\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 90, pp. 1--21.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/90\hfil Boundary-value problems] {Boundary-value problems for Hamiltonian systems and absolute minimizers \\ in calculus of variations} \author[V. N. Kolokol'tsov, A. E. Tyukov\hfil EJDE-2006/90\hfilneg] {Vassili N. Kolokol'tsov, Alexey E. Tyukov} % in alphabetical order \address{Vassili N. Kolokol'tsov \newline Dept of Statistics, University of Warwick, Coventry CV4 7AL, United Kingdom} \email{v.kolokoltsov@warwick.ac.uk} \address{Alexey E. Tyukov \newline Cardiff School of Mathematics, Cardiff University, Senghennydd Road, Cardiff, CF24 4AG, United Kingdom} \email{tyukovA@cardiff.ac.uk, tyukov@yahoo.co.uk} \date{} \thanks{Submitted November 14, 2005. Published August 15, 2006.} \subjclass[2000]{49J99, 34B15} \keywords{Hamiltonian systems; absolute minimizers; global field of extremals; \hfill\break\indent WKB method} \begin{abstract} We apply the method of Hamilton shooting to obtain the well-posedness of boundary value problems for certain Hamiltonian systems and some estimates for their solutions. The examples of Hamiltonian functions covered by the method include elliptic polynomials and exponentially growing functions. As a consequence we prove global existence, smoothness and almost everywhere uniqueness of absolute minimizers in the corresponding problem of calculus of variations and hence construct the global field of extremals. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newcommand{\wt}{\widetilde} \newcommand{\pa}{\partial} \section{Introduction} The classical problem of calculus of variations consists in finding a curve $\widetilde{y} (\tau)$ connecting $x_0$ and $x$ in time $t$ such that \begin{equation} \label{27} I_{\widetilde{y}} (t, x, x_0) = \min_{y(0)=x_0, y(t)=x} I_y (t, x, x_0), \end{equation} where \begin{equation} \label{25} I_y (t, x, x_0) = \int_0^t L(y (\tau) , \dot{y} (\tau)) \, d\tau . \end{equation} Here the function $L : \mathbb{R}^{2d} \to \mathbb{R}$ is called the Lagrangian of problem \eqref{27}. The value of minimum \eqref{27} is called the \emph{two-point function} corresponding to the Lagrangian $L$ and will be denoted $S(t, x, x_0)$. It is usually assumed to be convex with respect to the second variable. The function \begin{equation}\label{151} H(x, p) = \sup_{v \in \mathbb{R}^d} (pv -L (x, v)) \end{equation} is called the Hamiltonian of problem \eqref{27}. The celebrated Tonelli theorem is known to give the existence of $\widetilde{y} (\tau)$ under mild assumptions on $L$. The proof is based on the use of the so called direct methods of calculus of variations. However the minimizer $\widetilde{y} (\tau) $ given by Tonelli's theorem may be singular (see \cite{B} for an example and discussion). The aim of our paper is to single out some general enough class of Lagrangians (or Hamiltonians) having always non-singular minimizers and moreover, to prove the existence of the global field of smooth extremals for these classes. For a review on existence of non-singular minimizers see e.g. \cite{GH}, \cite{T}. In this paper we apply the method of Hamilton shooting to obtain existence, uniqueness of smooth solutions of the boundary value problems for systems \begin{equation}\label{4} \begin{gathered} \dot{x} = H_p (x, p) \\ \dot{p} = -H_x (x, p) \end{gathered} \end{equation} with a rather general class of Hamiltonians. This class of Hamiltonians includes the elliptic polynomials and functions of uniform exponential growth. In turn the well-posedness of boundary value problems and uniform estimates for the domain of uniqueness for boundary value problems imply the global existence, uniqueness and smoothness of absolute minimizers for problem \eqref{27} and hence the existence of a global field of extremals. It is proved that the smooth solutions always exist and are unique almost everywhere i.e. for any $x_0 \in \mathbb{R}^d, t >0$ the set of those $x \in \mathbb{R}^d$ for which a trajectory delivering the absolute minimum to functional \eqref{25} is not unique is a closed set of Lebesgue measure zero. Our method also yields estimates for the ``two-point function'' $S(t, x, x_0)$. \par As an important application of our results, let us mention the construction of local and global fields of extremals. It is well-known that the construction of global field of extremals corresponding to a Hamiltonian $H$ is the crucial step in the construction of WKB-type asymptotic for solutions of pseudo-differential equations $$ i \, { \pa u \over \pa t} = H \big(x, i \, {\pa \over \pa x}\big) \, u \quad \text{and} \quad {\pa u \over \pa t} = H \big(x, - {\pa \over \pa x}\big) \, u $$ (see e.g. \cite{K1},~\cite{M}). The method of Hamiltonian shooting can be applied also to some degenerate (non strictly convex) Hamiltonians. For example in \cite{K1} a rather general class of degenerate quadratic in momentum Hamiltonians was introduced (called regular in \cite{K1}) for which one can prove not only local uniqueness and global existence of solutions but also one can obtain exact asymptotic expansions of the two-point function that are quite similar to the case of nondegenerate quadratic Hamiltonians. It is worth mentioning that while developing the theory of stochastic Hamilton-Jacobi equations we used the method of Hamiltonian shooting to construct solutions for stochastic Hamiltonian systems driven by semimartingale noise (see \cite{K1,K2,KST}). \section{Main results}\label{s2} We denote by $\mathcal{F}$ the set of functions $h : \mathbb{R} \to \mathbb{R}$ represented by a series $h(z)=\sum_{n=0}^\infty a_n z^n$ with the radius of convergence equal to infinity such that for some constant $M=M(h)>0$ and all $z \geq M$ \begin{equation}\label{28} |h^{(n)} (z) h^{(m)} (z)| \leq |h^{(n-1)} (z) h^{(m+1)} (z)| \quad \forall \, n >m \geq 0 \, . \end{equation} Here $h^{(n)} (z)$ is the $n$-the derivative of $h(z)$. In particular, class $\mathcal{F}$ contains functions $h_1 (z) =\exp\{z \}$ and $h_2(z)= z^k$, $k \in {\mathbb N}$. \noindent{\bf Notation.