%From piccoli@tonelli.ma.sissa.it Tue Nov 17 09:38:26 1998

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  This is a plain TeX file, containing a REVISED VERSION of the paper
%
%    REGULAR SYNTHESIS AND SUFFICIENCY CONDITIONS FOR OPTIMALITY
%
%            by Benedetto Piccoli and Hector J. Sussmann 
%
%          The paper has been accepted for publication by the 
%                       SIAM Journal on Control 
%
%             The date of this revised version is October 17, 1998
%
%    This version is still a DRAFT:  DO NOT QUOTE WITHOUT PERMISSION
%
%            THIS VERSION SUPERSEDES ALL PREVIOUS VERSIONS 
%
%   This file actually contains TWO versions of the manuscript, one 
%   long and the other one short.  The long version is 50 pages long.
%   The short version is 40 pages long.
%  
%   As it is, the file will produce the LONG version. If you want to 
%   produce the SHORT version,  all you need to do is activate the third
%   line of the file (the one that ends with %SHORT)  by deleting the %
%   sign at the beginning of the line.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\def\myrunninghead{Regular synthesis}
\def\thegoodheadline{{\it Benedetto Piccoli and H\'ector J. Sussmann\hfill\myrunninghead}\hfill{\bf\folio}}
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{\mymagonebf
\pltwocen{Benedetto Piccoli}{H\'ector J. Sussmann$^{*}$}
}
\vskip 0.2true in
\pltwocen{SISSA--ISAS}{Department of Mathematics, Rutgers University}

\pltwocen{{\it via Beirut 2-4}}{{\it 110 Frelinghuysen Rd.}}
\pltwocen{{\it 34014 Trieste, Italy}}{{\it Piscataway, NJ 08854, USA}}
\vskip 0.2true in

\pltwocen{{\it piccoli@sissa.it}}{{\it sussmann@hilbert.rutgers.edu}}

\pltwocen{$\,\,\,$}{http://www.math.rutgers.edu/$\tilde{\,\,}$sussmann}
}
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 \newcount\footno%the footnote number
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%it's 0 while writing a letter and 1 for endfile and labels
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\def\thephoneson{Phone: (908)445-3072}
\def\theemail{sussmann@hamilton.rutgers.edu\enspace---
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\def\sighjsneru{\pltwo{Mahesh Nerurkar}{H. J. Sussmann}}
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\pltwo{\hbox to 5true cm{\hrulefill}}{\hbox to 5true cm{\hrulefill}}
\sighjswithmcclam\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
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\def\thedefaultps{ }
%END OF MACROS FOR LETTER
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\def\clearsubsectno{\global\subsectno=0}  
\def\cleartheono{\global\theono=1}
\def\cleartheotno{\global\theotno=1}
\def\cleartheononosec{\global\theononosec=1}
\def\clearlemmno{\global\lemmno=1}
\def\clearcorono{\global\corono=1}
\def\clearpropno{\global\propno=1}
\def\clearequano{\global\equano=1}
\def\clearequanonosec{\global\equanonosec=1}
\def\clearfiguno{\global\figuno=1}
\def\figcap#1#2{\medbreak
\centerline{ Figure {#1}:\ #2}
\vbox to 0.3true in{\vfill}}
\def\mybigbreak{\vskip 0.3true in}
\def\mynotsobigbreak{\vskip 0.2true in}
\def\equanosectno{\the\sectno .\the\equano}
\def\equanoapno{{\rm \appenlet}.\the\equano}
\def\figunosectno{\the\sectno .\the\figuno}
\def\adequano{\global\advance\equano by 1} 
\def\adequanonosec{\global\advance\equanonosec by 1} 
\def\adfiguno{\global\advance\figuno by 1} 
\def\numeq{\eqno(\equanosectno\adequano)}
\def\numeqnosec{\eqno(\the\equanonosec\adequanonosec)}
\def\numeqa{\eqno({\rm A}.\the\equano)\adequano}
\def\numeqap{\eqno({\rm \appenlet}.\the\equano)\adequano}
\def\numeqb{\eqno({\rm B}.\the\equano)\adequano}
\def\numeqalign{(\equanosectno)\adequano}
\def\numeqnoa{(\equanosectno.{\rm a})}
\def\numeqnob{(\equanosectno.{\rm b})}
\def\numeqnoc{(\equanosectno.{\rm c})}
\def\numeqcons{\eqno(\the\equano)\adequano}
%use \numeq for displayed eqs. (Includes when you are in \eqalign, but
%NOT \eqalignno.)  Use \numeqalign within \eqalignno.
\def\clearallcounters{\clearequano \cleartheono \clearlemmno \clearcorono 
    \clearfiguno \clearsubsectno}
\def\section#1\par{\global\advance\sectno by 1
    \bigbreak \noindent {\bf \S \the\sectno
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    \bigbreak \noindent {\bf \the\sectno.\the\subsectno
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\def\sectionnopar#1\par{\global\advance\sectno by 1
    \bigbreak \noindent {\bf \the\sectno
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    \bigbreak \noindent {\mymagonebf \the\sectno
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\def\appen#1#2\par{
    \bigbreak \noindent {\bf Appendix #1
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\def\nitem{{\global\advance\myitemno by 1}\item{\the\myitemno .}}
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\def\theoap#1\par{\medbreak \noindent {\bf Theorem \enspace 
   {\rm \appenlet}.\the\theono}. \enspace 
   {\sl#1}\medbreak \global\advance\theono by 1} 
\def\theot#1\par{\medbreak \noindent {\bf Theorem \enspace 
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   {\sl#1}\medbreak \global\advance\theotno by 1} 
\def\theonosec#1\par{\medbreak \noindent {\bf Theorem\enspace 
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   {\sl#1}\medbreak \global\advance\theononosec by 1} 
\def\propnosec#1\par{\medbreak \noindent {\bf Proposition\enspace 
   \the\theononosec}. \enspace 
   {\sl#1}\medbreak \global\advance\theononosec by 1} 
\def\lemm#1\par{\medbreak \noindent {\bf Lemma \enspace 
   \the\sectno .\the\lemmno}.\enspace 
   {\sl#1}\medbreak \global\advance\lemmno by 1} 
\def\lemmap#1\par{\medbreak \noindent {\bf Lemma \enspace 
   {\rm \appenlet}.\the\theono}.\enspace 
   {\sl#1}\medbreak \global\advance\theono by 1} 
\def\coro#1\par{\medbreak \noindent {\bf Corollary \enspace 
   \the\sectno .\the\corono}. \enspace 
   {\sl#1}\medbreak \global\advance\corono by 1} 
\def\prop#1\par{\medbreak \noindent {\bf Proposition \enspace 
   \the\sectno .\the\propno}. \enspace 
   {\sl#1}\medbreak \global\advance\propno by 1} 
\def\lemmtheo#1\par{\medbreak \noindent {\bf Lemma \enspace 
   \the\sectno .\the\theono}.\enspace 
   {\sl#1}\medbreak \global\advance\theono by 1} 
\def\corotheo#1\par{\medbreak \noindent {\bf Corollary \enspace 
   \the\sectno .\the\theono}. \enspace 
   {\sl#1}\medbreak \global\advance\theono by 1} 
\def\proptheo#1\par{\medbreak \noindent {\bf Proposition \enspace 
   \the\sectno .\the\theono}. \enspace 
   {\sl#1}\medbreak \global\advance\theono by 1} 
\def\defi#1\par{\medbreak \noindent {\bf Definition \enspace 
   \the\sectno .\the\theono}. \enspace 
   {#1}\medbreak \global\advance\theono by 1} 
\def\defid#1\par{\medbreak \noindent {\bf Definition \enspace 
    D\the\defidno}. \enspace 
   {#1}\medbreak \global\advance\defidno by 1} 
\def\rema#1\par{\medbreak \noindent {\bf Remark \enspace 
   \the\sectno .\the\theono}. \enspace 
   {#1}\medbreak \global\advance\theono by 1} 
\def\remap#1\par{\medbreak \noindent {\bf Remark \enspace 
   \appenlet .\the\remano}. \enspace 
   {#1}\medbreak \global\advance\remano by 1} 
\def\exam#1\par{\medbreak \noindent {\bf Example \enspace 
   \the\sectno .\the\theono}. \enspace 
   {#1}\medbreak \global\advance\theono by 1} 
\def\theoieeeinfo#1\par{\medbreak {\it Theorem \enspace 
   \the\theono:} \enspace 
   {#1}\medbreak \global\advance\theono by 1} 
\def\lemmieeeinfo#1\par{\medbreak {\it Lemma \enspace 
   \the\lemmno:}\enspace 
   {#1}\medbreak \global\advance\lemmno by 1} 
\def\coroieeeinfo#1\par{\medbreak {\it Corollary \enspace 
   \the\corono:} \enspace 
   {#1}\medbreak \global\advance\corono by 1} 
\def\remaieeeinfo#1\par{\medbreak {\it Remark:} \enspace 
   {#1}\medbreak }
\def\exam#1\par{\medbreak \noindent {\bf Example \enspace 
   \the\sectno .\the\theono}. \enspace 
   {#1}\medbreak \global\advance\theono by 1}
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\def\lint{\enspace\leftarrow\!\!\!\!\!\int}
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\def\sre{S_{\rho}(\eta)}
\def\vari{E=\{\eta_{\varepsilon}:\varepsilon\in C\}}
\def\varii{E_{i}=\{\eta_{i,\varepsilon_{i}}:\varepsilon_{i}\in C_{i}\}}
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\def\ckad{$C^{k}$-admissible}
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\def\rkveoet{\rho_{{\cal V},\eta_{1},\eta_{2}}^{k}}
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\def\deloveoet{\delta_{\cal V}^{1}(\eta_{1},\eta_{2})}
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\def\domet{{\rm Dom}(\eta_{2})}

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\newcount\bibno
\bibno=0

\global\advance\bibno by 1
\xdef\bardicount{[\the\bibno]}
\xdef\bardi{\vbox{\medbreak\item{\the\bibno.}
{Bardi, M., and I. Capuzzo-Dolcetta, ``Optimal control and viscosity
solutions of Hamilton-Jacobi-Bellman equation,''
Birk$\ddot{\rm a}$user Boston, Inc., Boston MA, 1997. 
}}}

\global\advance\bibno by 1
\xdef\boltcount{[\the\bibno]}
\xdef\bolt{\vbox{\medbreak\item{\the\bibno.}
{Boltyanskii, V. G., ``Sufficient conditions for optimality
and the justification of the dynamic programming principle,''
{\it SIAM J. Control Optim.} {\bf 4} (1966), pp. 326-361.
}}}

\global\advance\bibno by 1
\xdef\piccccount{[\the\bibno]}
\xdef\piccc{\vbox{\medbreak\item{\the\bibno.}
{Bressan, A., and B. Piccoli, ``Structural stability for time-optimal planar
syntheses,'' {\it Dynam. Contin. Discrete Impuls.  Systems} {\bf 3} (1997),
pp.~335--371.
}}}

\global\advance\bibno by 1
\xdef\piccdcount{[\the\bibno]}
\xdef\piccd{\vbox{\medbreak\item{\the\bibno.}
{Bressan, A., and B. Piccoli, ``A generic classification of time optimal
planar stabilizing feedbacks,'' {\it SIAM J. Control Optim.} {\bf 36}
(1998), no. 1, pp.~12-32. 
}}}

\global\advance\bibno by 1
\xdef\brunacount{[\the\bibno]}
\xdef\bruna{\vbox{\medbreak\item{\the\bibno.}
{Brunovsk\'y, P., ``Every normal linear system has a regular
time--optimal synthesis,'' {\it Math. Slovaca} {\bf 28} (1978),
pp. 81-100.
}}}

\global\advance\bibno by 1
\xdef\brunbcount{[\the\bibno]}
\xdef\brunb{\vbox{\medbreak\item{\the\bibno.}
{Brunovsk\'y, P., ``Existence of regular syntheses for general
problems,'' {\it J. Diff. Equations} {\bf 38} (1980),
pp. 317-343.
}}}

\global\advance\bibno by 1
\xdef\cesaricount{[\the\bibno]}
\xdef\cesari{\vbox{\medbreak\item{\the\bibno.}
{Cesari, L., ``Optimization -- Theory and applications,''
Springer--Verlag, New York, 1983.
}}}

\global\advance\bibno by 1
\xdef\sontagcount{[\the\bibno]}
\xdef\sontag{\vbox{\medbreak\item{\the\bibno.}
Clarke, F.~H., Yu.~S.~Ledyaev, A.~I.~Subbotin and E.~D.~Sontag,
``Asymptotic controllability implies feedback stabilization,''
{\it IEEE Trans. Autom. Control} {\bf 42} (1997), pp.~1394-1407.
}}


\longversion{
\global\advance\bibno by 1
\xdef\fedcount{[\the\bibno]}
\xdef\fed{\vbox{\medbreak\item{\the\bibno.}
{Federer, H., ``Geometric measure theory,'' Springer, Berlin, 1969.
}}}
}

\global\advance\bibno by 1
\xdef\filicount{[\the\bibno]}
\xdef\fili{\vbox{\medbreak\item{\the\bibno.}
{Filippov, A. F., ``Differential equations with discontinuous
right-hand side,'' Reidel, 1989.
}}}

\global\advance\bibno by 1
\xdef\flemcount{[\the\bibno]}
\xdef\flem{\vbox{\medbreak\item{\the\bibno.}
{Fleming, W.H., and  R. W. Rishel, ``Deterministic and stochastic
optimal control,'' Springer--Verlag, New York, 1975.
}}}

\global\advance\bibno by 1
\xdef\flemsonercount{[\the\bibno]}
\xdef\flemsoner{\vbox{\medbreak\item{\the\bibno.}
{Fleming, W.H., and  M. Soner, ``Controlled Markov processes and
viscosity solutions,'' Springer--Verlag, New York, 1993.
}}}


\global\advance\bibno by 1
\xdef\fullcount{[\the\bibno]}
\xdef\full{\vbox{\medbreak\item{\the\bibno.}
{Fuller, A. T., ``Relay control systems optimized for various
performance criteria,'' Automatic and remote control, Proc. first
world congress IFAC Moscow, vol. 1, Butterworths, 1961, pp. 510-519.
}}}

\longversion{
\global\advance\bibno by 1
\xdef\lojacount{[\the\bibno]}
\xdef\loja{\vbox{\medbreak\item{\the\bibno.}
{Lojasiewicz Jr., S., and H. J. Sussmann, 
``Some examples of reachable sets and
optimal cost functions that fail to be subanalytic,'' {\it SIAM J.
Control Optim.} {\bf 23} (1985), no. 4, pp. 584-598. 
}}}
}

\global\advance\bibno by 1
\xdef\marchalcount{[\the\bibno]}
\xdef\marchal{\vbox{\medbreak\item{\the\bibno.}
{Marchal, C., ``Chattering arcs and chattering controls,''
{\it J. Optim Theory Appl.} {\bf 11} (1973), pp. 441-468.
}}}

\global\advance\bibno by 1
\xdef\piccacount{[\the\bibno]}
\xdef\picca{\vbox{\medbreak\item{\the\bibno.}
{Piccoli, B., ``Regular time-optimal syntheses 
for smooth planar systems,'' {\it
Rend. Sem. Mat. Univ. Padova} {\bf 95} (1996), pp. 59-79.
}}}

\global\advance\bibno by 1
\xdef\piccbcount{[\the\bibno]}
\xdef\piccb{\vbox{\medbreak\item{\the\bibno.}
{Piccoli, B., ``Classification of generic singularities 
for the planar time optimal syntheses,'' {\it SIAM J.
Control Optim.} {\bf 34} (1996), no. 6, pp.~1914-1946. 
}}}

\global\advance\bibno by 1
\xdef\sussacount{[\the\bibno]}
\xdef\sussa{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J.,  ``Subanalytic sets and feedback control,'' {\it J. Diff.
Equations} {\bf 31} (1979), no. 1, pp. 31-52.
}}}

\global\advance\bibno by 1
\xdef\sussbcount{[\the\bibno]}
\xdef\sussb{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J.,  
``Piecewise analyticity of optimal cost functions and optimal
feedback,'' in {\it Proc. Joint Automatic Control Conference}, Denver,
Colo. (August 1979).  
}}}

\global\advance\bibno by 1
\xdef\sussccount{[\the\bibno]}
\xdef\sussc{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``Analytic stratifications and control theory,'' in {\it Proc. 1978
Int. Congress of Mathematicians}, Helsinki (1980), pp. 865-871.
}}}

\global\advance\bibno by 1
\xdef\sussdcount{[\the\bibno]}
\xdef\sussd{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Time-optimal control in the plane,'' in {\it Feedback
Control of Linear and Nonlinear Systems}, D. Hinrichsen and A. Isidori
Eds., Springer-Verlag, New York,  1982, pp. 244-260.    
}}}

\global\advance\bibno by 1
\xdef\sussecount{[\the\bibno]}
\xdef\susse{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``Subanalytic sets and optimal control in the plane,'' {\it Proc.
21st I.E.E.E. Conference on Decision and Control}, Orlando, Fl, (Dec.
1982), pp. 295-299. 
}}}

\global\advance\bibno by 1
\xdef\sussfcount{[\the\bibno]}
\xdef\sussf{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Lie brackets, real analyticity and geometric control
theory,'' in {\it Differential Geometric Control Theory}, R.~W. Brockett,
R.~S. Millman and H.J. Sussmann Eds., Birkh\"auser Boston Inc. (1983),
pp. 1-115.  
}}}

\longversion{
\global\advance\bibno by 1
\xdef\sussgcount{[\the\bibno]}
\xdef\sussg{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``Lie Brackets and real analyticity in control theory,'' in
{\it Mathematical Control Theory}, C. Olech
Ed., Banach Center Publications, Volume 14, PWN-Polish Scientific
Publishers, Warsaw, Poland, 1985, pp. 515-542.
}}}
}


\global\advance\bibno by 1
\xdef\susshcount{[\the\bibno]}
\xdef\sussh{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``The structure of time-optimal trajectories for single-input
systems in the plane : the $C^{\infty}$ nonsingular
case,'' {\it SIAM J. Control Optim.} {\bf 25} (1987), no. 2,
pp. 433-465. 
}}}

\global\advance\bibno by 1
\xdef\sussicount{[\the\bibno]}
\xdef\sussi{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``The structure of time-optimal trajectories for single-input
systems in the plane: the general real-analytic case,'' {\it  SIAM J.
Control Optim.} {\bf 25} (1987), no. 4, pp. 868-904.
}}}

\global\advance\bibno by 1
\xdef\susslcount{[\the\bibno]}
\xdef\sussl{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., 
``Regular synthesis for time-optimal control of single-input
real-analytic systems in the plane,'' {\it SIAM J. Control 
Optim.} {\bf 25} (1987), no. 5, pp. 1145-1162.
}}}

\global\advance\bibno by 1
\xdef\sussncount{[\the\bibno]}
\xdef\sussn{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Recent developments in the regularity theory of optimal
trajectories,'' in Linear and nonlinear mathematical control 
theory, {\it Rend. Sem. Mat. Univ. e Pol. Torino}, Fascicolo speciale, 1987,
pp.~149-182.
}}}

\longversion{
\global\advance\bibno by 1
\xdef\sussenvacount{[\the\bibno]}
\xdef\sussenva{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Envelopes,  high-order optimality conditions and
Lie brackets,'' in {\it Proc. 28th I.E.E.E. Conference on Decision and
Control}, Tampa, Fl, (Dec. 1989), pp.~1107-1112.
}}}
}

\global\advance\bibno by 1
\xdef\sussenvbcount{[\the\bibno]}
\xdef\sussenvb{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Envelopes, conjugate points and optimal bang-bang
extremals,'' in {\it Algebraic and Geometric Methods in Nonlinear Control
Theory}, M. Fliess and M. Hazewinkel Eds., Math. Appl., 29,
D.~Reidel Publishing Co., Dordrecht-Boston, Mass., 1986, pp.~325-346.
}}}

\global\advance\bibno by 1
\xdef\sussmcount{[\the\bibno]}
\xdef\sussm{\vbox{\medbreak\item{\the\bibno.}
{Sussmann, H. J., ``Synthesis, presynthesis, sufficient conditions for
optimality and subanalytic sets,'' in Nonlinear controllability and
optimal control, H. J. Sussmann ed., Marcel Dekker, New York, 1990,
pp. 1-19.
}}}

\global\advance\bibno by 1
\xdef\zelikincount{[\the\bibno]}
\xdef\zelikin{\vbox{\medbreak\item{\the\bibno.}
{Zelikin, M. I., and V. F. Borisov, 
{\it Theory of Chattering Control, with applications to 
astronautics, robotics, economics and engineering}, 
Birkh\"auser, Boston, 1994.}}}

\pageno=0
\null

\vskip 2em

\centerline{\bf REGULAR SYNTHESIS} 
\centerline{\bf AND SUFFICIENCY CONDITIONS FOR OPTIMALITY}

\vskip 4em

\picsus

\vfill\vfill

\centerline{\sl October 1998}

\vfill\vfill
\centerline{\bf ABSTRACT}

\vskip 1em

\noindent
We propose a definition of ``regular synthesis,'' more general than those
suggested by other authors such as Boltyanskii and Brunovsk\'y, and an even
more general notion of ``regular presynthesis.'' We give a complete proof
of the corresponding sufficiency theorem, a slightly weaker version of
which had been stated in an earlier article, with only a rough outline of
the proof.  We illustrate the strength of our result by showing that the
optimal synthesis for the famous Fuller problem satisfies our hypotheses.
We also compare our concept of synthesis with the simpler notion of a
``family of solutions of the closed-loop equation arising from an optimal
feedback law,'' and show by means of examples why the latter is inadequate,
and why the difficulty cannot be resolved by using other concepts of
solution---such as Filippov solutions, or the limits of sample-and-hold
solutions recently proposed as feedback solutions by Clarke, Ledyaev,
Subbotin and Sontag---for equations with a non-Lipschitz and possibly
discontinuous right-hand side.

\vskip 1.2em

\noindent
{\bf Keywords.} Optimal control, regular synthesis, sufficient conditions.

\vskip 2.5em
\centerline{Ref. SISSA 68/96/M}

\vskip 1em
\vfill\vfill\vfill
{\baselineskip=8pt

\noindent{\mysmall ${\,}^{*}$\ \ Supported in part by 
NSF Grant DMS95-00798 and AFOSR Grant 0923.  The final part of the
work was carried out between January and June 1998, while this author
was a visitor at the Institut Henri Poincar\'e (IHP), in Paris. He is
grateful to the IHP for its kind hospitality during this period.
}

}

\vfill\eject

\pageno=0


\null

\vfill\eject

\shortversion{\baselineskip=11.8pt}

\parnormal


\section{Introduction}

\noindent The purpose of the theory of ``regular synthesis'' is to turn into a
rigorous, precise theorem the vague assertion that ``if a collection
$\Gamma$ of extremals covers the whole space in a smooth enough way, then
all the members of $\Gamma$ are optimal.'' Naturally, a good definition
should at the same time be stringent enough to lead to a correct proof of
the sufficiency assertion, and broad enough to cover as large as possible a
class of optimal control problems. In particular, it would be desirable for
the theory to apply to well-known examples such as the famous ``Fuller's
problem'' (cf. Fuller \fullcount{}, Marchal \marchalcount{},
Zelikin-Borisov \zelikincount{}), where the optimal controls have infinitely
many switchings.

The three main goals of this paper are (a)~to propose a new definition of
regular synthesis, more general than those suggested by other authors
(e.~g.  Boltyanskii
\boltcount, Brunovsk\'y \brunacount, \brunbcount), as well as an even more
general notion of ``regular presynthesis,'' (b)~to give a detailed
statement and proof of the corresponding sufficiency theorem, a slightly
weaker version of which had been announced by one of us in an earlier
article (Sussmann \sussmcount), with only a rough sketch of the proof, and
(c)~to compare our definition with other concepts of synthesis such as
those of Boltyanskii and Brunovsk\'y, and with the simpler notion of a
``family of solutions of the closed-loop equation arising from an optimal
feedback law.''

Our main sufficiency result is Theorem 2.13.  The key to the proof turns
out to be a technical result---Theorem 2.14---on the relation between the
adjoint vector of the Maximum Principle and the gradient of the
cost function.  This in turn depends on results about differentiability
of trajectories with respect to a parameter under weak differentiability
assumptions on the dependence of the data on the parameter.  These results
are of interest in themselves and are presented separately in Appendix A.

Our definitions of regular presynthesis and regular synthesis are explained in
\S 2, cf. Definition 2.3 for the notion of ``presynthesis,'' 2.4 for that
of ``synthesis,'' and 2.12 for ``regularity.''  The strength of Theorem
2.13 is illustrated in \S 3 by showing that it applies to the synthesis for
Fuller's problem, where other sufficiency theorems such as those of
\boltcount{} and \brunbcount{} cannot be used because of the technical
problem arising from the occurrence of an infinite number of switchings.
(Not surprisingly, the optimality of Fuller's synthesis was originally
proved without resorting to regular synthesis arguments, and using instead
special symmetry properties of the problem, cf., e.~g., Fuller
\fullcount{}, Marchal \marchalcount{}, Zelikin-Borisov \zelikincount.)
Other definitions of synthesis---including those of Boltyanskii and
Brunovsk\'y---are reviewed and compared with ours in \S 4, where we also
discuss in detail why the concept of ``optimal synthesis'' cannot just be
defined by simply identifying it with that of ``optimal feedback law.''  In
particular, we explain in \S 4 that, if $x\to v(x)$ is an optimal feedback,
then in general the concept of ``solution'' of the closed-loop equation
$\dot x=f(x,v(x))$ is problematic (because $x\to f(x,v(x))$ may fail to be
Lipschitz-continuous) and compare the classical definition of solution with
two other notions: (i)~Filippov solutions, and (ii)~the limits of
sample-and-hold solutions recently proposed as feedback solutions by
Clarke, Ledyaev, Subbotin and Sontag in \sontagcount, which will be referred
to here as ``CLSS solutions.'' We argue that the classical notion of
solution fares better than the other two concepts, since there are
situations where the optimal trajectories are not Filippov solutions of the
optimal closed-loop equation, as well as cases where they are not CLSS
solutions.  The analysis of \S 4 is supplemented, in \S 5, with several
examples that clarify the relationship between the various definitions.
Finally, a number of technical points, including some arguments pertaining
to the proof of Theorem 2.13, are relegated to\shortversion{two}
\longversion{three} appendices.

As explained in \S 4, there are essentially two opposite starting points 
for  defining the notion of a ``regular'' optimal synthesis. One approach, 
following Boltyanskii \boltcount, Brunovsk\'y \brunacount-\brunbcount,
Cesari \cesaricount, Fleming and Rishel \flemcount, and Sussmann
\sussacount, is to take as the primary object a smooth or 
``piecewise smooth'' feedback control $v$, and then let the synthesis be the
collection $Traj(v)$ of all trajectories generated by $v$. If this
collection is too large (for example, if the closed-loop equation
determined by $v$ does not have unique solutions), then a synthesis would
be a suitably defined subclass $\Gamma$ of $Traj(v)$. The other strategy,
proposed in Piccoli \piccacount{} and Sussmann \sussmcount, is to take a
``synthesis'' to be just a collection of trajectories---or, more precisely,
trajectory-control pairs---not necessarily arising from a feedback control.
To carry out the first stategy, Boltyanskii and Brunovsk\'y were forced to
make precise sense of the notion of piecewise smoothness, which they did by
requiring that the set of interest---that is, the set of initial points of
optimal trajectories---be partitioned into submanifolds of various
dimensions on which a smooth feedback control is specified.  It turns out,
however, that with this definition the feedback control by itself is not
sufficient for the synthesis to be completely determined, since a
discontinuous feedback control may generate too many trajectories, some of
which could fail to be optimal, as shown in Example 5.3 below.  So, in any
case, the final outcome of the effort to implement the first strategy is a
result more consistent with the second one, since one ends up with a notion
of synthesis as a pair consisting of a feedback control $v$ together with a
selection $\Gamma$ of a family of trajectories generated by $v$, suggesting
that the specification of $\Gamma$ is needed but that of $v$ might not be.
This leads naturally to the approach of this paper, in which $v$ is
dispensed with altogether.

An important point that ought to be further investigated but will not
be touched upon here is the relation between the geometrical approaches of
\brunacount, \brunbcount, \sussacount, \sussbcount, \sussccount,
\sussecount, \sussfcount, \susslcount, \sussmcount, and the theory of
viscosity solutions (cf.  Bardi and Capuzzo Dolcetta \bardicount, Fleming
and Soner \flemsonercount).  We point out that the existing results of this
theory for deterministic optimal control do not appear to cover cases such
as Fuller's problem.  (Cf. for example \bardicount, where the main result
is for Lagrangians that are bounded away from zero.)

\vs

\section{Regular Presynthesis and Synthesis, and the Sufficiency Theorem}

\noindent We start with some definitions that will enable us to state and
prove a mild generalization of
the theorem whose proof was outlined in \sussmcount.  Our assumptions 
will be minimal, so as to achieve maximum generality.
%In later sections we will impose more restrictive conditions and apply
%our result to other definitions of synthesis given in the literature only
%for more regular control systems.

We consider a control system:
\xdef\system{\equanosectno}
$$\dot x=f(x,u)\,,\qquad x\in\Omega\,,\qquad u\in U\,,\numeq$$
and a family $\pmb{{\cal P}}=\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)=
\{{\cal P}(\Omega,U,f,L,{\cal T},\tau,x_0)\}_{x_0\in\Omega}$ 
of minimization
problems, where  ${\cal P}(\Omega,U,f,L,{\cal T},\tau,x_0)$ is 
the  problem: 
\xdef\minim{\equanosectno}
$${\rm minimize\,\,\,\,\,}\tau(x(b))+\int_a^b L(x(t),u(t))dt
\qquad{\rm subject\,\,\,to}\,\,\,(\system)\,,\,\,\,x(a)=x_0\,,\,\,\,
{\rm and}\,\,\,x(b)\in{\cal T}\,.\numeq$$
(A more precise formulation of the problems ${\cal P}(\Omega,U,f,L,{\cal
T},\tau,x)$, 
carefully specifying the space of curves to be considered, will be given
below. Since both $f$ and $L$ have no explicit dependence on $t$, 
we can always take $b=0$.) We assume that 

\medbreak
{\sl 
{\pari
\item{(A1)} $n$ is a positive integer and $\Omega$---the ``state
space''---is an open subset of $\R^n$;

\item{(A2)} $U$---the ``control set''---is a nonempty set;

\item{(A3)} $f:\Omega\times U\rightarrow \R^n$---the ``dynamics''---and
$L:\Omega\times U\rightarrow \R$---the ``Lagrangian''---are maps;

\item{(A4)} $f(\cdot,u)$ and $L(\cdot,u)$ are
maps of class ${\cal{C}}^1$  for every fixed $u\in U$;

\item{(A5)} ${\cal T}$---the ``target''---is a nonempty subset of $\Omega$;

\item{(A6)} $\tau:\Omega\to\R$ is a function of class ${\cal{C}}^1$.
}
}

\medbreak\noindent We use $T_x^cS$ to denote the {\sl contingent tangent
cone} to a set $S$ at a point $x\in S$, defined as follows: a vector $v$ is
in $T_x^cS$ iff there exist a sequence $\{x_j\}$ of points of $S$ and a
sequence $\{h_j\}$ of strictly positive numbers such that $h_j\to 0$ and
$x_j=x+h_jv+o(h_j)$ as $j\to\infty$.  We write
$$
\tilde T_xS=T_x^cS\cap(-T_x^cS)=\{v:v\in T_x^cS\quad{\rm and\quad}-v\in 
T_x^cS\}\,.\numeq
$$
 

\medbreak\noindent
As in Sussmann \sussmcount, $\tf$ will denote the map
$\Omega\times U\ni(x,u)\to \tf(x,u)\equaldef (f(x,u),L(x,u))\in\R^{n+1}$.

If $\Omega\times\R\ni(x,t)\to g(x,t)\in\R^m$ is a time-varying
vector-valued map on $\Omega$, then $Dg(x,t)$ denotes, for each
$(x,t)\in\Omega$, the Jacobian matrix at $x$ of the map $y\to g(y,t)$, so
$Dg(x,t)\in\R^{m\times n}$. 

If $\mu:A\to B$ is a map, we use $\Dom(\mu)$ to indicate the domain
of $\mu$, i.e. the set $A$. 

A {\sl control} is a $U$-valued map $\eta$ such that $\Dom(\eta)$ is a
compact subinterval $[a,b]$ of $\R$.  If $\eta$ is a control, we define
maps $f_{\eta}:\Omega\times\rr\to\R^{n}$, $L_{\eta}:\Omega\times\rr\to\R$,
$\tf_{\eta}:\Omega\times\rr\to\R^{n+1}$ by letting
$$
\left.\matrix{f_{\eta}(x,t)\!=\!f(x,\eta(t)),
&L_{\eta}(x,t)\!=\!L(x,\eta(t)),
&\tf_{\eta}(x,t)\!=\!\tf(x,\eta(t)),
&\,{\rm if}\,\,t\!\in\!\Dom(\eta)\,;\cr
f_{\eta}(x,t)\!=\!0,\phantom{f(,\eta(t))}
&L_{\eta}(x,t)\!=\!0,\phantom{L(,\eta(t))}
&\tf_{\eta}(x,t)\!=\!0,\phantom{\tf(,\eta(t))}
&\,{\rm if}\,\,t\!\notin\!\Dom(\eta)\,.\cr}\right.
$$

A {\sl trajectory} for a control $\e$ is an absolutely continuous map
$\g:\Dom(\e)\to\Omega$ with the property that $\dot\g(t)=f(\g(t),\e(t))$
for almost every $t\in\Dom(\e)$.  If $\Dom(\e)=[a,b]$ (so that
$\Dom(\g)=[a,b]$ as well), and $x=\gamma(a)$, $y=\gamma(b)$, we say that
$\gamma$ (or the pair $(\gamma,\eta)$) {\sl goes from $x$ to $y$}, and that
$\eta$ {\sl steers $x$ to $y$}, and we write $\gamma^-=x$, $\gamma^+=y$.
We denote by $\Range(\g)$ the range of the map $\g$, that is the set
$\{\g(t):t\in\Dom(\g)\}$.

\medbreak A control $\eta$ is ${\cal C}^1$-{\sl admissible} for $\tf$
if $\tf_{\eta}$ satisfies the following ${\cal{C}}^1$
Carath\'eodory conditions: 

\medbreak
{\pariim

\item{(C.i\phantom{i})} {\sl the map $\Dom(\e)\ni t\to
\tf_{\eta}(x,t)\in\R^{n+1}$ is measurable for each $x\in\Omega$;} 
\vs
\item{(C.ii)} {\sl for every compact subset $K$ of $\,\Omega$ there exists an
integrable function $\vp_K:\R\to\R$  such that for every $x\in K$ and
$t\in\Dom(\e)$:} 
\xdef\intbound{\equanosectno}
$$\|\tf_{\eta}(x,t)\|+\|D\tf_{\eta}(x,t)\|\leq \vp_K(t)\,.\numeq$$

}

\medbreak\noindent If $\eta$ is a ${\cal C}^1$-admissible control for
$\tf$,  and $\gamma$ is a trajectory corresponding to $\eta$, then we say
that $(\gamma,\eta)$ is a {\sl ${\cal C}^1$-admissible pair} for 
$\tf$.  We use $\Adm(\tf)$ to denote the set of all ${\cal C}^1$-admissible
pairs for $\tf$, and write $\Admo(\tf)$, $\Admo(\tf,{\cal T})$,
$\Admo_x(\tf,{\cal T})$, to denote, respectively, the set of all
$(\g,\e)\in\Adm(\tf)$ such 
that $\Dom(\e)$ is an interval of the form $[T,0]$ with $T\le 0$, the set
of those $(\g,\e)\in\Admo(\tf)$ such that $\gamma(0)\in{\cal T}$, and the
set of those $(\g,\e)\in\Admo(\tf,{\cal T})$ such that $\g^-=x$.  For a
pair $(\gamma,\eta)\in\Admo(\tf,{\cal T})$, we use
$J(\gamma,\eta)$ to denote the {\sl cost} of $(\gamma,\eta)$, given by 
$$
J(\g,\e)=\tau(\g^+)+\int_{\Dom(\g)} L(\g(t),\e(t))\,dt\,.\numeq
$$
If $\Gamma\!\subseteq\!\Admo(\tf,{\cal T})$ is an arbitrary set of 
${\cal C}^1$-admissible pairs for $\tf$ ending at the target ${\cal T}$ at
time $0$, we define a function
$V_{\Gamma}\!:\!\Omega\!\to\!\R\cup\{+\infty\}$, called the 
{\sl cost function of $\Gamma$}, by letting 
$$
V_{\Gamma}(x)=\inf\,\{J(\gamma,\eta):(\gamma,\eta)\in\Gamma,\,\gamma^-=x\}\numeq
$$ 
for $x\in\Omega$.  
(In particular, $V_{\Gamma}(x)=+\infty$ iff there is no
$(\gamma,\eta)\in\Gamma$ starting at $x$.)  If $\Gamma=\Admo(\tf,{\cal T})$,
then the cost function $V_{\Gamma}$ is the {\sl value function} of our
problem, and in that case we write $V_{\tf,{\cal T}}$ rather than
$V_{\Admo(\tf,{\cal T})}$.  Naturally, 
$V_{\Gamma}\ge V_{\tf,{\cal T}}$ pointwise for every subset $\Gamma$
of $\Admo(\tf,{\cal T})$. 