} We write $(X(t) , P(t))=(X(t, x_0, p_0) , P(t, x_0, p_0))$ for the solution of the system \eqref{4} with initial conditions $(x_0,p_0)$ at $t=0$ and call the $x$-projection $X(t)$ of a solution $(X (t), P(t))$ characteristic of the system \eqref{4}. Denote by \begin{equation}\label{94} \widetilde x(\tau) =\widetilde x(\tau; t,x,x_0) \end{equation} a characteristic of \eqref{4} with boundary conditions $\widetilde x(0)=x_0$, $\widetilde x(t)=x$. We will use these notations throughout the paper. \begin{theorem}\label{t1} Let $H \in C^4(\mathbb{R}^{2d})$, $d \geq 1$. We assume that for some $h \in \mathcal{F}$ and $\varepsilon >0$ the following conditions hold: \begin{itemize} \item[(i)] $ H_{pp} (x, p) \geq \varepsilon \big(1+|h''(|p|)| \big) \, E_d$ for all $x, p \in \mathbb{R}^d$, where $E_d \in \mathbb{R}^{d \times d}$ is identity matrix; \item[(ii)] $$ \sum_{|I|\leq 3, |J|=k} \Big| {\pa^{|I|+|J|} H (x, p) \over \pa x^I \pa p^J} \Big| \leq | h^{( k )} (|p|) |\chi_\varepsilon (p)+\varepsilon^{-1}(1-\chi_\varepsilon (p)) \quad $$ for all $ x, p \in \mathbb{R}^d$ such that $|p| \geq \varepsilon^{-1}$, where \[ \chi_\varepsilon (p) =\begin{cases} 1 & \text{if } |p| \geq \varepsilon^{-1} \\ 0 & \text{otherwise} \end{cases}, \] $I=(i_1,\dots, i_{|I|})$, $J=(j_1,\dots, j_{|J|})$ are multi-indexes, $k=0,1,2,3$. \end{itemize} Then there exist $r>0$ and $T >0$ such that for any $x_0 \in \mathbb{R}^d$, $x \in B_r (x_0) =\{y: |y-x_0| \leq r\}$ and $t \in (0, T)$ there exists $p_0 =p_0 (t,x,x_0) \in \mathbb{R}^d$ such that \begin{equation}\label{84} X(t, x_0, p_0 (t, x, x_0))\equiv x \, . \end{equation} Moreover, $p_0 (t,x,x_0)$ is continuously differentiable with respect to all variables. \end{theorem} In other words the theorem claims that the boundary value problem for the system \eqref{4} is well-posed in a neighborhood of any $x_0 \in \mathbb{R}^d$, i.e. for small $|x-x_0|$ and small $t$ there exists the unique solution of the system \eqref{4} such that $x(0)=x_0$, $x(t)=x$. The lengthy assumptions of Theorem \ref{t1} are designed to include the main examples of Hamiltonian functions $H(x, p)$, which are used in geometry and mathematical physics, more precisely they include: \noindent{(1)} Convex elliptic polynomials (which for instance represent the symbols of elliptic differential operators widely studied both in $\mathbb{R}^d$ and on Riemannian manifolds); the most well-studied particular case is surely given by quadratic polynomials, the corresponding Hamiltonians describing the energy of classical mechanical systems in $\mathbb{R}^d$ or on Riemannian manifolds. \noindent{(2)} Hamiltonians provided by the L\'evy-Khinchine formula, namely \[ H(x, p)={1 \over 2} \, (G(x) p, p)-(A(x), p) + \int_{\mathbb{R}^d\setminus \{0\}} \Big( e^{- (p, \xi)} -1 +\frac{(p, \xi)}{1+\xi^2} \Big) \, d \nu_x (\xi), \] where all elements of $G(x)\in \mathbb{R}^{d \times d}$, $A(x) \in \mathbb{R}^d$ and their derivatives up to order three are bounded, $\mu_1 E_d < G(x)< \mu_2 E_d$ for some $\mu_1 , \mu_2 > 0 $ and $\nu_x$ is a L\'evy measure on $\mathbb{R}^d\setminus \{0\}$ with bounded support and some mild regularity assumptions; these Hamiltonians represent the symbols of pseudo-differential operators, which describe generators of most general Feller processes; the solutions of boundary value problem for these Hamiltonians describe the quasi-classical (or small diffusion) asymptotics of trajectories of this processes. It is well known (see \cite{GH}, vol. 1, Proposition 2, p.330) that extremal $x(\cdot)$ connecting $x_0$ and $x$ is a strong minimizer, if it can be embedded into Meyer field covering domain $\Gamma \subset \mathbb{R}^{d+1}$ and \begin{equation}\label{21} \mathcal{E}_L (z,\mathcal{P} (\tau , z ), q )> 0 \end{equation} for $(t,z) \in \Gamma $ and $q \not = \mathcal{P} (t,z)$, where $$ \mathcal{E}_L (x,p,q):=L(x,q) -L(x,p)-(q-p)\cdot L_p(x,p) $$ is the Weierstrass excess function of the Lagrangian $L(x,p)$ and $\mathcal{P} (t, x)$ is the slope function of Meyer field. Theorem \ref{t1} implies the existence of the Meyer field in $\Gamma (x_0)=(0,T)\times B_{r} (x_0)$ for any $x_0 \in \mathbb{R}^d$ with slope function $\mathcal{P} (t, x)=P(t, x_0, p_0(t, x_0, x))$. Moreover, condition (i) Theorem \ref{t1} infer that $H_{pp}(x, p) \geq \varepsilon E_d$, which in turn yields \eqref{21}. Hence \begin{equation}\label{92} S(t, x, x_0) : = \min I_y (t, x, x_0) =I_{\widetilde x (\cdot)} (t, x, x_0) \, , \end{equation} where minimum is taken over all $y$ lying completely in $B_{r} (x_0)$ with the boundary conditions $y(0)=x_0$ and $y(t)=x$. \begin{lemma}\label{l3} Under the conditions of Theorem \ref{t1} there exist $r_1 \in (0,r]$, $T_1 \in (0,T]$\, ($r,T$ as in Theorem \ref{t1}) such that \begin{equation}\label{17} \max_{x, x_0 \in \mathbb{R}^d : |x-x_0|\leq r_1} S(t, x, x_0) < \min_{0 < \tau \leq t} \min_{x, x_0 \in \mathbb{R}^d : |x-x_0| = r} S(\tau, x, x_0) \, \end{equation} for all $0 < t < T_1$. Moreover \begin{equation}\label{112} \lim_{t \to 0+} \min_{x, x_0 \in \mathbb{R}^d : |x-x_0| = r} S(t, x, x_0) =+\infty \, . \end{equation} \end{lemma} The proofs of Theorem \ref{t1} and Lemma \ref{l3} will be given in Sections \ref{s3} and \ref{s4} respectively. Observe that (i) Theorem \ref{t1} implies that $H(x,p)$ is bounded from below and so is $L(x,v)$. Note that functionals \eqref{25} with Lagrangians $L(x,v)$ and $L(x,v)+c$ with $c \in \mathbb{R}$ being some constant have the same minimizing functions. Therefore without loss of generality we assume that $L(x,v) \geq 0$ and then $$ I_y (t,x,x_0) \geq 0 $$ for all piecewise smooth $y(\tau)$. \begin{corollary}\label{c2} Let $0 S (t, x,x_0) =I_{\widetilde x} (t, x,x_0) \, . $$ \end{proof} \begin{theorem}[Tonelli's theorem]\label{t2} Under the conditions of Theorem \ref{t1} for any $t>0$ and $x, x_0 \in \mathbb{R}^d$ there exists a characteristic $\widetilde x(\tau)$ with boundary conditions $\widetilde x(0)=x_0$, $\widetilde x(t)=x$ that provides an absolute minimum (probably not unique) for $I_y (t, x, x_0)$ over all piecewise smooth curves $y(\tau)$ connecting $x_0$ and $x$ in time $t$. \end{theorem} Our proof of the implication Lemma \ref{l3} $\Rightarrow$ Theorem \ref{t2} (and also the deduction of Theorem \ref{t3} given below) is similar to the one given in \cite[pp. 56-57]{K1}. We give it here for reader's convenience. \begin{proof}[Proof of Theorem \ref{t2}] Let $y_n=y_n(\tau)$ will be a minimizing sequence for $I_y (t, x, x_0)$, i.e. \begin{equation}\label{72} I_{y_n} (t, x, x_0) \to \inf_{y} I_{y} (t, x, x_0) \end{equation} and $y_n (0)=x_0$, $y_n (t)=x$. Let $k \in {\mathbb N}$ be the integer part of ${|x-x_0|/r_1}$, namely \begin{equation}\label{100} kr_1 \leq |x-x_0| < (k+1)r_1 \end{equation} and suppose first (in Steps 1 and 2) that \begin{equation}\label{87} t \in (0,T_1) \end{equation} ($T_1$, $r$, $r_1$ as in Lemma \ref{l3}). \noindent\emph{Step 1.} We show that for all $\tau \in (0,t)$ and $n \in {\mathbb N}$ \begin{equation}\label{37} y_n (\tau) \in \mathcal{K} \quad \text{for some compact} \quad \mathcal{K} \subset \mathbb{R}^d \, . \end{equation} We define $\tau_i=\tau_i (n)$ \, $i \in {\mathbb N}_0$ by the recurrent formula $$ \tau_0 =0; \quad \tau_i=\inf\{\tau\in (\tau_{i-1}, t]: |y_n (\tau)-y_n (\tau_{i-1})| =r\} $$ ($\inf \emptyset := +\infty$) and put $m=m(n)=\max\{i : \tau_i < \infty\}$. Since $y(0)=y(\tau_0)=x_0$ $$ |y_n (\tau)-x_0| \leq \sum_{j=1}^{i-1} \big|y_n (\tau_j)-y_n (\tau_{j-1}) \big| + \big|y_n (\tau)-y_n (\tau_{i-1}) \big| \leq i r \leq mr $$ for $\tau \in (\tau_{i-1}, \tau_i)$ and $$ |y_n (\tau)-x_0| \leq |y_n (\tau)-y_n (\tau_{m})|+|y_n (\tau_m)-x_0|\leq (m+1)r $$ for $\tau \in (\tau_m, t)$. Thus \begin{equation}\label{83} y_n (\tau) \in B_{(m+1) r} (x_0) \, . \end{equation} We show that \begin{equation}\label{82} m (n) \leq k+1 \end{equation} for all but finite number of $n$. Then combining \eqref{83} and \eqref{82} proves \eqref{37} with $\mathcal{K}= B_{(k+2) r} (x_0)$. To prove \eqref{82} we assume by contradiction that $m \geq k+2$. Let $\widehat y_n (\tau)$ be solution of \eqref{4} on each interval $[\tau_{i-1}, \tau_{i}]$ $i=1, \dots, k$ ($k$ as in \eqref{100}) and on the interval $[\tau_{k}, t]$ and $$ \widehat y_n (\tau_{i})=x_0+{i \over k+1}\, (x-x_0) \quad i=1, \dots, k \,, $$ $\widehat y_n (t)=x$. Since $|\widehat y_n (\tau_i)- \widehat y_n (\tau_{i-1})| < r_1$ and $|y_n (\tau_i)- y_n (\tau_{i-1})|=r$, an application of \eqref{17} gives \begin{equation} \label{80} \begin{aligned} S(\tau_i-\tau_{i-1},y_n (\tau_i), y_n (\tau_{i-1})) &> S(\tau_i-\tau_{i-1},\widehat y_n (\tau_i), \widehat y_n (\tau_{i-1} )) \\ &= I_{\widehat y_n} (\tau_i-\tau_{i-1},\widehat y_n (\tau_i), \widehat y_n (\tau_{i-1})) \, . \end{aligned} \end{equation} Summing \eqref{80} over $i=1, \dots, k$ and using $S(\tau_{k+1}-\tau_{k},y_n (\tau_{k+1}), y_n (\tau_{k})) \geq S(t-\tau_{k},x, \widehat y_n (\tau_{k}))$ we obtain \begin{align*} I_{y_n}(\tau_{k+1},y_n (\tau_{k+1}), x_0) &=\sum_{i=1}^{k+1} I_{y_n}(\tau_{i}-\tau_{i-1},y_n (\tau_{i}),y_n (\tau_{i-1})) \\ &\geq \sum_{i=1}^{k+1} S (\tau_{i}-\tau_{i-1},y_n (\tau_{i}),y_n (\tau_{i-1})) \\ &\geq I_{\widehat y_n}(t, x , x_0) \, , \end{align*} and so $$ \inf_{n} \, I_{y_n}(\tau_{k+1},y_n (\tau_{k+1}), x_0) \geq \inf_{n} \, I_{\widehat y_n}(t, x , x_0) \, . $$ On the other hand, by \eqref{112} we have $$ \inf_{n} \, I_{y_n}(\tau_{k+2}-\tau_{k+1} ,y_n (\tau_{k+2}), y_n (\tau_{k+1})) >0 \, . $$ Thus $$ \inf_{n} I_{y_n}(t,x, x_0) >\inf_{n} I_{\widehat y_n}(t,x, x_0) \, , $$ which contradicts \eqref{72}. This contradiction proves \eqref{82}. Without loss of generality we may assume that \eqref{82} holds for all $n$. \smallskip \noindent \emph{Step 2.} Since the sequence $(\tau_i,y_n (\tau_i))$ lies in the compact $[0,t] \times \mathcal{K}$ for any $i=1, \dots, m$, then without loss of generality we assume that $$ \tau_i (n) \to s_i\, , \quad y_n (\tau_i) \to b_i \quad\text{as}\quad n \to \infty $$ $i=1, \dots, m$ for some $s_i \in (0,t)$, $b_i \in \mathbb{R}^d$. Let $\widetilde{y}_n (\tau)$ be solution of \eqref{4} on each interval $[\tau_{i-1}, \tau_{i}]$ \, $i=1, \dots, m$ and on the interval $[\tau_{m}, t]$ and $$ \widetilde{y}_n (\tau_{i})=y_n (\tau_{i}) \quad i=1, \dots, k \, , $$ $\widetilde{y}_n (t)=x$. Clearly \begin{equation}\label{88} I_{y_n} (t,x,x_0) \geq I_{\widetilde{y}_n} (t,x,x_0) \, . \end{equation} Let $y (\tau)$ be solution of \eqref{4} on each interval $[s_{i-1}, s_{i}]$ \, $i=1, \dots, m$ and on the interval $[s_{m}, t]$ and $y (s_{i})= b_{i}$, $i=1, \dots, m$, $y (t)=x$. Theorem \ref{t1} implies \begin{equation}\label{85} \lim_{n \to \infty} \left\|y-\widetilde{y}_n \right\|_{C^1([s_{i-1}+\delta, \, s_i-\delta])} = 0 \end{equation} for any $\delta <(s_{i-1}-s_i)/2$, $i=1, \dots, m$. Note that for all but finite number $i$ we have $[s_{i-1}+\delta, s_i-\delta] \subset [ \tau_{i-1}, \tau_i ]$. Therefore, \begin{equation}\label{86} \lim_{n \to \infty} I_{\widetilde{y}_n} (t,x,x_0) =I_{ y} (t,x,x_0) \, . \end{equation} Combining \eqref{88} and \eqref{86} we get $$ \lim_{n \to \infty} I_{y_n} (t,x,x_0) \geq I_{ y} (t,x,x_0) \, . $$ It suffices to show smoothness of $y(\tau)$ at $\tau=s_i$ $i=1, \dots, n$. For each $i=1, \dots, m$ we take $s'_i 0$ the curve $X(\tau)$ $\tau \in [\tau_1, \tau_2+\delta]$ does not provide even a local minimum among curves joining $X(\tau_1)$ and $X(\tau_2+\delta)$ in time $\tau_2-\tau_1+\delta$. \end{proposition} The proof of the above Proposition can be found in \cite[p.77 Theorem 1]{A}. Following \cite{K1} we say that for some $x_0 \in \mathbb{R}^d$ the point $(t, x) \in \mathbb{R}_{+} \times \mathbb{R}^d$ is \emph{regular} and write $(t,x) \in \mathop{\rm Reg}(x_0)$ if ({\bf i}) points $x,x_0$ are not conjugate; ({\bf ii}) the absolute minimum is attained on a unique curve; ({\bf iii}) this curve is a characteristic $\widetilde x(\tau,t,x,x_0)$ with boundary conditions $\widetilde x(0)=x_0$, $\widetilde x(t)=x$. \begin{theorem}\label{t3} Under the conditions of Theorem \ref{t1} for any $x_0 \in \mathbb{R}^d$ the set of regular points is open, connected and everywhere dense in $\mathbb{R}_{+}\times \mathbb{R}^d$. For any fixed $t>0$ the set $\{x \in \mathbb{R}^d: (t, x) \in \mathop{\rm Reg}(x_0) \}$ is open and everywhere dense in $\mathbb{R}^d$. \end{theorem} \begin{proof} For any $t>0$, $x, x_0 \in \mathbb{R}^d$ we take characteristic $\widetilde x(\tau)=\widetilde x(\tau;t,x,x_0)$ which delivers absolute minimum for $I_y (\tau , x,x_0)$. Note that the existence of such $\widetilde x(\tau)$ is guaranteed by Theorem \ref{t2}. We show that for any $\tau \in (0, t)$ \, characteristic $\widetilde x(s)$, \, $s \in [0,\tau]$ delivers the unique absolute minimum to $I_y (\tau , \widetilde x(\tau) ,x_0)$. Clearly $\widetilde x(s)$ provides an absolute (a priori not unique) minimum to $I_y (\tau , \widetilde x(\tau),x_0)$. If another function $\widetilde{y}=\widetilde{y} (s)$, \, $s \in [0,\tau]$ with $\widetilde{y} (0)=x_0$, $\widetilde{y} (\tau)=\widetilde x (\tau)$ also provides an absolute minimum, then defining it on $(\tau,t]$ by $\widetilde{y} (s)=\widetilde x (s)$, $s \in [\tau,t]$ we deduce that $\widetilde{y}$ also provides a minimum for $I_y (t, x,x_0)$. Take $s' < \tau < s''$ such that $\widetilde{y}(s'), \widetilde{y}(s'') \in B_{2^{-1} r_1} (\widetilde{y} (\tau))$, and so $|\widetilde{y}(s')-\widetilde{y}(s'') | \leq r_1$. By Corollary \ref{c2} the characteristic connecting $\widetilde{y}(s')$ and $\widetilde{y}(s'')$ in time $s''-s'$ provides the unique absolute minimum. Therefore $\widetilde x(s)= \widetilde{y}(s)$ for $s \in [s', s'']$. Consequently we have $\widetilde x(s)=\widetilde{y}(s)$ for all $s \in [0, \tau]$. By Proposition \ref{p1} $\widetilde x(0)$ and $\widetilde x (\tau)$ are not conjugate. Hence $ (\tau, \widetilde x(\tau))$ is regular. This immediately implies that Reg $(x_0)$ is connected and everywhere dense in $\mathbb{R}_{+} \times \mathbb{R}^d$. Moreover, $x$-projection of Reg $(x_0)$ is everywhere dense in $\mathbb{R}^d$. To prove that Reg $(x_0)$ is open, note that if $(t, x)$ is regular then $\pa X(t, x_0,p_0) /\pa p_0 \not = 0$. An application of implicit function theorem gives that~$p_0=p_0 (t, x, x_0)$ is well-defined by \eqref{84} in some neighborhood $U(t,x) \subset \mathbb{R}^{d+1}$ of $(t,x)$. As we know the characteristic $\widetilde x (\tau)= \widetilde x(\tau;t', x', x_0)$ delivers local minimum for any $(t', x') \in U(t,x)$. Since the quantity $\min_y I_y (t', x', x_0)$ depends continuously on $(t', x')$ and for $(t', x')=(t, x)$ the unique absolute minimum is provided by $\widetilde x(\tau;t,x,x_0)$, then $x(\tau;t', x', x_0)$ provides an absolute minimum for any $(t',x') \in U_0(t,x)$ for some $U_0 (t,x) \subset U (t,x)$. Similarly fixing $t>0$ one shows that the set $\{x: (t,x) \in \mathop{\rm Reg} (x_0) \}$ is open. \end{proof} \section{Auxiliary results}\label{s5} In this section we will deduce some implications from \eqref{28} and (i), (ii) Theorem \ref{t1}. \noindent{\bf Notation.} Throughout this paper $\varepsilon>0$ will be from the conditions of Theorem \ref{t1} and $M=M(h)>0$ will be the constant from \eqref{28}. In sections \mbox{\ref{s5} - \ref{s4}} we will construct constants $M_1, \dots, M_{11}>0$, $\rho>0$, $$ 1>t_1 \geq t_2 \geq T \geq T_1 >0 \,, \quad 1 >c_1 \geq c_2 >0\, , \quad r \geq r_1>0 $$ which depend only on $\varepsilon$ and $h(z)$. \begin{lemma}\label{l5} The functions $h \in \mathcal{F}$ satisfy the following inequalities \begin{itemize} \item[(i)] $ |h^{(n)} (z) (h (z))^{n-1} | \leq | h' (z) |^n $ $n \in {\mathbb N}_0$, \item[(ii)] $\Big| h^{(n) } \Big( z + \lambda { {|h(z)| \over |h' (z)|}} \Big) \Big| \leq 3 |h^{(n) } (z)| $, $n \in {\mathbb N}_0$ \end{itemize} for all $z >M$ and $|\lambda | \leq 1$. \end{lemma} \begin{proof} The case $n=0$ in (i) is trivial. We assume that $n \geq 1$. By repeated application of \eqref{28} we obtain $$ |h^{(n)} (z) (h (z))^{n-1} | \leq |h^{(n-1)} (z) (h (z))^{n-2} h' (z) | \leq \dots \leq | h' (z)|^n $$ and (i) follows. We proceed with (ii). Using Taylor's development $h^{(n)} (z+a) = \sum^{\infty}_{m=0} h^{(n+m)} (z) a^m / m!$ with $ a= | \lambda h (z) /h^{ \prime} (z)|$ gives $$ h^{(n)} \Big( z + \lambda {|h(z) |\over |h' (z)|}\Big) = \sum_{m=0}^\infty |\lambda|^m {h^{(m+n)} (z) \over m!} \, {|h (z)|^{m} \over |h^{ \prime }(z)|^{m}} \, . $$ Since, by \eqref{28}, $$ |h^{(m+n)} (z)(h (z))^{m}| \leq |h^{(m+n-1)} (z) h' (z) (h (z))^{m-1}| \leq \dots \leq |h^{(n)} (z)| |h' (z)|^m\, , $$ it follows $$ \Big| h^{(n)} \Big( z + \lambda {|h(z) |\over |h' (z)|}\Big)\Big| \leq |h^{(n)} (z)| \sum_{m=0}^\infty {|\lambda|^m \over m!} \leq |h^{(n)} (z)|\exp\{1\}, $$ where we used $|\lambda| <1$. \end{proof} \begin{lemma}\label{l6} Let $h \in \mathcal{F}$, $H \in C^4 (\mathbb{R}^{2d})$ be such that conditions (i), (ii) of Theorem \ref{t1} hold. Then for some constants $M_1, M_2, M_3 >0$ \begin{itemize} \item[(i)] $ |h^{(m)} (z)|$ is monotone on $(M_1,+\infty)$ for all $m=0, \dots, 3 $, \item[(ii)] $ |h' (z)|$ increases on $(M_1,+\infty)$, \item[(iii)] $|H_p (x, p)| \geq \varepsilon |h' (|p|)| -M_2$ for all $x, p \in \mathbb{R}^d$, \item[(iv)] $(p, H_p (x, p)) \geq \varepsilon |h' (|p|)|\, |p| -M_3$ for all $x, p \in \mathbb{R}^d$ . \end{itemize} \end{lemma} \begin{proof} An application of \eqref{28} with $n=m+2$, $m \in {\mathbb N}_0$ implies $$ \big((g (z) )^2\big)'' = 2 g'' (z) g(z)+2(g' (z))^2 >0 $$ for $g(z)=h^{(m)} (z) $. Therefore, $g(z)$ has at most two intervals of monotonicity. Hence $g(z)$ is monotone for $z \geq M_{1, m} $ for some $M_{1, m}>0$. Take $$ M_1=\varepsilon^{-1}+\max_{m=0, \dots, \, 3} M_{1, m} $$ and (i) follows. By (i) Theorem \ref{t1} we have $H_{pp} (x,p) \geq \varepsilon E_d$. Hence \begin{align}\label{73} |p| \, |H_p(x, p)| \geq \big(p, H_{p} (x, p)\big)&= \big(p, H_{p} (x, 0)\big) + \int_0^1\big(p, H_{pp} (x, \tau p) p\big) \, d \tau \\ &\geq \big(p, H_{p} (x, 0)\big) + \varepsilon \, |p|^2 \\ &\geq \varepsilon \, |p|^2 -\varepsilon^{-1} |p| \, , \end{align} where we used that, by (ii) Theorem \ref{t1}, $|H_{p} (x, 0)| \leq \varepsilon^{-1}$ . Hence $$ |H_p(x, p)| \geq \varepsilon \, |p| -\varepsilon^{-1} \, . $$ Using (ii) Theorem \ref{t1} we get \begin{equation}\label{34} |h' (z) | \geq \varepsilon \, z -\varepsilon^{-1} \end{equation} for $ z \geq M_1$ (recall $M_1 \geq \varepsilon^{-1}$). This and (i) Lemma \ref{l6} imply (ii) Lemma \ref{l6}. We proceed with (iii) and (iv). Without loss of generality we may assume that $h' (z)$ increases for $z \geq M_1$. (Otherwise we replace $h$ by $-h$ noticing that the conditions of Theorem \ref{t1} for $-h$ and $H$ still hold.) So $$ h'' (z) > 0 \quad \text{for all} \quad z \geq M_1\, . $$ Hence for $|p| \geq M_1$ we get \[ \int_0^1|h''(\tau |p|) |\, |p| \, d\tau \geq \int_{{M_1 \over |p|}}^1 h''(\tau |p|) \, |p| \, d\tau =h'(|p|)-h'({M_1}) \, . \] Consequently, due to (i) Theorem \ref{t1}, \begin{align*} \int_0^1\big(p, H_{pp} (x, \tau p) p\big) \, d\tau &\geq \varepsilon |p|^2 \int_0^1(1+|h''(\tau |p|) ) \, d\tau \\ &\geq \varepsilon |p| \left(|p|+h'(|p|)-h'(M_1) \right) \, . \end{align*} Using this and the first line in \eqref{73} we have \begin{align}\label{53} (p, H_p(x, p)) &\geq - \varepsilon^{-1}|p| +\varepsilon |p| \left(|p|+h'(|p|)-h'(M_1) \right) \\ &\geq \varepsilon h'(|p|)\, |p|-M_3 \, , \end{align} where $$ M_3:=-\min_{z>0} \{- \varepsilon^{-1}z +\varepsilon z (z- h'(M_1) )\} \, . $$ Using again \eqref{53} we get \begin{equation}\label{48} |H_p(x,p)| \geq - \varepsilon^{-1} +\varepsilon \left(|p|+h'(|p|)-h'(M_1) \right) \geq \varepsilon h'(|p|) - \varepsilon^{-1} -\varepsilon h'(M_1) \, , \end{equation} where $M_2:=\varepsilon^{-1} +\varepsilon h'(M_1)$. \end{proof} \begin{lemma}\label{l1} Under the conditions of Theorem \ref{t1} there exist $t_1, M_4 >0$ such that for all $c \in (0, 1/2)$, $t \in (0, t_1)$, $x_0 \in \mathbb{R}^d$, \begin{equation}\label{6} p_0 \in V_{c, t}:= \{p \in \mathbb{R}^d : t |h' (|p|)| < c \} \end{equation} and $\tau \in [0, t]$ we have \begin{equation}\label{10} |h^{(n)} ( | P(\tau ) | )| \leq M_4 (|h^{(n)}( |p_0| )| +1)\quad n=0,1,2. \end{equation} Moreover, \begin{gather}\label{55} |X(t, x_0, p_0) -x_0 | \leq M_4 t (|h' ( |p_0| )| +1) \, , \\ \label{59} |P(t, x_0, p_0) -p_0 | \leq M_4 t (|h ( |p_0| )| +1) \,. \end{gather} \end{lemma} \begin{proof} We choose $0 < t_1 <1 $ such that the system \eqref{4} has a solution on $[-t_1,t_1]$. Let us first (in Steps 1 and 2) assume that \begin{equation}\label{39} |P (\tau)| \geq M_1 \quad \text{for all} \quad \tau \in [0,t_1] \, . \end{equation} \noindent \emph{Step 1.