We can now formulate ${\cal P}(\Omega,U,f,L,{\cal T},\tau,x_0)$ precisely:
${\cal P}(\Omega,U,f,L,{\cal T},\tau,x_0)$ is the problem of finding a
$(\g,\e)\in\Admo_{x_0}(\tf,{\cal T})$ 
such that $J(\g,\e)=V_{\tf,{\cal T}}(x_0)$.
Any such pair $(\gamma,\eta)$ is said to be a {\sl solution} of ${\cal
P}(\Omega,U,f,L,{\cal T},\tau,x_0)$.  
A pair $(\g,\e)\in\Admo(\tf,{\cal T})$
is {\sl optimal} if it is a solution of ${\cal P}(\Omega,U,f,L,{\cal
T},\tau,\g^-)$.  We use $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$---or,
simply, $\pmb{{\cal P}}$, in any context where the meaning of $\Omega$,
$U$, $f$, $L$, ${\cal T}$, $\tau$ is clear---to denote 
the family of problems 
$\Big\{{\cal P}(\Omega,U,f,L,{\cal T},\tau,x_0)\Big\}_{x_0\in\Omega}$.

\rema
We emphasize that in the optimal control problems studied  here {\it the
trajectories have to end at the target but could pass through it
before}. For example, consider the dynamical law $\dot x=1$ in $\rr$, with
Lagrangian $L(x)\equiv -1$ and target ${\cal T}=\{0,1\}$. Then the value
function $V$ is given by $V(x)=x-1$ for $x\le 1$ and $V(x)=+\infty$ for
$x>1$. If $x<1$, then the optimal trajectory $\g_x$ from $x$ to the target
is given by $\g_x(t)=t+1$ for $T_x\le t\le 0$, where $T_x=x-1$. Notice that
if $x<0$ then $t=0$ is not the first time when $\g_x(t)$ belongs to 
${\cal T}$. \proofend





Given $\la\in\R_n$ (the space of row $n$-vectors), 
$\l0\in\R$, $x\in\Omega$ and $u\in U$, we define:
\shortversion{
\xdef\hamil{\equanosectno}
$$H^c(\la,\l0,x,u)=\langle \la , f(x,u)\rangle +\l0
L(x,u)\,,\quad H(\la,\l0,x)=\inf\,\{H^c(\la,\l0,x,u):u\in
U\}\,.\numeq$$ 
}
\longversion{
\xdef\hamil{\equanosectno}
$$
H^c(\la,\l0,x,u)=\langle \la , f(x,u)\rangle +\l0
L(x,u)\,,\numeq
$$
$$
H(\la,\l0,x)=\inf\,\{H^c(\la,\l0,x,u):u\in
U\}\,.\numeq
$$ 
}
The functions $H^c:\R_n\times\R\times\Omega\times U\to\R$
and $H:\R_n\times\R\times\Omega\to\R\cup\{-\infty\}$
are known, respectively, as the {\sl control Hamiltonian} and the {\sl
minimized Hamiltonian} of $\tf$.

\defi
We say that the pair $(\gamma,\eta)\in \Admo(\tf,{\cal T})$ is {\sl
extremal} if there exist an absolutely continuous map $\la:{\rm
Dom}(\g)\to \R_n$ and a constant $\l0\geq 0$, that satisfy:

\medbreak


{\pariim
\item{(EX1)} {\sl $(\l0,\la(t))\neq (0,0)$ for some (and hence every) $t$;}

\midbreak

\item{(EX2)} {\sl $\dot\la = -{\partial H^c\over \partial x\,\,\,\,}
(\la(t),\l0,\gamma(t),\eta(t))$ for a.e. $t\in\Dom(\e)$;}

\midbreak

\item{(EX3)} $H(\la(t),\l0,\gamma(t))=H^c(\la(t),\l0,\gamma(t),\eta(t))=0$
{\sl for a.e.}  $t\in \Dom(\eta)$; 

\midbreak

\item{(EX4)} $\langle\la(0),v\rangle=\l0\langle \nabla\tau(\g(0)),v\rangle
\,\,\,{\sl for\,\,\,all}\,\,\,v\in \tilde T_{\g(0)}{\cal T}$. \proofend

}

\medbreak\noindent
Properties (EX1), (EX2), (EX3), (EX4) are, respectively,
the {\sl nontriviality condition}, the {\sl adjoint equation},
the {\sl Hamiltonian minimization condition}, and the {\sl transversality
condition}. A pair $(\la,\l0)$ such that (EX2) holds is called
an {\sl adjoint vector for} $(\g,\e)$.  An adjoint vector for which (EX3) 
holds is a {\sl minimizing adjoint vector for} $(\g,\e)$.  

The Maximum Principle says that, under some special conditions
on the target set ${\cal T}$, {\sl every optimal pair $(\g,\e)$
is extremal.}
(This is true, for example, if ${\cal T}$ is a submanifold of $\Omega$ 
of class ${\cal{C}}^1$, or a closed convex subset of $\Omega$, or
a set which is locally equivalent to a convex set via a ${\cal{C}}^1$
diffeomorphism.  More generally, it is true if the cone 
$\tilde T_{\g(0)}{\cal T}$ is convex---in which case it is of course a 
subspace, since $v\in\tilde T_{\g(0)}{\cal T}\Rightarrow
-v\in\tilde T_{\g(0)}{\cal T}$---and is an approximating cone 
to ${\cal T}$ at $\g(0)$.   Recall that a closed convex cone $C$ is an
{\sl approximating cone} to a set $S$ at a point $s\in S$ if there exist a
neighborhood $W$ of $0$, and a continuous map $F:W\cap C\to S$,  
such that $F(v)=s+v+o(||v||)$ as $v\to 0$ via values in $C$.) 

\defi
A {\sl presynthesis} for $\pmb{{\cal P}}$ is a subset
$\Gamma$ of $\Admo(\tf,{\cal T})$ such that,

\medbreak

{\pariim

\item{(PS)} {\sl Whenever $(\gamma_1,\eta_1)\in\Gamma$,
$(\gamma_2,\eta_2)\in\Gamma$, 
and $\g_1^-=\g_2^-$, it follows that $(\gamma_1,\eta_1)=(\gamma_2,\eta_2)$.}

}

\medbreak\noindent The set $\Dom(\Gamma)\equaldef\{\g^-:(\g,\e)\in\Gamma\}$
is called the {\sl domain} of $\Gamma$.  We say that $\Gamma$ is a {\sl
presynthesis on} $S$ if $\Gamma$ is a presynthesis and 
$S=\Dom(\Gamma)$. \proofend

\noindent Clearly, giving a presynthesis on $S$ amounts to choosing, for
each $x\in S$, a $(\g,\e)\in \Admo(\tf,{\cal T})$ such that $\g^-=x$.

If $\Gamma$ is a presynthesis such that $\Dom(\Gamma)$ consists of all
points that can be steered to a point of ${\cal T}$  by 
means of a pair of $\Admo(\tf,{\cal T})$, 
then we say that $\Gamma$ is {\sl total}.  

Given a presynthesis $\Gamma$ and a point $x\in\Dom(\Gamma)$,
we will always use $(\gx^\Gamma,\ex^\Gamma)$---or, simply, 
$(\gx,\ex)$ when the meaning of $\Gamma$ is clear from the 
context---to denote the unique pair ($\g,\e)\in\Gamma$ 
such that $\g^-=x$, and will write  $T_x=\min\,\Dom(\g_x)$, so 
$T_x\le 0$ and $\Dom(\g_x)=\Dom(\e_x)=[T_x,0]$.

\xdef\defsynthesis{\the\sectno.\the\theono}
\defi
A presynthesis on $S$ is {\sl memoryless} if
whenever $x\in S$ and $t\in \Dom(\ex)$ it follows that $y=\gx(t)$ belongs to
$S$ and $\eta_y$ is the restriction of $\ex$ to the interval $[t,0]$.  A
{\sl synthesis} is a memoryless presynthesis. \proofend

\defi
If each pair of a presynthesis $\Gamma$ is optimal (resp. extremal) then we
say that $\Gamma$ is {\sl optimal} (resp. {\sl extremal}). \proofend

\noindent  In particular, a presynthesis $\Gamma$
for $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$ is a total optimal
presynthesis if and only if $V_{\Gamma}\equiv V_{{\tf},{\cal T}}$.

\medbreak
As explained in the introduction, we wish to propose a definition of
``regular synthesis'' that will give rise to a sufficiency theorem.  The
theorem will say that a total extremal regular synthesis that satisfies the
appropriate boundary conditions is optimal.  We will actually define the
more general notion of ``regular presynthesis,'' and prove the sufficiency
theorem in this broader setting, so the desired optimality result will be
true in particular when the presynthesis is a true
synthesis.

The definition will say that a ``regular presynthesis''
is a presynthesis that satisfies suitable regularity assumptions.
So we begin by describing these regularity assumptions in detail.

\defi We call a function $V:\Omega\rightarrow \R\cup \{+\infty\}$ 
{\sl weakly upper semicontinuous} (w.u.s.c) at a point $x\in\Omega$ if
$\lim\inf_{y\to x} \lim\sup_{z\to y}V(z)\le V(x)$.  \proofend

Equivalently, $V$ is w.u.s.c. at $x$ if there exist sequences $\{x_j\}$,
$\{\ve_j\}$ and $\{\delta_j\}$ such that $x_j\in\Omega$, $\ve_j>0$, and
$\delta_j>0$ for all $j$, and $x_j\rightarrow x$ and $\delta_j\rightarrow
0$ as $j\to\infty$, having the property that $V(y)\le V(x)+\delta_j$
whenever $\|y-x_j\|<\ve_j$.

\defi 
Suppose we are given a Lipschitz-continuous vector field 
$X$ on $\Omega$ and a function
$V\!:\!\Omega\!\rightarrow\!\R\!\cup\!\{+\infty\}$. We say that $V$ has
the {\sl NDJ {\rm (``no downward jumps'')} property along $X$} if 

{\pariim

\item{(NDJ)} {\sl if $[a,b]\ni t\to\gamma(t)\in\Omega$ is an integral
curve of $X$, then $\lim\inf_{h\downarrow
0}V(\gamma(t-h))\leq V(\gamma(t))$ for every $t\in\,]a,b]$.} \proofend

}

\xdef\defweakc{\the\sectno.\the\theono}
\defi
We say that $V:\Omega\rightarrow \R\cup \{+\infty\}$ 
satisfies the {\sl weak continuity
conditions} for the control system (\system) if $V$ is lower semicontinuous,
weakly upper semicontinuous at every $x\in\Omega$, 
and has the NDJ property along the vector
field $x\to f(x,u)$ for every $u\in U$. \proofend

\def\rhofgx{\rho_{\tf,\Gamma,\xb,T}}

We now define the ``weak differentiability conditions'' for $\Gamma$.  In
this definition, we let $\rhofgx(v)$ be the
integrable $\R^{n+1}$-valued function on $[T,0]$
(see definition below) given by
$$
\rhofgx(v)(t)=\tf_{\eta_{\xb+v}}(\gxb(t),t)-\tf_{\exb}(\gxb(t),t)\,,
$$
and use the symbol $\rhofgx$ to denote the correspondence 
$v\rightarrow\rhofgx(v)$, regarded as a map into the space ${\rm
Bor}([T,0],\R^{n+1})$ of 
$\R^{n+1}$-valued Borel measures on $[T,0]$.  (Recall that ${\rm
Bor}([T,0],\R^{n+1})$ is the dual of the space $C^0([T,0],\R^{n+1})$ of
continuous $\R^{n+1}$-valued functions on $[T,0]$.)

\defi
A presynthesis $\Gamma=\{(\gamma_x,\eta_x):x\in S\}$
for $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$ is 
said to be {\sl $(f,L)$--differentiable} at a point $\xb\in\Omega$ if
there exist: (a)~a neighborhood $N$ of $\xb$ in $\Omega$
such that $N\subseteq \Dom(\Gamma)$, (b)~an interval $[T,0]$ such that
$\Dom(\ex)\subseteq [T,0]$ for all $x\in N$, and (c)~an $\bar\ve>0$, 
for which

{\pariim

{\sl 

\item{(DC1)} there are integrable functions $\pe :[T,0]\rightarrow
\R$, for $\ve\in\,]0,\bar\ve]$, with the property that 
$\lim_{\ve\to 0}\int_T^0 \pe(t)\, dt=0$, for which the inequality
$$\Bigl\|\tf_{\ex}(y,t)-\tf_{\ex}(\gxb(t),t)-D\tf_{\exb}(\gxb(t),t)\cdot
(y-\gxb(t))\Bigr\|\leq\pe(t).(||y-\gxb(t)||+||x-\bar x||)\numeq
$$
holds for every $y\in\Omega$, $x\in N$,  $t\in[T,0]$ such that 
$\|y-\gxb(t)\|\leq\ve$ and $||x-\bar x||\le\ve$;



\item{(DC2)} the map $\rhofgx$ is weak$^*$-differentiable at $v=0$, 
in the sense that, for every continuous function
$\alpha:[T,0]\rightarrow\R$, the map $\R^n\ni v\to \int_T^0
\alpha(t).\rhofgx(v)(t)\,dt \, \in\R^{n+1}$ is differentiable at
$v=0$. \proofend 

}
}


\xdef\defithin{\the\sectno.\the\theono}
\defi
A subset $A$ of $\Omega$ is {\sl thin} if there exist $A_1$, $A_2$
such that $A=A_1\cup A_2$, $A_1$ is a finite or countable union of
connected ${\cal{C}}^1$ submanifolds of positive codimension, and
${\cal{}H}^{n-1}(A_2)=0$, where ${\cal{}H}^k$ denotes $k$-dimensional
Hausdorff measure. \proofend

If $S$ is a subset of $\Omega$, we use $\Inte{S}$ to denote the interior 
of $S$ in $\Omega$. 


\defi
If $n$, $\Omega$, $U$, $f$, $L$, ${\cal T}$, $\tau$ are such that Assumptions
(A1) -- (A6) hold, $\Gamma$ is a presynthesis
for $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$, and $S={\rm
Dom}(\Gamma)$, we say that $\Gamma$ has the {\sl interior approximation
property} if 

{\pari

\medbreak
\item{(IAP)} {\sl If $(\g,\e)\in\Admo(\tf)$ 
is such that $\gamma(t)\in S$ for all $t\in{\rm Dom}(\g)$, then 
there exists a sequence $\{(\g_j,\e_j)\}_{j=1}^\infty$ 
in $\Admo(\tf)$ such that 

\medbreak
\itemitem{(IAP.1)} $\g_j(t)\in\Inte{S}$ for every $t\in{\rm Dom}(\g_j)$; 

\itemitem{(IAP.2)} $V_{\Gamma}(\gamma_j^+)\to V_{\Gamma}(\gamma^+)$; 

\itemitem{(IAP.3)} $\gamma_j^-\to\gamma^-$;

\itemitem{(IAP.4)} $\int_{\Dom(\g_j)} L(\g_j(t),\e_j(t))\,dt\to 
\int_{\Dom(\g)} L(\g(t),\e(t))\,dt$.}\proofend 

}

\noindent We are now ready to present our definition or ``regular
presynthesis'': 
\vs
\xdef\maindef{\the\sectno.\the\theono}
\defi
Let $n$, $\Omega$, $U$, $f$, $L$, ${\cal T}$, $\tau$ be such that Assumptions
(A1) -- (A6) hold.  Let $\Gamma$ be a presynthesis
for $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$. We say that $\Gamma$ is 
{\sl regular} if 

{\parix

\item{(\maindef.a)} {\sl the associated cost function $V_{\Gamma}$
satisfies the weak continuity conditions}; 

\midbreak

\item{(\maindef.b)}  {\sl there exists a thin subset $A$ of $\Omega$ such 
that $\Gamma$ is $(f,L)$--differentiable at all points of 
$\Dom(\Gamma)\backslash A$}. \proofend


}

With this definition, our main sufficiency theorem (a weaker version of
which was already stated in \sussmcount, with only an outline of the proof)
says:  
\vs
\xdef\maintheo{\the\sectno.\the\theono}
\theo 
Let $n$, $\Omega$, $U$, $f$, $L$, ${\cal T}$, $\tau$ be such that Assumptions
(A1) -- (A6) hold.  Let $\Gamma$ be an extremal regular presynthesis
for $\pmb{{\cal P}}(\Omega,U,f,L,{\cal T},\tau)$ and let $S={\rm
Dom}(\Gamma)$. Then:

\midbreak

{\parixx {\sl 

\item{(\maintheo.a)} The ``dynamic programming inequality''
\xdef\dpi{\equanosectno}
$$
V_{\Gamma}(\g^-)\le\int_{\Dom(\g)} L(\g(t),\e(t))\,dt+
V_{\Gamma}(\g^+)\numeq
$$
holds for every $(\g,\e)\in\Admo(\tf)$ such that 
$\Range(\g)\subset\Inte{S}$.

\item{(\maintheo.b)} If $V_\Gamma$ satisfies the boundary condition 
\xdef\bounc{\equanosectno}
$$
V_{\Gamma}(x)\le\tau(x)\quad{\rm for\quad every\quad}x\in{\cal T}\,,\numeq
$$
then $V_{\Gamma}(\g^-)\le J(\g,\e)$ for all $(\g,\e)\in\Admo(\tf,{\cal T})$
such that $\Range(\g)\subset\Inte{S}$.

\item{(\maintheo.c)} If $\Gamma$ has the interior approximation property,
then (\dpi) holds for all $(\g,\e)\in\Admo(\tf)$ such that 
$\Range(\g)\subset S$.

\item{(\maintheo.d)} If $\Gamma$ has the interior approximation property
and satisfies (\bounc)
then $V_{\Gamma}(\g^-)\le J(\g,\e)$ for all $(\g,\e)\in\Admo(\tf,{\cal T})$
such that $\Range(\g)\subset S$.

\item{(\maintheo.e)} If $\Gamma$ is total and satisfies
(\bounc) then $\Gamma$ is optimal. \proofend

}
}

%\medbreak

\beginproof We first establish the relation between the differential of the
cost function $V_{\Gamma}$ and the adjoint fields of vectors along the
extremal trajectories of the presynthesis.  This result is important in its
own right, so we state it as a separate theorem:

\xdef\maintheodiff{\the\sectno.\the\theono}
\theo 
Let $n$, $\Omega$, $U$, $f$, $L$, ${\cal T}$, $\tau$ be such that Assumptions
(A1) -- (A6) hold, and let $\Gamma=\{(\gamma_x,\eta_x):x\!\in\!S\}$ be a
presynthesis. Let $\bar x\!\in\! S$ be such that $\Gamma$ is
$(f,L)$--differentiable at $\bar x$.  Then

{\pariim
{\sl 
\item{(\maintheodiff.a)} the function $V_{\Gamma}$ is differentiable 
at $\bar x$;

\midbreak

\item{(\maintheodiff.b)} if the pair $(\gamma_{\bar x},\eta_{\bar x})$ is
extremal, then every minimizing adjoint vector $(\la,\l0)$ for 
$(\gamma_{\bar x},\eta_{\bar x})$ satisfies the identity
\shortversion{$\l0 (\nabla V_{\Gamma})(\xb)= \la(\txb)$. In particular,} 
\longversion{$$
\l0 (\nabla V_{\Gamma})(\xb)= \la(\txb)\,.
$$
In particular,}
$\l0\neq 0$, and $(\la,\l0)$ is unique up to multiplication by a positive 
constant. 

}

}

\medbreak\noindent{\sl PROOF OF THEOREM \maintheodiff.\enspace}
We apply Theorem A.11 in Appendix A, taking $P$ to be $\R^n$
and letting $f_x=f_{\ex}$, $L_x=L_{\ex}$, $\tf_x=\tf_{\ex}$. 
Then, if $y\in\Omega$ and $x\in N$, we have 
$f_x(y,t)=f(y,\ex(t))$, $L_x(y,t)=L(y,\ex(t))$, 
$\tf_x(y,t)=\tf(y,\ex(t))$ if $t\in[T_x,0]$, and 
$f_x(y,t)=0$, $L_x(y,t)=0$, $\tf_x(y,t)=0$ if $t\notin[T_x,0]$.

We let $\hat V_\Gamma(x)=\int_{T_x}^0L(\gx(t),\ex(t))\,dt$, so 
$\hat V_\Gamma$ is the ``Lagrangian part'' of our cost function.
We extend the curve $\gx$ to the interval 
$[T,0]$ by letting $\gx(t)=x$ for $T\le t<T_x$, so $\gx$ 
still is an integral curve of $f_{\ex}$ on $[T,0]$, and 
$\hat V_{\Gamma}(x)=\int_{T}^0L_x(\gx(t),t)\,dt$.

The map $p\to \bar x_p$ of Theorem A.11 is the identity map, which is
certainly differentiable at $\bar x$.  So Theorem A.11 tells us that the
``endpoint map'' $x\to{\cal{}E}(x)\equaldef \gx(0)$ and the function $\hat
V_\Gamma$ are differentiable at $\bar x$. This implies that the function
$V_\Gamma$ is differentiable at $\bar x$ proving (\maintheodiff.a).

Assume now that $(\gxb,\exb)$ is extremal and $(\lambda,\l0)$ is a minimizing
adjoint vector for $(\gxb,\exb)$.
We extend the adjoint vector $\la$ to $[T,0]$ by letting
$\la(t)=\la(T_{\xb})$ for $T\le t<T_{\xb}$. We then define 
\shortversion{$H^c_x(p,p^0,y,t)=\langle p,f_x(y,t)\rangle+p^0L_x(y,t)$.}
\longversion{$$H^c_x(p,p^0,y,t)=\langle p,f_x(y,t)\rangle+p^0L_x(y,t)\,.$$}
Then the adjoint equation 
$\dot\la(t)=-\nabla_yH^c_{\xb}(\la(t),\la^0,\gxb(t),t)$ holds
a.e. on $[T,0]$.  Moreover, \shortversion{it is clear that}
\xdef\moreover{\equanosectno}
$$
H^c_{x}(\la(t),\l0,\gxb(t),t)\ge H^c_{\xb}(\la(t),\l0,\gxb(t),t)
=0\,\,\,{\rm for\,\,\,a.\,\,e.\,\,\,}t\in[T,0]\,.\numeq
$$
\longversion{
(To see this, we first observe that $H^c_{\xb}(\la(t),\l0,\gxb(t),t)=0$
a.~e. on $[T,0]$, because (a)~the equality holds a.~e. when $t\in[T_x,0]$
thanks to (EX3), and (b)~it holds everywhere when $t\in[T,T_x\,[\,$, because in
this case $f_{\xb}(\gxb(t),t)=0$ and $L_{\xb}(\gxb(t),t)=0$.  
Next, we show that, if $x\in N$, then $H^c_{x}(\la(t),\l0,\gxb(t),t)\ge
0$. The inequality follows from (EX3) for almost every $t$ 
when $t\ge T_{\xb}$ and $t\ge T_{x}$, because in this case
$H^c_{x}(\la(t),\l0,\gxb(t),t)\!=\!H^c(\la(t),\la^0,\gxb(t),\ex(t))
\!\ge\! H(\la(t),\la^0,\gxb(t))\!=\!0$ 
for almost every $t$.  
When $t<T_x$, $H^c_{x}(\la(t),\l0,\gxb(t),t)$ vanishes,  because
$f_x(\gxb(t),t)=0$ and $L_x(\gxb(t),t)=0$. Finally, we must consider the
possibility that $T_x\le t<T_{\xb}$. For this purpose, we observe that, if
$u\in U$ is arbitrary, then 
$H^c(\la(s),\la^0,\gxb(s),u)\ge H(\la(s),\la^0,\gxb(s))\ge 0$ 
for almost every $s\in[T_{\xb},0]$. Since
$s\to H^c(\la(s),\la^0,\gxb(s),u)$ is continuous, we conclude that
$H^c(\la(s),\la^0,\gxb(s),u)\ge 0$ for {\sl all} $s\in[T_{\xb},0]$. In
particular, $H^c(\la(T_{\xb}),\la^0,\xb,u)\ge 0$.  Since $\la$ is constant
on $[T,T_{\xb}]$, we see that $H^c(\la(s),\la^0,\xb,u)\ge 0$ for all
$s\in[T,T_{\xb}]$ and all $u\in U$. Now, if $T_x\le t<T_{\xb}$, then
$H^c_{x}(\la(t),\l0,\gxb(t),t)=H^c(\la(t),\la^0,\xb,\ex(t))\ge 0$, and
(\moreover) is proved.)}

{}From Theorem A.11 it follows the identity
\xdef\identi{\equanosectno}
$$
\eqalign{
&\,\lambda(0).D{\cal E}(\bar x).v+\lambda^0 \nabla\hat V_{\Gamma}(\bar x).v
-\lambda(T_{\bar x})\cdot v\cr
&\,\phantom{\lambda(0).D{\cal E}(\bar x).v+}=\lim_{\ve\downarrow
0}\,\,\ve^{-1}\!\int_T^0\!\Big(H^c_{\xb +\ve v}(\la(t),\l0,\gxb(t),t)- 
H^c_{\xb}(\la(t),\l0,\gxb(t),t)\Big)\,dt\cr}\numeq
$$
for every vector $v$. Then (\moreover) tells us that 
the right-hand side of inequality (\identi) is nonnegative. So
$$
\lambda(0).D{\cal E}(\bar x).v+\lambda^0\nabla\hat V_{\Gamma}(\bar x).v
-\lambda(T_{\bar x})
\cdot v\ge 0\qquad{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Given $v$, the curve $r\to \gamma_{\bar x+rv}(0)$ is defined
for $r$ in a neighborhood of $0$ in $\R$, and is differentiable 
at $0$ with derivative $D{\cal E}(\bar x).v$.  On the other hand, this
curve is contained in ${\cal T}$, so the vector 
$w=D{\cal E}(\bar x).v$ belongs to  $\tilde T_{\g_{\bar x}(0)}{\cal T}$.  
It then follows from the transversality condition (EX4) that  
$\langle\la(0),w\rangle=\l0\langle \nabla\tau(\g(0)),w\rangle$.
Therefore 
\xdef\thisineq{\equanosectno}
$$
\l0\biggl(\nabla\tau(\g(0)).D{\cal E}(\bar x).v+\nabla\hat V_{\Gamma}(\bar
x).v\biggr) 
-\lambda(T_{\bar x})
\cdot v\ge 0\qquad{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Since the left-hand side of (\thisineq) is linear in $v$, it has to vanish
for all $v$. So  
\xdef\thiseq{\equanosectno}
$$
\l0 \nabla V_\Gamma(\bar x).v=
\l0\biggl(\nabla\tau(\g(0)).D{\cal E}(\bar x).v+\nabla\hat V_{\Gamma}(\bar
x).v\biggr)=\lambda(T_{\bar x})\cdot v\qquad
{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Therefore $\l0\nabla V_\Gamma(\bar x)=\lambda(T_{\bar x})$. 

In particular, $\l0$ cannot vanish, because $\l0=0$ would imply
$\lambda(T_{\bar x})=0$, contradicting (EX1).  So we may, after
multiplication by a positive constant, assume that $\l0=1$, and then
$\nabla V_{\Gamma}(\xb)= \la(\txb)$. This proves, in particular, the
uniqueness of $(\lambda(\cdot),\l0)$ up to multiplication by a positive
constant, since the Cauchy problem for the adjoint equation has unique
solutions. \proofend 

\medbreak\noindent{\sl CONTINUATION OF THE PROOF OF THEOREM \maintheo.\enspace}
We first prove (\maintheodiff.a). For any given $x\in S\setminus A$, let
$\la_x$ be the unique $\la$ such 
that $(\la,1)$ is a minimizing adjoint vector along $(\gx,\ex)$. 
(The existence and uniqueness
of $\la_x$ follows from Theorem \maintheodiff.) The minimization condition
(EX4) of Maximum Principle then 
implies that, for any given $u\in U$, the inequality
$$
\langle \la_x(t),f(\gx(t),u)\rangle +L(\gx(t),u)\geq 0
$$
holds for almost all $t\in [T_x,0]$.
Since $t\to \langle \la_x(t),f(\gx(t),u)\rangle +L(\gx(t),u)$
is continuous on $[T_x,0]$, the 
inequality is valid at time $\tx$. In view of 
Theorem \maintheodiff{}, 
\xdef\difvalf{\equanosectno}
$$
\langle\nabla V_{\Gamma}(x),f(x,u)\rangle +L(x,u)\geq 0\,.
\numeq
$$
Now consider an arbitrary $(\g,\eta)\in\Admo(\tf)$, with domain
$[T,0]$, such that $\Range(\g)\subset \Inte{S}$.   We want
to prove that (\dpi) holds.   
It follows from Lemma~4.1 of \sussncount{} that
there exists a sequence of admissible
pairs $(\gamma_j,\eta_j)\in \Admo(\tf)$, with domain $[T,0]$, 
such that $\gamma_j(0)=\g(0)$,
$\eta_j$ is piecewise constant, $\gamma_j\rightarrow\g$ uniformly 
on $[T,0]$, and
$\int_{T}^0 L(\g_j(t),\e_j(t))\,dt\to\int_{T}^0 L(\g(t),\e(t))\,dt$.

Since $\Range(\g)\subset \Inte{S}$, we can clearly assume that the
$\Range(\g_j)\subset\Inte{S}$. Since $V_{\Gamma}$ is lower
semicontinuous, it follows that, if 
$V_{\Gamma}(\gamma_j(T))+\int_{T}^0 L(\g_j(t),\e_j(t))\,dt
\leq V_{\Gamma}(\gamma(0))$ for each $j$, then 
$V_{\Gamma}(\gamma(T))+\int_{T}^0 L(\g(t),\e(t))\,dt
\leq V_{\Gamma}(\gamma(0))$. 
Hence it suffices to assume that  the control $\eta$ is piecewise
constant.

If (\dpi) holds for two pairs
$(\g_i,\e_i)\in\Adm(\tf)$ such that ${\rm Dom}(\g_1)=[a,b]$,
${\rm Dom}(\g_2)=[b,c]$, and $\g_1(b)=\g_2(b)$ for some $a,b,c$, 
then it also holds for the concatenated pair  $(\g,\e)$ defined in an 
obvious way. 
It therefore suffices to prove (\dpi) for all $(\g,\e)\in\Adm(\tf)$
corresponding to a constant control.
Since $L$ does not depend on time we can assume
$(\g,\e)\in\Admo(\tf)$ and $\e$ constant, 
so we make this assumption from now on. 

Let $u\in U$ be such that $\e(t)=u$ for $t\in[T,0]$.  For each
$y\in\Omega$, let $t\to\zeta^y(t)$ be the trajectory of $\eta$ that
satisfies the terminal condition $\zeta^y(0)=y$. Let $W$ be an open subset
of $\Omega$ such that $\g(0)\in W$, $\zeta^y(T)$ is defined for all $y\in
W$, and $\zeta^y([T,0])$ is contained in $\Inte{S}$ whenever $y\in W$. We will
show that

\medbreak
\itemitem{(\#)} {\sl for almost every $y\in W$ the set
$B^y=\{t\in [T,0]:\zeta^y(t)\in A\}$ is finite or countable.}

\medbreak\noindent Since $A$ is thin, we can write $A=A_1\cup A_2$,
where $A_1=\bigcup_{j}M_j$ and $\{M_j\}_{j\in J}$ is a finite or
countable family of connected submanifolds of $\Omega$ of codimension
$d_j>0$, while ${\cal{}H}^{n-1}(A_2)=0$.  After replacing each $M_j$
by a finite or countable family of open submanifolds of $M_j$, we may
assume that the $M_j$ are embedded.

Define $\tilde W=W\times\,]\,T,0\,[\,$, and let
$\Phi$ be the map $\tilde W\ni(y,t)\to \zeta^y(t)\in\Omega$. Then $\Phi$ is
a ${\cal{C}}^1$ submersion, because for each fixed $t$ the partial map
$y\to\Phi(y,t)$ is a ${\cal{C}}^1$ diffeomorphism. Therefore each set
$\tilde M_{j}=\Phi^{-1}(M_j)$ is an embedded submanifold of $\tilde W$ of
codimension $d_j$, and ${\cal{}H}^{n}(\Phi^{-1}(A_2))=0$.

Let $\pi$ be the projection $\tilde W\ni(y,t)\to y\in W$.
Let $S_{j}$ be the set of points $s\in\tilde M_{j}$ such that
the restriction $\pi\restr\tilde M_{j}$ of $\pi$ to 
$\tilde M_{j}$ is not regular at $s$. (Recall that, if
$S_1$, $S_2$ are ${\cal{C}}^1$ manifolds, then  a ${\cal{C}}^1$ map
$F:S_1\to S_2$ is {\sl regular} at $s\in S_1$ if the differential $d\pi(s)$
maps the tangent space $T_sS_1$ surjectively onto $T_{F(s)}S_2$.)
Then ${\cal{}H}^{n}(\pi(\Phi^{-1}(A_2)))=0$ \longversion{(see for example
Federer \fedcount)}, and  ${\cal{}H}^{n}(\pi(S_{j}))=0$ for each $j$, by
Sard's Theorem. So the ``bad'' set 
${\cal{}B}=\pi(\Phi^{-1}(A_2))\cup\Bigl(\bigcup_{j}\pi(S_{j})\Bigr)$
has Lebesgue measure zero. 

Now let $y\in W\backslash{\cal{}B}$. Then $\zeta^y(t)\notin A_2$ if
$T<t<0$. (Otherwise, if $\zeta^y(t)\in A_2$, $T<t<0$, it
would follow that $(y,t)\in\Phi^{-1}(A_2)$, so
$y\in\pi(\Phi^{-1}(A_2))$, and then $y\in{\cal{}B}$.)  Also, we claim
that the set $\{t\in [T,0]:\zeta^y(t)\in A_1\}$ is at most countable. To see this,
it suffices to show that, for each $j$, the set 
$E_{j}=\{t:T<t<0\,,\,\zeta^y(t)\in M_j\}$ is finite or
countable.  Fix $j$, and suppose
$t\in E_{j}$. The manifold $\tilde M_{j}$ has codimension
$d_j$ in $\tilde W$. Since $d_j>0$, the dimension $\nu_{j}$
of $\tilde M_{j}$ is $\le n$. Moreover, 
$d\pi(y,t)$ maps $T_{y,t}\tilde M_{j}$ surjectively onto 
$\R^n$ (the tangent space to $\Omega$ at $y$),
because $y\notin{\cal{}B}$. So $\nu_{j}=n$, and $d\pi(y,t)$ is injective
on $T_{y,t}\tilde M_{j}$.  Since 
$d\pi(y,t)\Big({\partial\,\,\over\partial t}\Big)$ is
obviously equal to 
$0$, it follows that ${\partial\,\,\over\partial t}\notin T_{y,t}\tilde
M_{j}$.  Since $\tilde M_{j}$ is embedded, $(y,t')\notin\tilde
M_{j}$ if $0<|t'-t|\le\varepsilon$, provided that $\varepsilon$ is
sufficiently small. Therefore $t$ is an isolated point of $E_{j}$.
So $E_{j}$ is a discrete subset of the open interval $\,]\,T,0\,[\,$. 
Therefore $E_{j}$ is finite or countable, as desired.

We have therefore proved that, if $y\in W\backslash{\cal{}B}$, then 
$\zeta^y(t)\notin A_2$ for all $t\in ]T,0[$, and 
the set $\{t\in [T,0]:\zeta^y(t)\in A_1\}$ is at most countable.  Since $A=A_1\cup
A_2$, and ${\cal{B}}$ has measure zero, ($\#$) follows. 

\medbreak
{}In view of the weak upper semicontinuity of $V_{\Gamma}$, there exist $x_j$,
$\ve_j>0$, $\delta_j>0$ such that $x_j\rightarrow \gamma(0)$ and
$\delta_j\rightarrow 0$ as $j\to\infty$, and $V_{\Gamma}(y)\leq
V_{\Gamma}(\gamma(0))+\delta_j$ whenever $\|y-x_j\|\leq \ve_j$.  Choose the
$\ve_j$'s so that $\|y-x_j\|\leq\ve_j$ implies $y\in W$. Then, by ($\#$), we
can choose for each $j$ a $y_j$ such that $\|y_j-x_j\|\leq \ve_j$ and the
bad set $B^{y_j}$ is at most finite or countable.  Since 
$\zeta^{y_j}(t)\in\Inte{S}$ for every $t\in[T,0]$, 
(\difvalf) implies that the function
$[T,0]\ni t\to\varphi_j(t)=V_{\Gamma}(\zeta^{y_j}(t))
+\int_T^t L(\zeta^{y_j}(s),\eta(s)) ds\in\R$
is differentiable at every $t$, except for a finite or countable number 
of values, with a nonnegative derivative.
Therefore $\varphi_j$ satisfies assumption (B.1.a) of Lemma~B.1 (Appendix~B).
Moreover, the lower semicontinuity of $V_{\Gamma}$ ensures assumption
(B.1.b) of Lemma~B.1 for $\varphi_j$, and the (NDJ) condition ensures
assumption (B.1.c). Hence we can apply Lemma~B.1 to $\varphi_j$, and
conclude that $\varphi_j(T)\le\varphi_j(0)$.  So
$$
V_{\Gamma}(\zeta^{y_j}(T))\leq \int_T^0
L(\zeta^{y_j}(t),\eta(t))dt +V_{\Gamma}(y_j)\leq
\int_T^0 L(\zeta^{y_j}(t),\eta(t))dt +V_{\Gamma}(\gamma(0))+\delta_j.
$$ 
If we let $j\to\infty$ we find, using the lower
semicontinuity of $V_{\Gamma}$, that (\dpi) holds
for $(\g,\e)$, as desired.  
So (\dpi) holds when $\eta$ is constant and, as explained earlier, this
completes the proof of (\maintheo.a).  

\longversion{
\medbreak
To prove (\maintheo.b), we pick $(\g,\e)\in\Admo(\tf,{\cal T})$ 
such that $\g$ is contained in $\Inte{S}$, 
and apply (\maintheo.a) to conclude that 
$V_{\Gamma}(\g^-)\leq \int_{{\rm Dom}(\g)}L(\g(t),\e(t))dt
+V_{\Gamma}(\g(0))$.  
Then (\bounc) implies the inequality $V_{\Gamma}(\g(0))\le\tau(\g(0))$, so 
$V_{\Gamma}(\g^-)\leq J(\g,\e)$ and (\maintheo.b) is proved. 
}

\longversion{
\medbreak
To prove (\maintheo.c), we pick $(\g,\e)\in\Admo(\tf)$ 
such that $\g$ is contained in $S$, and find pairs 
$(\g_j,\e_j)$ such that conditions (IAP.1,2,3,4) hold. Then 
(\maintheo.a) implies that the inequality 
$V_{\Gamma}(\g_j^-)\leq \int_{{\rm Dom}(\g_j)}
L(\g_j(t),\e_j(t))dt +V_{\Gamma}(\g_j(0))$ holds. Using (IAP.2.3.4) and 
the lower semicontinuity of $V_{\Gamma}$, we conclude that 
(\dpi) holds, completing the proof of (\maintheo.c). 
}

\longversion{
The proof of (\maintheo.d) is identical to that of 
(\maintheo.b), using (\maintheo.c) instead of (\maintheo.a).
}

\shortversion{
Statements (\maintheodiff.b),  (\maintheodiff.c), and (\maintheodiff.d)
are immediate consequences of (\maintheodiff.a).
}

\longversion{Finally, to prove}\shortversion{To prove} 
(\maintheo.e), we assume that $\Gamma$ is total and prove that it
satisfies the interior approximation property.  So now $S$ is the set of
all points of $\Omega$ that can be steered to a point of ${\cal T}$
by a pair of $\Admo(\tf)$. 
Let $(\g,\e)\in\Admo(\tf)$ be a trajectory-control pair such
that $\g$ is entirely contained in $S$. Then $\g(0)\in S$, so $\g(0)\in{\rm
Dom}(\Gamma)$ and then $V_{\Gamma}(\g(0))<+\infty$.  The weak upper
semicontinuity of $V_{\Gamma}$ implies that there exist $x_j$, $\ve_j>0$,
$\delta_j>0$ such that $x_j\rightarrow \gamma(0)$ and $\delta_j\rightarrow
0$ as $j\to\infty$, and $V_{\Gamma}(y)\leq V_{\Gamma}(\gamma(0))+\delta_j$
whenever $\|y-x_j\|\leq \ve_j$.  It then follows that $V_{\Gamma}(x_j)\to
V_\Gamma(\g(0))$, since $V_{\Gamma}$ is lower semicontinuous. Let ${\rm
Dom}(\g)=[T,0]$.  Let $\g_j$ be the maximally defined trajectory for the
control $\e$ that satisfies $\g_j(0)=x_j$. Then $\g_j$ is defined on
$[T,0]$ if $j$ is large enough, and conditions (IAP.2,3,4) hold with
$\e_j=\e$.  We may then assume that $\g_j$ is defined on $[T,0]$ for all
$j$. To prove that (IAP.1) holds as well, we observe that $y\in S$ whenever
$\|y-x_j\|\leq \ve_j$, because in that case $V_{\Gamma}(y)<+\infty$, since
$V_{\Gamma}(y)\leq V_{\Gamma}(\gamma(0))+\delta_j$, and
$V_{\Gamma}(\gamma(0))<+\infty$ because $\g(0)\in S$.  Let $\zeta^y$ be the
trajectory for the control $\e$ that satisfies $\zeta^y(0)=y$. Then for
each $j$ there exists a $\tilde\ve_j$ such that $0<\tilde\ve_j\le\ve_j$,
with the property that the curve $\zeta^y$ is defined on $[T,0]$ for all
$y$ such that $\|y-x_j\|\le\tilde\ve_j$. Moreover, it is clear that $y$ is
reachable from $\zeta^y(t)$ for every $t\in[T,0]$, so
$\zeta^y([T,0])\subseteq S$ if $\|y-x_j\|\le\tilde\ve_j$, because $y\in S$.
Now, if $T\le t\le 0$, the map $y\to\zeta^y(t)$ is a ${\cal C}^1$
diffeomorphism from the neighborhood $\{x:\|x-x_j\|<\tilde\ve_j\}$ of $x_j$
onto a neighborhood $N_j(t)$ of $\g_j(t)$. Clearly, $N_j(t)\subseteq S$, so
$\g_j(t)\in\Inte{S}$. This completes the proof of
(\maintheo.e). \proofend

\vs


\rema
\shortversion{It is not hard to prove that {\it the function $V_{\Gamma}$
is weakly upper semicontinuous  everywhere if and only if it is weakly
upper semicontinuous  at all the points of the target ${\cal T}$.} \proofend}
\longversion{{\it The function $V_{\Gamma}$ is weakly upper semicontinuous 
everywhere if and only if it is weakly upper semicontinuous 
at all the points of the target ${\cal T}$.}
Indeed, let $x\in\Omega$.  If $x\notin S$, then 
$V_{\Gamma}(x)=+\infty$, and the conclusion follows. Next, suppose 
$x\in S$, and let  $(\gx,\ex)$ be its associated
admissible pair.  Assume that $V_{\Gamma}$ is weakly upper semicontinuous 
at $\gx(0)$. Let
$y_j$, $\ve_j$, $\delta_j$ be such that $y_j\to \gx(0)$, $\ve_j>0$,
$\delta_j>0$, $\delta_j\to 0$, and $V_{\Gamma}(y)\le
V_{\Gamma}(\gx(0))+\delta_j$  whenever $\|y-y_j\|\le\ve_j$. Let $\zeta^{y_j}$
be the trajectory corresponding to the control $\ex$ and such that
$\zeta^{y_j}(0)=y_j$.
}
\longversion{
Since $y_j\to \gx(0)$, we may assume, by picking $j$ large enough,
that $\zeta^{y_j}$ is defined on $[\tx,0]$ for all $j$, 
and let $x_j=\zeta^{y_j}(\tx)$. Let
$\tilde\ve_j$ be such that $0<\tilde\ve_j<{1\over j}$ and, if
$\|z-x_j\|\le\tilde\ve_j$, then the trajectory $\theta^z$ of $\ex$
with initial condition $\theta^z(\tx)=z$ is defined on $[\tx,0]$ and
satisfies $\|\theta^z(0)-y_j\|<\ve_j$. Then, using the same reasoning
as in the proof of Theorem \maintheo{}, we can show that 
\xdef\whatever{\equanosectno}
$$V_\Gamma(z)\le\int_{\tx}^0L(\theta^z(t),\ex(t))\,dt+V_{\Gamma}(\gx(0))+\delta_j\numeq
$$
whenever $\|z-x_j\|\le\tilde\ve_j$. (First approximate $\ex$ by
piecewise constant controls, $\eta^k$, and let $\theta^{z,k}$ be the
corresponding trajectories with terminal condition 
$\theta^{z,k}(0)=\theta^z(0)$.  Then for each $k$ use Statement ($\#$)
of the proof of Theorem \maintheo{} to approximate the terminal point 
$\theta^{z}(0)$ by points $w^{z,k,i}$ with the property that, if 
$\tilde\theta^{z,k,i}$ is the trajectory for $\eta^k$ for which 
$\tilde\theta^{z,k,i}(0)=w^{z,k,i}$, it follows that 
$V_\Gamma(\tilde\theta^{z,k,i}(\tx))
\le\int_{\tx}^0L(\tilde\theta^{z,k,i}(t),\eta^k(t))\,dt+
V_{\Gamma}(w^{z,k,i})$. Since $\|\theta^z(0)-y_j\|<\ve_j$, 
the $w^{z,k,i}$ satisfy $\|w^{z,k,i}-y_j\|<\ve_j$ if $i$ is large
enough, so $V_\Gamma(\tilde\theta^{z,k,i}(\tx))\le
\int_{\tx}^0L(\tilde\theta^{z,k,i}(t),\eta^k(t))\,dt+
V_\Gamma(\gx(0))+\delta_j$ when $i$ is large enough. If we let
$i\to\infty$ and then let $k\to\infty$, and use the lower
semicontinuity of $V_\Gamma$, we find that (\whatever) holds.)
}

\longversion{
Let
$\tilde\delta_j=\sup\Bigl\{\|\int_{\tx}^0L(\theta^z(t),\ex(t))\,dt-
\int_{\tx}^0L(\gx(t),\ex(t))\,dt\|:\|z-x_j\|\le\tilde\ve_j\Bigr\}$.
Then $\tilde\delta_j\to 0$, because $x_j\to x$ and $\tilde\ve_j\to
0$. On the other hand, $V_\Gamma(z)\le
V_{\Gamma}(x)+\delta_j+\tilde\delta_j$ whenever
$\|z-x_j\|\le\tilde\ve_j$. So $V_{\Gamma}$ is weakly upper
semicontinuous at $x$. \proofend
}

\vs


\section{The Fuller Phenomenon}

\medbreak

\noindent In this section we show that the famous minimization
problem considered by Fuller in \fullcount---and extensively studied
by other authors, e.g. Zelikin-Borisov \zelikincount---presents an optimal
synthesis that satisfies the
hypotheses of Theorem \maintheo. 

Consider the system:
%\longversion{
%\xdef\constr{\equanosectno}
%$$
%\cases{ \dot x_1 =  x_2  &$\,,$\cr
%          \dot x_2 =  u  &$\,,$\cr
%            |u|  \leq 1  &$\,,$\cr}\numeq
%$$
%}
%\shortversion{
\xdef\constr{\equanosectno}
$$
\dot x_1 =  x_2\,,\qquad\,\,\,\dot x_2 =  u\,,\qquad\,\,\,|u|  \leq 1\,,\numeq
$$%}
and the family $\hat{\pmb{{\cal P}}}=\{\hat{{\cal P}}_x\}_{x\in\R^2}$
of minimization problems:
\xdef\minimfulinfhor{\equanosectno}
$$\hat{\cal P}_x\,:\qquad{\rm minimize}\,\,\,
\int_0^{+\infty} x_1^2(t)\,dt\,\,\,{\rm subject\,\,\,to}
\quad(x_1(0),x_2(0))=x\,.\numeq
$$
We proceed to describe the optimal synthesis for this problem, determined
by Fuller, referring the reader to Zelikin-Borisov \zelikincount{}, Ch.~2,
for further details. First define the ``switching locus''
\shortversion{$\zeta=\{(x_1,x_2):|x_1|=C x_2^2\,,\,\,x_1x_2\leq 0\}$,}
\longversion{$$\zeta=\{(x_1,x_2):|x_1|=C x_2^2\,,\,\,x_1x_2\leq 0\}\,,$$}
where $C$ is a constant
whose precise definition will be given later.  Write $\zeta=\zeta^+
\cup\zeta^-\cup\{(0,0)\}$, where $\zeta^{\pm}=\{(x_1,x_2)\in\zeta:\pm
x_1> 0\}$. 
Then the piecewise smooth curve $\zeta$ divides the plane into two regions
$A^-$ and $A^+$ consisting, respectively, of the points that lie above and
below $\zeta$. We define a discontinuous feedback control $k:\R^2\to[-1,1]$
by letting $k(x_1,x_2)=-1$ if $(x_1,x_2)\in A^-\cup\zeta^-$, 
$k(x_1,x_2)=1$ if $(x_1,x_2)\in A^+\cup\zeta^+$, and $k(0,0)=0$.

Then for every $x\in\R^2$ there exists a unique solution 
$\hat\gamma_x:[0,\infty\,[\,\to\R^2$ of the Cauchy problem
$\dot x_1=x_2$, $\dot x_2=k(x_1,x_2)$, $(x_1(0),x_2(0))=x$.
The corresponding open-loop control $\hat\eta_x$ is related to
$\hat\gamma_x$ by $\hat\eta_x(t)=+1$ when $\hat\gamma_x(t)\in
A^+$, $\hat\eta_x(t)=-1$ when $\hat\gamma_x(t)\in A^-$, and
$\hat\eta_x(t)=0$ when $\hat\gamma_x(t)=0$.

The switchings occur when $\hat\gamma(t)\in\zeta$. The curve $\hat\gamma_x$
reaches the origin in a finite time $\hat T_x$, and then stays
there. Therefore $\hat\eta_x(t)=0$ when $\hat\gamma_x(t)=0$, which happens
iff $t\ge\hat T_x$ (see Fig.~1).

\vskip 1em

\vbox{


\centerline{\hbox{\psfig{figure=fuller.ps,width=3in,height=3in}}}

\medbreak
\centerline{Figure 1}

}

\vskip 1em

\noindent Therefore the admissible pairs $(\hat\gamma_x,\hat\eta_x)$
actually solve the family $\pmb{{\cal P}}=\{{\cal P}_x\}_{x\in\R^2}$
of minimization problems
\xdef\minimful{\equanosectno}
$$
{\cal P}_x\,:\quad{\rm minimize}\,\,\int_a^{b} x_1^2(t) dt\,\,{\rm
subject\,\,to}\,\,(x_1(a),x_2(a))=x\,\,{\rm and}\,\,
(x_1(b),x_2(b))=(0,0)\,,\numeq 
$$
i.e., the problems of reaching the origin in finite time, subject
to the dynamical constraints (\constr), and minimizing the
cost given by (\minimful). 

So far, $C$ was an arbitrary positive constant. It turns out that, if $C$
is chosen to be equal to $C^*$, where $C^*$ is the unique positive root of
the polynomial $x^4+x^2/12-1/18$, then the pairs $(\hat\g_x,\hat\eta_x)$
are extremal.  (This is proved in \zelikincount{}. In page~26, it is shown
that $C$ must be chosen so that $(1-2C)^{1/2}(1+2C)^{-1/2}=\mu$, where
$\mu$ is such that, if $\sigma=\mu+\mu^{-1}$, then
$\sigma^2-3\sigma-6=0$. These equations easily imply $C^4+C^2/12-1/18$=0.)

We now choose $C=C^*$, so in particular $0.4<C<0.5$. 
Then $\{\hat\gamma_x\}_{x\in\R^2}$ is a family of
extremals reaching the target, and one would like to prove that the
$\hat\gamma_x$ are optimal.  The sufficiency theorems of \boltcount{} and
\brunbcount{} do not apply, since the controls $\hat\eta_x$ 
have an infinite number of switchings.  For this reason, the optimality of
the synthesis obtained by Fuller is usually established by other means.
For example, one can prove optimality using the {\sl dilation
symmetry properties}
of the problem.  Precisely, there exists a one parameter family
$\Delta=\{\Delta_{\rho}\}_{\rho>0}$ of maps from $\R^2$ to $\R^2$ such
that, if $(\gamma,\eta)$ is an admissible pair having cost $c$, and we let
$\gamma_\rho(t)=\Delta_\rho(\gamma({t\over\rho}))$ and
$\eta_\rho(t)=\eta({t\over\rho})$,
then $(\gamma_\rho,\eta_\rho)$ is an admissible pair having cost $\rho^5c$.
It then follows that the $\Delta_\rho$ map optimal trajectories to optimal
trajectories.  

The proof of optimality using the dilation symmetry is
given, for example, in
\zelikincount{}, and will not be repeated here. It will turn out, however,
that the dilations $\Delta_\rho$ will play a role in our arguments, so we
recall their definition and one simple identity that will be useful later:

\medbreak
\itemitem{(1)}  By definition, 
\shortversion{$\Delta_\rho(x_1,x_2)=(\rho^2x_1,\rho x_2)$.} 
\longversion{$$\Delta_\rho(x_1,x_2)=(\rho^2x_1,\rho x_2)\,.$$} 
\itemitem{(2)} It follows from (1), in particular, that 
\shortversion{$\Delta_\rho(x_1,x_2)-x=(\r-1)\, (2x_1,x_2)+(\r-1)^2\,
(x_1,0)$}
\longversion{$$\Delta_\rho(x_1,x_2)-x=(\r-1)\, (2x_1,x_2)+(\r-1)^2\, (x_1,0)$$}
for all $\rho,x_1,x_2.$

\medbreak\noindent
In order to apply Theorem \maintheo, we first make a time translation. We 
define $T_x=-\hat T_x$,  and then let 
\shortversion{$\gamma_x(t)=\hat\gamma_x(t+\hat T_x)$,
$\eta_x(t)=\hat\eta_x(t+\hat T_x)$,} 
\longversion{$$
\gamma_x(t)=\hat\gamma_x(t+\hat T_x)\,,\qquad
\eta_x(t)=\hat\eta_x(t+\hat T_x)\,,$$} 
for $T_x\le t\le 0$.  Then
$\Gamma=\{(\gamma_x,\eta_x)\}_{x\in\R^2}$ is a memoryless family of
${\cal C}^1$-admissible extremal pairs such that $\gamma_x$ is defined on $[T_x,0]$,
$\gamma_x(T_x)=x$, and $\gamma_x(0)=0$. 

We want to prove that $\Gamma$ is an optimal synthesis.  For this purpose,
we will verify the hypotheses of Theorem \maintheo, and in particular the
$(f,L)$--differentiability of $\Gamma$ at every point
$x\in\R^2\backslash\zeta$.  Since it is very easy to check Property (DC1),
we will only verify (DC2).

\longversion{
As in \zelikincount{}, page~26, we let
$$
\mu=\sqrt{ {1-2C \over 1+2C}}\,. 
$$
Then $0<\mu<1$. 
}
\shortversion{
As in \zelikincount{}, page~26, we let $\mu=\sqrt{(1-2C)(1+2C)^{-1}}$.
Then $0<\mu<1$. 
}

\longversion{
Fix a point $\xb=(\xb_1,\xb_2)$ belonging to
$\R^2\backslash\{(0,0)\}$. Then there exist 
$s_0(\xb)$ and $s_1(\xb)$ such that 
$T_{\xb}=s_0(\xb)<s_1(\xb)\le\mu s_0(\xb)<0$, and  
$$
\exb=\pm\sum_{k=0}^{+\infty} (-1)^k \chi_{_{_{I_k(\xb)}}}
$$
where  
\medbreak
\itemitem{(a)}$I_0(\xb)=[s_0(\xb),s_1(\xb)]\,$; 
\itemitem{(b)}$I_k(\xb)=[\mu^{k-1}s_1(\xb),\mu^ks_{1}(\xb)]$ for $k=1,2,\ldots\,\,$;
\itemitem{(c)}$\chi_{_{_{S}}}$ is the indicator function of a set $S$;
\medbreak\noindent and 
\medbreak\itemitem{(d)}the $\pm$ sign  depends on $\xb$ as follows: it is 
$+$ if $\xb\in A^+\cup\zeta^+$, and $-$ if $\xb\in A^-\cup\zeta^-$.
}
\shortversion{
Fix a point $\xb=(\xb_1,\xb_2)$ belonging to
$\R^2\backslash\{(0,0)\}$. Then there exist 
$s_0(\xb),s_1(\xb)$ such that 
$T_{\xb}\!=\!s_0(\xb)\!<\!s_1(\xb)\!\le\!\mu s_0(\xb)\!<\!0$, and 
$\exb=\pm\sum_{k=0}^{+\infty} (-1)^k \chi_{_{_{I_k(\xb)}}}$,
where  (a)~$I_0(\xb)=[s_0(\xb),s_1(\xb)]$, 
(b)~$I_k(\xb)=[\mu^{k-1}s_1(\xb),\mu^ks_{1}(\xb)]$ for
$k=1,2,\ldots$, (c)~$\chi_{_{_{S}}}$ is the indicator function of a set $S$, 
and (d)~the $\pm$ sign  depends on $\xb$ as follows: it is 
$+$ if $\xb\in A^+\cup\zeta^+$, and $-$ if $\xb\in A^-\cup\zeta^-$.
}

\longversion{\medbreak\noindent}Let $T$ be such that $T<0$ and the
interval $[T,0]$ contains  
$\Dom(\ex)$---i.e. $T\le T_x$---for $x$ near $\bar x$.  To prove (DC2),
we must 
pick a point $\xb\in\R^2\setminus\zeta$ and a continuous function
$\a:[T,0]\rightarrow\R$, and show that the map 
$$
\R^2\,\,\ni\,\, x \,\,\,\longrightarrow \,\,\,J^\alpha(x)
\,\,\equaldef\,\,\int_T^0\a(t).\tilde
f_{\eta_{x}}(\gxb(t),t)\,dt\,\,\in\,\,\R^3\,,\numeq 
$$
is differentiable at $x=\xb$. We will do this by showing that, for two
linearly independent directions $v_1(x)$, $v_2(x)$, depending smoothly on
$x$, the directional derivatives $D_{v_1(x)}J^\alpha$ and
$D_{v_2(x)}J^\alpha$ exist and are continuous with respect to $x$.  We
choose $v_1(x)$ to be the direction of the curve $\rho\to \Delta_\rho(x)$
at $\rho=1$, and take $v_2(x)=f(x,\eta_x(T_x))$.
\longversion{The determinant of $v_1(x)$ and $v_2(x)$ is then equal to 
$\pm(2x_1-x_2^2)$ (the sign being as in (d)), 
which never vanishes on $A^{\pm}$, since $C<1/2$. So 
$v_1(x)$ and $v_2(x)$ are linearly independent for all $x\in A^+$.}
\shortversion{It is then easy to see that
$v_1(x)$ and $v_2(x)$ are linearly independent for all $x\in A^+$.}
We will just
determine the directional derivatives at $\xb$, and the result will
make it obvious that they depend continuously on $\xb$, so
differentiability will follow.   

Let us assume, for simplicity, that $\xb\in A^+$. (The case when 
$\xb\in A^-$ is similar.)   We first
differentiate $J^\alpha$ in the direction of $v_1(\xb)$.  
If $\rho>0$, we have (letting $\bar s_0=s_0(\xb)$,
$\bar s_k=\mu^{k-1} s_1(\xb)$ for $k=1,2,\ldots$,
$I_k=[\bar s_k,\bar s_{k+1}]$,  and 
$I_k^\rho=[\r \bar s_k,\r \bar s_{k+1}]$ for $k=0,1,\ldots\,$, 
and writing $\xb_\rho=\Delta_\rho(\xb)$)$\,$:
\shortversion{
\xdef\tildef{\equanosectno}
$$
T_{_{\xb_\rho}}\!=\!\rho T_{\xb}\,,\quad
\eta_{_{\xb_\rho}}\!=\!\sum_{k=0}^{+\infty}(-1)^k\chi_{_{_{I_k^\rho}}}\,,
\quad{\rm and}\quad\tilde f_{\eta_{_{\xb_\rho}}}(\gxb(t),t)\!-\!\tilde
f_{\exb}(\gxb(t),t)\!=\! 
\left( \matrix{0\cr \eta_{_{\xb_\rho}}(t)\!-\!\exb(t) \cr 0\cr}
\right)\!+\!R_\rho(t),
$$ 
where $R_\rho(t)$ is a term arising because of the possibility that 
$T_{\xb}\neq T_{_{\xb_\rho}}$.
}
\longversion{
$$
T_{_{\xb_\rho}}=\rho T_{\xb}\,,\quad\quad\quad
\eta_{_{\xb_\rho}}=\sum_{k=0}^{+\infty}(-1)^k\chi_{_{_{I_k^\rho}}}\,,
$$
and
\xdef\tildef{\equanosectno}
$$\tilde f_{\eta_{_{\xb_\rho}}}(\gxb(t),t)-\tilde f_{\exb}(\gxb(t),t)=
\left( \matrix{0\cr \eta_{_{\xb_\rho}}(t)-\exb(t) \cr 0\cr}
\right)+R_\rho(t)\,,\numeq$$ 
where $R_\rho(t)$ is a term arising because of the possibility that 
$T_{\xb}\neq T_{_{\xb_\rho}}$.
}

\longversion{
Then, if $\rho\ge 1$ and $\rho-1$ is sufficiently small,
$$
\eqalign{
\int_T^0 (\eta_{_{\xb_\r}}(t)-\exb(t)) \a(t)\, dt&=
\int_{\rho\bar s_0}^0 \sum_{k=0}^{+\infty} (-1)^k
\Big(\chi_{_{_{I_k^\rho}}}(t)-\chi_{_{_{I_k}}}(t)\Big)\a(t)\,dt\cr
&=\sum_{k=0}^{+\infty} (-1)^k\Biggl(\int_{\rho\bar s_k}^{\rho\bar s_{k+1}}
\a(t)\,dt-\int_{\bar s_k}^{\bar s_{k+1}}\a(t)\,dt\Biggr)\cr
&=\sum_{k=0}^{+\infty} (-1)^k\Biggl(\int_{\rho\bar s_k}^{\bar s_k}\a(t)\,dt
-\int_{\rho\bar s_{k+1}}^{\bar s_{k+1}}\a(t)\,dt\Biggr)\cr
&=\int_{\rho\bar s_0}^{\bar s_0}\a(t)\,dt+2\sum_{k=1}^{+\infty}
(-1)^k\Biggl(\int_{\rho\bar s_k}^{\bar s_k}\a(t)\,dt\Biggr)\,.\cr}
$$
The numbers $(\rho-1)^{-1}\Big\|\int_{\rho\bar s_k}^{\bar s_k}\a(t)\,dt\Big\|$ are
bounded by $\mu^{k-1}|\bar s_1|.\|\a\|_{L^\infty}$, which is the general
term of a convergent series, since $\mu<1$. So we can divide by
$\rho-1$, let $\rho\to 1$, and  take the limit of each term separately.
We then conclude that 
\xdef\therefore{\equanosectno}
$$
\lim_{\r\downarrow 1} {1\over \r-1}\int_T^0  (\eta_{\xb_{\r}}(t)-\exb(t)) \a(t)
dt=-\bar s_0\a(\bar s_0)-2\bar s_1\sum_{k=1}^{+\infty}(-1)^k
\mu^{k-1}\a(\mu^{k-1}\bar s_1)\,.\numeq
$$ 
}

\longversion{
We now find the limit of $(\rho-1)^{-1}\int_T^0\a(t).R_\rho(t)\,dt$
as $\rho\downarrow 1$.  Clearly, 
$T_{_{\xb_\r}}=\rho T_{\xb}<T_{\xb}$ if $\rho>1$. So
$$
\int_T^0\a(t).R_\rho(t)\,dt=\int_{\rho T_{\xb}}^{T_{\xb}}
\a(t).\left(\matrix{\gamma_{_{\xb_\r}}^2(t)\cr 0
\cr\Bigl( \gamma_{_{\xb_\r}}^1(t)\Bigr)^2}\right)\,dt\,,
$$
where $ \gamma_{_{\xb_\r}}^i(t)$, for $i=1,2$, is the 
$i$-th component of  $\gamma_{_{\xb_\r}}(t)$.  Therefore
\xdef\thereforeb{\equanosectno}
$$
\lim_{\rho\downarrow 1}\,(\rho-1)^{-1}
\int_T^0\a(t).R_\rho(t)\,dt=-T_{\xb}\a(T_{\xb}).
\left(\matrix{\xb_2\cr 0\cr \xb_1^2}\right)\,.\numeq
$$
Similar computations show that (\therefore) and (\thereforeb) hold as well
for the limits as $\rho\uparrow 1$.  So 
the directional derivative of $J^\a$ at
$x=\xb$  in the direction of $v_1(\xb)$ exists and is given by the sum
of the right-hand sides of (\therefore) and (\thereforeb), 
which depend continuously on $\xb$,
since $\bar s_0$, $\bar s_1$ and $T_{\xb}$ (which is none other than 
$\bar s_0$) 
depend continuously on $\xb$ as long as $\xb\in A^+$.   
}

\shortversion{
Then a somewhat tedious but straightforward calculation shows that}

\shortversion{
\xdef\therefore{\equanosectno}
\medbreak
\line{\noindent \hfill\hfill $\quad\quad\lim_{\r\to 1} {1\over \r-1}\int_T^0
(\eta_{\xb_{\r}}(t)-\exb(t)) \a(t) 
\,dt=-s_0\a(s_0)-2s_1\sum_{k=1}^{+\infty}(-1)^k
\mu^{k-1}\a(\mu^{k-1}s_1)\,,$\adequano\hfill(\equanosectno)}}

\shortversion{
\xdef\thereforeb{\equanosectno}
\line{\noindent {\rm and}\hfill$\lim_{\rho\to 1}(\rho-1)^{-1}
\int_T^0\a(t).R_\rho(t)\,dt=-T_{\xb}\a(T_{\xb}).
\left(\matrix{\xb_2\cr 0\cr \xb_1^2}\right)$.\adequano\hfill\hfill\hfill(\equanosectno)}}

\shortversion{
\noindent So the directional derivative of $J^\a$ at
$x=\xb$  in the direction of $v_1(\xb)$ exists and is given by the sums
of the right-hand sides of (\therefore) and (\thereforeb), 
which depend continuously on $\xb$,
since $s_0$, $s_1$ and $T_{\xb}$ (which is none other than $s_0(\xb)$) 
depend continuously on $\xb$ as long as $\xb\in A^+$.   
}

A much simpler argument verifies the existence of the derivative of
$J^\a$ in the direction of $v_2(\xb)$---i.e., of $f(\xb,1)$---whose
value turns out to be the vector with components
$\a(T_{\xb})\xb_2$, $\a(T_{\xb})$, and $\a(T_{\xb})\xb_1^2$.  Once again,
this expression depends continuously on $\xb$, so the proof of (DC2)
is complete.

The other assumptions of  Theorem \maintheo{} are easily checked. So the
theorem applies, and the optimality of the synthesis described above
follows.\proofend 

\vs

\section{Comparison with Other Definitions of Synthesis}

\medbreak\noindent The purpose of this section is to compare our concept 
of regular presynthesis with other definitions of synthesis that either
have been proposed by other authors or are sufficiently natural to be worth
considering as possible ways to define this notion.  For simplicity, we
will only consider the case of a single-point target set ${\cal T}$, and
take this point to be the origin.  (Both Boltyanskii and Brunovsk\'y,
whose definitions will be discussed in detail, do the same, so 
this will facilitate the comparison.)  Moreover, we will only consider 
nonnegative Lagrangians, so a trajectory containing a loop can never be
optimal, and an optimal trajectory necessarily terminates when it hits the
origin. 

Perhaps the simplest conceivable notion of a ``regular synthesis'' is that
of a ``feedback control.''  In principle, one could define a
``feedback control law'' on a set $S$ to be a map $v:S\to U$.  If $v$ is
such that the vector field $S\ni x\to f(x,v(x))\in\R^n$ is
Lipschitz-continuous, and we assume for simplicity that $S$ is open, then
$v$ gives rise to unique maximally defined trajectories $\hat\g_x$ starting
at each $x\in S$, and corresponding open-loop controls $\hat\e_x$, given by
$\hat\e_x(t)=v(\hat\g_x(t))$.  If we make the additional assumption that
every trajectory $\hat\g_x$ obtained in this way reaches the target at a
time $\hat T_x$, define $T_x=-\hat T_x$, and let
$\g_x(t)=\hat\g(t+\hat T_x)$ and $\e_x(t)=\hat\e(t+\hat T_x)$ for $T_x\le
t\le 0$, then we will have constructed a synthesis
$\Gamma=\{(\g_x,\e_x)\}_{x\in S}$---in the sense of Definition
\defsynthesis---with domain $S$.
 
The main drawback of such a definition is that, as is well known, for most
reasonable optimal control problems there does not exist an optimal
feedback that renders the map $x\to f(x,v(x))$ Lipschitz-continuous, or
even continuous.  So it is absolutely essential to allow ``discontinuous
feedback laws,'' and when this is done one immediately runs into the
problems of 

\medbreak
\itemitem{(a)} the lack of a truly satisfactory notion of solution

\noindent and

\itemitem{(b)} the lack of good theorems guaranteeing existence of solutions.

\medbreak\noindent Difficulty (b) can be handled in at least two ways,
namely, by 

\medbreak\itemitem{(A)} incorporating into the definition 
requirements that imply existence and uniqueness of solutions of the
closed-loop equation (CLE)  $\dot x=f(x,v(x))\,$; 

\medbreak\itemitem{(B)} incorporating into the definition
requirements that imply existence---but not necessarily uniqueness---of
solutions of the CLE, and adding to the specification of
$v$ a definite prescription for choosing a solution when uniqueness fails,
so that the ``feedback law'' is no longer the closed-loop control $v$
alone, but the {\it pair} $(v,\Gamma)$, where $\Gamma=\{(\g_x,\e_x)\}$ 
is a family that selects one solution of the CLE for each initial condition
$x$. 

\medbreak\noindent Moreover, whether we choose (A) or (B), one has to be
precise about the concept of ``solution.''  Here we will consider three 
such concepts, namely, 

\medbreak\itemitem{(1)} classical solutions (i.~e., 
absolutely continuous curves $t\to x(t)$ having the property that the
equality $\dot x(t)=f(x(t),v(x(t)))$ holds for almost every $t$),

\medbreak\itemitem{(2)} Filippov solutions (cf. \filicount),

\medbreak\itemitem{(3)} ``CLSS solutions,'' that is, the ``feedback solutions''
defined by Clarke, Ledyaev, Subbotin and Sontag in \sontagcount.