} Notice that (i), (ii) Lemma \ref{l6} imply that \begin{equation}\label{111} |h(z)| \quad \text{increases for} \quad z \geq M_1 \, . \end{equation} In particular $h(z)$ preserves sign for $z \geq M_1$. Without loss of generality we may assume (in Steps 1 and 2) that $h(z) > 0$ for $z \geq M_1$. Using (ii) Theorem \ref{t1} and $M_1 >\varepsilon^{-1}$ we find from \eqref{39} and the system \eqref{4} $$ |P(t)-p_0| \leq \int_{0}^t |H_x (X(\tau), P(\tau)) | \, d \tau \leq \int_0^t h (|P(\tau)|) \, d \tau $$ and so $$ |p_0|-\int_0^t h (|P(\tau)|) \, d \tau\leq |P(t)| \leq |p_0|+\int_0^t h (|P(\tau)|) \, d \tau \, . $$ Due to \eqref{111} and Gronwall Lemma, $$ y_1(\tau) \leq | P(\tau) | \leq y_2(\tau) , $$ where $y_i(\tau)$ $i=1,2$ solve the equation $$ \dot{y}_i (\tau) = (-1)^i \, h(y_i (\tau)) \, , \quad y_i(0)=|p_0| . $$ One readily sees that $y_i(\tau)= \Phi^{-1} \left( \Phi (|p_0|) +(-1)^i\tau \right)$, $i=1,2$, where $$ \Phi (\lambda) = \int_{M_1}^\lambda {d \zeta \over h(\zeta)} \, . $$ Hence \begin{equation}\label{16} \Phi^{-1} \left( \Phi (|p_0|) -\tau \right) \leq | P(\tau ) | \leq \Phi^{-1} \left( \Phi (|p_0|) +\tau \right) . \end{equation} \emph{Step 2.} An application of Taylor's formula on $\Phi^{-1}$ yields \begin{equation}\label{90} \Phi^{-1} \left( \Phi (z) +\tau \right)= z+ \sum_{n=1}^\infty {a_n (z) \over n !} \, \tau^n , \end{equation} where $$ a_n(z) = {d^n \over d \tau^n} \, \Phi^{-1} \left( \Phi (z) +\tau \right) \big|_{\tau=0} = (\Phi^{-1})^{(n)} (\Phi (z)) $$ or \begin{equation}\label{60} a_n (\Phi^{-1} (\kappa)) = (\Phi^{-1} )^{(n)} (\kappa) \, \end{equation} with $\kappa = \Phi (z)$. We differentiate \eqref{60} in $\kappa$ to get \begin{equation}\label{49} a_{n}' (\Phi^{-1} (\kappa)) (\Phi^{-1} (\kappa))' = (\Phi^{-1} )^{(n+1)} (\kappa) =a_{n+1} (\Phi^{-1} (\kappa)) \, . \end{equation} Consequently, \eqref{49} and $(\Phi^{-1} (\kappa))'= dz /d \kappa = (\Phi' (z))^{-1}= h (z)$ imply \begin{equation}\label{61} a_1 (z) = h(z) \quad \text{and} \quad a_{n+1} (z)=a_{n}' (z) h(z) \, . \end{equation} Using induction we deduce from \eqref{61} \begin{equation}\label{62} a_n (z) = {\sum}' h^{(m_1)} (z) \dots h^{(m_{n-1})} (z) h(z) \quad n \geq 2 \, , \end{equation} where the sum $\sum'$ is taken over some $m_1, \dots , m_{n-1} \in {\mathbb N}_0$ such that $m_1+\dots+m_{n-1}=n-1$ and contains $(n-1)!$ terms. It follows $$ | a_n (z) | \leq (n-1)! \max_{m_1+\dots+m_{n-1}=n-1} \big| h^{(m_1)} (z) \dots h^{(m_{n-1})} (z) h(z) \big| \, . $$ By (i) Lemma \ref{l5} and because of $\sum_{k=1}^{n-1}(m_k-1)=0$ we have $$ | \, h^{(m_1)} (z) \dots h^{(m_{n-1})} (z) | =\prod_{k=1}^{n-1}| \, h^{(m_k)} (z) (h(z))^{m_k-1} | \leq |h' (z)|^{n-1} . $$ Consequently, $$ |a_n (z) \tau^n | \leq (n-1)! \, | \tau h' (z) |^{n-1} \, \tau |h (z)| . $$ Since, by $\eqref{6}$, $|\tau h' (|p_0|) | \leq c $, it follows that \begin{equation}\label{93} \big| {a_n (|p_0|) \tau^n \over n!} \big| < c^{n-1} \, \tau |h(|p_0|)| . \end{equation} Substituting \eqref{93} in \eqref{90} we get \begin{equation}\label{15} \Phi^{-1} \left( \Phi (|p_0|) +\tau \right) \leq |p_0| + { 1 \over 1- c }\, \tau |h(|p_0|)| \leq |p_0| + 2 \tau |h(|p_0|)| \end{equation} for $c \in (0, 1/2)$. Similarly we obtain \begin{equation}\label{150} \Phi^{-1} \left( \Phi (|p_0|) -\tau \right) \geq |p_0| - 2 \tau |h(|p_0|)| \end{equation} for $c \in (0, 1/2)$. Combining \eqref{16}, \eqref{15} and \eqref{150} we obtain $$ |p_0| - 2 \tau |h(|p_0|)| \leq | P(\tau) | \leq |p_0| + 2 \tau |h(|p_0|)| $$ for $c \in (0, 1/2)$. Using $p_0 \in V_{c, t}$ gives $\tau \leq t \leq c/|h' (|p_0|)|$ and so \begin{equation}\label{26} |p_0| - 2 c \, {|h(|p_0|)| \over |h' (|p_0|)|} \leq | P(\tau) | \leq |p_0| + 2 c \, {|h(|p_0|)| \over |h' (|p_0|)|}. \end{equation} An application of (ii) Lemma \ref{l5} with $\lambda= 2 c \leq 1$ gives \begin{equation} \label{101} \begin{aligned} | h^{(n) } ( |P(\tau)| ) | &\leq \max\big\{ h\big(|p_0| + 2 c \, {|h(|p_0|)| \over |h' (|p_0|)|}\big) , h\big(|p_0| -2 c \, {|h(|p_0|)| \over |h' (|p_0|)|}\big) \big\} \\ &\leq 3 |h^{(n)} (|p_0|)| \quad n=0,1,2. \end{aligned} \end{equation} Similarly one can check (we will need this in the proof of Lemma \ref{l7}) that \begin{equation}\label{33} |h^{(n)} (|P(\tau_1)|)| \leq 3 |h^{(n)} (|P(\tau_2)|)| \quad n=0,1,2 \end{equation} for any $\tau_1, \tau_2 \in [0,t]$. \noindent\emph{Step 3.} Finally we suppose that \eqref{39} does not hold. If $|P(\tau)| \leq M_1$ for all $\tau \in [0,t_1]$ then taking \begin{equation}\label{45} M_4 :=3+3 \max_{z\in [0, \, M_1], n=0,1,2} |h^{(n)}(z)| \, \end{equation} we have \begin{equation}\label{40} | h^{(n) } ( |P(\tau)| ) |\leq {1 \over 3} \, M_4 \quad n=0,1,2 \end{equation} for all $\tau$. If $|P(s_0)|=M_1$ and $|P(\tau)| \geq M_1$ on $[s_0, s_1] \subset [0, t_1]$ for some $s_0 0$ such that for all $t \in (0,t_1)$, $c \in (0,c_1)$, $x_0 \in \mathbb{R}^d$, $p_0 \in V_{c, \, t}$ one has \begin{itemize} \item[(i)] $$|p_0| \geq \omega \big(M_4^{-1} \, {|x-x_0| \over t} -M_5\big) \quad \text{provided that } M_4^{-1} {|x-x_0|\over t} -M_5>M_4 \, , $$ \item[(ii)] $$ |p_0| \leq \omega \big( {2 \over \varepsilon} \, {|x-x_0| \over t} +M_5\big) \, , $$ \item[(iii)] $$ \big(p_0 , {x-x_0 \over t} \big)\geq {\varepsilon \over 2}\, |p_0| \big(M_4^{-1}{|x-x_0| \over t } -M_5\big)- M_6 \, , $$ \end{itemize} where we write for short $x=X(t, x_0, p_0)$. \end{lemma} \begin{remark} \label{rmk1} \rm Notice that (ii) Lemma \ref{l6} and the fact that $M_4>|h' (M_1)|$ imply the correctness of definition \eqref{64}. \end{remark} \begin{proof} \emph{Step 1.