\medbreak\noindent The examples of \S 5 will show that the concepts 
of CLSS solution and Filippov solution are not adequate. Indeed, 
Example 5.4 shows an optimal synthesis whose trajectories 
are not Filippov solutions of the optimal closed-loop equation.
In this example, the optimal trajectories are CLSS solutions, but 
the optimal closed-loop equation has many other CLSS  solutions that 
are not optimal (cf. Remark 5.5). In Example 5.6, we exhibit an optimal
control problem whose optimal trajectories are not CLSS solutions
of the optimal closed-loop equation. 

This leaves the notion of classical solution as the only viable candidate
for the concept of solution to be used in the definition of optimal
synthesis.  Example 5.3 shows that it may happen that the set of classical
solutions of the optimal closed-loop equation contains, in addition to all
the optimal trajectories, some other arcs that are not optimal. 

It then follows that (A) is not an adequate way of handling difficulty (b),
and we are left with (B).  This was indeed the strategy followed by
Boltyanskii in \boltcount{}, Brunovsk\'y in \brunbcount{}, and Sussmann in
\sussacount{}.  In all these cases, a ``regular synthesis'' is defined by
first specifying a---not necessarily continuous---feedback control law $x\to
v(x)$. The feedback $v$ is supposed to be ``piecewise smooth'' in
a technical sense that, among other things, guarantees the existence of
trajectories for every initial condition. Uniqueness is not
assumed, but the specification of $v$ is supplemented with
a prescription for selecting a trajectory when non-uniqueness occurs. So, a
``synthesis'' in the sense of \boltcount{}, \brunbcount{}, and
\sussacount{} is more than just an optimal feedback: it is really a pair
$(v,\Gamma)$ of the kind discussed in (B).

\medbreak We now review the three concepts of regular 
synthesis proposed in \boltcount{}, \brunbcount{}, and \sussacount{},
starting with Brunovsk\'y's definition as stated in \brunbcount{}, and then
explaining how to modify this idea to obtain the alternative
formulations suggested by Boltyanskii in {}\boltcount{} and Sussmann in
\sussacount{}.  Since the notations used in \boltcount{}, \brunbcount{},
and \sussacount{} are different, we will give a unified account of the
three definitions using a single set of symbols, and indicating which
notations of \boltcount{} and {}\brunbcount{} they correspond to.
Furthermore, our accounts of \boltcount{} and {}\brunbcount{} will be
slightly modified versions of the text of the published papers, correcting
what we believe are some minor imprecisions or typographical
errors, and introducing some additional notations of our own for extra
clarity.

Both Boltyanskii and Brunovsk\'y work with a system defined on an open
subset $\Omega$ of a finite-dimensional real space (called $X$ in
\boltcount, equal to $\rr^n$ in \brunbcount). Both assume that the
dynamical behavior of the controlled system is given by a law $\dot
x=f(x,u)$, where $U$ is a subset of $\rr^m$, and the maps $f$ and $L$ are
of class ${\cal C}^1$, in the sense that they can be extended to maps of
class ${\cal C}^1$ on an open subset of $\Omega\times\R^m$ that contains
$\Omega\times clos_{\R^m}(U)$.  Both assume that the synthesis is defined
on an open subset $S$ of $\Omega$ (called $V$ in \boltcount, $G$ in
\brunbcount).

Brunovsk\'y's definition uses the concept of a stratification, so we review
this notion first.

%Consider the control system:
%$$\dot x=f(x,u),\qquad x\in\R^n,\qquad u\in U\subset\R^m,$$
%$f:\R^n\times\R^m\rightarrow\R^n$ being ${\C}^1$, and the cost
%function
%$$J=\int_0^T f^0(x,u)dt$$
%where $f_0:\R^n\times\R^m\rightarrow\R$ is ${\C}^1$. Given an initial
%point $x_0$ and target point $\hat x$, by an admissible control we
%understand a piecewise continuous function $u:[0,T]\rightarrow U$ such
%that the solution $x(t)$ of the equation
%$$\dot x=f(x,u(t))$$
%satisfying $x(0)=x_0$ exists on $[0,T]$.

%definition of stratification 

\defi
Given a $k\in\{1,2,\ldots\}\cup\{+\infty,\omega\}$, and a manifold $M$ of
class ${\cal{C}}^k$, a {\sl ${\cal{C}}^k$ stratification in $M$} of a
subset $S$ of $M$ is a partition $\P$ of $S$ into nonempty connected
embedded submanifolds of $M$ of class ${\cal{C}}^k$, such that $\P$ is
locally finite in $M$ (i.e.,  every compact subset of $M$ intersects finitely
many members of $\P$) and the following ``frontier axiom'' holds:

\medbreak
\itemitem{(FA)} {\sl if $P_1,P_2\in\P$, $P_1\neq P_2$ (so that $P_1\cap
P_2=\emptyset$), and $P_1\cap clos_M(P_2)\not=\emptyset$, then $P_1\subseteq
clos_M(P_2)$ and  $dim(P_1)<dim(P_2)$}.  
 
\medbreak\noindent A {\sl $\C^k$-stratified subset of $M$} is a
pair $(S,\P)$ having the property that $S\subseteq M$ and ${\cal P}$
is a $C^k$ stratification of $S$ in $M$. \proofend

We are now ready to present Brunovsk\'y's definition, inserting some
comments of our own---labeled  ``BP\&{}HS''---in square brackets.  

\defi 
Let $S\subseteq\Omega$ be an open subset such that the origin belongs to
$S$. A {\sl Brunovsk\'y regular synthesis on $S$ for the control problem
(\system), (\minim), with target the origin}, is a 6--tuple
$\Xi=(\P,\P_1,\P_2,\Pi,\Sigma,v)$ (called $({\cal S},{\cal S}_1,{\cal
S}_2,\Pi,\Sigma,v)$ in \brunbcount) such that

{\sl 


\midbreak
\itemitem{(Br.1)} $\P$ is the union of $\{\{0\}\}$ 
%$\{\hx\}$ 
and a locally finite (in $S$) partition $\P\backslash\{\{0\}\}$ of 
$S\setminus \{0\}$ into nonempty connected embedded ${\C}^1$ submanifolds
of $S$  (called ``cells''),

\midbreak
\itemitem{(Br.2)} $\P\backslash\{\{0\}\}$ is the disjoint union of $\P_1$ (the
set of ``type I cells'') and
$\P_2$ (the set of ``type II cells''), 

\midbreak
\itemitem{(Br.3)} $v:S\rightarrow U$ (the ``closed loop
control''), $\Pi:\P_1\rightarrow\P$, and $\Sigma:\P_2\rightarrow\P_1$ 
are maps such that the following properties are satisfied:

\midbreak

{\parindent=27pt


\itemitem{(Br.3.A)}  If  $\P'=\{P\in\P:{\rm dim}\;P<n\,,\,P\neq\{0\}\}$, and 
$S'=\bigcup \{P\in\P'\}$, then 
$(S',\P')$ is a stratified subset of $S$ of dimension $<n$.  
{\rm \quad [BP\&{}HS: This turns out to be equivalent to the much simpler
statement that {\sl ${\cal P}$ is a ${\cal C}^1$ stratification 
of $S$}.\shortversion{]}
\longversion{The proof is somewhat delicate, and will be given in full
detail in Appendix C.]}} 

\midbreak\itemitem{(Br.3.B)} The function $v$ is of class $\C^1$ on each cell.

\midbreak\itemitem{(Br.3.C)}  If $P_1\in\P_1$ then $f(x,v(x))\in T_x P_1$ (the
tangent space to $P_1$ at $x$) for every $x\in P_1$. In addition, for each
$x\in P_1$, if we let $\xi_x$ be the maximally defined solution of the
initial value problem
\xdef\initval{\equanosectno}
$$
\dot\xi=f(\xi,v(\xi))\,,\,\,\,\xi(0)=x\,,\,\,\,\xi\in P_1\,,\numeq
$$
and define $t_x=\sup\,\Dom(\xi_x)$, then 
the limit $\xi_x(t_x-)\,\equaldef\,\lim_{t\uparrow t_x}\xi_x(t)$ exists (in
$\Omega$)  and belongs to $\Pi(P_1)$. {\rm \quad [BP\&{}HS: 
\longversion{The limit 
$\xi_x(t_x-)$ cannot belong to $P_1$, because if it did 
then it would be equal to the limit of  $\xi_x(t)$ in $P_1$
(because $P_1$ is embedded), so $\xi_x(t)$ would have a limit in $P_1$ as
$t\uparrow t_x$, and then $\xi_x$ would be extendable to a solution 
of (\initval) on an interval containing $[0,t_x+\ve\,[\,$ for some
positive $\ve$, contradicting the choice of $t_x$.}
\shortversion{The limit 
$\xi_x(t_x-)$ cannot belong to $P_1$.}   It then follows
that {\sl $P_1\neq\Pi(P_1)\subseteq clos(P_1)$ and, moreover, 
${\rm dim}(\Pi(P_1))<{\rm
dim}(P_1)$ for all $P_1\in{\cal P}_1$}.\shortversion{]}
\longversion{Indeed, if $P_1\in{\cal P}_1$, then 
we can pick $x\in P_1$ and conclude that $\xi_x(t_x-)$ belongs to 
$\Pi(P_1)\backslash 
P_1$. So $\Pi(P_1)\cap clos(P_1)\neq\emptyset$ and $\Pi(P_1)\neq P_1$.
Therefore  $\Pi(P_1)\subseteq clos(P_1)$ and 
${\rm dim}(\Pi(P_1))<{\rm dim}(P_1)$, since $P_1\in{\cal P}$, 
$\Pi(P_1)\in{\cal P}$, and ${\cal P}$ is a stratification.]}}

\midbreak\itemitem{(Br.3.D)}   If $P_2\in\P_2$, then the control
$v$ is continuous on $P_2\cup \Sigma(P_2)$, and for each $x\in P_2$
there exists a unique curve $\xi_x:[0,t_x\,[\,\to\Omega$ such that
the restriction  
$\xi_x\restr\,]\,0,t_x\,[\,$ is a maximally defined integral curve 
of the vector field $f(\cdot,v(\cdot))$ on $\Sigma(P_2)$,
and $\xi_x(0)=x$. {\rm \quad [BP\&{}HS: This implies that 
$\Sigma(P_2)\neq P_2$, $P_2\subseteq clos(\Sigma(P_2))$, and 
${\rm dim}\Big(\Sigma(P_2)\Big)>{\rm dim}(P_2)$ for every $P_2\in{\cal
P}_2$.]} 

\midbreak\itemitem{(Br.3.E)} On every cell $P$, $x\to t_x$ is a
continuously differentiable function, and $(t,x)\to\xi_x(t)$, 
$(t,x)\to u_x(t)\equaldef v(\xi_x(t))$ are
continuously differentiable maps on the set
\xdef\defofep{\equanosectno}
$$
E(P)\equaldef\{(t,x):x\in P\,,\,t\in[0,t_x]\}\numeq
$$
in the sense that they can be prolonged to
maps of class $\C^1$ on
some open subset of $\R\times P$ containing $E(P)$. 
\midbreak
\itemitem{(Br.3.F)} For every $x\in S\backslash\{0\}$, if we let
$\tilde\xi_x$ denote the curve obtained in an obvious
way by piecing together the trajectories on every single cell, 
and write $\tilde\eta_x(t)=v(\tilde\xi_x(t))$, then 
the admissible pair $(\tilde\xi_x,\tilde\eta_x)$
ends at the origin after passing from one cell to another
a finite number of times,
and is extremal. 
\midbreak
\itemitem{(Br.3.G)} The cost function $V^\Xi\equaldef V_{\Gamma(\Xi)}$
corresponding to the synthesis
$\Gamma(\Xi)=\{(\tilde\xi_x,\tilde\eta_x)\}_{x\in S}$  
is continuous. \proofend

}
}

\midbreak


\noindent Boltyanskii's definition, as given in \boltcount{}, includes an
extra ingredient, namely, a subset $N$ of $S$ where the synthesis is
allowed to have a more singular behavior.  The definition is formulated in
terms of a sequence ${\bf P}=(P^0,\ldots,P^n)$ of subsets of $S$ 
such that 
\vskip -0.3cm
\xdef\incrunion{\equanosectno}
$$
P^0\subseteq P^1\subseteq\ldots\subseteq P^n=S\,,\numeq
$$
and the object that we 
called ${\cal P}$ here (that is, the set of ``cells'') is the union
\vskip -0.3cm
\xdef\pisunion{\equanosectno}
$$
{\cal P}=\bigcup_{i=0}^n{\cal P}^i\,,\numeq
$$
where 

\midbreak
\itemitem{(Bo.*)} {\sl ${\cal P}^i$, for $i\geq 0$,
is the set of all connected components of $P^i\setminus(P^{i-1}\cup N)$,
where $P^{-1}=\emptyset$} 

\medbreak\noindent so now ${\cal P}$ is a partition of 
$(S\setminus N)\cup\{0\}$
rather than of $S$. Boltyanskii does not explicitly require that this
partition be locally finite in $S$. He only asks that the sets $P^i$ and
$N$ be ``piecewise smooth'' (that is, locally finite unions of
``curvilinear polyhedra,'' i.~e.  of sets that are ${\cal
C}^1$-diffeomorphic to closed bounded polyhedra in finite-dimensional
Euclidean spaces), but this does not imply that ${\cal P}$ is locally
finite. (In fact, in \S 5, Example 5.2, we exhibit a Boltyanskii regular
synthesis whose set of cells is not locally finite.)

In addition, Boltyanskii imposes some extra requirements, which imply
some of the Brunovsk\'y conditions.  In order to facilitate
the comparison, we will include in our definition of a Boltyanskii
synthesis all the Brunovsk\'y conditions that necessarily follow from the
Boltyanskii assumptions, even though this will introduce a number of
redundancies.

\defi 
A {\it Boltyanskii regular synthesis} on the open subset $S$ of the open
set $\Omega\subseteq\rr^n$, for the control problem (\system), (\minim),
with target the origin, is an 8-tuple 
$\Xi=({\bf P},\P_1,\P_2,\Pi,\Sigma,v,N,\Xi^N)$ such that

\midbreak

{\sl 

{\parindent=27pt

\itemitem{(Bo.1)} ${\bf P}=(P^0,\ldots,P^n)$ is a sequence of
piecewise smooth subsets of $S$ such that (\incrunion) holds.

\midbreak
\itemitem{(Bo.2)} $N$ is a piecewise smooth subset of $S$ such that ${\rm
dim}(N)<n$. 

\midbreak
\itemitem{(Bo.3)} If ${\cal P}$---the set of ``cells''---is defined by
(\pisunion), with the ${\cal P}^i$ defined by (Bo.*), then all the cells
are embedded submanifolds of $S$, of class ${\C}^1$, such that ${\rm
dim}(P)=i$ whenever $P\in{\cal P}^i$.  {\rm \quad [BP\&{}HS: The cells are
obviously connected, by definition, and form a partition of $(S\setminus
N)\cup\{0\}$. But ${\cal P}$ need not be locally finite and {\it a
fortiori} ${\cal P}$ need not be a stratification.]}

\midbreak
\itemitem{(Bo.4)} $dim(\Pi(P))=\dim(P)-1$ whenever $P\in\P_1$.

\midbreak
\itemitem{(Bo.5)} The only zero-dimensional cell is $\{0\}$.

\midbreak
\itemitem{(Bo.6)} $\Xi^N$ is a family of pairs in 
$\Admo(\tf,\{0\})$ such that (a)~if $(\xi,\e)\in\Xi^N$ then 
$\xi^-\in N$, (b)~for every $x\in N$ there is a $(\xi,\e)\in\Xi^N$ such 
that $\xi^-=x$, and (c)~if $(\xi_1,\e_1)$ and $(\xi_2,\e_2)$ belong to 
$\Xi^N$, and $\xi_1^-=\xi_2^-$, then $J(\xi_1,\e_1)=J(\xi_2,\e_2)$. 

\midbreak
\itemitem{(Bo.7)} Conditions (Br.2) and (Br.3) of Brunovsk\'y's
definition hold, with the following modifications:  (a)~$v$ is defined
on $S\setminus N$ rather than on $S$, (b)~the stratification 
requirement (Br3.A) is dropped, and (c)~$V^{\Xi}(x)$ is defined to be 
equal to $J(\tilde\xi_x,\tilde\e_x)$ if $x\in S\setminus N$, and 
to $J(\xi,\e)$ if $x\in N$ and $(\xi,\e)$ is any member of $\Xi^N$ such
that $\xi^-=x$.  

\midbreak
\itemitem{(Bo.8)} Whenever the trajectory $\tilde\xi_x$ enters a new cell,
it does so ``at a nonzero angle''. (That is, more precisely:
whenever $P\in\P_1$ and $x\in P$, the tangent vector
to $\tilde\xi_x$ at time $t_x$ is not tangent to $\Pi(P)$ at 
$\tilde\xi_x(t_x)$.) \proofend

}
}

\medbreak\noindent It is then clear that the two concepts of a regular
synthesis defined by Boltyanskii and Brunovsk\'y are not comparable.  A
Brunovsk\'y synthesis can fail to be a Boltyanskii synthesis because

\medbreak
\item{a.} it violates the ``nonzero angle'' condition 
(Bo.8), as in the example given by Brunovsk\'y in 
\brunbcount{}, 
\medbreak\noindent or 
\item{b.} it violates the dimension condition (Bo.4), as in our Example 5.4
below,  
\medbreak\noindent or 
\item{c.} it violates condition (Bo.5), as in Example 5.1 below.

\medbreak\noindent
On the other hand, a Boltyanskii synthesis can fail to be a Brunovsk\'y
synthesis because, for example, the set of cells could fail to be locally
finite, as in our Example 5.2. Also, the Brunovsk\'y definition, as stated
in \brunbcount, does not allow for an extra ``singular set'' $N$.

\medbreak\noindent
The uniqueness requirement of $(Br.3.D)$ is rather strong.  In 
\sussacount{} Sussmann proposed an even less restrictive
formulation, as we now explain.  For a $P\in{\cal P}_2$, we not only
specify a cell $\Sigma(P)\in{\cal P}_1$ as in (Br.3.D), but also give a
continuous ``exiting map'' $\E_P$, defined on
the set $\{(x,t):x\in P,0\le
t<\ve(x)\}$ and with values in $P\cup\Sigma(P)$. Here $\ve:P\to\R$ is a
continuous strictly positive function, and the map $\E_P$ is required to be
such that, for every $x\in P$, (a)~$\E_P(x,0)=x$, and (b)~the map
$\,]\,0,\ve(x)\,[\,\ni t\to\E_P(x,t)$ takes values in $\Sigma(P)$ and is an
integral curve of $f(\cdot,v(\cdot))$. So in
\sussacount{} a {\sl synthesis} is not just a 6-tuple 
$\Gamma=(\P,\P_1,\P_2,\Pi,\Sigma,v)$ as in the Brunovsk\'y
definition, but a 7-tuple
$\Gamma=(\P,\P_1,\P_2,\Pi,\Sigma,v,\E)$, where the extra ingredient
is the family $\E=\{\E_P\}_{P\in{\cal P}_2}$ of exiting maps. 

\rema
The map $\E_P$ can be thought of as way of selecting in a continuous 
fashion, for each $x\in P$, an initial piece $\,]\,0,\ve(x)\,[\,\ni t\to
\E_P(x,t)\in\Sigma(P)$ of an integral curve of the feedback 
vector field $\Sigma(P)\ni y\to f(y,v(y))\in T_y\Sigma(P)$
such that $\lim_{t\downarrow 0}\E_{P,x}(t)=x$.
Naturally, if (Br.3.D) was satisfied we would just choose 
this curve to be the restriction of $\xi_x$ to $]0,t_x[$. \proofend

It is not hard to produce a concept of regular synthesis that
contains as special cases the three definitions of 
\boltcount, \brunbcount, and \sussacount{} but does not differ 
too much from any of them. This can be done by removing from 
the definitions of \boltcount, \brunbcount, and \sussacount{} 
conditions that not really needed, such as the local finiteness
requirements and Boltyanskii's conditions (Bo.4), (Bo.5) 
and (Bo.8). One can also eliminate the requirement 
that the control space $U$ be a subset of some Euclidean space $\R^m$, 
and take $U$ to be an arbitrary set with no extra structure, provided 
only that all the conditions involving differentiability of the 
feedback control $x\to v(x)$ are replaced by 
a differentiability requirement for the 
map $x\to\tf(x,v(x))$. One possible result is the following
concept, that we will call ``BB-regular synthesis,'' using 
the letters BB to stand for ``Boltyanskii and Brunovsk\'y'':

\xdef\defibibi{\the\sectno.\the\theono}
\defi A {\sl BB-regular synthesis}
on the open subset $S$ of $\Omega$, for the control problem (\system),
(\minim), with target $\{0\}$, is a $9$-tuple
$\Xi=(\P,\P_1,\P_2,\Pi,\Sigma,\E,v,N,\Xi^N)$ such that

{\sl 
\midbreak
\itemitem{(BB.1)} $0\in N\subseteq S$, and 
$\P$ is a finite or countable partition of $S\setminus N$ into nonempty
connected embedded submanifolds of $S$ of class ${\C}^1$(called ``cells'');

\midbreak
\itemitem{(BB.2)} $\P$ is the disjoint union of 
$\P_1$ (the set of ``type I cells'') and
$\P_2$ (the set of ``type II cells''), 

\midbreak
\itemitem{(BB.3)} $v:\bigcup\{P:P\in\P_1\}\rightarrow U$ (the ``closed loop
control''), $\Pi:\P_1\rightarrow\P\cup\{\{0\}\}$, 
and $\Sigma:\P_2\rightarrow\P_1$, are maps, 
and $\E$ is a family $\{\E_P\}_{P\in\P_2}$, 
such that the following properties are satisfied:

\midbreak

{\parindent=27pt


\itemitem{(BB.3.A)} If $P\in\P_1$, then the map $P\ni x\to \tf(x,v(x))$ is
of class $\C^1$, and such that $f(x,v(x))$ belongs to $T_x P$ (the tangent
space to $P$ at $x$) for every $x\in P$.

\midbreak
\itemitem{(BB.3.B)}  If $P\in\P_2$, then $\Sigma(P)\in\P_1$,
and $\E_P$ is a map that assigns to each $x\in P$ a maximally defined 
integral curve $\E_{P,x}$ of the vector field 
$\Sigma(P)\ni y\to f(y,v(y))\in T_y\Sigma(P)$,  
in such a way that $0=\inf\,\Dom(\E_{P,x})$ and 
$\lim_{t\downarrow 0}\E_{P,x}(t)=x$. 

\midbreak
\itemitem{(BB.3.C)}  For $x\in P\in\P_2$, define a map
$\xi_x:\{0\}\cup\Dom(\E_{P,x})\to\Omega$ by letting $\xi_x(0)=x$ and
$\xi_x(t)=\E_{P,x}(t)$ for $t\in\Dom(\E_{P,x})$.  For $x\in P_1\in\P_1$,
let $\xi_x$ be the maximally defined solution of the initial value problem
(\initval).  Define $t_x=\sup\,\Dom(\xi_x)$ for $x\in P\in\P$.  Then for
every cell $P\in\P$ the function $P\ni x\to t_x$ is of class ${\cal C}^1$, and
the maps $\hat E(P)\ni(t,x)\to\xi_x(t)\in\Omega$ and $\hat
E(P)\ni(t,x)\to\tf(\xi_x(t),v(\xi_x(t)))\in\R^{n+1}$ have continuously
differentiable extensions to a neighborhood of $E(P)$ in $\R\times P$, where
$E(P)$ is the set defined by (\defofep), and $\hat E(P)=\{(t,x)\in
E(P):0<t<t_x\}$.

\midbreak
\itemitem{(BB.3.D)}  If $x\in P\in\P_1$, then the limit 
$\xi_x(t_x-)=\lim_{t\uparrow t_x}\xi_x(t)$---which exists 
because of (BB.3.C)---belongs to $\Pi(P)$. 

}


\midbreak
\itemitem{(BB.4)} Boltyanskii's condition (Bo.6) holds. 

\midbreak
\itemitem{(BB.5)} Brunovsk\'y's condition (Br.3.F) holds for
$x\in S\setminus N$. 

\midbreak
\itemitem{(BB.6)} Brunovsk\'y's condition (Br.3.G) holds, with $V^\Xi$
defined as in (Bo.7.c). \proofend

}


It is then easy to show that if $\Xi$ is a regular synthesis in the sense
of Boltyanskii \boltcount,  Brunovsk\'y \brunbcount, or Sussmann
\sussacount, then it is possible to associate to $\Xi$ 
a BB-regular synthesis in the sense of
Definition \defibibi.  Moreover, a BB-regular synthesis gives rise in a
natural way to a regular presynthesis $\Gamma(\Xi)$ in the sense of our
Definition \maindef{}, which is a synthesis if $N=\{0\}$.

Our Definition \maindef{}---a slightly different version
of which was already proposed in
\sussmcount{}---is even more general, because:

\medbreak
\item{a.} We do not require that $V_\Gamma$ be continuous,
and only assume the weak continuity conditions.

\medbreak
\item{b.} We do not require that the domain of the synthesis be an open
set. This is quite important because one wants the theory to apply to
systems that are not locally controllable, and in those cases one usually
wants to take $S$ to be the set $\hat S$ of all points of $\Omega$ that can
be steered to the target, and in general $\hat S$ is not open.

\medbreak
\item{c.} We dot require the existence of a partition into ``cells.'' 
>From our point of view, such a partition is only needed to guarantee the
weak differentiability condition (\maindef.b).  

\medbreak
\item{d.} Even when a good partition into ``cells'' exists, satisfying 
the most restrictive conditions of both definitions (that is, 
Brunovsk\'y's stratification condition (Br.3.A) and 
Boltyanskii's conditions (Bo.4), (Bo.5), and (Bo.8)), it may still happen
that the crucial ``finite  number of steps'' condition 
(Br.3.F) is violated, as in Fuller's example, but our requirements for a
regular presynthesis are still met. 

\section{Examples}

\noindent We now present several examples of optimal
regular syntheses to illustrate the differences between the 
various definitions. 

One example showing how our notion of regular synthesis is more general
than those of Boltyanskii and Brunovsk\'y's was already discussed in \S 3,
where we showed that our theory applies to Fuller's problem while the other
ones clearly do not, since they do not allow the trajectories of the
synthesis to have infinitely many switchings.  Another important example,
relevant for the comparison of the Boltyanskii and Brunovsk\'y 
definitions, is the one given by Brunovsk\'y in \brunbcount, showing 
how the ``nonzero angle'' condition (Bo.8) can be violated. 

We now show by means of an example how Boltyanskii's condition (Bo.5) can
be violated. In principle, it is quite easy to produce examples of
syntheses that satisfy all of Boltyanskii's conditions except for
(Bo.5). It turns out, however, that in most such examples one can
construct a new synthesis where (Bo.5) holds, for example by
incorporating the zero-dimensional cells in the singular set $N$, or by
drawing a one-dimensional arc $A_q$ through every $q$ such that $\{q\}$ is
a cell and making the $A_q$'s cells of the new synthesis. To produce an
example where no synthesis can possibly satisfy (Bo.5)  more
work is needed, but it can be done, as we now show.

\exam Let 

\medbreak
\line{\hfill\hbox to 4cm{$S_1=\{(x,0):x\le 0\}\,,$\hfill}\hfill
\hbox to 4cm{$S_2=\{(x,x):0\le x\le 1\}\,,$\hfill}\hfill}

\line{\hfil\hfill\hbox to 4cm{$S_3=\{(x,2x-1):x\ge 1\}\,.$\hfill}\hfill
\hbox to 4cm{$S_4=\{(x,0):x\ge 0\}\,,$\hfill}\hfill\hfil}


\medbreak\noindent Let $\Lambda_1=S_1\cup S_2\cup S_3$,
$\Lambda_2=S_1\cup S_4$. Let 
$A$ be the set of points $(x,y)$ such that $x\ge 0$ and either $0\le y\le
x\le 1$ or $0\le y\le 2x-1$, and let $B$ be the set of those $(x,y)$ such that 
either $y\le 0$, or $x\le 0$, or $0\le x\le 1$ and $x\le y$, or 
$x\ge 1$ and $y\ge 2x-1$.     Let  
$\varphi,\psi:\rr^2\to\rr$ be smooth nonnegative functions such that 
$\varphi(x,y)=0$ if and only if $(x,y)\in A$,
and $\psi(x,y)=0$ if and only if $(x,y)\in B$. 

We consider the minimum time optimal control problem in $\rr^2$ with 
target $\{(0,0)\}$, and dynamics given by 
$$
\dot x={u_1\over 1+\varphi(x,y)}\,,\qquad\qquad
\dot y={u_2 \over 1+\psi(x,y)}\,,
$$
where the control constraint is $|u_1|+|u_2|\le 1$. 

If $q=(\bar x,\bar y)$ is any point in $\rr^2$, and $[a,b]\ni
t\to\xi(t)=(x(t),y(t))\in\rr^2$  
is any trajectory from $q$ to the target, corresponding to a control
$[a,b]\ni t\to\eta(t)=(u_1(t),u_2(t))$, then 
$$
b-a\ge\int_a^b\Big(|u_1(t)|+|u_2(t)|\Big)\,dt\ge\int_a^b\Big(|\dot
x(t)|+|\dot y(t)|\Big)\,dt\ge|\bar x|+|\bar y|\,.
$$
Moreover,  a trajectory $\xi_q$ from $q$ to the origin whose cost 
is exactly $|\bar x|+|\bar y|$ can be constructed as follows:

\medbreak
\item{1.}  If $q$ belongs to $\Lambda_1$,  then the point $\xi_q(t)$ moves
along $\Lambda_1$ towards the origin, with a control vector 
$\eta_q(t)=(u_1(t),u_2(t))$ such that $|u_1(t)|+|u_2(t)|=1$ (so that
$\eta_q(t)=(1,0)$ if $q\in S_1$ and $\eta_q(t)=(-{1\over 2},-{1\over 2})$ 
if $q\in  S_2$;  if $q\in S_3$, then 
$\eta_q(t)=(-{1\over 3},-{2\over 3})$  until $\xi_q(t)=(1,1)$, and 
$\eta_q(t)=(-{1\over 2},-{1\over 2})$ from then on). 
\item{2.}  If $q\in A$, then $\xi_q(t)$ moves horizontally to the left 
(with control $(u_1,u_2)=(-1,0)$) until it reaches $\Lambda_1$, and then 
follows the trajectory described in item 1.
\item{3.}  If $q\in B$ and  $\bar y>0$, then $\xi_q(t)$ moves vertically
down (with control $(u_1,u_2)=(0,-1)$) until it reaches
$\Lambda_1$, and then  follows the trajectory described in item 1.
\item{4.}  If $q\in B$ and $\bar y<0$, then $\xi_q(t)$ moves vertically up
(with control $(u_1,u_2)=(0,1)$) until it reaches $\Lambda_2$, and then 
moves horizontally towards the origin with control $(1,0)$ 
or $(-1,0)$. 

\medbreak\noindent The family $\Xi=\{\xi_q\}_{q\in\rr^2}$ is an optimal
synthesis.  If we define $P^0$ to be the two-point set consisting of
$(0,0)$ and $(1,1)$, and let $P^1=\cup_{i=1}^4S_i$, $P^2=\rr^2$, and
$N=\emptyset$, then all the conditions of a Boltyanskii regular synthesis
are satisfied (with $17$ cells), except for (Bo.5).  Moreover, it is easy
to show that {\it this problem does not admit a regular synthesis that
satisfies all the Boltyanskii conditions, including (Bo.5)}.
\longversion{(The proof is as follows.  The optimal trajectories for our
problem are clearly unique for all initial points $q=(\bar x,\bar y)$. So
for any optimal synthesis the trajectories are the $\xi_q$. Assume we have
a regular synthesis satisfying all of Boltyanskii's conditions.  Then the
point $(1,1)$ cannot belong to the singular set $N$, because the definition
of a Boltyanskii synthesis implies that every marked trajectory starting at
a point of $\R^2\setminus N$ is entirely contained in $\R^2\setminus N$, so
the set $U=\{(x,y):x\ge 1\,,\,y\ge 1\}$ must be entirely contained in
$N$---because if $q\in U$ then $\xi_q$ goes through $(1,1)$---and this
contradicts the fact that the dimension of $N$ is at most $1$.  Now, let
$C$ be the cell containing $(1,1)$. Since every optimal trajectory going
through $(1,1)$ has a discontinuous velocity at $(1,1)$, the cell $C$
cannot be of type I. In particular, $C$ is not two-dimensional, so (Bo.5)
implies that it is one-dimensional.  It follows from the definition of a
Boltyanskii synthesis that the direction $\nu(q)$ of the marked trajectory
$\xi_q$ at its starting point $q$ must be a continuous function of $q$ as
long as $q\in C$.  But $\nu(1,1)=(-{1\over 2},-{1\over 2})$, and there is no
direction along which $\nu(q)$ is continuous as $q\to (1,1)$,
so it is impossible for $C$ to be
one-dimensional.)}\proofend

\medbreak In our second example, we present a Boltyanskii
optimal synthesis which is not a Brunovsk\'y synthesis,
because the set of cells is not locally finite. 

\exam Let $\psi:\rr\to\rr$ be a function of class ${\cal C}^\infty$ 
such that (a)~$\psi(x)=0$ whenever $x\le 0$, (b)~$0\le\psi(x)<x$ when 
$x>0$, (c)~the set of zeros of $\psi$ is the union of $\{0\}$ and the
points  of a decreasing infinite sequence $\{x_k\}_{k=1}^\infty$ having an
accumulation point at $0$, (d)~for every $k\geq 1$ there exists
$\xb_k\in ]x_{k+1},x_k[$ such that $\psi$ is strictly increasing
on $[x_{k+1},\xb_k]$ and strictly decreasing on $[\xb_k,x_k]$
and (e) $\psi$ is strictly increasing on $[x_1,+\infty[$. 
Let
$$
\eqalign{
A&=\{(x,y)\in\rr^2:x\ge 0\,\,\,{\rm and}\,\,\,-\psi(x)\le y\le\psi(x)\}\,,\cr
B&=\{(x,y)\in\rr^2:x\le|y|\}\,,\cr
E&=\Big\{(x,y)\in\rr^2:x\ge 0\,\,{\rm and}\,\,\Big(\psi(x)\le y\le
x\,\,{\rm or}\,\,-x\le y\le-\psi(x)\Big)\Big\}\,.\cr}
$$
Let $L$ be the $y$ axis. 
Then $A$, $B$, $E$, and $L$ are closed subsets
of $\rr^2$. Let $\sigma:\rr^2\to\rr$ be a function of class
${\cal{}C}^\infty$ such that $\sigma\equiv 0$ on $A\cup B$ and $\sigma>0$
on $\rr^2\setminus(A\cup B)$. Let $\tau:\rr^2\to\rr$ be a function of class 
${\cal C}^\infty$ such that $\tau\equiv 0$ on $E\cup L$ and $\tau>0$ on 
$\rr^2\setminus(E\cup L)$. 

Consider the optimal control problem in $\rr^2$ with target $\{(0,0)\}$, 
in which the dynamics is given by
$$
\dot x=u_1\,,\qquad\dot y=u_2\,,
$$
the control constraints are $u_1\in\{-1,0,1\}$ and $|u_1|+|u_2|=1$, and the
cost functional  to be minimized is
$$
J=\int_a^b\biggl(u_1(t)^2\Big(1+\sigma(x(t),y(t))\Big)
+u_2(t)^2\tau(x(t),y(t))\biggr)\,dt\,.
$$
If $q=(\bar x,\bar y)\in\rr^2$, and $[a,b]\ni t\to\xi(t)=(x(t),y(t))$ 
is any trajectory going from $q$ to the origin, 
then $J(\xi)\ge\int_a^bu_1(t)^2\,dt=\int_a^b|u_1(t)|\,dt\ge
|\int_a^b\dot x(t)\,dt|=|\bar x|$.  On the other hand, there exists a
trajectory $\xi_q$ for which $J(\xi_q)=|\bar x|$.  To see this, observe that
purely vertical motion---i.~e. $|u_2|=1$ and $u_1=0$---costs nothing 
as long as it takes place in $E\cup L$, while 
horizontal motion---i.~e. $|u_1|=1$ and $u_2=0$---has a cost per unit time 
equal to $1$ on $A\cup B$.  If $q\in B$, then we can
move horizontally towards the $y$ 
axis---using $u_1=-{\rm sgn}(\bar x)$---and then move vertically along the
$y$ axis and end up at  
the origin, thereby obtaining a trajectory whose cost is exactly $|\bar x|$.  
If $q\in E$, then we can move
vertically up or down---using $u_2={\rm sgn}(\bar y)$---until we 
hit the boundary of $B$. Once we are in $B$, 
we follow the optimal trajectory already described for points of $B$, and 
end up producing a trajectory from $q$ to the origin 
whose cost is $|\bar x|$. If $q\in A$ and $\bar y=0$, then
the constant control  
$u_1=-1$, $u_2=0$ steers $q$ to the origin with cost $\bar x$.
If $q\in A$ and $\bar y\neq 0$, then we can use 
the constant control $u_1=-1$, $u_2=0$ until we reach a point $(\hat x,\hat y)$
such that $|\hat y|=\psi(\hat x)$.  The cost of this is clearly 
$\bar x-\hat x$, and $(\hat x,\hat y)$ belongs to $E$, so  
we know how to go from $(\hat x,\hat y)$ to the origin with cost $\hat x$. So 
we can go from $q$ to the origin with cost $\bar x$.

Therefore the value function $V$ for this problem is given by
$V(x,y)=|x|$. Moreover, we have given an explicit description of a
family $\Xi=\{\xi_q\}_{q\in\rr^2}$ which is a total optimal
synthesis. 

We now show that {\it $\Xi$ is the family of trajectories
of a Boltyanskii regular synthesis whose set of cells is not locally
finite.}  To see this, we define $N$ to be the set
$[0,\infty\,[\,\times\{0\}$, so $N$ is clearly piecewise smooth. We
take $P^0=\{(0,0)\}$, and let $P^1$ be the union 
of the two coordinate axes, the two half-lines
$L_1=\{(x,x):x\ge 0\}$ and $L_2=\{(x,-x):x\ge 0\}$, and 
the graphs of $\psi$ and  $-\psi$. 
Finally, we let $P^2=\rr^2$. It is then clear that $P^0$, $P^1$,
and $P^2$ are piecewise smooth.   The connected components 
of $P^1\setminus(P^0\cup N)$ are the open half-lines 
\shortversion{$\Lambda_1=\{(x,x):x>0\}$, $\Lambda_2=\{(0,y):y>0\}$,
$\Lambda_3=\{(x,0):x<0\}$, $\Lambda_4=\{(0,y):y<0\}$, 
and $\Lambda_5=\{(x,-x):x> 0\}$, and the arcs
$A_k^+=\{(x,\psi(x)):x_{k+1}<x<x_k\}$, 
$A_k^-=\{(x,-\psi(x)):x_{k+1}<x<x_k\}$.}
\longversion{
$$
\left.\matrix{\boxa{\Lambda_1=\{(x,x):x>0\}\,,}&
\boxa{\Lambda_2=\{(0,y):y>0\}\,,}&
\boxa{\Lambda_3=\{(x,0):x<0\}\,,}\cr
\boxa{\Lambda_4=\{(0,y):y<0\}\,,}&
\boxa{\Lambda_5=\{(x,-x):x> 0\}\,,}&
\boxa{\,\,}\cr}\right.
$$
and the arcs
$$
A_0^+=\{(x,\psi(x)):x>x_1\}\,,\qquad
A_0^-=\{(x,-\psi(x)):x>x_1\}\,.
$$
$$
A_k^+=\{(x,\psi(x)):x_{k+1}<x<x_k\}\,,\qquad
A_k^-=\{(x,-\psi(x)):x_{k+1}<x<x_k\}\,.
$$}
The connected components of 
$P^2\setminus(P^1\cup N)$ are the sets
\shortversion{$U_1=\{(x,y):x>0\,,\,\psi(x)<y<x\}$, 
$U_2=\{(x,y):0<x<y\}$, $\qquad U_3=\{(x,y):x<0<y\}$, 
$\qquad U_4=\{(x,y):x<0\,,\,y<0\}$, $U_5=\{(x,y):x>0\,,\,y<-x\}$, 
and $U_6=\{(x,y):x>0\,,\,-x<y<-\psi(x)\}$, and the regions
$W_k^+=\{(x,y):x_{k+1}<x<x_k\,,\,0<y<\psi(x)\}$, 
$W_k^-=\{(x,y):x_{k+1}<x<x_k\,,\,0>y>-\psi(x)\}$.}
\longversion{
$$
\left.\matrix{\boxo{U_1=\{(x,y):x>0\,,\,\psi(x)<y<x\}\,,}&
\boxo{U_2=\{(x,y):0<x<y\}\,,}\cr
\boxo{U_3=\{(x,y):x<0<y\}\,,}&\boxo{U_4=\{(x,y):x<0\,,\,y<0\}\,,}\cr
\boxo{U_5=\{(x,y):x>0\,,\,y<-x\}\,,}&
\boxo{U_6=\{(x,y):x>0\,,\,-x<y<-\psi(x)\}\,,}\cr}\right.
$$
and the bounded regions
$$
W_0^+=\{(x,y):x>x_1\,,\,0<y<\psi(x)\}\,,\,\,\,
W_0^-=\{(x,y):x>x_1\,,\,0>y>-\psi(x)\}\,.
$$
$$
W_k^+=\{(x,y):x_{k+1}<x<x_k\,,\,0<y<\psi(x)\}\,,\,\,\,
W_k^-=\{(x,y):x_{k+1}<x<x_k\,,\,0>y>-\psi(x)\}\,.
$$}
The arcs $A_0^+$, $A_0^-$,
$A_k^+$ and $A_k^-$ and the half-lines 
$\Lambda_1$, $\Lambda_5$ are declared to be type II cells, and
\shortversion{we choose $\Sigma(A_k^+)=U_1$, $\Sigma(A_k^-)=U_6$, 
$\Sigma(\Lambda_1)=U_2$, $\Sigma(\Lambda_5)=U_5$.}   
\longversion{we choose 
$$
\Sigma(A_0^+)=\Sigma(A_k^+)=U_1\,,\quad 
\Sigma(A_0^-)=\Sigma(A_k^-)=U_6\,, 
\quad\Sigma(\Lambda_1)=U_2\,,\quad \Sigma(\Lambda_5)=U_5\,.
$$}   
The remaining cells (that is, $\Lambda_2$, $\Lambda_3$, $\Lambda_4$,  
the $U_i$, $W_0^+$, $W_0^-$, the $W_k^+$ and the $W_k^-$) are type I cells. 
The feedback $v$ is defined by letting 
\shortversion{$v\equiv(-1,0)$ on $N\setminus\{(0,0)\}$, the $W_k^+$, the
$W_k^-$, $U_2$ 
and $U_5$,  $v\equiv(1,0)$ on $U_3$ and $U_4$, 
$v\equiv(0,1)$ on $U_1$ and $\Lambda_4$,  
$v\equiv(0,-1)$ on $U_5$ and $\Lambda_2$.}
\longversion{
$$
v\equiv\left\{\matrix{\boxir{(-1,0)}&{\rm on}&\boxu{(N\!\setminus\!\{(0,0)\})
\bigcup\,(W_0^+\cup W_0^-) \bigcup\,(\cup_{k}W_k^+)\bigcup\,(\cup_{k}W_k^-)
\bigcup U_2\bigcup U_5\,,}\cr
\boxir{(1,0)}&{\rm on}&\boxu{U_3\cup U_4\,,}\cr
\boxir{(0,1)}&{\rm on}&\boxu{U_1\cup \Lambda_4\,,}\cr
\boxir{(0,-1)}&{\rm on}&\boxu{U_6\cup \Lambda_2\,.}\cr}\right.
$$}
The definitions of $\Pi$ and $\Xi^N$ are the obvious ones.
It is then clear 
that all the conditions of the Boltyanskii definition are
satisfied, but the set of cells is not locally finite.  \proofend

\exam  We now present an example of a synthesis that satisfies all the 
Brunovsk\'y conditions except for the continuity requirement 
(Br.3.G). This example will, in addition, exhibit the phenomenon of
nonuniqueness of trajectories for the closed-loop equation arising from 
the optimal feedback, thereby providing a concrete illustration of
the reasons for including as an extra ingredient  a selection of 
trajectories, i.~e., for following strategy (B) of \S 4. 

We construct and study in detail the time-optimal synthesis for the
planar system: 
\xdef\eqoo{\equanosectno} 
$$
\dot q= F(q)+ u\,G(q)\,,\qquad |u|\leq 1\,, \numeq
$$
where
\xdef\eqot{\equanosectno} 
$$
q\equaldef\left(\matrix{x\cr \cr y\cr}\right)\,\,\in\,\,\R^2\,,\qquad\,\,\,
F(q)\,\equaldef \left(\matrix{1-{y\over 2}\cr \cr {x+1\over
2}\cr}\right)\,,\qquad\,\,\,
G(q)\,\equaldef\left(\matrix{-{y\over 2}\cr \cr {x+1\over 2}\cr}\right)\,,
\numeq
$$
and the target is the origin of $\R^2$.

The trajectories of (\eqoo) corresponding to the constant control $u\equiv
-1$ are straight horizontal lines going from left to right, while those
corresponding to $u\equiv +1$ are circles centered at the point (-1,1),
running counterclockwise.  The optimal synthesis, to be determined below,
is shown in Fig.2.

\shortversion{
An arc $\gamma:[a,b]\to\R^2$ is {\it simple} if $\gamma$ is an injective
map. Clearly, all optimal trajectories are simple.} 

\shortversion{
Since $G(q)=0$ only for $q=(-1,0)$, it is clear that a trajectory
$\gamma:[a,b]\to\R^2$ uniquely determines the corresponding control
$u(\cdot)$, unless the set $\{t:\gamma(t)=(-1,0)\}$ is of positive
measure. In particular, for simple trajectories $\gamma$---the only ones
that will be of interest to us---the control is uniquely determined by
$\gamma$.}

\vskip 0.5em
\vbox{

\centerline{\hbox{\psfig{figure=benny.ps,width=3in,height=3in}}}

\medbreak
\centerline{Figure 2}

}

\vskip 1em

%A {\it control} is a measurable function $u:[a,b]\rightarrow [-1,1]$ for
%some $a,b\in \R$ such that $a\le b$.  (As usual, we identify measurable
%functions that are equal almost everywhere.)  A {\it trajectory} for a
%control $u$ is an absolutely continuous function
%$\gamma:[a,b]\rightarrow\R^2$ verifying
%$\dot\gamma(t)=F(\gamma(t))+u(t)G(\gamma(t))$ for almost every $t\in[a,b]$.
%A trajectory $\gamma:[a,b]\to\R^2$ is {\it optimal} if there is no
%trajectory $\gamma':[a',b']\to \R^2$ such that $\gamma'(a')=\gamma(T)$,
%$\gamma'(b')=\gamma(b)$, and $b'-a'<b-a$.

\longversion{
An arc $\gamma:[a,b]\to\R^2$ is {\it simple} if $\gamma$ is an injective
map. Clearly, all optimal trajectories are simple.} 

\longversion{
Since $G(q)=0$ only for $q=(-1,0)$, it is clear that a trajectory
$\gamma:[a,b]\to\R^2$ uniquely determines the corresponding control
$u(\cdot)$, unless the set $\{t:\gamma(t)=(-1,0)\}$ is of positive
measure. In particular, for simple trajectories $\gamma$---the only ones
that will be of interest to us---the control is uniquely determined by
$\gamma$.}

As in previous sections, we use $\R_k$ to denote the space of real row
$k$-vectors. If $\Omega$ is an open subset of $\R^k$ and $X,Y:\Omega\to\R^k$
are vector fields of class $C^1$, then $[X,Y]$ will denote the Lie bracket
of $X$ and $Y$, so $[X,Y]=(DY).X-(DX).Y$.  
%A {\it variational covector field} (or ``adjoint vector'') along a
%trajectory $\gamma$ of (\eqoo)---corresponding to a control 
%$u(\cdot)$---is an absolutely continuous function
%$\lambda:[a,b]\rightarrow\R^2_*$  that verifies the ``adjoint equation''
%$$
%\dot\lambda(t)=-\lambda(t)\cdot\Bigl(\nabla F(\gamma(t))+
%u(t)\nabla G(\gamma(t))\Bigr) \numeq
%$$
%for almost every $t\in[a,b]$. We call $\lambda$ {\it nontrivial}
%if $\lambda(t)\not\equiv 0$. We define the {\it Hamiltonian}
%$H^c:\R^2_*\times\R^2\times\R\rightarrow \R$ by:
%$$
%H^c(\lambda,q,u)=\lambda\cdot\Bigl(F(q)+uG(q)\Bigr)\,.\numeq
%$$
%A covector field $\lambda$ along $\gamma$ is said to be {\it minimizing}
%if:
%\xdef\mineq{\equanosectno} 
%$$
%H^c\Bigl(\lambda(t),\gamma(t),u(t)\Bigr)=
%\min\{H^c(\lambda(t),\gamma(t),w):|w|\leq 1\}\,\,\,\,{\rm for\,\,\,
%a.e.}\,\,\,t\in[a,b]\,.\numeq
%$$
%If a trajectory $\gamma$ is optimal, then $\gamma$ is simple, and has to be
%{\it extremal}, meaning that, if $u(\cdot)$ is the corresponding control,
%then the pair $(\gamma,u(\cdot))$ satisfies the necessary condition given
%by the Pontryagin Maximum Principle:
%
%\medbreak
%\item{(E)} {\it there exist a nontrivial
%minimizing variational covector field
%$\lambda$ and a nonnegative constant $\lambda_0$ such that}
%$$
%\lambda_0+H^c(\lambda(t),\gamma(t),u(t))=0
%\,\,\,{\rm for\,\,\,almost\,\,\,every}\,\,\,t\in[a,b]\,.\numeq
%$$

\medbreak\noindent If $X,Y$ are $\R^2$-valued functions on a subset of
$\R^2$, then  $(X,Y)$ will denote the $2\times 2$ matrix with columns
$X,Y$. In particular, following Sussmann \susshcount, we  define two 
real-valued functions $\Delta_A$, $\Delta_B$ on $\R^2$, by letting
$$
\Delta_{A}=det\,(F,G)\qquad\Delta_{B}=det\,(G,[F,G])\numeq
$$
where ``{\it det\/}'' stands for determinant.  
{}From the results of \susshcount, we know that every extremal trajectory 
of (\eqoo) is a finite concatenation of ``bang'' and ``singular'' arcs.
An arc is a {\it bang arc} if it is a $B_-$ arc---i.~e. a trajectory for
the control $u=-1$---or a $B_+$ arc---i.~e. a trajectory for $u=1$. 
Singular arcs lie in the set of zeros of the function $\Delta_{B}$.
In our case,
\xdef\eqdelab{\equanosectno} 
$$
\Delta_{A}={x+1\over 2}\,\,\,\,\,{\rm \,\,\,and\,\,\,}\,\,\,\,\,
\Delta_{B}=-{y\over 4}\numeq
$$
so the only singular arcs are those entirely contained in the $x$ axis,
and these correspond to the control $-1$, so they actually are  bang.
So {\it all the extremals of our problem are bang-bang, that
is finite concatenations of bang arcs}.

If $\Delta_A(q)\not= 0$ then the vectors $F(q)$ and $G(q)$ are linearly
independent, and we can define the numbers $f(q)$ and $g(q)$ as the
coefficients of the linear combination:
\xdef\lieequal{\equanosectno}
$$
[F,G](q)=f(q)F(q)+g(q)G(q).\numeq
$$
Let $\Omega_A$,  $\Omega_B$ be, respectively, the sets 
$\{q:\Delta_A(q)\neq 0\}$, $\{q:\Delta_B(q)\neq 0\}$, and define
$\Omega_{AB}=\Omega_A\cap\Omega_B$. 
It was proved in \susshcount{} that, if $\gamma$ is an extremal trajectory
and $\gamma(t)\in\Omega_{AB}$ for every $t$, then $\gamma$ is bang-bang with at
most one switching, and this switching can 
only be from the control value $-1$ to the value $+1$ if $f>0$,
and from $+1$ to $-1$ if $f<0$. In our case, it follows from (\eqdelab) that
$$
\Omega_A=\{(x,y):x\neq-1\}\,,\quad\,\,\,\,\,\,\Omega_B=\{(x,y):y\neq
0\}\,,\numeq 
$$
so $\Omega_{AB}$  is the union of four open quadrants $\Omega_{AB,i}$,
$i=1,2,3,4$, defined by
\longversion{
$$\eqalign{
\Omega_{AB,1}&=\{(x,y):x>-1\,,\,y>0\}\,,\cr
\,\,\,\Omega_{AB,2}&=\{(x,y):x<-1\,,\,y>0\}\,,\cr
\,\,\,\Omega_{AB,3}&=\{(x,y):(x,-y)\in\Omega_{AB,2}\}\,,\cr
\,\,\,\Omega_{AB,4}&=\{(x,y):(x,-y)\in\Omega_{AB,1}\}\,.\cr}\numeq
$$
}
\shortversion{
$$\left.\matrix{\Omega_{AB,1}=\{(x,y):x>-1\,,\,y>0\}\,,\,\,\,\,&\,\,&
\Omega_{AB,2}=\{(x,y):x<-1\,,\,y>0\}\,,\,\,\,\,\cr
\Omega_{AB,3}=\{(x,y):(x,-y)\in\Omega_{AB,2}\}\,,&\,\,&
\Omega_{AB,4}=\{(x,y):(x,-y)\in\Omega_{AB,1}\}\,.}\right.\numeq
$$
}
Moreover, 
$$
[F,G]= \left(\matrix{0\cr {1\over 2}\cr}\right)=
{y\over 2(x+1)}F+{2-y\over 2(x+1)}G\,,\numeq
$$
so $f>0$ on $\Omega_{AB,1}\cup\Omega_{AB,3}$ and 
$f<0$ on $\Omega_{AB,2}\cup\Omega_{AB,4}$.  So the control switchings
are from $-1$ to $1$ on $\Omega_{AB,1}\cup\Omega_{AB,3}$ and 
{}from $1$ to $-1$ on $\Omega_{AB,2}\cup\Omega_{AB,4}$.

%Assume now to have two trajectories $\gamma_1$, $\gamma_2$, having the
%same initial point and the same terminal point. Following [], we have
%that if $\gamma_2^{-1}*\gamma_1$, that is the concatenation of
%$\gamma_1$ with $\gamma_2$ run backwards, is a simple curve enclosing
%the region $\cal R$ then:
%$$T(\gamma_1)-T(\gamma_2)=\int \int_{\cal R}
%{\Delta_B\over(\Delta_A)^2}$$
%where $T$ is the time along a trajectory.

To determine the optimal synthesis, we will first construct a ``sufficient
family'' of trajectories for our problem, i.~e. a collection ${\cal F}$ of
simple trajectories such that every optimal trajectory ending at the target
is in ${\cal F}$. We will then find, for every initial point $q$, a
$\gamma_q\in{\cal F}$ such that if $\delta\in{\cal F}$ goes from $q$ to $0$
and $\delta\neq\gamma_q$ then $\delta$ is not optimal. Since the existence
of optimal trajectories {}from any initial point follows by elementary
arguments, we will be able to conclude that $\gamma_q$ is optimal.

Suppose that $\gamma:[a,b]\to\R^2$ is time-optimal, $\gamma(b)=(0,0)$,
$a<b$, and $\gamma$ arises from an open-loop control $t\to u(t)$.  Then
$\gamma$ is simple, and is a concatenation of finitely many maximal bang
arcs.  Since $\gamma$ is an extremal, we can fix a nontrivial minimizing
adjoint vector $\lambda$ along $\gamma$ and a constant $\lambda_0\ge 0$ for
which (EX4) holds. Define the {\it switching function}
\longversion{$$\varphi(t)=\lambda(t)\cdot G(\gamma(t))\,.$$}
\shortversion{$\varphi(t)=\lambda(t)\cdot G(\gamma(t))$.}
{}The Hamiltonian minimization condition implies that
$$
\varphi(t)\not= 0\,\, \Longrightarrow \,\,u(t)=-sgn(\varphi(t))\,.\numeq
$$
Let $\delta:[c,d]\to\R^2$ be a maximal $B_-$ piece of $\gamma$, so
$c<d$ and $\delta$ is of the form: 
$$
\delta(t)=(x_0+t,y_0)\,.\numeq
$$
%Since $\delta$ is maximal,
%
%\medbreak
%\item{(i)}~either 
%\itemitem{(i.1)} $c=a$,
%\item{}~or 
%\itemitem{(i.2)} $c>a$ and a control switching from $1$ to $-1$ occurs at $c$,
%\item{}~and
%\item{(ii)}~either 
%\itemitem{(ii.1)} $d=b$,
%\item{}~or 
%\itemitem{(i.2)}$d<b$ and a control switching from $-1$ to $1$ occurs at $c$.

%\medbreak
\noindent For a.e. $t\in[c,d]$, we have
$\dot \lambda=-\lambda\cdot D(F-G)=0$.
So $\lambda=(\lambda_1,\lambda_2)$ is constant,
$(\lambda_1,\lambda_2)\neq(0,0)$, and:
$$
\varphi(t)=-\lambda_1 {y_0\over 2}+\lambda_2 {x_0+t+1\over 2}\numeq
$$
is a linear function of $t$.  Since a linear function cannot have more than
one zero unless it vanishes identically, we conclude that
$a<c<d<b$ can only be true if $\lambda_2=y_0=0$. 
If $d=b$, then of course $y_0=0$, since $\gamma(b)=(0,0)$. 
So one of the following two possibilities occurs:

\medbreak
\itemitem{(I)\phantom{I}}~$\delta$ is contained in the $x$ axis,
\itemitem{(II)}~$a=c$. 
\medbreak

This shows that all the maximal $B_-$ pieces of $\gamma$ are entirely 
contained in the $x$ axis, with the only possible exception that the domain
$[a,b]$ of $\gamma$ may contain an initial segment $[a,d]$ 
such that $\g([a,d])$ is contained in a horizontal line for
which $y\neq 0$. 

%In particular there are extremal trajectories that switch from
%control $+1$ to control $-1$ at a point $(-L,0)$ with $L<0$, and then
%reach the origin with control $-1$.

Let us now analyze the adjoint equation for a maximal $B_+$ piece
$\delta:[c,d]\to\R^2$ of $\gamma$, such that $c<d$. Clearly, $\delta$ is of
the form: 
\xdef\eqxycircle{\equanosectno} 
$$
x(t)=-1+r \cos(t+\theta)\,,\qquad y(t)=1+r \sin(t+\theta)\,,
\,\qquad c\le t\le d\,,\numeq
$$
where $r, \theta$ are constants, $r>0$, and $\theta\in [0,2\pi]$. (The
possibility 
that $r=0$ is obviously excluded, because $\gamma$ is time-optimal and
$c<d$.)  The  components $\lambda_1,\lambda_2$ of $\lambda$ satisfy
$\dot\lambda_1=-\lambda_2$ and $\dot\lambda_2=\lambda_1$, 
so there exist constants $A>0$, $\theta_1\in[0,2\pi]$, such that
$$
\lambda_1(t)=A\cos(t+\theta_1)\,,\qquad\lambda_2(t)=A\sin(t+\theta_1)
\,,\,\qquad c\le t\le d\,.\numeq
$$
Therefore, up to multiplication by a positive constant,
$$
\varphi(t)= -\cos(t+\theta_1)+r\sin(\theta_1-\theta)\,.\numeq
$$
Now suppose that one of the two endpoints $c,d$ of the domain 
of $\delta$ is a switching time and the corresponding point
in $\R^2$ lies in the $x$ axis.  That is, we assume that
$h\in\{c,d\}$ is such that $\varphi(h)=0$ and $y(h)=0$.  Then 
\xdef\rsintheta{\equanosectno} 
$$
r\sin(h+\theta)=-1\,,\numeq
$$
so 
$$
\eqalign{\varphi(t)&=-\cos(t+\theta_1)
+r\Bigl(\sin(h+\theta_1)\cos(h+\theta)
-\cos(h+\theta_1)\sin(h+\theta)\Bigr)\cr
&=-\cos(t+\theta_1)+\cos(h+\theta_1)+r\sin(h+\theta_1)\cos(h+\theta)
\,.}\numeq
$$
In particular, 
\xdef\iszero{\equanosectno} 
$$
0=\varphi(h)=r\sin(h+\theta_1)\cos(h+\theta)\,,\numeq
$$
so $\varphi(t)= -\cos(t+\theta_1)+\cos(h+\theta_1)$. If $r\neq 1$
then (\rsintheta) implies $0\ge \sin(h+\theta)\neq -1$, so 
$\cos(h+\theta)\neq 0$.  But then (\iszero) implies 
$\sin(h+\theta_1)=0$, so 
$\cos(h+\theta_1)=\pm 1$.  The possibility that 
$\cos(h+\theta_1)=1$ is excluded, for in that case
$\varphi$ would be $>0$ throughout $\,]\,c,d\,[\,$, contradicting the
minimization condition. So 
$\varphi(t)= -1-\cos(t+\theta_1)$. This shows that 
$\varphi(t)<0$ for $t\in[c,d]\setminus\{h\}$, unless $d-c=2\pi$.  But 
if $d-c=2\pi$ then $\gamma$ would not be simple.
So $d-c<2\pi$, and we have shown that the other endpoint 
$h'$ of $[c,d]$ cannot be a switching point, and therefore $h'\in\{a,b\}$.

It then follows that one of the following possibilities must occur:

\medbreak
\itemitem{(i)\phantom{ii}} $d=b\,$;
\itemitem{(ii)\phantom{i}} $a=c<d<b$ and $\gamma(d)$ lies in the $x$ axis;
\itemitem{(iii)} $a<c<d<b$ and $\gamma(d)=(-1,0)$.

\medbreak\noindent
Indeed, if $d<b$ but $\gamma(d)$ is not in the $x$ axis, then we could let
$\delta'$ be the maximal $B_-$ piece of $\gamma$ that starts at time $d$,
and our analysis of the $B_-$ pieces of $\gamma$, applied to $\delta'$,
would show that $\delta'$ cannot in fact occur, since $\delta'$ is not an
initial segment of $\gamma$ and is not contained in the $x$ axis.  So, if
$d<b$ then $\gamma(d)$ lies in the $x$ axis, and $d$ is a switching time of
$\gamma$. Our previous argument shows that $c=a$, unless $r=1$.  But if
$r=1$, and $\gamma(d)$ is in the $x$ axis, then $\gamma(d)=(-1,0)$.  So
$d<b$ implies that either (ii) or (iii) must hold.

We now combine the results about the two types of pieces and provide a
complete description of the optimal trajectories ending at the origin. 
Let us use $L$, $L_0$, $C$ to denote, respectively, the trajectory
types ``$B_-$ and not contained in the $x$ axis,''
``$B_-$ and contained in the $x$ axis,'' and ``$B_+$.'' (The symbols
$L$ and $C$ stand for ``line'' and ``circle.'')  Let us 
use $T_1T_2\ldots T_m$ to denote the trajectory type 
``$T_1$ followed by $T_2\,\,\ldots\,\,\,$followed by $T_m$.''

With this notation, we now show that {\sl a nontrivial optimal
trajectory $\gamma$ ending at the origin is of one of the following seven
types}: $L_0$, $C$, $L_0C$, $LC$, $CL_0$, $LCL_0$, $CL_0C$.
First, concatenations of six or more types are ruled out as follows: 
any such concatenation would have to contain at least 
three different $C$ pieces; at most one of them can satisfy (i) and at most
one can satisfy (ii);  if both of these possibilities occurred, then 
we would have a concatenation of the form $C*C*C$, i.~e. five pieces rather
than six; so there must be two different pieces satisfying
(iii), and then $\gamma$ would go at least twice through the point
$(-1,0)$, and would not be simple.  %Concatenations of the form
% $LCLC$, $L_0CLC$, $LCL_0C$ and $L_0CL_0C$ are excluded  for the same
%reason, and this implies that all concatenations of five types
%that start with a $C$ are excluded. 
Next, all concatenations containing an $L_0CL_0$ sequence are excluded,
because if 
$\gamma$ is any such concatenation then the $C$ piece that lies between the
two $L_0$'s would have to satisfy (iii), so its last switching would have to
happen at $(-1,0)$; then the $C$ piece would have to be an arc of 
the circle with center $(-1,1)$ and radius $1$, which intersects the $x$
axis only at $(-1,0)$; then the other switching would also happen at
$(-1,0)$, and $\gamma$ would not be simple. A five-piece concatenation
must be $CL_0CL_0C$, $LCL_0CL_0$ or $L_0CL_0CL_0$, 
because every $B_-$ piece must satisfy (I) or (II), so it cannot be an $L$
unless it is  the first piece. So all such concatenations contain an
$L_0CL_0$ piece and are therefore excluded. A four-piece concatenation 
must be $CL_0CL_0$ or $L_0CL_0C$ or $LCL_0C$. The types 
$CL_0CL_0$ and $L_0CL_0C$ are ruled out because they contain an 
$L_0CL_0$ triple. The type $LCL_0C$ is excluded because the first $C$ would
have to satisfy  (iii), so the switching from  $C$ to $L_0$ would have to
happen at $(-1,0)$; then the $L_0$ piece following 
the switching has to be a segment going from $(-1,0)$ to a point $(\al,0)$ and
$\al$ must satisfy $-1<\al<0$, for if $\al$ was $\ge 0$ then the $L_0$
piece would already reach the origin before switching to the last $C$;
and if $-1<\al<0$ then the $C$ arc starting at $(\al,0)$ does not go through
the origin. So all four-piece concatenations are excluded.  
Of the three-piece possibilities, $CLC$, $LCL$ and $L_0CL$ are excluded 
because an $L$ piece can only occur at the beginning of the sequence,
and we also know that $L_0CL_0$ is excluded. So only $CL_0C$ and 
$LCL_0$ are left.  Of the two and one-piece concatenations, 
$CL$ and $L$ are obviously excluded, and we are left with
$L_0$, $C$, $CL_0$, $L_0C$ and $LC$.

The $LCL_0$ trajectories are further restricted by the condition that the
switching from $C$ to $L_0$ must happen at the point $(-1,0)$.  Let us call
those $LCL_0$ trajectories that satisfy this extra condition ``good.''

Let ${\cal F}$ be the set of all simple trajectories that end at the
origin and are either of one of the six types $L_0$, $C$, $L_0C$, $LC$,
$CL_0$, $CL_0C$, or of type $LCL_0$ and good. We have shown that every
optimal trajectory belongs to ${\cal F}$.  It will turn out that the
optimal synthesis involves all seven types listed above, but not all
trajectories of one of these types are optimal, so we need a finer
analysis.

To begin with, observe that $\dot x=1-y$ along a $B_+$ arc, so $\dot x>1$
along a $B_+$ arc as long as $y<0$. Therefore, if $q_1=(x_1,y_1)$ and
$q_2=(x_2,y_2)$ satisfy $y_1=y_2\le 0$, $x_1<x_2$, and $x_1+x_2=-2$, so that
$q_1$ and $q_2$ can be joined both by a $C$ arc and an $L$ (or $L_0$) arc,
then the $C$ arc is contained in the lower half-plane, and is faster than
the $L$ or $L_0$ arc.  In particular, if we let $L_0(\beta)$ denote, for
$\beta>0$, the $L_0$ arc going from $(-\beta,0)$ to $(0,0)$, we see that
{\it $L_0(\beta)$ is not optimal if $1<\beta\le 2$}, because the piece of
$L_0$ going from $(-\beta,0)$ to $(-2+\beta,0)$ can be replaced by a faster
type $C$ arc.  Then the principle of optimality implies that 
{\it $L_0(\beta)$ is not optimal if $1<\beta$}, because if $\beta>2$ then 
$L_0(\beta)$ contains $L_0(2)$, which is not optimal. 

Next we show that {\it $L_0(1)$ is optimal}. It suffices to notice that
there is no trajectory from $(-1,0)$ to $(0,0)$ of any of the types $C$,
$LC$, $LCL_0$, and exactly one trajectory of each of the types $L_0$,
$L_0C$, $CL_0$, and $CL_0C$, but the last three are not simple, so $L_0(1)$
is left as the only possible candidate. It then follows from the principle
of optimality that {\it $L_0(\beta)$ is optimal for $0\le\beta\le 1$}.

We now study the optimal trajectories that are of one of the types $C$,
$L_0C$, $LC$, and $CL_0C$.  Let $\gamma:[a,b]\to\R^2$ be such a trajectory,
and take $b=0$, so $\gamma(0)=(0,0)$.  

If $\gamma$ is a $C$ arc, then $a>-2\pi$, for otherwise $\gamma$ would
contain a full loop. For $0<a<2\pi$, we let $C(a)$ be the $C$ arc
$\gamma:[-a,0]\to\R^2$ such that $\gamma(0)=(0,0)$.  It will be shown below
that the arcs $C(a)$ are all optimal.

Next suppose that $\gamma$ is $LC$.  Since the $C$ piece satisfies
(\eqxycircle), we must have $r=\sqrt{2}$ and $\theta={7\pi\over 4}$. Let 
$c$ be the time when the switching from $L$ to $C$ occurs, so 
$-2\pi<c<0$, for if $c\le -2\pi$ then $\gamma$ would not be simple.
On $[c,0]$, the switching function $\varphi$ is given, up to multiplication 
by a positive constant, by:
$$
\varphi(t)=\sqrt{2}\sin\Bigl(\theta_1-{7\pi\over 4}\Bigr)-\cos (t+\theta_1)
=\sin{\theta_1}+\cos{\theta_1}-\cos (t+\theta_1).
$$
So
$$
\varphi(t)=(1+\sin t)\sin{\theta_1}+(1-\cos t).\cos{\theta_1}
$$
Suppose that $\sin{\theta_1}=0$. Then $\cos{\theta_1}=\pm 1$, so
$\cos{\theta_1}=-1$, because $\varphi(t)<0$  when $t$ is negative and close
to $0$, since $u(t)=+1$ for such $t$.  
So $\varphi(t)=\cos t-1$, and then $\varphi(c)\neq 0$, since 
$-2\pi<c<0$. Since $\varphi(c)=0$, we conclude that $\sin{\theta_1}\neq 0$.  
We can then define
$$
\psi(t)={1+\sin t\over 1-\cos t}\,,\,\,\,\,\quad
\nu={\cos{\theta_1}\over\sin\theta_1}\,.
$$
Then $\psi$ is well defined and smooth on $[c,0\,[\,$, 
and $\psi(t)$ goes to $+\infty$ as $t\uparrow 0$. Moreover, $\varphi$ can
be expressed as 
$\varphi(t)=\sin{\theta_1}.(1-\cos{t}).(\nu+\psi(t))$ for 
$t\in[c,\,0\,[\,$. Therefore $\sin{\theta_1}<0$, and
$\psi(c)=-\nu$, since $1-\cos{c}\neq 0$, because $-2\pi<c<0$. Clearly, 
$\psi(t)\ge 0$ for all $t\in\,]\!-\!2\pi,0\,[\,$, 
and $\psi$ attains the value $0$ on
$\,]\!-\!2\pi,0\,[\,$  at $t=-{\pi\over 2}$ and 
nowhere else.  A simple calculation shows that $\psi$ is strictly 
increasing on $\,]\!-\!{\pi\over 2},0\,[\,$. 
If $\nu>0$, then $\psi+\nu$ never vanishes on
$\,]\!-\!2\pi,\,0[\,$, so $\varphi(c)\neq 0$.  So
$\nu\le 0$. If $\nu=0$, then $\psi(c)=-\nu$ implies $c=-{\pi\over 2}$. 
If $\nu<0$, then $\psi$ must take the value 
$-\nu$ at some point $t(\nu)$ such that $-{\pi\over 2}<t(\nu)<0$, 
because $\psi({\pi\over 2})=0$ and $\psi(0-)=+\infty$. Since $\psi$ is
strictly increasing on $\,]\!-\!{\pi\over 2},0\,[\,$, we see that 
$t(\nu)$ is unique, and $\varphi$ changes sign at $t(\nu)$. If $c<t(\nu)$,
we would  contradict the fact that $\varphi$ has constant sign on
$[c,0]$. On the other hand, it is impossible that $c>t(\nu)$, because
then the equation $\psi(t)=-\nu$ would  have two solutions on
$\,]\!-\!{\pi\over 2},0\,[\,$. So $c=t(\nu)$, and we have proved that 
$-{\pi\over 2}\le c<0$.   If ${\pi\over 4}<c$, then 
$\gamma(c)\in \Omega_{AB,4}$, and the switching is not permitted.
So the only possibility left to us is $-{\pi\over 2}\le c\le-{\pi\over 4}$.
If $c=-{\pi\over 2}$, then we really have an $L_0C$ arc rather than 
an $LC$ arc. So the only $LC$ arcs that could be optimal 
are those for which $-{\pi\over 2}<c\le-{\pi\over 4}$.