} In this step we show that \begin{equation}\label{105} \left|\left(p_0, H_p (X(\tau), P(\tau)) -H_p (x_0, p_0) \right) \right| \leq 2^{-1} \varepsilon \, |h'(|p_0|)| |p_0| +2 \omega (M_4) ( \varepsilon^{-1}+M_4) \, . \end{equation} Let us first assume that \begin{equation}\label{13} |P(\tau)|>M_1 \quad \forall \tau \in [0,t] \, . \end{equation} From \eqref{4}, \eqref{33} and (ii) Theorem \ref{t1} (recall $M_1> \varepsilon^{-1}$) we deduce \begin{align*} \sum_{j=1}^d |(X(\tau)-x_0)_j| &\leq \sum_{j=1}^d \int_0^t \left| {\pa H \over \pa p_j}\left( X(\tau), P(\tau) \right) \right| \, d\tau \\ &\leq t \max_{0 \leq \tau \leq t} |h'(|P(\tau)|)| \leq 3 t \big|h' (|p_0|)\big| \end{align*} and $$ \sum_{j=1}^d |(P(\tau)-p_0)_j| \leq t \max_{0 \leq \tau \leq t} |h(|P(\tau)|)| \leq 3 t \big|h (|p_0|)\big| \, . $$ Denote by \begin{equation}\label{1} R:=H_p (X(\tau), P(\tau)) -H_p (x_0, p_0) \, , \end{equation} $R= (R_1, \dots, R_d)$. Using the mean value theorem we get \begin{align*} R_i&=\sum_{j=1}^d H_{p_i x_j} (\nu_{i}, \xi_{i})(X(t)-x_0)_j+\sum_{j=1}^d H_{p_i p_j} (\nu_{i}, \xi_{i})(P(t)-p_0)_j \, \\ &=:R_i^1+R_i^2 \end{align*} for some $\nu_{i} \in [x_0, X(\tau)]$, $\xi_{i} \in [p_0, P(\tau)]$, \, \, $i=1, \dots, d$. It follows that $$ |R_i^2| \leq \max_{i,j=1, \dots, d} \left| H_{p_i p_j} (\nu_{i}, \xi_{i}) \right| \Big( \sum_{j=1}^d |(P(\tau)-p_0)_j| \Big) \, . $$ Due to $|\xi_i|> M_1 > \varepsilon^{-1}$ and (ii) Theorem \ref{t1} we get $$ \left| H_{p_i p_j} (\nu_{i}, \xi_{i}) \right| \leq \big|h'' (|\xi_{i} |)\big| \leq \max \{ |h'' (|p_0|)| , |h'' (|P(\tau)|)| \} \leq 3 \big|h'' (|p_0|)\big| \quad $$ $i,j=1, \dots, d$ and so $$ |R_i^2| \leq 9t \big|h'' (|p_0|) h (|p_0|)\big| \leq 9 t \big|h' (|p_0|)\big|^2 \, . $$ Similarly $|R_i^1| \leq 9 t \big|h' (|p_0|)\big|^2 $. Consequently, $$ |R| \leq \sum_{i=1}^d |R_i| \leq 18 \, d t \big|h' (|p_0|)\big|^2 \, . $$ Setting $c_1= \varepsilon /(36 \, d )$ and using $t \leq |h'(|p_0|)|^{-1} \, c$ we get (under assumption \eqref{13}) \begin{equation}\label{41} \left|H_p (X(\tau), P(\tau)) -H_p (x_0, p_0) \right| \leq { \varepsilon \over 2} \, |h'(|p_0|)| \, . \end{equation} If \eqref{13} does not hold then, by \eqref{46} and (ii) Theorem \ref{t1}, $$ |H_p(X(\tau), P(\tau))| \leq \varepsilon^{-1}+|h'(|P(\tau)|)| \leq \varepsilon^{-1}+ M_4\, . $$ Using, by \eqref{46}, $|h' (|p_0|)|\leq M_4$ and the fact that $z \geq \omega(M_4)$ implies $|h' (z)|> M_4$ we obtain $|p_0| \leq \omega(M_4)$. Hence $$ \left|\left(p_0, H_p (X(\tau), P(\tau)) -H_p (x_0, p_0) \right) \right| \leq 2 \omega(M_4) ( \varepsilon^{-1}+M_4)\, . $$ Combining this and \eqref{45} we arrive at \eqref{105}. \noindent\emph{Step 2.} If \eqref{13} holds then, by \eqref{41} \begin{align*} |x-x_0| -t |H_p(x_0, p_0)| &\geq -\left| x-x_0 - t H_p(x_0, p_0)\right|\\ &= -\Big| \int_0^t \left(H_p (X(\tau), P(\tau)) -H_p (x_0, p_0) \right) \, d \tau \Big| \\ &\geq -{ t \varepsilon \over 2} \, |h'(|p_0|)| \, , \end{align*} and so using (iii) in Lemma \ref{l6}, we get \begin{equation}\label{43} |x-x_0| \geq {\varepsilon t \over 2} \, |h'(|p_0|)| -t M_2 \,. \end{equation} If \eqref{13} does not hold, then, due to \eqref{46}, \begin{equation}\label{42} |x-x_0| \geq 0 \geq t |h' (|p_0|)|- M_4 t \, . \end{equation} Combining \eqref{43} and \eqref{42} (recall $\varepsilon <1$) we arrive at $$ {|x-x_0| \over t} \geq { \varepsilon \over 2} \, |h' (|p_0|)|-M_2 - M_4 \, . $$ This and \eqref{55} yield \begin{equation}\label{104} M_4^{-1}{|x-x_0| \over t}-M_5 \leq |h' (|p_0|)| \leq {2 \over \varepsilon }\, {|x-x_0| \over t} +M_5 \, , \end{equation} where $M_5:= 2 \varepsilon^{-1} \, ( M_2+M_4)$. Applying $\omega (\cdot)$ to both sides of \eqref{104} proves (i) and (ii) Lemma \ref{l7}. \noindent\emph{Step 3.} An application of \eqref{105} and (iv) Lemma \ref{l6} give \begin{align} \left(p_0 , x-x_0\right) &= t \left(p_0 , H_p (x_0, p_0)\right)+\int_0^t \big(p_0 , H_p (X(\tau), P(\tau))- H_{p}(x_0, p_0) \big) \, d\tau \\ &\geq t \left(p_0 , H(x_0, p_0)\right) -{\varepsilon t \over 2}\, |h' (|p_0|)| \, |p_0| - 2 \omega (M_4) ( \varepsilon^{-1}+M_4)t \\ &\geq {\varepsilon t \over 2}\, |h' (|p_0|)| \, |p_0| - M_3 t- 2 \omega(M_4) ( \varepsilon^{-1}+M_4)t \, . \end{align} Using the first inequality in \eqref{104} and setting $M_6:=M_3+2 \omega(M_4) ( \varepsilon^{-1}+M_4)$ we complete the proof. \end{proof} \section{Proof of Theorem \ref{t1}} \label{s3} Let \begin{equation}\label{96} \|B\| := \max_{1 \leq i \leq n,\, 1 \leq j \leq m} |(B)_{ij}| \end{equation} for any $B \in \mathbb{R}^{n \times m}$. We say that $R(t)= { O} (r(t))$ \, $t \in {\bf T} \subset \mathbb{R}$, where $R(t) \in \mathbb{R}^{n \times m}$, $r (t) \in \mathbb{R}_{+}$ if \begin{equation}\label{76} \|R (t) \| \leq C r (t) \end{equation} for some constant $C >0$ and all $t \in {\bf T}$. \begin{lemma}\label{l2} Let $c_1, t_1$ be as in Lemma \ref{l7} and the conditions of Theorem \ref{t1} be satisfied. Then for all $t \in (0, t_1)$, $c \in (0, c_1)$, $x_0 \in \mathbb{R}^d$, $ p_0 \in V_{c, t}$ one has \begin{gather} \label{7} {\pa X (s) \over \pa p_0} = s H_{pp} (x_0, p_0 )+ s \gamma { O} (c +t) , \\ \label{12} {\pa P (s) \over \pa p_0} = E_d + { O} (c+t) , \end{gather} where $s \in [0,t]$, $E_d$ is the identity matrix, \begin{equation}\label{97} \gamma=|h'' (|p_0|)|+1 \end{equation} and ${ O}(\cdot)$ is taken uniformly in $x_0, p_0$ and $s$. \end{lemma} \begin{proof} \emph{Step 1.} From the system \eqref{4} we find \begin{equation}\label{9} \ddot{X} (\tau)= A (X(\tau), P(\tau)), \quad A (x, p) = H_{px} H_p-H_{pp} H_x. \end{equation} Here we omit $x, p$ in arguments of the derivatives of $H$, the terms $H_{px} H_p$, $H_{pp} H_x$ are understood as matrix products and $A(x, p) \in \mathbb{R}^d$. Differentiating with respect to $p_0$ the Taylor's developments \begin{equation} \label{18} \begin{gathered} X(s) = x_0+ \dot{X} (0) s + \int_0^s (s-\tau) \, \ddot{X} (\tau) \, d \tau , \\ P(s) = p_0 + \int_0^s \dot{P} (\tau) \, d \tau \end{gathered} \end{equation} and using $\pa \dot{X} (0)/ \pa p_0 = H_{pp} (x_0, p_0 )$ gives \begin{equation}\label{29} {\pa X (s) \over \pa p_0 } = H_{pp} (x_0, p_0 ) s + \int_0^s (s-\tau) \, {\pa \ddot{X} (\tau) \over \pa p_0 } \, d \tau \end{equation} and \begin{equation}\label{31} {\pa P (s) \over \pa p_0} = E_d + \int_0^s {\pa \dot{P} (\tau) \over \pa p_0 } \, d \tau \end{equation} respectively, where \begin{gather}\label{35} {\pa \ddot{X} (\tau) \over \pa p_0 } = A_x (X(\tau), P(\tau)) \, {\pa X (\tau) \over \pa p_0} + A_p (X(\tau), P(\tau))\, {\pa P (\tau) \over \pa p_0} \, , \\ \label{38} {\pa \dot{P} (\tau) \over \pa p_0 } =- H_{xx} (X( \tau), P( \tau)) {\pa X (\tau) \over \pa p_0} -H_{xp} (X( \tau), P( \tau)) {\pa P (\tau) \over \pa p_0} \, . \end{gather} We take $2d \times 2d $ matrix $$ \ell (s, \tau) = \begin{pmatrix} s^{-1}(s-\tau) \tau A_x (X(\tau), P(\tau)) & s^{-1}(s-\tau) \gamma^{-1} A_p (X( \tau), P( \tau)) \\ -\tau \gamma H_{xx} (X( \tau), P(\tau)) &- H_{xp} (X( \tau), P( \tau)) \end{pmatrix} \, , $$ where $\gamma$ is given by \eqref{97}. Define a mapping $L :C([0, t], \mathbb{R}^{ 2 d \times d}) \to C([0, t], \mathbb{R}^{ 2 d \times d}) $ by $$ [L M] (s) := \int_0^s \ell(s, \tau) M(\tau) \, d\tau \, , $$ $M (\cdot) \in C([0, t], \mathbb{R}^{ 2d \times d}) $. We rewrite \eqref{29}-\eqref{31} as $\mathcal{M}=\mathcal{M}_0+L \mathcal{M}$, where $$ \mathcal{M}= \begin{pmatrix}{ {1 \over s} \, {\pa X (s) \over \pa p_0 }} \\ { \gamma \, {\pa P (s) \over \pa p_0 } } \end{pmatrix} \quad \text{and} \quad \mathcal{M}_0= \begin{pmatrix} H_{pp} (x_0, p_0 ) \\ \gamma E_d \end{pmatrix} \, . $$ Using (ii) in Theorem \ref{t1}, we get $\| \mathcal{M}_0 \| =O( \gamma)$. Thus \begin{equation}\label{8} \mathcal{M} = \sum_{n=0}^\infty { L}^n \mathcal{M}_0 = \mathcal{M}_0 + \|{L}\| { O} (\gamma) \end{equation} provided $\|{L}\| <1$. \noindent\emph{Step 2.