For $0<\sigma<2\pi$, $s>0$, let $LC(s,\sigma)$ denote the trajectory
defined on the interval $[-s-\sigma,0]$ that ends at the origin and
corresponds to a $u=-1$ control on $[-s-\sigma,-\sigma]$ followed by a
$u=1$ control on $[-\sigma,0]$. We have shown that {\it $LC(s,\sigma)$ is
not optimal unless ${\pi\over 4}\le\sigma<{\pi\over 2}$.}  We now show that
{\it if ${\pi\over 4}\le\sigma<{\pi\over 2}$ then $LC(s,\sigma)$ is
optimal.}  To see this, let $q=(x,y)$ be the starting point of
$LC(s,\sigma)$, for $s>0$, ${\pi\over 4}\le\sigma<{\pi\over 2}$. Then 
$1-\sqrt{2}leq y<0$. It is clear that 
$LC(s,\sigma)$ is the only simple $LC$ trajectory from $q$ to $(0,0)$.  Also, it
is easy to see that there are no trajectories from $q$ to $(0,0)$ of the
types $L_0$, $C$, $L_0C$, or good $LCL_0$.  There is exactly one $CL_0$
trajectory, but it is easy to see that the time $\hat\tau$ along this
trajectory is larger than the time $\tau$ along $LC(s,\sigma)$.  (Actually,
$\tau=s+\sigma<|x|<\hat\tau-{3\pi\over 2}$.) Finally, there is exactly one
$CL_0C$ trajectory $\gamma$, which can also be ruled out. (Let 
$\bar x=-1-\sqrt{2-(y-1)^2}$, so $\bar q=(\bar x,y)$ is the point where
$LC(s,\sigma)$ switches from $L$ to $C$. Then both $LC(s,\sigma)$ and
$\gamma$ go through $\bar q$ and coincide from that point on. So it
suffices to compare the parts of $LC(s,\sigma)$ and $\gamma$ up to $\bar
q$. The time along $LC(s,\sigma)$ up to $\bar q$ is $s=\bar x-x$, and that
along $\gamma$ is $>\sqrt{(1+x)^2+(1-y)^2}-\sqrt{2}+{3\pi\over 2}$. So we 
\shortversion{have to show that $\alpha>0$, where 
\hbox{$\alpha=\sqrt{(1\!+\!x)^2\!+\!(1\!-\!y)^2}-\!\sqrt{2}+\!
{3\pi\over 2}\!-(\bar x\!-\!x)=\sqrt{(1\!+\!x)^2\!+\!(1\!-\!y)^2}-\!
\sqrt{2}+\!{3\pi\over 2}\!+\!x\!+\!1\!+\!\sqrt{2\!-\!(y\!-\!1)^2}$}.} 
\longversion{have to show that $\alpha>0$, where 
$$
\eqalign{\alpha&=\sqrt{(1\!+\!x)^2\!+\!(1\!-\!y)^2}-\!\sqrt{2}+\!
{3\pi\over 2}\!-(\bar x\!-\!x)\cr
&=\sqrt{(1\!+\!x)^2\!+\!(1\!-\!y)^2}-\!
\sqrt{2}+\!{3\pi\over 2}\!+\!x\!+\!1\!+\!\sqrt{2\!-\!(y\!-\!1)^2}\,.\cr}
$$}

\shortversion{\noindent Then, using the fact that 
$\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$  when $a\ge 0$ and $b\ge 0$, we get 
the inequalities $\alpha\ge\sqrt{(1+x)^2+2}-\sqrt{2}+{3\pi\over 2}+x+1
>\sqrt{(1+x)^2}+x+1=|1+x|+1+x\ge 0$, as desired.)}

\longversion{\noindent Then, using the fact that 
$\sqrt{a}+\sqrt{b}\ge \sqrt{a+b}$  when $a\ge 0$ and $b\ge 0$, we get 
the inequalities 
$$
\eqalign{\alpha&\ge\sqrt{(1+x)^2+2}-\sqrt{2}+{3\pi\over 2}+x+1\cr
&>\sqrt{(1+x)^2}+x+1\cr
&=|1+x|+1+x\ge 0\,,\cr}
$$
as desired.)}

For $s>0$, we let $L_0C(s)$ denote the trajectory that consists of a $B_-$
arc during time $s$ followed by a $B_+$ arc during time ${\pi\over 2}$, and
ending at the origin.  Then it is easily shown that {\it $L_0C(s)$ is
optimal for every $s>0$.}  (Indeed, if $q=(x,0)$ is the starting point of
$L_0C(s)$, for $s>0$, then $L_0C(s)$ is the only simple $L_0C$ trajectory from $q$
to $(0,0)$, and there are no simple 
trajectories from $q$ to $(0,0)$ of the types
$C$,  $LC$, $CL_0$, $LCL_0$ or $CL_0C$.  There is an $L_0$ arc, namely,
$L_0(|x|)$, but we know that this arc is not optimal, because $|x|>1$.)

\shortversion{Now define $R_1=\{(0,0)\}$, $R_2=\,]\!-\!1,0\,[\,\times\{0\}$,
$R_3=\{(-1,0)\}$. For $q=(0,0)$, let $\gamma_q$ be the trajectory
$\gamma$ with domain $\{0\}$ such that $\gamma(0)=(0,0)$. For $q=(x,0)\in
R_2\cup R_3$, let $\gamma_q=L_0(|x|)$.  Then the $\gamma_q$, for
$q\in R_1\cup R_3\cup R_3$, are optimal.}
\longversion{Now define 
$$
R_1=\{(0,0)\}\,,\quad\,\,\,R_2=\,]\!-\!1,0\,[\,\times\{0\}\,,\,\,\,\quad
R_3=\{(-1,0)\}\,.
$$
For $q=(0,0)$, let $\gamma_q$ be the trajectory
$\gamma$ with domain $\{0\}$ such that $\gamma(0)=(0,0)$. If $q=(x,0)$
belongs to $R_2\cup R_3$, let $\gamma_q=L_0(|x|)$.  Then the $\gamma_q$,
for $q\in R_1\cup R_2\cup R_3$, are optimal.}

For $r>0$, let $D(r)$ be the open disc with center $(-1,1)$ and radius $r$,
and let $\partial D(r)$ denote the boundary of $D(r)$.  Let 
\shortversion{$R_4=D(1)$, $R_5=\partial D(1)\backslash\{(-1,0)\}$.}
\longversion{
$$
R_4=D(1)\,,\quad\quad R_5=\partial D(1)\backslash\{(-1,0)\}\,.
$$}
Then every point $q\in R_4$ can be 
joined to $(0,0)$ by a unique good $LCL_0$ trajectory. We use $\gamma_q$
to denote this trajectory.  Then {\it the arcs $\gamma_q$, for $q\in R_4$, are
optimal}. (To see this, observe that if $q\in R_4$ then the only
trajectories other than $\gamma_q$ that are of one of our seven types 
and go from $q$ to the origin are either $LCL_0$ but not good
or not simple, or 
$LC$ with a switching from $L$ to $C$ at a time $c$ that violates the
requirement that $-{\pi\over 2}<c$. So all the alternative candidates are
excluded, and $\gamma_q$ must 
be the optimal trajectory.)

A similar argument shows that {\sl if $q\in R_5$ and $\gamma_q$ is the
unique $CL_0$ trajectory going from $q$ to $(0,0)$ then $\gamma_q$ is
optimal}.

\shortversion{Next, let $R_6=D(\sqrt{2})\backslash (R_2\cup R_3\cup R_4\cup
R_5)$.}
\longversion{Next, let 
$$
R_6=D(\sqrt{2})\backslash (R_2\cup R_3\cup R_4\cup R_5)\,.
$$}
If $q\in R_6$, then the obvious candidate for optimality is the arc $\gamma_q$
of type $CL_0$ obtained by following the $C$ arc that starts at $q$ until
$R_2$ is reached, and then continuing along $R_2$ towards the origin.  We
now prove that {\it the arcs $\gamma_q$, $q\in R_6$, are optimal}.  To see
this, observe that if $q=(x,y)\in R_6$ then there are no simple arcs from
$q$ to $(0,0)$ of the types $C$, $L_0C$, $CL_0C$.  If $y\neq 0$ then there
are no $L_0$ arcs either. If $y=0$ then there is an $L_0$ arc, but it is
$L_0(\beta)$ for $\beta=-|x|>1$, so it is not optimal.  If $y\neq 0$ then
there is an $LC$ arc, but (i)~if $y<0$ then this arc switches from $L$ to
$C$ in $\Omega_{AB,4}$, so it is not optimal; (ii)~if $y>0$, then the arc
is $LC(s,\sigma)$ for some $\sigma\in\,]\,5\pi/4,2\pi\,[\,$, so it is not
optimal either.  If $y\le 0$ then $\gamma_q$ is the only simple $CL_0$ arc, and if
$y>0$ then there are two such arcs, but the other one is not optimal,
because it contains $L_0(\beta)$ for some $\beta>1$.  So we have ruled out
all the alternative candidates to $\gamma_q$ other than good $LCL_0$ arcs.
We now exclude this possibility too, by doing it first for the case when
$y\neq 2$.  If $y\le 0$ or $y>2$ or $x\geq -1$, then there is no good $LCL_0$
arc. If $0<y<2$ and $x<-1$ then there is a good $LCL_0$ arc, but it
switches from $L$ to $C$ at a point in $\Omega_{AB,2}$, so this arc is not
optimal.  So now all the alternative possibilities have been ruled out if
$q=(x,y)\in R_6$ is such that $y\neq 2$ or $y=2$ and $x>-1$.  So $\gamma_q$
is optimal for all such $q$'s.  Suppose now that $y=2$ and $x<-1$. Then
there is a good $LCL_0$ arc $\gamma$, which switches from $L$ to $C$ at the
point $p=(-1,2)$.  This arc can be ruled out in a number of ways, of which
the following qualitative argument appears to be the shortest. Let $\gamma$
be defined on $[T,0]$. Let $\alpha$ be a small positive number---actually,
any $\alpha$ such that $0<\alpha<\pi$ will do---and let
$\tilde\gamma:[T-\alpha,0]\to\R^2$ be the trajectory such that
$\tilde\gamma(t)=\gamma(t)$ for $t\in[T,0]$ and $\tilde\gamma$ is a $B_+$
arc on $[T-\alpha,T]$.  Let 
$(\tilde x,\tilde y)=\tilde q=\tilde\gamma(T-\alpha)$. Then $\tilde q$ also
belongs to $R_{6}$, but 
it's not true that $\tilde y=2$ and $\tilde x<-1$.  So we already know that
$\gamma_{\tilde q}$ is the unique optimal arc from $\tilde q$ to the
origin.  Since $\gamma_q$ is a cofinal piece of $\gamma_{\tilde q}$, the
principle of optimality implies that $\gamma_q$ is optimal as well. If
$\gamma$ was optimal, then $\gamma_q$ and $\gamma$ would go from $q$ to $0$
in the same time, so $\gamma_q(T)=\gamma_{\tilde q}(T)=q$. Let $[\tilde
T,0]$ be the domain of $\gamma_{\tilde q}$, and define $\delta:[\tilde
T,0]\to\R^2$ by letting $\delta$ agree with $\gamma_{\tilde q}$ on $[\tilde
T,T]$ and with $\gamma$ on $[T,0]$. Then $\delta$ is an optimal arc from
$\tilde q$ to $0$, and $\delta$ is $CLCL_0$. This, however, is impossible,
because we know that $CLCL_0$ is excluded. So we have now completed the
proof that {\sl for each $q\in R_6$, $\gamma_q$ is the unique optimal arc
from $q$ to the origin}.

\shortversion{Next, we define five sets $R_7$, $R_8$, $R_9$, $R_{10}$,
$R_{11}$,  by letting $R_7=\partial
D(\sqrt{2})\cap\Big(\R\times\,]\,0,+\infty\,[\,\Big)$,  
$R_8=\Big\{(-2,0)\Big\}$, $R_9=\Big\{(x,y)\in\partial
D(\sqrt{2}):-2<x<-1\Big\}$, $R_{10}=\Big\{(-1,1-\sqrt{2})\Big\}$ and,
finally, $R_{11}=\Big\{(x,y)\in\partial D(\sqrt{2}):-1<x<0\Big\}$.}
\longversion{Next, we define five sets $R_7$, $R_8$, $R_9$, $R_{10}$,
$R_{11}$,  by letting 
$$
\eqalign{
R_7&=\partial D(\sqrt{2})\cap\Big(\R\times\,]\,0,+\infty\,[\,\Big)\,,\cr
R_8&=\Big\{(-2,0)\Big\}\,,\cr
R_9&=\Big\{(x,y)\in\partial D(\sqrt{2}):-2<x<-1,\,y<0\Big\}\,,\cr
R_{10}&=\Big\{(-1,1-\sqrt{2})\Big\}\,,\cr
R_{11}&=\Big\{(x,y)\in\partial D(\sqrt{2}):-1<x<0,\,y<0\Big\}\,.\cr}
$$}
Then 
the sets $R_1$, $R_7$, $R_8$, $R_9$, $R_{10}$ and $R_{11}$ are pairwise
disjoint and 
\shortversion{$D(\sqrt{2})=R_1\cup R_7\cup R_8\cup R_9\cup R_{10}\cup
R_{11}$.}  
\longversion{
$$
D(\sqrt{2})=R_1\cup R_7\cup R_8\cup R_9\cup R_{10}\cup R_{11}\,.
$$} 
For $q\in R_7\cup R_8\cup R_9\cup R_{10}\cup R_{11}$ 
there is a unique trajectory $\gamma_q$ of type $C$ going from $q$ to
$(0,0)$. The optimality of $\gamma_q$ for $q\in R_7\cup R_8\cup R_9\cup
R_{10}\cup R_{11}$ is proved by
considerations similar to those used for $R_6$.

\shortversion{We let $R_{12}=\,]\!-\!\infty,-2\,[\,\times\{0\}$, 
$R_{13}=\{(x,y):1-\sqrt{2}<y<0\,,\,
x<-1\,,\,(x+1)^2+(1-y)^2>2\}$, $R_{14}=\,]\!-\!\infty,-1\,[\,\times\{
1-\sqrt{2}\}$.} 
\longversion{We let 
$$
\eqalign{
R_{12}&=\,]\!-\!\infty,-2\,[\,\times\{0\}\,,\cr
R_{13}&=\{(x,y):1-\sqrt{2}<y<0\,,\,x<-1\,,\,(x+1)^2+(1-y)^2>2\}\,,\cr
R_{14}&=\,]\!-\!\infty,-1\,[\,\times\{1-\sqrt{2}\}\,.\cr}
$$}
Then $R_{13}\cup R_{14}$ is exactly the set of starting points $q$ of the
trajectories $LC(s,\sigma)$, for $s>0$, ${\pi\over 4}\le\sigma<{\pi\over
2}$, and we know that for each $q\in R_{13}\cup R_{14}$
the $LC(s,\sigma)$ trajectory is unique and optimal. We define $\gamma_q$ 
to be this trajectory. A similar fact is true for $R_{12}$, with 
$L_0C(s)$ rather than $LC(s,\sigma)$ in the role of $\gamma_q$.

\shortversion{Finally, we let %gnung
$R_{15}=\R^2\backslash(R_1\cup\ldots\cup R_{14})$.}
\longversion{Finally, we let 
$$
R_{15}=\R^2\backslash(R_1\cup\ldots\cup R_{14})\,.
$$}
For $q\in R_{15}$, we let $\gamma_q$ be the unique simple $CL_0C$ trajectory
from $q$ to $(0,0)$. We show that {\it $\gamma_q$ is optimal for all $q\in
R_{15}$}.  Let $q\in R_{15}$, and let $\gamma\in {\cal F}$ be an extremal
trajectory starting at $q$ and verifying $\gamma(0)=0$.  It is then obvious
that $\gamma$ is not of type $L_0$, $C$ or $L_0C$. The case $LC$ is
impossible, because in that case $\gamma$ would have a switching at a time
$c$ such that $c <-\pi/2$.  If $\gamma$ is of type $CL_0$ then it contains
an $L_0(\beta)$ piece, with $\beta>1$, and we know that any such piece is
nonoptimal, so $\gamma$ is not optimal.  Finally, if $\gamma$ is of type
$LCL_0$ and good then $\gamma(t)=p=(x_p,y_p)\in R_5$ for some $t$. It is
clear that $y_p>0$. If $x_p>-1$, then $\gamma$ is not simple, so this
possibility is excluded. If $x_p\le-1$, then we can pick a positive $\ve$
such that $\hat q=\gamma(t-\ve)$ is in $R_6$. Then the restriction
$\hat\gamma$  of  $\gamma$ to the interval  $[t-\ve,0]$ is a good $LCL_0$
trajectory  from $\hat q$ to $0$. Since $\hat q\in\R_6$, we know that 
$\hat\gamma$ is not optimal. So $\gamma$ is not optimal either, by the
principle of optimality.

We have now completely analyzed the minimum time problem with target set
$\{(0,0)\}$ for every initial condition $q$, and proved in all possible
cases that $\gamma_q$ is the unique solution whose terminal time is $0$.
So the synthesis $\Gamma=\{\gamma_q\}_{q\in\R^2}$ shown in Figure 2 is
optimal, and is the unique optimal presynthesis.  The partition ${\cal
P}=(R_1,\ldots,R_{15})$ is a $C^1$---and, actually,
$C^\omega$---stratification of $\R^2$. 

\shortversion{Let us define the feedback control $(x,y)\to v(x,y)$ by letting 
$v=0$ on  $R_1$, $v=-1$ on the set
$R_2\cup R_3\cup R_4\cup R_{12}\cup R_{13}\cup R_{14}$,  
and $v=+1$ on $R_5\cup R_6\cup R_7\cup R_8\cup R_9\cup R_{10}\cup R_{11}
\cup R_{15}$.  We let ${\cal P}_1=\{R_2,R_4,R_5,R_6,R_7,R_9,R_{11},R_{12}, 
R_{13},R_{14},R_{15}\}$, ${\cal P}_2=\{R_3,R_{8},R_{10}\}$. We define 
$\Pi(R_2)=R_1$, $\Pi(R_4)=R_5$, $\Pi(R_5)=R_3$, 
$\Pi(R_6)=R_2$, $\Pi(R_7)=R_8$, $\Pi(R_9)=R_{10}$, 
$\Pi(R_{11})=R_1$, $\Pi(R_{12})=R_8$, $\Pi(R_{13})=R_9$, 
$\Pi(R_{14})=R_{10}$, $\Pi(R_{15})=R_{12}$, $\Sigma(R_3)=R_2$,
$\Sigma(R_8)=R_9$, $\Sigma(R_{10})=R_{11}$.}

\longversion{Let us define the feedback control $(x,y)\to v(x,y)$ by letting 
$$
v=\left\{\matrix{\phantom{+}0&{\rm on}&\boxo{R_1\,,}\cr
-1&{\rm on}&\boxo{R_2\cup R_3\cup R_4\cup R_{12}\cup R_{13}\cup R_{14}\,,}\cr
+1&{\rm on}&\boxo{R_5\cup R_6\cup R_7\cup R_8\cup R_9\cup R_{10}\cup R_{11}
\cup R_{15}\,.}\cr}\right.
$$
We let 
$$
\eqalign{
{\cal{}P}_1&=\{R_2,R_4,R_5,R_6,R_7,R_9,R_{11},
R_{12},R_{13},R_{14},R_{15}\}\,,\cr 
{\cal P}_2&=\{R_3,R_{8},R_{10}\}\,.\cr}
$$
We then define  
$$
\left.\matrix{
\,\,\Pi(R_2)=R_1\,,&\Pi(R_4)=R_5\,,&\,\,\Pi(R_5)=R_3\,,&\,\,\Pi(R_6)=R_2\,,\cr
\,\,\Pi(R_7)=R_8\,,&\,\,\,\Pi(R_9)=R_{10}\,,
&\Pi(R_{11})=R_1\,,&\Pi(R_{12})=R_8\,,\cr
\Pi(R_{13})=R_9\,,&\Pi(R_{14})=R_{10}\,,&\,\,\Pi(R_{15})=R_{12}\,,&
\,\,\Sigma(R_3)=R_2\,,\cr
\,\,\Sigma(R_8)=R_9\,,&\,\Sigma(R_{10})=R_{11}\,.&\,\,\cr}\right.
$$}

Then $({\cal P},{\cal P}_1,{\cal P}_2,\Pi,\Sigma,v)$ is a regular synthesis
that satisfies all the conditions of Brunovsk\'y's definition, except for
the continuity requirement (Br.3.G) on the cost function $V_{\Gamma}$.  On
the other hand, it is easy to verify that {\sl $V_\Gamma$ satisfies the
weak continuity conditions}. So $({\cal P},{\cal P}_1,{\cal
P}_2,\Pi,\Sigma,v)$ gives rise to a ``regular synthesis'' in the sense of our 
definition.

Moreover, it is clear that {\sl the $+1$ trajectory starting from the point
$(-1,0)$ is an admissible classical trajectory for the optimal feedback control but
is not optimal}.  So this example illustrates a very important positive
aspect of the Boltyanskii-Brunovsk\'y's definition, namely, the role of the
``instantaneous exit map'' $\Sigma$. 
%This map assigns to each Type II cell
%$P$ a Type I cell $\Sigma(P)$, in such a way that, if $x\in P$, then the
%trajectory of the synthesis starting at $x$ enters $\Sigma(P)$
%instantaneously. Our example shows that, if we had only used the
%discontinuous feedback $v$ in order to generate trajectories, then some
%non-optimal trajectories would be admissible solutions of the corresponding
%equation $\dot x=f(x,v(x))$. In the Boltyanskii-Brunovsk\'y's setup, these
%bad trajectories are excluded by specifying $\Sigma$, and insisting that,
%if we start at time $0$ at an $x\in P\in{\cal P}_2$, then only those
%trajectories $\xi$ such that $\xi(t)\in\Sigma(P)$ for small positive $t$
%are admissible, among all $\xi$'s such that $\xi(0)=x$.
To understand how $\Sigma$ matters, let us suppose we
did not specify $\Sigma$, and allowed all the classical trajectories of the
discontinuous feedback $v$.  Then we could, for example, start from the
Type II cell $R_3=\{(-1,0)\}$ and follow the trajectory corresponding to
the control $+1$ up to time $2\pi$, returning to the starting point, after
which we could go to the origin using the control $-1$. The curve
$\gamma_1$ defined in this way is also an admissible classical trajectory of $v$, but
it is clear that $\gamma_1$ is not time optimal.  Even worse, we
can define a trajectory $\gamma_2$ of $v$, with domain $[0,+\infty[$,
corresponding to the constant control $+1$ and starting from $(-1,0)$.
This trajectory neither reaches the origin in finite time nor approaches it
as time goes to infinity.  Notice that the trajectories $\gamma_1$ and $
\gamma_2$ satisfy the equation $\dot\gamma(t)=f(\gamma(t),v(\gamma(t)))$
for every $t$ in their domain, so they satisfy the equation corresponding
to the discontinuous feedback $v$ in calssical sense.  
However, if we regard it as part of the
specification of our synthesis that $\Sigma$ has to be given as well, then
the difficulty disappears.\proofend

\xdef\nofili{\the\sectno.\the\theono}
\exam
We now exhibit an optimal synthesis in Brunovsk\'y's sense 
that does not satisfy Condition (Bo.4) of Boltyanskii's definition.
We consider the minimum time problem for the system:
$$
\dot x = u\,,\,\,\,\qquad
\dot y = v\,,\qquad 
\dot z = {w\over \sqrt{1+x^2+y^2}}\,,
$$
with target $\{(0,0,0)\}$ and control constraint $(u,v,w)\in U$, where $U$
is the set of those $(u,v,w)\in\rr^3$ such that $(u^2+v^2)^{1/2}+|w|\leq
1$, i.~e. the solid of revolution obtained by rotating the square
$\{(u,w):|u|+|w|\le 1\}$ in the $u,w$-plane about the $w$ axis.
Define the sets:
$$
\eqalign{R_0&=\{(0,0,0)\}\,,\phantom{:x=y=0 \, x\, z>0}
\qquad R_3=\{(x,y,z):x^2+y^2>0,\, z=0\}\,,\cr
R_1&=\{(x,y,z): x=y=0, \, z>0\}\,,\qquad R_4=\{(x,y,z):x^2+y^2>0,\,z>0\}\,,\cr
R_2&=\{(x,y,z):x=y=0,\, z<0\}\,,\qquad R_5=\{(x,y,z):x^2+y^2>0,\,z<0\}\,.\cr}
$$  
We $\,\,$then $\,\,$define $\,\,$a $\,\,$feedback $\,\,$control $\,\,$law 
$\,\,(u,v,w):\rr^3\to U\,\,$ by $\,\,$specifying $\,\,$its $\,\,$restriction 
$R_i\ni(x,y,z)\to(u_i(x,y,z),v_i(x,y,z),w_i(x,y,z))\in\R^3$ for 
every $R_i$ as follows:
\xdef\defioptlaw{\equanosectno}
$$\eqalign{u_0&=v_0=w_0=0\,,\qquad\qquad\qquad
u_3=u_4=u_5=-{x\over\sqrt{x^2+y^2}}\,,\cr
u_1&=v_1=0,\;w_1=-1\,,\qquad\qquad v_3=v_4=v_5=-{y\over\sqrt{x^2+y^2}}\,,\cr
u_2&=v_2=0,\;w_2=1\,,\qquad\qquad\,\,\,\,\, w_3=w_4=w_5=0\,.\cr}\numeq
$$
If we follow the trajectories of this feedback starting from a point
$(x,y,z)\in R_3\cup R_4\cup R_5$, we move horizontally towards the $z$ axis
with speed $1$, and then, if we are not yet at the origin, we move to the
origin along the $z$ axis with speed $1$. This obviously gives rise to a
synthesis $\Gamma$, and it is possible to prove that $\Gamma$ is optimal by
applying Theorem \maintheo. However, in this case it is easy to verify
directly that the associated cost function
$$
V_{\Gamma}=|z|+\sqrt{x^2+y^2}
$$
is the value function of our problem.  Indeed, since $V_{\Gamma}$ is the
cost function arising from the synthesis $\Gamma$, all we need is to show
that the cost $b-a$ of any trajectory
$[a,b]\ni{}t\to\xi(t)=(x(t),y(t),z(t))$ from a point $\bar q=(\bar x,\bar
y,\bar z)$ to the origin is bounded below by $V_{\Gamma}(\bar q)$.  

Assume first that $x(t)^2+y(t)^2>0$ for all $t\in\,]\,a,b\,[\,$, and let
$t\to(u(t),v(t),w(t))$ be the corresponding control.  Write
$\rho(t)=\sqrt{u(t)^2+v(t)^2}$.  Then an elementary calculation shows that
$$
\eqalign{
{d\,\,\over dt}\Big(V_{\Gamma}(\xi(t))\Big)&=
{u(t).x(t)+v(t).y(t)\over\sqrt{x(t)^2+y(t)^2}}+
{\rm sgn}(z(t)).{w(t)\over\sqrt{1+x(t)^2+y(t)^2}}\cr
&\ge -\rho(t)-|w(t)|\cr
&\ge -1\,.\cr}
$$
Therefore, since the function $t\to V_{\Gamma}(\xi(t))$ is Lipschitz,
and hence absolutely continuous, we can conclude that
$$
V_{\Gamma}(\bar q)=-\int_a^b
{d\,\,\over dt}\Big(V_{\Gamma}(\xi(t))\Big)\,dt\le b-a\,.
$$
Now, if $\xi:[a,b]\to\rr^3$ is an arbitrary trajectory from $\bar q$ to the
origin, then $\xi$ can be approximated by a sequence of trajectories
$\xi_j:[a_j,b_j]\to\rr^3$ from $\bar q$ to $(0,0,0)$ such that $\xi_j(t)$
is not in the $z$ axis for $a_j<t<b_j$. Then $b_j-a_j\ge V_{\Gamma}(\bar
q)$ for all $j$, so $b-a\ge V_{\Gamma}(\bar q)$. Therefore $\Gamma$ is an
optimal synthesis.

Notice that in this case $\Pi(R_4)=R_1$, $\Pi(R_5)=R_2$, and
$\Pi(R_3)=R_0$. So the optimal synthesis $\Gamma$ does not satisfy 
the conditions of Boltyanskii's definition, because (Bo.4) is violated,
since the dimension of the exit manifold from $R_i$ is equal to
${\rm dim}(R_i)-2$ for $i=3,4,5$. 

\rema
In Example \nofili{} the optimal pair $(\g_q,\e_q)$ starting at $q$ is
unique for every $q$, and is a classical solution of the closed-loop
equation corresponding to the optimal feedback law defined by \defioptlaw.

On the other hand, it is easy to see that {\it the optimal trajectories are
not Filippov solutions of the optimal closed-loop equation}.  Finally, we
point out that, {\it although the optimal trajectories are CLSS solutions,
it is not true that all CLSS solutions of the optimal closed-loop equation
are optimal trajectories.} For example, if the initial condition is a point
$q$ which is not in the $z$ axis and also not in the plane $z=0$, then most
CLSS solutions $t\to\g(t)$ will never switch at a time $t$ such that
$\g(t)$ lies in the $z$ axis, and will therefore be entirely contained in a
plane $z={\rm constant}$, without ever reaching	 the target. \proofend


\exam
We now exhibit an example of an optimal synthesis 
for which it is not true that all the optimal trajectories are CLSS
solutions of the optimal closed-loop equation. 

Consider the optimal control problem in $\rr$, with control space $U=\rr$,
dynamics given by $\dot x=u$, Lagrangian $L=(u^4-x^2)^2$, and target 
$\{1\}$. The value function $V$ is given by $V(x)\equiv 0$. The
feedback control law $v$ defined by 
$$
v(x)=\left\{\matrix{\sqrt{|x|}&\quad&{\rm if}&x<1\,,\cr
-\sqrt{x}&\quad&{\rm if}&x>1\,,\cr
0&\quad&{\rm if}&x=1\cr}\right.
$$
leads to trajectories $\g_x:[T_x,0]\to\rr$, corresponding to
open-loop controls $\e_x:[T_x,0]\to\rr$, given by the formulas
$$
\eqalign{
T_x&=\,\,2.{\rm sgn}(1-x).\Big({\rm sgn}(x).\sqrt{|x|}-1\Big)\,,\cr
\g_x(t)&=\left\{\matrix{
{(2+t)^2\over 4}&\quad\quad\,\,\,\,\quad\quad\,\,\,\,{\rm if}&x\le 1&{\rm and}&\max(T_x,-2)\le t\le
0\,,\cr 
-{(2+t)^2\over 4}&\quad\quad\,\,\,\,\quad\quad\,\,\,\,{\rm if}&x\le 1&{\rm and}&T_x\le
t\le-2\,,\qquad\qquad\cr  
{(2-t)^2\over 4}&\quad\quad\,\,\,\,\quad\quad\,\,\,\,{\rm if}&x\ge 1&{\rm and}&T_x\le t\le
0\,,\qquad\qquad\,\,\,\,\cr}\right.\cr 
\e_x(t)&=\left\{\matrix{
{1\over 2}|2+t|&\quad\quad\,\,\,\,\quad\quad\,\,\,\,{\rm if}&x\le 1&{\rm and}&T_x\le t\le 0\,,\cr
{1\over 2}(2-t)&\quad\quad\,\,\,\,\quad\quad\,\,\,\,{\rm if}&x\ge1&{\rm and}&T_x\le t\le 0\,.\cr}\right.\cr}
$$
It is easy to see that the family $\Gamma=\{(\g_x,\e_x)\}_{x\in\rr}$ is a
regular synthesis in the sense of our definition. Moreover, although the
optimal trajectories are not unique, it is not hard to show that {\sl
$\Gamma$ is the only optimal synthesis for our 
problem}. \longversion{(Indeed, suppose that
$\tilde\Gamma=\{(\tilde\g_x,\tilde\e_x)\}_{x\in\rr}$ was another optimal synthesis,
and let ${\rm Dom}(\g_x)=[\tilde T_x,0]$ for $x\in\rr$.  Let
$x\in\rr$. Then the cost of $(\tilde\g_x,\tilde\e_x)$ must be zero, so
$|\tilde\e_x(t)|=\sqrt{|\tilde\g_x(t)|}$ for a.~e. $t$.  Suppose
$x<1$. Then $\tilde\g_x(t)<1$ for all $t\in[\tilde T_x,0\,[\,$, because if
$\tilde\g_x(t)\ge 1$ for some $t\in[\tilde T_x,0\,[\,$ then the map
$\tilde\g_x$ would not be injective, and this would contradict the fact
that $\Gamma$ is a synthesis.  If $\tilde\e_x(t)<0$ for all $t$ in a subset
$E$ of $[\tilde T_x,0]$ of positive measure, then once again the map
$\tilde\g_x$ would not be injective. So $\tilde\e_x(t)\ge 0$ a.~e., and
then $\tilde\e_x(t)=v(\tilde\g_x(t))$ for a.~e. $t$. A similar argument
shows that $\tilde\e_x(t)=v(\tilde\g_x(t))$ for a.~e. $t$ if $x>1$. So the
$\tilde\g_x$ are solutions of the equation $\dot\g=v(\g)$. This equation
has unique solutions except for the possibility that, starting at some $x<0$, a
trajectory reaches the origin and 
stays at $0$ for an arbitrary length of time before it
resumes its motion to the right.  This possibility, however is clearly
excluded for a trajectory of a synthesis.  So, for each $x$, 
$\tilde\g_x$ is the unique solution of $\dot\g=v(\g)$ that 
starts at $x$, ends at $1$ at time $0$, and is one-to-one.  Therefore
$\tilde\g_x=\g_x$ for all $x$, so $\tilde\Gamma=\Gamma$.)}

If we let $f(x,u)=u$, then \longversion{we have established that}
the arcs $\g_x$ are classical solutions
of the closed-loop equation $\dot x=f(x,v(x))$.  On the other hand, {\sl
the trajectory $\g_{0}$ corresponding to the initial condition $x=0$ is not
a CLSS solution of the closed-loop equation $\dot x=f(x,v(x))$} because, on
an interval $[a,b]$, the only CLSS solution $\g$ of this equation with
initial condition $\g(a)=0$ is $\g(t)\equiv 0$. \proofend


\bigbreak
\vbox{
\appen{A}{Differentiation of trajectories with respect to a parameter} 

\noindent The purpose of this appendix is to 
\longversion{prove}  \shortversion{present}
a general theorem on differentiation with respect to a parameter $p$,
taking values in a normed space $P$, of a family of trajectories
$x_p:[a,b]\to\R^n$ of time-varying vector fields $f_p$.  We will state 
\longversion{and prove} the
theorem under minimal hypotheses, much weaker than what is actually needed
for the main results of this paper, because this result has other
applications, such as the theory of envelopes (cf. Sussmann
\longversion{\sussenvacount,}
\sussenvbcount), where the more general statement is useful. 

}


If $\dot x_p(t)=f_p(x_p(t),t)$, then formal differentiation with respect
to $p$ at $p=p_0$ in the direction of a vector $v$ yields the variational
equation 
\xdef\varformal{\equanoapno}
$$
\dot y_{p_0,v}(t)={\partial f_{p_0}\over \partial{x}\,\,\,}
\Bigl(x_{p_0}(t),t\Bigr).y_{p_0,v}(t) 
+\lim_{\ve\downarrow 0}{W_{p_0+\ve v}(t)-W_{p_0}(t)\over \ve}\,,\numeqap
$$
where $y_{p_0,v}(t)$ is the directional derivative of $x_p(t)$ at $p_0$ in
the $v$-direction, and 
$W_p(t)=f_p(x_{p_0}(t),t)$.   

The technical problem that will concern
us here is to make (\varformal) rigorous even when the limit in the
right-hand side only exists in a weak sense. 
\shortversion{It can then happen that the variational equation is a linear
equation $\dot y(t)=A(t)y(t)+w(t)$ where the ``input'' $w$ is the formal
derivative of a function $W$ which is not necessarily absolutely continuous
or even continuous. So our first task will be to study the solutions of
linear systems with ``generalized inputs'' $w$.}


\longversion{
For example, suppose that $n=1$ and we are looking at the control system
$\dot x=u$, and a family of ``bang-bang'' controls $u_s$ depending on
$s\in[0,\infty[$, such that $u_s(t)=1$ for $t<s$, and $u_s(t)=-1$ for $t\ge
s$.  Let $x_s:[0,\infty[\,\to\R$ be the solution of $\dot x=u_s$,
$x(0)=0$. Then $x_s(t)=t$ for $t\le s$, and $x_s(t)=2s-t$ for $t\ge
s$. Given any $T>0$, the limit $\lim_{\ve\to
0}{u_{s+\ve}^T-u_s^T\over\ve}$---where $u_s^T$ is the restriction of $u_s$
to $[0,T]$---does not exist pointwise, but exists in the weak$^*$
sense---regarding the $u_s$ as members of the dual of the space of
continuous functions on $[0,T]$---and equals $2\delta_s$, where $\delta_s$
is the Delta function at $s$.  If we let $y_s(t)$ be the derivative of
$x_s(t)$ with respect to $s$---so $y_{s}(t)=y_{s,1}(t)$---then $y_s$
satisfies the differential equation $\dot y_s(t)=2\delta_s(t)$, which makes
better sense in integrated form: $y_s(t)=2\chi_{[s,\infty[}(t)$.  This
says, in particular, that $y_s(T)=2$ for $s<T$, and $y_s(T)=0$ for $s>T$.}

\longversion{As this example shows, it can happen that
the variational equation is a linear equation of the form
$\dot y(t)=A(t)y(t)+w(t)$
where the ``input'' $w$ is the formal derivative of a function $W$ which is
not necessarily absolutely continuous or even continuous. So our first task
will be to study the solutions of linear systems with ``generalized
inputs'' $w$.}

\medbreak
Let $-\infty<a<b<\infty$, and let $n$ be a positive integer.  
If $A\in L^1([a,b],\R^{n\times n})$, and $\bar Y\in\R^n$,
then it is well known that the initial value problem
\xdef\inhom{\equanoapno}
$$
\dot Y(t)=A(t)\cdot Y(t)+w(t)\,,\,\,\,\,\qquad Y(a)=\bar Y\numeqap
$$
has a unique solution for every initial condition and every map 
$w\in L^1([a,b],\R^n)$.  Since (\inhom) can also be written formally as
\xdef\inhomfor{\equanoapno}
$$
dY(t)=A(t)\cdot Y(t)\,dt+w(t)\,dt\,,\,\,\,\,\qquad Y(a)=\bar Y\,,\numeqap
$$
we will use $Y_{A,w\,dt,\bar Y}$ to denote the unique solution of (\inhom). 

As explained before, we want to solve
(\inhom) for ``inputs'' $w$ more general than integrable functions. The
appropriate class of inputs turns out to be that of {\sl formal derivatives
of bounded measurable functions}.  Precisely, we let $BM([a,b],\R^n)$
denote the space of all bounded Lebesgue measurable functions
$W:[a,b]\to\R^n$, and use $BM_0([a,b],\R^n)$ to denote the set of those
$W$ in $BM([a,b],\R^n)$ such that $W(a)=0$.  (In $BM([a,b],\R^n)$ we do {\it
not} identify two functions that are equal almost everywhere.)  

\longversion{We endow $BM([a,b],\R^n)$ with the norm 
$$
||W||_{BM}\equaldef \sup\{||W(t)||:t\in[a,b]\}\,.\numeqap
$$
Then $BM([a,b],\R^n)$ is a Banach space, and $BM_0([a,b],\R^n)$ is a closed
subspace of $BM([a,b],\R^n)$.  The space $C^0([a,b],\R^n)$ of continuous
maps $W:[a,b]\to\R^n$ is of course a closed subspace of $BM([a,b],\R^n)$,
but $L^\infty([a,b],\R^n)$ is {\sl not} a subspace of $BM([a,b],\R^n)$,
because in $L^\infty([a,b],\R^n)$ we identify functions that are equal
a.e., but in $BM([a,b],\R^n)$ we do not.
} 

\shortversion{We endow $\,\,BM([a,b],\R^n)\,\,$ with the norm 
$\quad||W||_{BM}\,\equaldef\,\sup\{||W(t)||:t\in[a,b]\}\,.\quad$
Then $BM([a,b],\R^n)$ is a Banach space, and $BM_0([a,b],\R^n)$ is a closed
subspace of $BM([a,b],\R^n)$.  The space $C^0([a,b],\R^n)$ of continuous
maps $W:[a,b]\to\R^n$ is of course a closed subspace of $BM([a,b],\R^n)$,
but $L^\infty([a,b],\R^n)$ is {\sl not} a subspace of $BM([a,b],\R^n)$,
because in $L^\infty([a,b],\R^n)$ we identify functions that are equal
a.e., but in $BM([a,b],\R^n)$ we do not.}

If $W\in BM_0([a,b],\R^n)$ is absolutely continuous, and $w=\dot W$, 
so that, formally,  $dW=w\,dt$,  then we can rewrite (\inhom) formally as
\xdef\inhomint{\equanoapno}
$$
dY(t)=A(t)\cdot Y(t)\, dt+dW(t)\,\,\,\,\,\,\,\,\,\,Y(a)=\bar Y\,.\numeqap
$$
A solution of (\inhomint) is then a continuous map $t\to Y(t)$ such 
that 
\xdef\inhomintsol{\equanoapno}
$$
Y(t)=\bar Y+\int_a^tA(s)\cdot Y(s)\,ds+W(t)\,\qquad{\rm for
\quad all\quad} t\in[a,b]\,.\numeqap
$$ 
Now (\inhomintsol) makes sense for arbitrary $Y\in BM([a,b],\R^n)$,
$W\in BM_0([a,b],\R^n)$, even if $W$ is not absolutely continuous.
So we turn it into a definition:  for a pair $(W,\bar Y)\in
BM_0([a,b],\R^n)\times\R^n$, a {\sl solution} of (\inhomint) is a map
$Y\in BM([a,b],R^n)$ such that (\inhomintsol) holds.     It then follows that 

\medbreak


{\pariim

\item{(EU)} {\sl for every $W\!\!\in\!\!BM_0([a,b],\R^n)$  and every 
$\bar Y\!\!\in\!\!\R^n$ there exists a unique solution $Y$ of (\inhomint).}

}

\medbreak\noindent
Indeed, uniqueness is trivial, since the difference $Y=Y^1-Y^2$ of two solutions
is a solution of the homogeneous problem (i.e. of (\inhom) with $w\equiv 0$
and $\bar Y=0$),  so $Y^1-Y^2\equiv 0$.  To prove existence, we observe that 
(\inhomintsol) is equivalent to 
\xdef\inhomintsolequiv{\equanoapno}
$$
Y(t)-W(t)=\bar Y+\int_a^tA(s)\cdot (Y(s)-W(s))\,ds+\int_a^tA(s)\cdot W(s)
\,ds\,\qquad{\rm for\quad all\quad} t\in[a,b]\,,\numeqap
$$ 
i.~e. to $dZ=AZ\,dt+AW\,dt$, $Z(a)=\bar Y$, with $Z=Y-W$.  Since $AW\in
L^1$, because  
$A\in L^1$ and $W$ is bounded and measurable, the equation
$\dot Z(t)=A(t)\cdot Z(t)+A(t)W(t)$ has a unique solution $Z^*$ for which 
$Z^*(a)=\bar Y$.  If we let $Y(t)=Z^*(t)+W(t)$,  and recall that 
$W(a)=0$,  we see that $Y$ is the desired solution of (\inhomint).

We will use $Y_{A, dW,\bar Y}$ to denote the solution of (\inhomint),
whose existence  and uniqueness has just been established.   
In addition to proving (EU), the preceding argument also gives an explicit 
formula for $Y_{A,dW,\bar Y}$, namely:
\xdef\inhomintsolfla{\equanoapno}
$$
Y_{A,dW,\bar Y}=W+Y_{A,AW\,dt,\bar Y}\,.\numeqap
$$
Clearly, the function $Y_{A,AW\,dt,\bar Y}$ is absolutely continuous,
so the regularity properties of $Y_{A,dW,\bar Y}$---modulo absolutely
continuous  functions---are exactly the same as those of $W$. 
(For example: $Y$ is continuous,  right-continuous, or left-continuous 
at a point $t\in[a,b]$ if and only if $W$ is; $Y$ is of bounded variation 
iff $W$ is; $Y$ is absolutely continuous iff $W$ is.)

Gronwall's inequality, applied to (\inhomintsolequiv), tells us that
$$
||Y_{A,dW,\bar Y}(t)-W(t)||\le e^{||A||_{L^1}}\Bigl(||\bar Y||+
||A||_{L^1}||W||_{BM}\Bigr)\,.\numeqap
$$
Therefore
\xdef\ineqyw{\equanoapno}
$$
||Y_{A,dW,\bar Y}||_{BM}\le ||W||_{BM}+e^{||A||_{L^1}}\Bigl(||\bar Y||+
||A||_{L^1}||W||_{BM}\Bigr)\,.\numeqap
$$
On the other hand, (\inhomintsol) also implies that
\xdef\ineqwy{\equanoapno}
$$
||W||_{BM}\le||\bar Y||+||Y_{A,dW,\bar Y}||_{BM}\Bigl(1+||A||_{L^1}\Bigr)\,.\numeqap
$$   
\shortversion{It is not hard to prove the following technical fact:}
\xdef\theolin{{\rm \appenlet}.\the\theono}
\theoap
Let $\{W_j\}$,  $\{\bar Y_j\}$, $\{A_j\}$ be sequences in 
$BM_0([a,b],\R^n)$, $\R^n$, $L^1([a,b],\R^{n\times n})$, respectively.
Assume that $\{A_j\}$ is bounded in $L^1([a,b],\R^{n\times n})$. 
Let $Y_j=Y_{A_j,dW_j,\bar Y_j}$. Then

{\sl 

\itemitem{(\theolin.1)}  $\{Y_j\}$ is bounded in $BM([a,b],\R^{n})$ if 
and only if $\{W_j\}$ is bounded in $BM([a,b],\R^{n})$ and $\{\bar Y_j\}$ 
is bounded in $\R^{n}$. 

\itemitem{(\theolin.2)} If $\{A_j\}$ converges in $L^1([a,b],\R^n)$
to a limit  $A$, $\bar Y\in\R^n$, $\bar Y_j\to\bar Y$, 
$\{W_j\}$ is bounded in $BM([a,b],\R^n)$, 
$W\in BM_0([a,b],\R^n)$, 
and $Y=Y_{A,dW,\bar Y}$, then $Y_j\to Y$ almost everywhere iff $W_j\to W$ 
almost everywhere.  In that case,  (a)~$Y_j-W_j$ converges to $Y-W$
uniformly on  $[a,b]$ and, in particular, (b)~for every $t\in[a,b]$,
$Y_j(t)\to Y(t)$ iff  $W_j(t)\to W(t)$. \shortversion{\proofend}

} 

\longversion{
\beginproof
Statement (\theolin.1) follows from the bounds (\ineqyw) and (\ineqwy).  To
prove (\theolin.2), we first observe that under the hypotheses of
(\theolin.2) we can apply (\theolin.1) and conclude that $\{Y_j\}$ is
bounded in $BM([a,b],\R^n)$.
}

\longversion{
Let $\tilde Y_j=Y_j-W_j=Y_{A_j,A_jW_j\,dt,\bar Y_j}$, 
$\tilde Y=Y-W=Y_{A,AW\,dt,\bar Y_j}$.  Then 
$$
\tilde Y_j(t)=\bar Y_j+\int_a^tA_j(s)Y_j(s)\,ds
=\bar Y_j+\int_a^tA_j(s)\tilde Y_j(s)\,ds+\int_a^tA_j(s)W_j(s)\,ds\,,\numeqap
$$ 
so
\xdef\forgrony{\equanoapno}
$$
\tilde Y_j(t)-\tilde Y(t)=\int_a^t A(s)\Big(\tilde Y_j(s)-\tilde
Y(s)\Big)\,ds+ R_j(t)\,,\numeqap
$$
where 
$$
R_j(t)=\bar Y_j-\bar Y+\int_a^t\Bigl(A_j(s)-A(s)\Bigr) Y_j(s)\,ds
+\int_a^t A(s)\Bigl(W_j(s)-W(s)\Bigr)\,ds\,,\numeqap
$$
and also
\xdef\forgronw{\equanoapno}
$$
\tilde Y_j(t)-\tilde Y(t)=\bar Y_j-\bar Y+\int_a^t\Bigl(A_j(s)-A(s)\Bigr)
Y_j(s)\,ds
+\int_a^t A(s)\Bigl(Y_j(s)-Y(s)\Big)\,ds\,.\numeqap
$$
The assumption that the sequence $\{W_j\}$ is uniformly bounded
and $\{\bar Y_j\}$ converges to $\bar Y$ implies that 
the sequences $\{\tilde Y_j\}$ and  $\{Y_j\}$
are uniformly bounded as well. Since $A_j\to A$ in $L^1$, it is clear
that the first two terms of the sum defining $R_j(t)$ go to zero uniformly
as $j\to\infty$.  Also,
$$
\biggl\|\int_a^t A(s)\Big(W_j(s)-W(s)\Big)\,ds\biggr\|\le \int_a^b||A(s)||\cdot
||W_j(s)-W(s)||\,ds\,,\numeqap
$$
which goes to zero by the dominated convergence theorem if
$s\to ||A(s)||$ is integrable and the sequence $\{W_j-W\}$ is uniformly
bounded and goes to zero almost everywhere.  So, if we let 
$K_j=\sup\{||R_j(t)||:a\le t\le b\}$, we see that $K_j\to 0$ as
$j\to\infty$ if the assumptions of (\theolin.2) hold and 
$W_j(t)\to W(t)$ for a.e. $t$. 
It then follows from Gronwall's inequality, applied to (\forgrony), that
$||\tilde Y_j(t)-\tilde Y(t)||\le K_je^{||A||_{L^1}}$, 
so $\tilde Y_j-\tilde Y\to 0$  uniformly.
}

\longversion{
If $Y_j-Y\to 0$ a.e., then the first two terms of the right-hand side of (\forgronw) 
go to zero uniformly, and the third one  is bounded by 
$\int_a^b||A(s)||\cdot||Y_j(s)-Y(s)||\,ds$, which also goes to zero. So
$\tilde Y_j-\tilde Y\to 0$  uniformly.
}

\longversion{
We have thus shown that $\tilde Y_j-\tilde Y\to 0$  uniformly if either
$W_j\to W$ a.e. or $Y_j\to Y$ a.e. This clearly implies  all the conclusions
of (\theolin.2).\proofend
}
 
\shortversion{\noindent} A very important subspace of
$\,\,BM([a,b],\R^n)\,\,$ is $\,\,BV([a,b],\R^n)\,\,$, the space of 
all functions \hbox{$W:[a,b]\to\R^n$ such that} (a)~$W$ is of bounded
variation, (b)~$W$ is right-continuous at every point of $]a,b]$, and
(c)~$W(a)=0$.  There is a canonical correspondence between $BV([a,b],\R^n)$ and
the dual space $C^0([a,b],\R_n)^*$ of the Banach space $C^0([a,b],\R_n)$ of
continuous functions $[a,b]\to\R_n$. (We write $R_n$ rather than $R^n$
because we want to think of the members of $C^0([a,b],\R_n)$ as {\sl
row-vector-valued functions}, cf. below.)

The members of $C^0([a,b],\R_n)^*$ are the {\sl $\R^n$-valued Borel
measures on $[a,b]$}.  When $n=1$, they are the {\sl finite signed Borel
measures on $[a,b]$}.  
%Clearly, $C^0([a,b],\R_n)^*$ is canonically
%identified with $\Bigl(C^0([a,b],\R)^*\Bigr)^n$, so the members of
%$C^0([a,b],\R_n)^*$ can be thought of as $n$-tuples of finite signed Borel
%measures.  
The identification map from $BV([a,b],\R^n)$ to
$C^0([a,b],\R_n)^*$ is the one that assigns to a function $W\in
BV([a,b],\R^n)$ the unique Borel measure $\mu_W$ such that
$\mu_W([a,t])=W(t)$ for $a<t\le b$. 
%(The measure of a single-point set
%$\{t\}$ is then $\mu_W(t+)-\mu_W(t-)$, which is equal to
%$\mu_W(t)-\mu_W(t-)$ if $t>a$ and to $\mu_W(a+)$ if $t=a$.  In particular,
%$\mu_W(\{t\})=0$ iff $\mu_W$ is continuous at $t$.)

{}From now on we identify a function $W\in BV([a,b],\R^n)$ with its
corresponding measure $\mu_W$, and write $\int\alpha\cdot dW$ for
$\int\alpha\cdot d\mu_W$.  
%As explained above, it is convenient to think of
%the members of $C^0([a,b],\R_n)$ as row-vector-valued functions
%$\alpha=(\alpha_1,\ldots,\alpha_n)$, and of those of $BV([a,b],\R^n)$ as
%column-vector-valued functions $W=(W^1,\ldots,W^n)$, so the expression
%$\alpha\cdot dW$ stands for the inner product $\sum_k\alpha_k\,dW^k$.
Clearly, $\int\alpha\cdot dW$ is defined, more generally, for an arbitrary
bounded Borel measurable $\R_n$-valued map $\alpha$ on $[a,b]$.

Naturally,  the integral $\int\alpha\,dW$ of a scalar bounded Borel measurable
function $\alpha:[a,b]\to\R$ is also well defined, and is a
vector in $\R^n$.   
%We emphasize that, if
%$W$ is not continuous, then for an integral such as $\int_s^tdW$ to be well
%defined we have to be careful to specify whether the integral is to be
%interpreted as $\int \chi_{[s,t]}\,dW$ or $\int \chi_{]s,t]}\,dW$ or $\int
%\chi_{[s,t[}\,dW$ or $\int \chi_{]s,t[}\,dW$, so we will never write
%$\int_s^t\,dW$ unless $W$ is continuous at $s$ and $t$.
 
On $BV([a,b],R^n)$ there are at least three important topologies: 
(a)~the one induced by the norm $||\ldots||_{BM}$, (b)~the strong topology
of $BV([a,b],\R^n)$ as the dual of the Banach space $C^0([a,b],\R_n)$,
(c)~the weak$^*$ topology, also corresponding to the duality with
$C^0([a,b],\R_n)$.  
%The first one will not be of direct interest to us,
%except for the fact that, in order to apply Theorem \theolin{}, we will
%often need to know that a set is bounded in the norm $||\ldots||_{BM}$. 
The second one is induced by the {\sl total variation norm}:
$$
||W||_{TV}=\sup\biggl\{\Bigl|\int \alpha\cdot dW\Bigr|:\alpha\in
C^0([a,b],\R_n)\,,\,||\alpha||_{BM} \le 1\biggr\}\,.\numeqap
$$ 
%It is well known that $||W||_{TV}$ is the supremum of the sums
%$S_{\pi}(W)$, ranging over all partitions
%$\pi\in\Pi([a,b])=\cup_{m}\Pi_m([a,b])$, where $\Pi_m([a,b])$ is the set of
%all $m+1$-tuples $(t_0,t_1,\ldots,t_m)$ of members of $[a,b]$ such that
%$a=t_0<t_1<\ldots<t_m=b$ and, for $\pi=(t_0,t_1,\ldots,t_m)\in \Pi([a,b])$,
%$S_{\pi}(W)\equaldef\sum_{i=1}^m||W(t_i)-W(t_{i-1})||$.  In particular, 
It is clear that $||W||_{BM}\le ||W||_{TV}$ for all $W$ in $BV([a,b],\R^n)$.
Therefore, {\sl if a subset $S$ of $BV([a,b],\R^n)$ is bounded in $TV$ norm
then $S$ is also bounded in $BM$ norm}.

We will need the following \shortversion{result, whose proof
is carried out using the uniform boundedness theorem and standard 
compactness properties:} 
\xdef\lemweakconv{{\rm \appenlet}.\the\theono}
\lemmap
If a sequence $\{W_j\}$ in $BV([a,b],\R^n)$ weak$^*$-converges to a $W\in
BV([a,b],\R^n)$, then (a)~$\{W_j\}$ is bounded in $TV$ norm, (b)~$W_j(a)\to
W(a)$ and $W_j(b)\to W(b)$, (c)~every subsequence of $\{W_j\}$ has a
subsequence that converges to $W$ at all but at most countably many points
of $[a,b]$, (d)~$\int_a^b\beta(s)||W_j(t)-W(t)||\,dt\to 0$ for every
function $\beta\in L^1([a,b],\R)$. \shortversion{\proofend}

%gnung


\longversion{
\beginproof
Statement (a) follows from the uniform boundedness theorem, and Statement
(b) is trivial, since $W_j(a)=0=W(a)$, $W_j(b)=\int 1\,dW_j$, and
$W(b)=\int 1\,dW$. Statement (c) for $n=1$ is a consequence of Helly's
theorem, and the statement for arbitrary $n$ then follows easily from 
the case $n=1$.  Statement (d) then clearly follows from (c), using the
dominated convergence theorem together with the fact that boundedness in
$TV$ norm implies boundedness in $BM$ norm. \proofend
}

As a corollary of Lemma \lemweakconv{}, we get the integration by parts 
formula:
\xdef\lemparts{{\rm \appenlet}.\the\theono}
\lemmap
If $\lambda:[a,b]\to\R_n$ is absolutely continuous, and $W\in
BV([a,b],\R^n)$, then 
\xdef\forparts{\equanoapno}
$$
\lambda(b)\cdot W(b)=\int \lambda\cdot dW+\int_a^b \dot\lambda(t)
\cdot W(t)\,dt\,.\numeqap
$$

\beginproof
The result is obviously true, because of the standard product rule for
derivatives, if $W$ is also absolutely continuous. In the general case, we
use the fact that an arbitrary $W\in BV([a,b],\R^n)$ is the weak$^*$-limit
of a sequence $\{W_j\}$ of absolutely continuous functions.  It then
follows that
(i)~$\lambda(b)\cdot W_j(b)\to \lambda(b)\cdot W(b)$, because $W_j(b)\to
W(b)$, (ii)~$\int \lambda\cdot dW_j\to\int \lambda\cdot dW$, because
$\{W_j\}$ weak$^*$-converges to $W$, and $\lambda$ is continuous, and
(iii)~$\int_a^b \dot\lambda(t)\cdot W_j(t)\,dt\to
\int_a^b \dot\lambda(t)\cdot W(t)\,dt$, because Part (d) of Lemma
\lemweakconv{} implies that 
$\int_a^b ||\dot\lambda(t)||.||W_j(t)-W(t)||\,dt\to 0$,  since
$\dot\lambda\in L^1$.   \proofend

An important consequence of Lemma \lemparts{} is the following
\xdef\lemadjo{{\rm \appenlet}.\the\theono}
\lemmap
Let $W\in BV([a,b],\R^n)$, $A\in L^1([a,b],\R^{n\times n})$,
$\bar Y\in\R^n$, and write $Y=Y_{A,dW,\bar Y}$. Suppose
$\lambda:[a,b]\to\R_n$ is absolutely continuous and satisfies
$\dot\lambda(t)=-\lambda(t)A(t)$ for almost all $t$.  Then
\xdef\foradjo{\equanoapno}
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=\int\lambda\cdot dW\,.\numeqap
$$

\beginproof
Write $Y(t)=V(t)+W(t)$, where $V(t)=\bar Y+\int_a^tA(s).Y(s)\,ds$.  
Then 
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=
\lambda(b)\cdot V(b)-\lambda(a)\cdot V(a)+\lambda(b)\cdot W(b)\,.\numeqap
$$
The function $t\to\lambda(t)\cdot V(t)$ is absolutely continuous
and its derivative is $\dot\lambda(t)\cdot V(t)+\lambda(t)\cdot\dot V(t)$,
which equals $-\lambda(t)A(t)Y(t)+\lambda(t) A(t)Y(t)-\dot\lambda(t)\cdot W(t)$,
i.e. $-\dot\lambda(t)\cdot W(t)$. So 
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=\lambda(b)\cdot W(b)
-\int_a^b\dot\lambda(t)\cdot W(t)\,dt\,.\numeqap
$$
Equation (\foradjo) follows from this and the integration by parts formula
(\forparts). \proofend

It is clear that, if $A\in L^1([a,b],\R^{n\times n})$, $\bar Y\in\R^n$, and 
$W\in BM_0([a,b],\R^n)$, then $Y_{A,dW,\bar Y}-W$ is absolutely continuous,
so $W\in BV([a,b],\R^n)$ iff $Y_{A,dW,\bar Y}$ is of bounded variation,
which happens iff $Y_{A,dW,\bar Y}-\bar Y\in BV([a,b],\R^n)$.

\shortversion{The following theorem is proved by repeated use of Gronwall's
inequality:} 
\xdef\theolinbv{{\rm \appenlet}.\the\theono}
\theoap
Let $\{W_j\}$,  $\{\bar Y_j\}$, $\{A_j\}$ be sequences in 
$BV([a,b],\R^n)$, $\R^n$, $L^1([a,b],\R^{n\times n})$, respectively.
Assume that $\{A_j\}$ is bounded in $L^1([a,b],\R^{n\times n})$  and 
$\{\bar Y_j\}$ is bounded in $\R^{n}$. Let $Y_j=Y_{A_j,dW_j,\bar Y_j}$ . Then

\itemitem{(\theolinbv.1)}  {\sl $\{Y_j-\bar Y_j\}$ is bounded in 
$BV([a,b],\R^{n})$ if and only if $\{W_j\}$ is bounded in
$BV([a,b],\R^{n})$.}

\itemitem{(\theolinbv.2)} {\sl If $\{A_j\}$ converges in $L^1([a,b],\R^n)$
to a limit $A$, $\bar Y_j\to\bar Y$, $W\in BV([a,b],\R^n)$, and we write
$Y=Y_{A,dW,\bar Y}$, then $\{W_j\}$ weak$^*$-converges to $W$ iff
$\{Y_j-\bar Y_j\}$ weak$^*$-converges to $Y-\bar Y$.  In that case,
$\{Y_j-W_j-\bar Y_j\}$ converges to $Y-W-\bar Y$ strongly in
$BV([a,b],\R^n)$. In particular, (a)~$W_j(b)\to W(b)$, (b)~$Y_j(b)\to
Y(b)$, and (c)~for every $t\in[a,b]$, $Y_j(t)\to Y(t)$ iff 
$W_j(t)\to W(t)$. \shortversion{\proofend}}

\longversion{
\beginproof
We know from theorem \theolin{} that $\{Y_j-\bar Y_j\}$ is bounded in $BM$
norm iff $\{W_j\}$ is.  Since boundedness in $BV$ norm implies boundedness
in $BM$ norm, we see that boundedness of one of the two sequences in $BV$
norm implies that there is a constant $C$ such that $||Y_j(t)||\le C$ for
all $j,t$, and then the integral $I_j=\int_a^b||A_j(t)Y_j(t)||\,dt$ is
bounded by a fixed constant $K$.  But $I_j=||Y_j-\bar Y_j-W_j||_{BV}$.  So
boundedness in $BV$ of one of the sequences $\{W_j\}$, $\{Y_j\}$ implies
boundedness in $BV$ of the other one.
}

\longversion{
If either $\{W_j\}$ weak$^*$-converges to $W$, or $\{Y_j-\bar Y_j\}$
weak$^*$-converges to $Y-\bar Y$, then the uniform boundedness theorem
implies that either $\{W_j\}$ or $\{Y_j-\bar Y_j\}$ is bounded in $BV$
norm, so both sequences are bounded in $BV$ norm---and hence also in
$BM$ norm---by a constant $C$.  Writing $Z_j=Y_j-\bar Y_j-W_j$, $Z=Y-\bar
Y-W$, we have then $Z_j-Z=R_j^1+R_j^2$, where
$$
R_j^1(t)=\int_a^t\Bigl(A_j(s)-A(s)\Bigr)Y_j(s)\,ds\,,\qquad
R_j^2(t)=\int_a^tA(s)\Bigl(Y_j(s)-Y(s)\Bigr)\,ds\,.\numeqap
$$
Then $||R_j^1||_{BV}\le C||A_j-A||_{L^1}$, so $||R_j^1||_{BV}\to 0$.  Also,
$$||R_j^2||_{BV}\le||A||_{L^1}.||\bar Y_j-\bar Y||+
\int_a^b||A(t)||.||Y_j(t)-\bar Y_j-(Y(t)-\bar Y)||\,dt\,.\numeqap
$$  
}

\longversion{
If $\{Y_j-\bar Y_j\}$ weak$^*$-converges to $Y-\bar Y$, then Lemma
\lemweakconv{} implies that $||R_j^2||_{BV}\to 0$, so $Z_j\to Z$ in
$BV([a,b],\R^n)$.
}

\longversion{
If $\{W_j\}$ weak$^*$-converges to $W$, then we rewrite $R_j^2$ as
$$
R_j^2(t)=\int_a^tA(s)\Bigl(Z_j(s)-Z(s)\Bigr)\,ds+R_j^3(t)+R_j^4(t)\,,\numeqap
$$
where  $R_j^3(t)=\int_a^tA(s)\Bigl(\bar Y_j-\bar Y\Bigr)\,ds$ and
$R_j^4(t)=\int_a^tA(s)\Bigl(W_j(s)-W(s)\Bigr)\,ds$.  We then 
have $||R_j^3||_{BV}\le||A||_{L^1}.||\bar Y_j-\bar Y||$ and
$||R_j^4||_{BV}\le\int_a^b||A(s)||.||W_j(s)-W(s)||\,ds$.  Lemma
\lemweakconv{} then implies that $||R_j^4||_{BV}\to 0$.  So
$Z_j(t)-Z(t)=\int_a^tA(s).(Z_j(s)-Z(s))\,ds+R_j(t)$, where
$R_j=R_j^1+R_j^3+R_j^4$, so $||R_j||_{BV}\to 0$.  Then Gronwall's
inequality implies that $||Z_j-Z||_{BM}\le e^{||A||_{L^1}}||R_j||_{BV}$.
If $Q_j(t)=\int_a^tA(s).(Z_j(s)-Z(s))\,ds$, then $||Q_j||_{BV}\le
||A||_{L^1}||Z_j-Z||_{BM}$, so $||Q_j||_{BV}\to 0$.  Since
$Z_j-Z=Q_j+R_j$, we see that $||Z_j-Z||_{BV}\to 0$.
}

\longversion{
So we have shown that if one of the two sequences $\{W_j-W\}$,
$\{Y_j-\bar{Y}_j-(Y-\bar{Y})\}$ weak$^*$-converges to $0$, then
$Y_j-\bar{Y}_j-W_j\to Y-\bar{Y}-W$ strongly in $BV([a,b],\R^n)$.  This
implies all the conclusions of (\theolinbv.2).\proofend
}

\medbreak
Recall that, if $P,Q$ are normed spaces, and $F:\Omega\to Q$ is a map
defined on a neighborhood $\Omega$ of a point $p$ of $P$, then $F$ is said
to be {\sl Fr\'echet differentiable} at $p$ if there exists a bounded
linear map $L$---called the {\sl Fr\'echet differential} of $F$ at
$p$---such that $F(p')=F(p)+L(p'-p)+o(||p'-p||)$ as $p'\to p$.  If $X$ is a
Banach space and $Q=X^*$, then we call $F$ {\sl weak$^*$ Fr\'echet
differentiable} at $p$ if for every $x\in X$ the map $\Omega\ni
p'\to\langle F(p'),x\rangle\in\R$ is Fr\'echet differentiable at $p$.  (We
use $\langle x^*,x\rangle$, for $x\in X$, $x^*\in X^*$, as an alternative
notation for $x^*(x)$.)

%It is easy to see that in general weak$^*$ Fr\'echet differentiability does
%not imply Fr\'echet differentiability.  (For example, if $X=C^0([0,1],\R)$,
%$P=\R$, $\Omega=[0,1]$, and $F(t)=\chi_{[0,t]}$ for $t\in\Omega$, then for
%every $\alpha\in X$ the map $t\to \langle F(t),\alpha\rangle$ is just
%the map $t\to\int_a^t\alpha(s)\,ds$, which is differentiable at every
%$t\in]0,1[$.  The weak$^*$-limit $\lim_{h\to 0}h^{-1}(F(t+h)-F(t))$ is
%$\delta_t$, the delta function at $t$ (i.e. the functional
%$\alpha\to\alpha(t)$), but $||h^{-1}(F(t+h)-F(t))-\delta_t||_{BV}=2$ if
%$h\neq 0$ and $t,t+h\in[0,1]$.)

Suppose $P$ is a normed space, $X$ is a Banach space, $p\in P$,
$F:\Omega\to X^*$ is a map defined on a neighborhood $\Omega$ of $p$, and
$F$ is weak$^*$ Fr\'echet differentiable at $p$.  Then for each $x\in X$
there is a bounded linear functional $\theta_x$ on $P$ such that $\langle
F(p+v),x\rangle=\langle F(p),x\rangle+\theta_x(v)+o(||v||)$ as $v\to
0$ in $P$. Clearly, $\theta_x$ is unique, and is given by the formula
$$
\theta_x(v)=\lim_{h\to 0}h^{-1}\langle F(p+hv)-F(p),x\rangle\,.\numeqap
$$
For any given $v\in P$, use $\theta^v$ to denote the map
$x\to\theta_x(v)$. Then $\theta^v$ is a limit of linear maps, so it is
linear.  If $\{h_j\}$ is any sequence going to $0$, then the sequence of
linear functionals $\nu_j$ on $X$ given by $\nu_j(x)=h_j^{-1}\langle
F(p+h_jv)-F(p),x\rangle$ converges pointwise to $\theta^v$. Since each
$\nu_j$ is a bounded linear functional on $X$, it follows for the uniform
boundedness theorem that $\theta^v$ is bounded.  Since
$\theta^v(x)=\theta_x(v)$, which is linear with respect to $v$, we can
conclude that $v\to \theta^v$ is a linear map from $P$ to $X^*$. We use
$D^w\!F(p)$ to denote this map, and $D^w\!F(p).v$ to denote its value for a
particular $v$, so $D^w\!F(p).v=\theta^v$, and 
\xdef\defofdwf{\equanoapno}
$$
\langle D^w\!F(p).v,x\rangle=\lim_{h\to 0}h^{-1}\langle
F(p+hv)-F(p),x\rangle\,.\numeqap
$$
\shortversion{The following lemma follows by applying the uniform
boundedness principle:} 
\xdef\isbounded{{\rm \appenlet}.\the\theono}
\lemmap
If $P$ is a normed space, $X$ is a Banach space, $p\in P$,
$F:\Omega\to X^*$ is a map defined on a neighborhood $\Omega$ of $p$,
and $F$ is weak$^*$ Fr\'echet differentiable at $p$.  Then: 

{\pari
\itemitem{(\isbounded.i)} {\sl there is a $C>0$ such that} 
\xdef\bdforf{\equanoapno}
$$
||F(p+v)-F(p)||\le C||v|| \qquad{\sl for \,\,\,all\,\,\,sufficiently
\,\,\,small \qquad} v\in P\,,\numeqap
$$

\itemitem{(\isbounded.ii)} {\sl the linear map $D^w\!F(p):P\to X^*$ 
defined by (\defofdwf) is bounded. \shortversion{\proofend}} 

}

\longversion{
\beginproof
Suppose (\isbounded.i) is not true. Then there exists a sequence
$\{v_j\}$ in $P$, converging to $0$ and such that
$||F(p+v_j)-F(p)||>j||v_j||$. Define, for each $j$, a  
member $\sigma_j$ of $X^*$,  by 
$$
\sigma_j={F(p+v_j)-F(p)-\theta^{v_j}\over ||v_j||}\,.\numeqap
$$
Then $\sigma_j(x)\to 0$ for each $x$, so the sequence $\{\sigma_j\}$ is
pointwise bounded and  hence uniformly bounded.   
}

\longversion{    
On the other hand, for each $x\in X$, the sequence $\Big\{{\theta^{v_j}(x)\over
||v_j||}\Big\}$ is bounded, because $\theta^{v_j}(x)=\theta_x(v_j)$, and
$\theta_x$ is a bounded linear functional on $P$,  so
$|\theta^{v_j}(x)|\le||\theta_x||.||v_j||$. So, using the uniform
boundedness theorem once again, we conclude that the sequence
$\Bigl\{{\theta^{v_j}\over ||v_j||}\Bigr\}$ is bounded in $X^*$.  Since
$\{\sigma_j\}$ is 
bounded, we conclude that the sequence
$\Bigl\{{F(p+v_j)-F(p)\over ||v_j||}\Bigr\}$ is
bounded, contradicting the fact that $||F(p+v_j)-F(p)||>j||v_j||$.
So (\bdforf) holds for some $C>0$, and we have proved (\isbounded.i). 
}

\longversion{
It follows from  (\isbounded.i) that $|\langle
F(p+v)-F(p),x\rangle|\le C||x||.||v||$ for all $x\in X$, if $v\in P$ is
small enough.  Applying this with $hv$ in the role of $v$, and letting
$h\downarrow 0$, we find that
$$
|\langle D^w\!F(p).v,x\rangle|\le C||x||.||v||\qquad{\rm for\,\,\,all\qquad}
x\in X\,,\,\,\,v\in P\,.\numeqap
$$
Therefore $||D^w\!F(p).v||\le C||v||$, so $D^w\!F(p)$ is a bounded linear 
map from $P$ to $X^*$.  \proofend
}

\shortversion{\noindent} We are now ready to study the differentiability
properties with respect to 
a parameter $p$, belonging to a normed space $P$, of a family of
trajectories $x_p:[a,b]\to\R^n$ of time-varying vector fields $f_p$.  We
 will prove, under minimal technical hypotheses, that if the initial
condition map $p\to x_p(a)$ is differentiable at $p=0$, and the map
$p\to f_p$ satisfies a ``weak differentiability condition'' at the
trajectory $x_0$, then the map $p\to x_p$ is weak$^*$- differentiable
at $p=0$, as a map into $BV([a,b],\R^n)$, and in particular the endpoint
map $p\to x_p(b)$ is differentiable at $p=0$.