} We now estimate $\|{L}\|$. Using the elementary formula \begin{equation}\label{77} \|B_1 B_2\| \leq \nu \|B_1\| \|B_2\| \quad \forall B_1 \in \mathbb{R}^{n \times \nu}, B_2 \in \mathbb{R}^{\nu \times m} \end{equation} we obtain $$ \max_{0 \leq s \leq t} \left\| \int_0^s \ell(s, \tau) M(\tau) \, d\tau \right\| \leq s d \max_{0 \leq \tau \leq s \leq t} \| \ell(s, \tau) \| \, \, \max_{0 \leq \tau \leq t} \| M(\tau) \| \,. $$ Hence $$ \|L\| \leq s d \max_{0 \leq \tau \leq s \leq t} \| \ell(s, \tau) \| \, . $$ Since $\tau \leq s$ and $s^{-1}(s-\tau) \leq 1$, it follows that \begin{equation} \label{108} \begin{aligned} \max_{0 \leq \tau \leq s \leq t} \| \ell(s, \tau) \| &\leq s \max_{0 \leq \tau \leq t} \| A_x (X(\tau), P( \tau)) \| + \gamma^{-1} \max_{0 \leq \tau \leq t} \| A_p (X( \tau), P( \tau)) \| \\ &\quad + s \gamma \max_{0 \leq \tau \leq t} \| H_{xx} (X( \tau), P( \tau))\| + \max_{0 \leq \tau \leq t} \| H_{xp} (X( \tau), P( \tau))\| \, . \end{aligned} \end{equation} Applying \eqref{28} and (ii) Theorem \ref{t1} to \eqref{9}, we have \begin{equation} \label{68} \begin{aligned} \|A ( x, p)\| &\leq d \|H_{xp}\|\, \|H_p\|+ d \|H_{pp}\| \, \|H_x\| \\ &\leq d |h' (|p|)|^2+ d \left|h'' (|p|) h' (|p|)\right| \\ &\leq 2 d \, |h' (| p|)|^2 , \end{aligned} \end{equation} provided $|p| \geq M_1 $. Similarly we obtain \begin{gather*} \| A_x ( x, p) \| \leq 4d \, |h' (|p|)|^2 \, , \\ \begin{aligned} \|A_p (x,p)\| &\leq 3 d \big|h'' (|p|) \, h' (|p|)\big| +d \big|h^{\prime \prime \prime} (|p|) \, h (|p|)\big| \\ &\leq 4 d \big|h'' (|p|) \, h' (|p|)\big| \end{aligned} \end{gather*} for $|p| \geq M_1$. Consequently, \begin{gather}\label{56} A_x ( x, p) = (h' (|p|))^2 { O}(1)+{ O} (1), \\ \label{22} A_p (x,p) = \big|h'' (|p|) \, h' (|p|)\big| { O}(1)+{ O} (1) \end{gather} for all $p \in \mathbb{R}^{d}$. Hence $$ s^2 A_x (X (\tau) , P (\tau))= { O}([s h' (|P(\tau)|)]^2) +{ O} (s^2) \, . $$ By \eqref{10}, we get \begin{equation}\label{65} s |h' (|P(\tau)|)| \leq M_4 (s |h' (|p_0|)|+s) \leq M_4 (c+s) \end{equation} and so \begin{equation}\label{19} s^2 A_x (X (\tau) , P (\tau)) = { O}(c^2+t^2) . \end{equation} Since, by \eqref{10}, $\gamma^{-1} |h''(|P(\tau)|)|= O(1)$, then using \eqref{22}, \eqref{65} and $\gamma^{-1} \leq 1$ we find \begin{equation}\label{2} s \gamma^{-1} A_p (X (\tau) , P (\tau)) ={ O} (|sh' (P(\tau)) |) + \gamma^{-1}{ O} (s)= { O} (c+t). \end{equation} Piecing together \eqref{19} and \eqref{2} gives \begin{equation}\label{110} s^2 \max_{0 \leq \tau \leq t} \| A_x (X(\tau), P( \tau)) \| + s \gamma^{-1} \max_{0 \leq \tau \leq t} \| A_p (X( \tau), P( \tau)) \| = O(c+t) \, . \end{equation} Similarly we get \begin{equation}\label{109} s^2 \gamma \max_{0 \leq \tau \leq t} \| H_{xx} (X( \tau), P( \tau))\| + s \max_{0 \leq \tau \leq t} \| H_{xp} (X( \tau), P( \tau))\| = O(c+t) \, . \end{equation} Combining \eqref{108}, \eqref{110}, \eqref{109} we arrive at $s \max_{0 \leq \tau \leq s \leq t} \| \ell(s, \tau) \|=O(c+t) $. Hence $\|{L}\| = O(c+t)$. Consequently we rewrite $\eqref{8}$ as \begin{gather} \label{32} {1 \over s} {\pa X (s) \over \pa p_0 } = H_{pp} (x_0, p_0 ) + \gamma { O} (c +t), \\ \label{36} \gamma {\pa P (s) \over \pa p_0 } = \gamma (E_d + { O} (c +t)). \end{gather} Multiplying \eqref{32} and \eqref{36} by $s$ and $\gamma^{-1}$ respectively we complete the proof. \end{proof} \begin{corollary}\label{c1} Under the conditions of Theorem \ref{t1} there exist $c_2 \in (0,c_1)$, $t_2 \in (0, t_1)$ such that for all $t \in (0, t_2)$, $c \in (0, c_2) $ and $ p_1, p_2 \in V_{c, t}$ one has \begin{equation}\label{70} |X(t, x_0, p_2)-X(t, x_0, p_1)|^2 \geq {t^2 \varepsilon^2 \over 2} \, |p_1-p_2|^2 \, . \end{equation} \end{corollary} \begin{proof} One has \begin{equation} \label{44} \begin{aligned} & \left|X(t,x_0,p_2)-X(t,x_0,p_1) \right|^2 \\ &= \int_{0}^1 \int_{0}^1 {\left(p_2-p_1\right)}^T {\Big({\pa X \over \pa p_0 } (t,x_0, p_1+s(p_2-p_1)) \Big)}^T \\ &\quad \times \Big({\pa X \over \pa p_0 } (t,x_0, p_1 + \tau (p_2-p_1))\Big) \left(p_2-p_1\right) \, d\tau \,ds \,, \end{aligned} \end{equation} where $({\pa X /\pa p_0})^T$ is transposed matrix to ${\pa X /\pa p_0}$. We denote by $$ q_1=p_1+s (p_2-p_1) \, , \quad q_2=p_1+\tau (p_2-p_1) \, .$$ From \eqref{7}, we obtain \begin{equation} \label{69} \begin{aligned} \Big({\pa X (t, x_0, q_1) \over \pa p_0}\Big)^T {\pa X (t, x_0, q_2) \over \pa p_0} &= t^2( H_{pp} (x_0,q_1)+ \gamma_1 R_1) ( H_{pp} (x_0,q_2)+ \gamma_2 R_2)\\ &=: t^2 H_{pp} (x_0,q_1) H_{pp} (x_0,q_2)+t^2 {R}, \end{aligned} \end{equation} where $\gamma_i=|h'' (|q_i|)|+\varepsilon^{-1}$, $i=1, 2$ and $\|R_i\|\leq (c+t) \kappa$ for some constant $\kappa > 1$. Using \eqref{77} we get $$ \| {R} \| \leq d\left( \gamma_1 \|H_{pp} (x_0,q_2)\|\, \|R_1\|+ \gamma_2 \|H_{pp} (x_0,q_1)\| \, \|R_2\| +\gamma_1 \gamma_2 \, \|R_1\| \, \|R_2\|\right) $$ Let us take $c_21$, $c+t0$ is from the conditions of Theorem \ref{t1}. \par Let us take \begin{gather}\label{78} M_7:=\max_{|p| \leq M_1} \left[\left( \varepsilon^{-1}+|h'(|p|)|\right)^2 + \left( \varepsilon^{-1}+|h''(|p|)|\right) \left( \varepsilon^{-1}+|h(|p|)|\right) \right] \, , \\ \rho:=\min\big\{ {\varepsilon \over 8 d (4 M_4^2+ M_7) }\, , \, \, c_2 \big\} \, , \\ \label{91} T := \min\big\{ {\varepsilon \rho \over 4 M_4 (|h' (0)|^2+1)} \, , \, \, {\varepsilon \rho \over 2 M_2 } \, \, , \rho, t_2 \big\} \, , \\ r:= {\varepsilon \rho \over 2} \, . \end{gather} Let $W_t:=V_{\rho, t}$ (recall \eqref{6}) and \begin{equation}\label{30} \mathfrak{D}: p_0 \to X(t, x_0, p_0) , \quad \mathfrak{D}: W_t \to \mathbb{R}^d \end{equation} \emph{Step 1.} We will show that $B_r (x_0)\subset \mathfrak{D} (W_t)$. Using \eqref{9}, \eqref{18}, (iii) in Lemma \ref{l6} and the fact that $\dot{X} (0)=H_p (x_0,p_0) $, we have \begin{equation} \label{58} \begin{aligned} |X(t) -x_0| &\geq t |\dot{X} (0)|- \int_0^t (t-\tau) \,| \ddot{X} (\tau) | \, d \tau \\ &\geq t | H_p (x_0, p_0)| - \int_0^t (t-\tau) \, |A (X(\tau), P(\tau) ) | \, d \tau \\ &\geq \varepsilon t |h' (|p_0|)|-t M_2 - t^2 \max_{ 0 \leq \tau \leq t} \, |A (X(\tau), P(\tau) ) |\, . \end{aligned} \end{equation} Due to \eqref{78}, \begin{align*} \|A ( x, p)\| &\leq d \|H_{xp}\|\, \|H_p\|+ d \|H_{pp}\| \, \|H_x\| \\ &\leq d \, \big[\left( \varepsilon^{-1}+|h'(|p|)|\right)^2+ \left( \varepsilon^{-1}+|h''(|p|)|\right) \left( \varepsilon^{-1}+|h(|p|)|\right) \big]\\ &\leq d M_7 \end{align*} for any $x \in \mathbb{R}^d$, $|p| \leq M_1$. Piecing together this and \eqref{68} we obtain $$ \|A ( x, p)\| \leq 2 d \, |h' (| p|)|^2 +dM_7 $$ for all $x, p \in \mathbb{R}^n$. This and \eqref{10} imply \begin{equation} \label{67} \begin{aligned} |A (X(\tau), P(\tau) ) | &\leq 2 d \, M_4^2 (|h' (| p_0|)|+1)^2 +dM_7 \\ &\leq 4 d \, M_4^2 (|h' (| p_0|)|^2+1) +d M_7 \\ &\leq d \, (4 M_4^2 +M_7)(|h' (| p_0|)|^2+1)\, . \end{aligned} \end{equation} Substituting \eqref{67} into \eqref{58} and using $2 t M_2 <\varepsilon \rho$, \, $t |h' (|p_0|)|=\rho$ we obtain \begin{align*} |X(t) -x_0| &\geq \varepsilon t |h' (|p_0|)|-t M_2 -t^2 d (4 M_4^2+ M_7) (|h' (|p_0|)|^2+1) \\ &\geq {\varepsilon \rho \over 2} - d (4 M_4^2+ M_7) (\rho^2+t^2)\, \end{align*} for $p_0 \in \pa W_t$, $x_0 \in \mathbb{R}^d$. Since $t < T < \rho $ and $ \varepsilon > 8 d (4 M_4^2+ M_7) \rho $, it follows that $$ |X(t) -x_0| > {\varepsilon \rho \over 2} - 2 d (4 M_4^2+ M_7) \rho^2 > {\varepsilon \rho \over 4} \, \, $$ or \begin{equation}\label{98} \min_{p_0 \in \pa W_t} |X(t, x_0, p_0) -x_0| > r \end{equation} for any $x_0 \in \mathbb{R}^d$. On the other hand, using \eqref{55} with $p_0=0$ and \eqref{91} we have \begin{equation}\label{99} |X(t, x_0, 0)-x_0| \leq t M_4 (|h'(0)|+1) < T M_4 (|h'(0)|+1) < {\varepsilon \rho \over 4} = r \, . \end{equation} Combining \eqref{98}, \eqref{99} shows that $ x_0 \in \mathfrak{D} (W_t)$. Moreover, $B_r (x_0) \subset \mathfrak{D} ( W_t)$. \par \emph{Step 2.} Corollary \ref{c1} implies that $\mathfrak{D}$ is injective. It suffices to notice that, by \eqref{7} and (i) Theorem \ref{t1}, ${\pa X /\pa p_0} \not = 0$ and so the implicit function theorem implies that $\mathfrak{D}$ is a local diffeomorphism for any $t < T$. \end{proof} \section{Proof of Lemma \ref{l3}}\label{s4} For all $x_0, x \in B_r(x_0)$ and $t 0$. \end{lemma} \begin{proof} Using \eqref{70} with $p_1=0$ and $p_2 =p_0 (t, x_0, x_0)$ gives \begin{equation} \label{57} \begin{aligned} |X(t,x_0,0)- x_0 |^2 &=|X(t,x_0,0)- X(t,x_0,p_0 (t, x_0, x_0))|^2 \\ &\geq {t^2 \varepsilon^2 \over 2} \, {|p_0 (t, x_0, x_0)|}^2 \, . \end{aligned} \end{equation} Due to \eqref{55}, $$ \left|X(t,x_0, 0)- x_0 \right|^2 \leq t^2 M_4^2 \big(1+|h' (|0|)|\big)^2 \, . $$ Combining this and \eqref{57} gives \begin{equation}\label{54} |p_0 (t, x_0, x_0)|^2 \leq {2 \over \varepsilon^2}\, M_4^2 \big(1+|h' (|0|)|\big)^2 . \end{equation} On the other hand, an application of (ii) in Theorem \ref{t1} and \eqref{26} to \eqref{11} shows that $\sigma(t,x_0, p_0 )$ is bounded if $|p_0|$ is bounded. This remark together with \eqref{54} completes the proof. \end{proof} \begin{proof}[Proof of Lemma \ref{l3}] Let us choose $M_9 >0 $ such that \begin{gather}\label{23} {M_9 \over M_4}-M_5 \geq |h' (M_1)|, \\ \label{5} M_{10}:= {\varepsilon \over 2}\, \omega \big({M_9 \over M_4 }- M_5 \big) \big({M_9 \over M_4}-M_5 \big)-M_6> 0 \, . \end{gather} We take $t \in (0, T)$, $y, y_0 \in \mathbb{R}^d$ such that $|y-y_0|=r$. Using $$ {\pa S (t, y, y_0) \over \pa y_0} =-p_0(t, y ,y_0) $$ and applying the elementary formula $\phi(1)=\phi(0)+\int_0^1 \phi' (\tau)\, d\tau$ to \[ \phi(\tau) := S(t, y,y+\tau(y_0-y)), \] $\tau \in [0,1]$ we get \begin{equation} \label{63} \begin{aligned} S(t, y, y_0) &=S(t,y,y)+\int_0^1 \left(p_0 (t,y,y+\tau(y_0-y)), y-y_0\right) \, d \tau\\ &=S(t,y,y)+\int_0^1 \left( p_0^\tau , y-\xi_\tau \right) \, {d \tau \over \tau} \, , \end{aligned} \end{equation} where $\xi_\tau=y+\tau(y_0-y)$ and $p_0^\tau=p_0 (t,y,\xi_\tau)$. By (i) and (iii) in Lemma \ref{l7}, we have \begin{gather} \label{24} |p^\tau_0| \geq \omega \big( {r\tau \over t M_4} - M_5\big) \quad \text{provided that } {r\tau \over t M_4} - M_5>M_4 \, , \\ \label{107} \left( p_0^\tau , y-\xi_\tau \right) \geq {\varepsilon \over 2}\, |p^\tau_0| \big({r \tau \over M_4}-t M_5 \big)-t M_6 \, , \end{gather} where we have used $|y-\xi_\tau |=\tau r$. Setting $s =t M_9 / r$ and using \eqref{24}-\eqref{107} we get \begin{align*} \left( p_0^\tau , y-\xi_\tau \right) &\geq {\varepsilon \over 2}\, |p^\tau_0| \big(t \, {M_9 \over M_4}-t M_5 \big)-t M_6 > t M_{10} > 0 \quad \text{for } \tau \in [s, 1] \, , \end{align*} where we have used $\tau r \geq t M_9 $. Notice that due to \eqref{23} the condition in \eqref{24} is satisfied. If we assume that $t \leq r /(2M_9)$, then $s \leq 1/2$ and so \begin{equation}\label{20} \int_0^1 \left( p_0^\tau , y-\xi_\tau \right) \, {d \tau \over \tau} \geq \int_{1/2}^1 -\Big|\int^{s}_0 \Big| \, . \end{equation} Using again \eqref{107} and, due to (i) Lemma \ref{l7}, $$ |p_0^\tau| \geq \omega \big({\tau r \over t M_4 }- M_5 \big) $$ we obtain \begin{align}\label{102} ( p_0^\tau , y-\xi_\tau) &\geq {\varepsilon \over 2}|p^\tau_0| \big({r \over 2 M_4}-t M_5 \big) -t M_6 \\ &\geq {\varepsilon \over 2} \omega \big({r \over 2 t M_4 }- M_5 \big) \big({r \over 2 M_4}-t M_5 \big)-t M_6 \quad \text{for} \, \, \tau \in [{1 \over 2}, 1] \, . \end{align} On the other hand, due to (ii) Lemma \ref{l7} for $0 \leq \tau \leq s$, we have $$ |p^\tau_0| \leq \omega \big( {2 \over \varepsilon} \, {r \tau \over t} +M_5\big) \leq \omega \big({2 \over \varepsilon } \, M_9+M_5 \big) =:M_{11} $$ and so \begin{equation}\label{103} \Big| \int_0^s \left( p_0^\tau , y-\xi_\tau \right) \, {d \tau \over \tau} \Big| \leq r \int_0^s |p_0^\tau| \, d \tau \leq r M_{11} \, . \end{equation} Combining \eqref{63}, \eqref{20}, \eqref{102}, \eqref{103} we arrive at \begin{equation} \label{3} \begin{aligned} &\min_{|y-y_0|=r} S(t, y, y_0) \\ &\geq {-M_8+ \Big[ {\varepsilon \over 2} \omega \big({r \over 2 t M_4 } -M_5 \big) \big({r \over 2 M_4}-t M_5 \big)- t M_6 \Big] \ln 2 - r M_{11} } \end{aligned} \end{equation} for $t < r/(2M_9)$. \noindent\emph{Step 2.} We choose $r_1