A {\sl curve in $\R^n$} is a continuous map $\xi:I\to\R^n$, whose
domain $I={\rm Dom}(\xi)$ is a nonempty interval. A curve $\xi$ is an {\sl
arc} if ${\rm Dom}(\xi)$ is compact.  The {\sl graph} $G(\xi)$ of a curve
$\xi$ is the set $G(\xi)=\{(\xi(t),t):t\in{\rm Dom}(\xi)\}$.  If
$\xi:[a,b]\to\R^n$ is an arc in $\R^n$, and $\ve>0$, then the
$\ve$-{\sl tube about $\xi$} is the set ${\cal
T}(\xi,\ve)=\{(x,t):x\in\R^n,a\le t\le b,||x-\xi(t)||\le\ve\}$.

A {\sl time-varying vector field} on $\R^n$ is an $\R^n$-valued map $f$
whose domain ${\rm Dom}(f)$ is a---possibly empty---subset $S$ of the
product $\R^n\times\R$. We use $TVVF(\R^n)$ to denote the set of all
time-varying vector fields on $\R^n$.  A {\sl trajectory}, or {\sl integral
curve}, of an $f\in TVVF(\R^n)$ is a locally absolutely continuous curve
$\xi$ in $\R^n$ such that $G(\xi)\subseteq{\rm Dom}(f)$ and
$\dot\xi(t)=f(\xi(t),t)$ for almost all $t\in{\rm Dom}(\xi)$.  We use
$\traj(f)$ to denote the set of all trajectories of $f$, and $\traj_c(f)$
to denote the set of all $\xi\in\traj(f)$ such that ${\rm Dom}(\xi)$ is
compact.

If $f\in TVVF(\R^n)$ and $\xi\in\traj(f)$, we say that $\xi$ is a {\sl
maximal} trajectory of $f$ if it cannot be extended to a trajectory
$\tilde\xi:\tilde I\to\R^n$ of $f$ defined on an interval $\tilde I$
such that $I\subseteq\tilde I$ but $I\neq\tilde I$. We use $\mtraj(f)$ to
denote the set of all maximal trajectories of $f$.  Given $\bar x\in\R^n$,
$\bar t\in\R$, we use $\traj(f,\bar x,\bar t)$, $\mtraj(f,\bar x,\bar t)$
to denote, respectively, the set of all $\xi\in\traj(f)$ such that
$\xi(\bar t)=\bar x$, and the set $\traj(f,\bar x,\bar t)\cap\mtraj(f)$.
Zorn's lemma implies that every $\xi\in\traj(f,\bar x,\bar t)$ can be
extended to a $\tilde\xi\in\mtraj(f,\bar x,\bar t)$, so
$\traj(f,\bar{x},\bar t)\neq\emptyset$ iff
$\mtraj(f,\bar{x},\bar{t})\neq\emptyset$. Clearly,
$\traj(f,\bar{x},\bar{t})\neq\emptyset$ iff
$(\bar{x},\bar{t})\in{\rm{}Dom}(f)$.  (Indeed, if
$(\bar{x},\bar{t})\in{\rm{}Dom}(f)$, then the map $\xi$ with domain
$\{\bar{t}\}$ such that $\xi(\bar{t})=\bar x$ is in $\traj(f)$.)
 


We say that an $f\in TVVF(\R^n)$ has the {\sl forward existence property}
at a point $(\bar x,\bar t)$ if for some $\ve>0$ there exists a
$\xi\in\traj(f,\bar x,\bar t)$ with domain $[\bar t,\bar t+\ve]$.  We say
that $f$ has the {\sl forward limit property} on a subset $S$ of
${\rm{}Dom}(f)$ if, whenever $\xi:[a,b[\,\to\R^n$ is a trajectory of $f$
which is contained in a compact subset of $S$, it follows that
$\lim_{t\uparrow b}\xi(t)$ exists.

Suppose we are given the following data:

\medbreak

{\pari

\item{(D.1)} a  normed space $P$,

\item{(D.2)} a family ${\bf f}=\{f_p\}_{p\in P}$ of
time-varying vector fields on $\R^n$, depending on a parameter $p\in P$,

\item{(D.3)} a $p_0\in P$, 

\item{(D.4)} a trajectory $\xi:[a,b]\to\R^n$ of $f_{p_0}$.

}

\medbreak\noindent We then define functions $w_p:[a,b]\to\R^n$  by 
\xdef\defofwp{\equanoapno}
$$
w_p(t)=f_p(\xi(t),t)-f_{p_0}(\xi(t),t)\quad{\rm for\quad}t\in[a,b]\,,\numeqap
$$
so $w_p$ is defined on $[a,b]$ whenever $G(\xi)\subseteq{\rm Dom}(f_p)$.
If $w_p$ is integrable, we let 
\xdef\defofWp{\equanoapno}
$$
W_p(t)=\int_a^tw_p(s)\,ds\qquad{\rm for\qquad}t\in[a,b]\,.\numeqap
$$
We then let ${\cal W}$ be the map  that sends $p$ to ${\cal W}(p)=W_p$,
with domain  
\xdef\defofdomw{\equanoapno}
$$
{\rm Dom}({\cal W})=\{p\in P: G(\xi)\subseteq{\rm Dom}(f_p),
\,\,w_p\in L^1([a,b],\R^n)\}\,.\numeqap
$$
We regard ${\cal W}$ as a map from ${\rm Dom}({\cal W})$ to 
$BV([a,b],\R^n)$. 

\medbreak\noindent  We then say that ${\bf f}$ is {\sl weakly
differentiable at $p_0$ along $\xi$} if there exists $\bar\ve>0$ such that

\medbreak
{\pariim 

\item{(WD1)} {\sl ${\cal T}(\xi,\bar\ve)\subseteq{\rm Dom}(f_p)$ for every 
$p$ such that $||p-p_0||\le\bar\ve$.}

\item{(WD2)} {\sl For every $p$ such that $||p-p_0||\le\bar\ve$,  $f_p$ has
the forward limit property on ${\cal T}(\xi,\bar\ve)$ and the forward
existence property at every $(x,t)$ such that $a\le t<b$ and
$||x-\xi(t)||<\bar\ve$.}

}

{\pariim

\item{(WD3)}  {\sl There exist $A\!\in\!L^1([a,b],\R^{n\times n})$, 
and functions $\psi_\ve\!\in\!L^1([a,b],\R)$,  for 
$\ve\!\in ]0,\bar\ve]$, such that}
\xdef\fpxt{\equanoapno}
$$
||f_p(x,t)-f_p(\xi(t),t)-A(t).(x-\xi(t))||\le\psi_\ve(t)(||x-\xi(t)||
+||p-p_0||)\numeqap
$$
{\sl whenever $a\le t\le b$, $||x-\xi(t)||\le\ve$,   $||p-p_0||\le\ve$,  and} 
\xdef\limpsive{\equanoapno}
$$
\lim_{\ve\downarrow 0}\int_a^b\psi_\ve(t)\,dt=0\,.\numeqap
$$


\item{(WD4)}  {\sl The function $w_p$ defined by (\defofwp)
is integrable for every $p$ such that $||p-p_0||\le\bar\ve$ and, if
${\cal{}W}$ is defined as above, then ${\cal W}$ is weak$^*$ Fr\'echet
differentiable at $p_0$.}

}


\parnormal

\medbreak\noindent 
The second part of condition (WD4) says that for each $\alpha\in
C^0([a,b],\R_n)$ the map $p\to \int\alpha\cdot dW_p$ is Fr\'echet
differentiable at $p_0$.  Lemma \isbounded{} then implies---since
$W_{p_0}=0$---that there exists a bounded linear map $D^w{\cal 
W}(p_0):P\to BV([a,b],\R^n)$ having the property that
\xdef\flafordifwp{\equanoapno}
$$
\int\alpha\cdot dW_p=\langle{\cal W}(p)-{\cal W}(p_0),\alpha\rangle=
\int\alpha\cdot d\Bigl(D^w{\cal W}(p_0).(p-p_0)\Bigr)+o(||p-p_0||)
\qquad{\rm as}\qquad p\to p_0\numeqap
$$
for each $\alpha\in C^0([a,b],\R_n)$, and also that there is a constant
$C$ such that
\xdef\constantc{\equanoapno}
$$
||W_p||_{TV}\le C||p-p_0||\qquad{\rm for\,\,\,all\,\,\,}p\in P{\rm
\,\,\,such\,\,\,that\,\,\,}\|p-p_0\|{\rm \,\,\,is\,\,\,
sufficiently \,\,\,small}.\numeqap
$$

The following result is the main theorem on differentiation of trajectories 
with respect to a parameter. \shortversion{The proof involves a 
number of estimates using Gronwall's inequality.}

\xdef\theodiftraj{{\rm  \appenlet}.\the\theono}
\theoap
Let $P$, ${\bf f}$, $p_0$, $\xi$ be data as in (D.1,2,3,4). 
Assume that ${\bf f}$ is weakly differentiable at $p_0$ along $\xi$.  Let
$\bar\ve>0$ be such that (WD1) to (WD4) hold.  Then

{\parix

\item{(\theodiftraj.a)} {\sl  For every $\ve\in\,]\,0,\bar\ve]$ there
exists a $\delta\in\,]\,0,\ve[$ such that, whenever $||p-p_0||\le\delta$,
$||\bar x-\xi(a)||\le\delta$, and $\zeta\in\mtraj(f_p,\bar x,a)$, it
follows that $[a,b]\subseteq{\rm Dom}(\zeta)$ and
$||\zeta(t)-\xi(t)||\le\ve$ for all $t\in[a,b]$.}

\item{(\theodiftraj.b)} {\sl If $\{\bar x_p\}_{||p-p_0||\le\bar\ve}$ is a
family of points of $\R^n$ such that the map $p\to \bar x_p$ is
Fr\'echet differentiable at $p_0$ and $\bar x_{p_0}=\xi(a)$, then, if
$\{\xi_p\}_{||p-p_0||\le\bar\ve}$ is a family of curves such that
$\xi_p\in\mtraj(f_p,\bar x_p,a)$, and $x_p$ denotes the restriction of
$\xi_p$ to $[a,b]$---which is well defined for small enough $||p-p_0||$ by
(\theodiftraj.a)---then the map ${\cal X}$ that sends each $p$ to the
function $x_p$ is weak$^*$ Fr\'echet differentiable at $p_0$ as a map from
$P$ to $BV([a,b],\R^n)$.  Moreover, $D^w{\cal X}(p_0)$ can be computed as
follows.  For $v\in P$, 
$$
D^w{\cal{}X}(p_0).v=Y_{A,\,d(D^w{\cal{}W}(0).v),
\,L.v}\,,\numeqap
$$
where $L:P\to\R^n$ is the differential at $p_0$
of the map $p\to \bar x_p$. \shortversion{\proofend}} 

}

\longversion{
\beginproof
We will assume, as we clearly may without loss of generality, that
$p_0=0$. Also, we know that there is a $C>0$ such that the bound
(\constantc) holds.  So we can assume, by making $\bar\ve$ smaller, if
necessary, that $||W_p||_{TV}\le C||p||$ whenever $||p||\le\bar\ve$.  By
making $C$ larger and $\bar\ve$ smaller, if necessary, we may also assume
that $C\ge 1$, and
\xdef\alsoassume{\equanoapno}
$$
8e^{||A||_{L^1}}||\psi_\ve||_{L^1}<1 \qquad{\rm for\qquad}
\ve\in]0,\bar\ve]\,. \numeqap
$$ 
Pick $\ve\in]0,\bar\ve[$, and choose $\delta>0$ such that  
\xdef\choicea{\equanoapno}
$$
C\delta<\ve\,.\numeqap
$$  
Notice that $\delta<\ve$, since $C\ge 1$. 
}

\longversion{
Let $p$, $\bar x$ be such that $||p||\le\delta$ and
$||\bar{x}-\xi(a)||\le\delta$.  Let $\zeta\in\mtraj(f_p,\bar x,a)$, and let
$I={\rm Dom}(\zeta)$.  Then $a\in I$. Let $c$ be the supremum of those $t$
such that $a\le t\le b$ and $||\zeta(s)-\xi(s)||\le\ve$ for $a\le s\le t$.
Then $a<c\le b$, $c\in I$, and either $c=b$ or $||\zeta(c)-\xi(c)||=\ve$.
(The fact that $a<c$ follows from the local existence property, since
$||\bar x-\xi(a)||\le\delta<\ve<\bar\ve$, and $\zeta$ is continuous.  The
fact that $c\in I$ follows from the forward limit property, since the graph
of the restriction of $\zeta$ to $[a,c[$ is contained in ${\cal
T}(\xi,\ve)$.  If $c<b$ and $||\zeta(c)-\xi(c)||<\ve$, then the forward
existence property implies that $\zeta$ is not maximal.)
}

\longversion{
We will estimate $||\zeta(t)-\xi(t)||$ for $t\in[a,c]$, and show that---if
$\ve$ is small enough, and $\delta$ is suitably
chosen---$||\zeta(t)-\xi(t)||<\ve$ for all such $t$, including $t=c$.  This
will imply that $c=b$.
}
 
\longversion{
{}From now on, we work on the interval $[a,c]$.  We have  
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+\int_a^t\Bigl(f_p(\zeta(s),s)-f_0(\xi(s),s)\Bigr)
\,ds\,,\numeqap
$$
so 
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+W_p(t)+
\int_a^t\Bigl(f_p(\zeta(s),s)-f_p(\xi(s),s)\Bigr)\,ds\,.\numeqap
$$
Let $r_p(t)=||f_p(\zeta(t),t)-f_p(\xi(t),t)-A(t).(\zeta(t)-\xi(t))||$.
Then 
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+W_p(t)+\int_a^tA(s).(\zeta(s)-\xi(s))\,ds
+\int_a^tr_p(s)\,ds\,,\numeqap
$$
so
$$
||\zeta(t)-\xi(t)||\le||\bar x-\xi(a)||+
||W_p||_{sup}+\int_a^t||A(s)||.||\zeta(s)-\xi(s)||\,ds+\int_a^t
r_p(s)\,ds\,,\numeqap
$$
and then 
$$
||\zeta(t)-\xi(t)||\le||\bar x-\xi(a)||+||W_p||_{sup}+
2\ve||\psi_\ve||_{L^1}+\int_a^t||A(s)||.||\zeta(s)-\xi(s)||\,ds\,,\numeqap
$$
because $r_p(s)\le 2\ve\psi_{\ve}(s)$ for $s\in[a,c]$, since 
$||\zeta(s)-\xi(s)||\le\ve$ and $||p||\le\ve$.
}

\longversion{
Since $||\bar x-\xi(a)||\le\delta$, and  
$||W_p||_{sup}\le||W||_{TV}\le C||p||\le C\delta$, Gronwall's inequality
implies  
$$
||\zeta(t)-\xi(t)||\le
e^{||A||_{L^1}}\Bigl((C+1)\delta+2\ve||\psi_\ve||_{L^1}\Bigr)\,.\numeqap
$$
In view of (\alsoassume), we have  
$$
2\ve e^{||A||_{L^1}}||\psi_\ve||_{L^1}<{\ve\over 4}\,.\numeqap
$$
Now suppose that 
\xdef\choiceb{\equanoapno}
$$
\delta\le\kappa\ve\,,\qquad{\rm where}\qquad\kappa=(4C+4)^{-1}e^{-||A||_{L^1}}
\,.\numeqap
$$
Then  
$$
||\zeta(t)-\xi(t)||\le{\ve\over 2}\qquad{\rm for\qquad all}\qquad
t\in[a,c]\,.\numeqap
$$   
This is our desired estimate, from which it follows, as explained above, 
that $c=b$, and then $||\zeta(t)-\xi(t)||\le\ve$ for all $t\in[a,b]$.  This 
completes the proof of (\theodiftraj.a).  
}

\longversion{
Besides proving (\theodiftraj.a), we have also shown that, if $\kappa$ is
the constant given by (\choiceb), then the conclusion of (\theodiftraj.a)
holds for every $\delta$ such that $0<\delta\le\kappa\ve$.  (Notice that if
$\delta\le\kappa\ve$ then (\choicea) holds.)
}

\longversion{
Now suppose we are given a map $P\ni p\to\bar x_p\in\R^n$ which is
defined for $||p||\le\bar\delta$ for some $\bar\delta>0$.  Assume that this
map is Fr\'echet differentiable at $0$, with differential $L$, and is such
that $\bar x_0=\xi(a)$.  Write $z_p=Y_{A,dV_p,L.p}$, where $V_p=D^w{\cal
W}(0).p$.
}

\longversion{
It then follows that there exist a constant $C'\ge 1$ and a 
$\bar\delta'\in\,]\,0,\bar\delta]$ such that
(i)~$C'\bar\delta'\le\kappa\bar\ve$, 
and (ii)~$||\bar x_p-\xi(a)||\le C'||p||$ for  $||p||\le\bar\delta'$. 
}

\longversion{
Pick in an arbitrary fashion $\xi_p\in\mtraj(f_p,\bar x_p,a)$ for
$||p||\le\bar\delta'$. Then, if $||p||\le\bar\delta'$, it follows that
(i)~$\xi_p$ is defined, (ii)~$\xi_p\in\mtraj(f_p,\bar x,a)$ for an $\bar
x$---namely, $\bar x_p$---such that $||\bar x-\xi(a)||\le C'\bar\delta' 
\le\kappa\bar\ve$.    Since $C'\ge 1$, we also have 
$||p||\le\kappa\bar\ve$. Therefore we can conclude---using $\ve=\bar\ve$ and
$\delta=\kappa\bar\ve$---that $[a,b]\subseteq{\rm Dom}(\xi_p)$ 
and $(\xi_p(t),t)\in{\cal T}(\xi,\bar\ve)$ for $t\in[a,b]$. 
}

\longversion{
Moreover, using (\fpxt) with $p=p_0$ it is easy to see that
$\xi_0(t)=\xi(t)$ for every $t\in [a,b]$,
since both trajectories satisfy the same initial condition.
}

\longversion{
Let $x_p$ be the restriction of $\xi_p$ to $[a,b]$.  For
$||p||\le\bar\delta'$, we have
$$
x_p(t)-\xi(t)-z_p(t)=\bar
x_p-\xi(a)-L.p+\int_a^tA(s).(x_p(s)-\xi(s)-z_p(s))\,ds 
+W_p(t)-V_p(t)+R_p(t)\,,\numeqap
$$ 
where $R_p(t)=\int_a^t \tilde r_p(s)\,ds$ and
$\tilde r_p(s)=f_p(x_p(s),s)-f_p(\xi(s),s)-A(s).(x_p(s)-\xi(s))$.  
If $p\neq 0$, let 
$$
Y_p=||p||^{-1}(x_p-x_0-z_p)\,,\qquad U_p=||p||^{-1}(W_p-V_p+R_p)\,,
\qquad \bar Y_p=||p||^{-1}(\bar x_p-x_0(a)-L.p)\,.\numeqap
$$
Then $Y_p=Y_{A,dU_p,\bar Y_p}$.
}

\longversion{
We have to prove that $Y_p$ weak$^*$ converges to $0$ as $p\to 0$.  Since
$\bar Y_p\to 0$, using Theorem~{\theolinbv}
our conclusion will follow if we show that $U_p$ weak$^*$
converges to $0$.  Since $V_p=D^w{\cal W}(0).p$, it is clear that
$||p||^{-1}(W_p-V_p)$ weak$^*$-converges to $0$.  So all we need is to
prove that $||p||^{-1}R_p$ weak$^*$-converges to $0$.  If $||p||$ is
sufficiently small, then we can apply the previous estimates using
$\delta=||p||$, $\ve={\delta\over\kappa}$.  We conclude that $||r_p(s)||\le
2\ve\psi_{\ve}(s)$ for all $s\in[a,b]$. This implies that
$$
||\tilde r_p||_{L^1}\le{2||p||\over \kappa}||\psi_{\kappa^{-1}||p||}||_{L^1}\,.\numeqap
$$
This shows that $||R_p||_{TV}=||r_p||_{L^1}=o(||p||)$ as $p\to 0$.
So $||p||^{-1}R_p$ converges to $0$ strongly---and 
{\it a fortiori} weakly$^*$---in $BV([a,b],\R^n)$. This completes 
our proof. \proofend
}

\shortversion{\noindent} If $f\in TVVF(\R^n)$ and  $\xi:[a,b]\to\R^n$ is a
trajectory of $f$,  
we say that {\sl $f$ is differentiable along $\xi$} if the map 
$x\to f(x,t)$ is differentiable at $\xi(t)$ for almost every 
$t$.  In that case, we write 
$D_{\xi}f(t)\equaldef{\partial f\over\partial x}(\xi(t),t)$, and refer 
to the map $[a,b]\!\ni\! t\!\to\!D_\xi f(t)\!\in\!\R^{n\times n}$ as the {\sl differential
of $f$ along $\xi$.}

\xdef\lemdifisadj{{\rm \appenlet}.\the\theono}
\lemmap
If ${\bf f}$ is weakly differentiable at $p_0$ along a trajectory 
$\xi:[a,b]\to\R^n$ of $f_{p_0}$, then $f_{p_0}$ is differentiable along 
$\xi$ and the map $A$ of (WD3) is such that $A(t)=D_\xi f_{p_0}(t)$ 
for almost all $t\in[a,b]$. 

\beginproof
We begin by showing that the functions $\psi_\ve$ for which (WD3) holds can
be chosen so that $\psi_\ve(t)$ is a monotonically nonincreasing function
of $\ve$ for each $t$.  To see this, start with some family
$\{\psi_\ve\}_{\ve\in]0,\bar\ve]}$ of functions for which (WD3) holds, and
define $\hat\psi_{\ve}=\psi_{2^{-k}\bar\ve}$ whenever
$2^{-k-1}\bar\ve<\ve\le 2^{-k}\bar\ve$ and $k$ is a nonnegative integer.
Then (WD3) also holds if the $\hat\psi_\ve$ are substituted for the
$\psi_\ve$.  Next define
$\bar\psi_\ve(t)=\min\{\hat\psi_{\ve'}(t):\ve\le\ve'\le\bar\ve\}$.  Then
the $\bar\psi_\ve$ are well defined integrable functions of $t$, because
for each $\ve$ the set of functions
$\{\hat\psi_{\ve'}:\ve\le\ve'\le\bar\ve\}$ is finite.  It is clear that
(WD3) also holds with the $\bar\psi_\ve$ instead of the $\hat\psi_\ve$, and
in addition $\hat\psi_\ve(t)$ is nonincreasing as a function of $\ve$.

Now, if we pick the $\psi_\ve$ to be monotonically nonincreasing with
respect to $\ve$, it follows that the limit $\rho(t)=\lim_{\ve\downarrow
0}\psi_\ve(t)$ exists for every $t$.  By the dominated convergence theorem,
$\int_a^b\rho(t)dt=0$.  Since $\rho(t)\ge 0$ for all $t$, it follows that
$\rho(t)=0$ almost everywhere.

Let $t$ be such that $\rho(t)=0$.  Let $\theta(x)=\psi_{||x-\xi(t)||}(t)$.
Then $\lim_{x\to\xi(t)}\theta(x)=0$.  On the other hand, (WD3) implies that
\xdef\saysprec{\equanoapno}
$$
||f_{p_0}(x,t)-f_{p_0}(\xi(t),t)-A(t).(x-\xi(t))||\le\theta(x)
||x-\xi(t)||\,.\numeqap
$$
Since $\theta(x)\to 0$ as $x\to\xi(t)$,  (\saysprec) says precisely
that $A(t)$ is the differential at $\xi(t)$ of the map 
$x\to f_{p_0}(x,t)$, completing our proof. \proofend
 
If $f\in TVVF(\R^n)$ is differentiable along a trajectory 
$\xi:[a,b]\to\R^n$ of $f$, and the matrix-valued function $D_\xi f$ 
is integrable, then the ordinary differential equation 
\xdef\adjequa{\equanoapno}
$$
\dot\lambda(t)=-\lambda(t).D_{\xi}f(t)\,,\numeqap
$$
for an $\R_n$-valued function $\lambda$ on $[a,b]$, is the {\sl adjoint
variational equation}---or, simply, the {\sl adjoint equation}---for $f$
along $\xi$.  A solution of the adjoint equation (i.e. an absolutely
continuous function $\lambda:[a,b]\to\R_n$ such that (\adjequa) holds
for almost all $t$) is known as an {\sl adjoint vector for $f$ along
$\xi$.}

\xdef\theodifwithlambda{{\rm \appenlet}.\the\theono}
\theoap  
Let $P$, ${\bf f}=\{f_p\}_{p\in P}$, $p_0\in P$, $\xi:[a,b]\to\R^n$ be as
in Theorem \theodiftraj. Let $P\ni p\to\bar x_p\i\R^n$ be a map which
is Fr\'echet differentiable at $p_0$ and such that $\bar x_{p_0}=\xi(a)$,
and let $L_a$ be its differential at $p_0$.  Let $\xi_p$ be, for $p$ close
enough to $p_0$, maximal trajectories of $f_p$ such that
$\xi_p(a)=\bar{x}_p$, and let $x_p$ be the restriction of $\xi_p$ to
$[a,b]$. Then the map $p\to x_p(b)$ is differentiable at $p_0$. Let
$L_b$ be its differential at $p_0$.  Let $\lambda:[a,b]\to\R^n$ be an
adjoint vector for $f_{p_0}$ along $\xi$.  Then
$$
\lambda(b)\cdot L_b(v)-\lambda(a)\cdot L_a(v)
=\lim_{\ve\downarrow 0}\,\ve^{-1}\!\int_a^b\Bigl\langle\lambda(t),f_{p_0+\ve v}
(\xi(t),t)-f_{p_0}(\xi(t),t)\Bigr\rangle dt\,\numeqap
$$
for every $v\in\R^n$.

\beginproof
Assume, without loss of generality, that $p_0=0$. According to Theorem 
\theodiftraj{} and Lemma \lemdifisadj, the map $p\to x_p$ is weak$^*$
Fr\'echet differentiable at $0$---as a map from $P$ to
$BV([a,b],\R^n)$---and its differential at $0$ is the map that sends $v\in
P$ to $Y_{A,d(U.v),L_a(v)}$, where $U=D^w{\cal W}(0)$, and ${\cal W}$ is
the map $p\to W_p$, where
$W_p(t)=\int_a^t\Bigl(f_{p}(\xi(s),s)-f_{0}(\xi(s),s)\Bigr)dt$, and
$A=D_{\xi}f$.  Then $p\to x_p(b)$ is Fr\'echet differentiable at $0$,
and the differential $L_b$ is given by $L_b(v)=Y_{A,d(U.v),L_a(v)}(b)$.
Clearly, $L_a(v)=Y_{A,d(U.v),L_a(v)}(a)$.  Since $\lambda$ satisfies
$\dot\lambda=-\lambda.A$, Lemma \lemadjo{} tells us that
$$
\lambda(b)\cdot L_b(v)-\lambda(a)\cdot L_a(v)=\int_a^b\lambda.d(U.v)\,.
$$
Since $U.v$ is the weak$^*$-limit of $\ve^{-1}W_{\ve v}$ as $\ve\downarrow
0$, our conclusion follows. \proofend

We now turn to the differentiation of a Lagrangian cost functional.  

If $f\in TVVF(\R^n)$, a {\sl Lagrangian} for $f$ in $\R^n$ is a real-valued 
function $L$  such that

\medbreak

{\pari

\item{(L.1)}  ${\rm Dom}(f)\subseteq{\rm Dom}(L)$;

\item{(L.2)} the function $t\to L(\xi(t),t)$ 
is locally integrable for every $\xi\in\traj(f)$.

}

\medbreak\noindent
If $f\in TVVF(\R^n)$ has components $f^1,\ldots,f^n$, and $L$ is a
Lagrangian for $f$, then we can consider the time-varying vector field
$[L;f]$ in $\R^{n+1}$ whose components are $(f^0,f^1,\ldots,f^n)$, where
$f^0\equiv L$.  A trajectory of $[L;f]$ can then be regarded a pair
$(\xi^0,\xi)$, where $\xi\in\traj(f)$ and $\xi^0$ is a real-valued locally
absolutely continuous function on ${\rm Dom}(\xi)$ such that
$\dot\xi^0(t)=L(\xi(t),t)$ for a.e. $t$.  Any such function $\xi^0$ will be
called a {\sl running $L$-cost along $\xi$}.  Our definition of a
Lagrangian implies that a running $L$-cost along $\xi$ exists for all
$\xi\in\traj(f)$, and it is clear that any two running $L$-costs along
$\xi$ must differ by a constant. If $\xi$ is such that ${\rm Dom}(\xi)$ is
of the form $[a,b]$ or $[a,b[$ or $[a,\infty[$, then there is a canonical
choice of running $L$-cost along $\xi$, obtained by letting
$\xi^0(t)=\int_a^tL(\xi(s),s)\,ds$.  In that case, we will write $\xi^L$ to
denote the curve $t\to(\xi^0(t),\xi(t))$, where $\xi^0$ is the canonical
running $L$-cost.

Now suppose we are given data
$P$, ${\bf f}$, $p_0$, $\xi$  as in (D.1,2,3,4), as well as  

\medbreak

{\pari

\item{(D.5)} a family ${\bf L}=\{L_p\}_{p\in P}$ such that each $L_p$
is a Lagrangian for $f_p$.  

}

\medbreak\noindent
Given a trajectory $\xi:[a,b]\to\R^n$ of
$f_{p_0}$, we say that the pair $({\bf f},{\bf L})$ is {\sl weakly
differentiable at $p_0$ along $\xi$} if the family $\{[L_p;f_p]\}_{p\in P}$
is weakly differentiable along $\xi^{L_{p_0}}$.

It is easy to see that $({\bf f},{\bf L})$ is weakly differentiable at
$p_0$ along $\xi$ iff there exists $\bar\ve>0$ such that  (WD1,2,3,4)
hold, and in addition

\medbreak
{\pariim

\item{(WD5)}  {\sl There exists $h\in L^1([a,b],\R_n)$ such that
\xdef\lpxt{\equanoapno}
$$
||L_p(x,t)-L_p(\xi(t),t)-h(t).(x-\xi(t))||\le\psi_\ve(t)(||x-\xi(t)||
+||p-p_0||)\numeqap
$$
whenever $a\le t\le b$, $||x-\xi(t)||\le\ve$ and
$||p-p_0||\le\ve$.}  

\item{(WD6)}  {\sl If we let $w_p^0(t)=L_p(\xi(t),t)-L_{p_0}(\xi(t),t)$,
$W_p^0(t)=\int_a^tw_p^0(s)\,ds$, then the map
${\cal W}^0$ that sends $p$ to ${\cal W}^0(p)=W_p^0\in BV([a,b],\R)$ is
weak$^*$ Fr\'echet differentiable at $p_0$.} 

}

\medbreak\noindent
If $f\in TVVF(\R^n)$, $\xi:[a,b]\to\R^n$ is a trajectory of $f$, and
$L$ is a Lagrangian for $f$, we say that {\sl $L$ is differentiable along
$\xi$} if the map $x\to L(x,t)$ is differentiable at $\xi(t)$ for almost
every $t$.  In that case, we write $\nabla_{\xi}L(t)\equaldef
\nabla_xL(\xi(t),t)$, and refer to the map $[a,b]\ni t\to \nabla_\xi
L(t)\in\R^{n}$ as the {\sl differential of $L$ along $\xi$.}
 
Lemma \lemdifisadj{}, applied to the curve $\xi^{L_{p_0}}$,  yields

\xdef\lemdifisadjwithl{{\rm \appenlet}.\the\theono}
\lemmap
If $({\bf f},{\bf L})$ is weakly differentiable at $p_0$ along a trajectory 
$\xi:[a,b]\to\R^n$ of $f_{p_0}$, then $f_{p_0}$ and $L_{p_0}$ are
differentiable along $\xi$ and the maps $A$, $h$ of (WD3,5) are such that 
$A(t)=D_\xi f_{p_0}(t)$ and $h(t)=\nabla_\xi L_{p_0}(\xi(t),t)$ for almost all 
$t\in[a,b]$.\proofend

Under the hypothesis of Lemma \lemdifisadjwithl{}, we can apply Theorem
\theodifwithlambda{} to the family $\{[L_p;f_p]\}_{p\in P}$ and the curve
$\xi^{L_{p_0}}$.   Define
$$
J_p(\zeta)=\int_a^b L_p(\zeta(t),t)\,dt\,,\numeqap
$$
if $\zeta:[a,b]\to\R^n$ is a curve such that the integral exists.  (We will
only be using this when $\zeta\in\traj(f_p)$, in which case $J_p(\zeta)$ is
well defined.)  An adjoint vector for $[L_{p_0};f_{p_0}]$ along
$\xi^{L_{p_0}}$ is then a pair $(\lambda^0,\lambda)$ such that
$\lambda^0\in\R$, $\lambda:[a,b]\to\R_n$ is absolutely continuous, and
\xdef\adjeqwithlzero{\equanoapno}
$$
\dot\lambda(t)=-\lambda(t).A(t)-\lambda^0h(t)\qquad{\rm for\qquad a.e.\qquad}
t\in[a,b]\,.\numeqap
$$
Theorem \theodifwithlambda{} then implies:
\xdef\theodifwithlambdalag{{\rm \appenlet}.\the\theono}
\theoap
Assume that we are given data 
$P$, ${\bf f}$, $p_0$, $\xi$, ${\bf L}$  as in (D.1,2,3,4,5).
Assume that $({\bf f},{\bf L})$ is weakly differentiable at
$p_0$ along $\xi$.  Let $P\ni p\to\bar x_p\in\R^n$ be a map which is
Fr\'echet differentiable at $p_0$ and such that $\bar x_{p_0}=\xi(a)$, and
let $L_a$ be its differential at $p_0$.  Let $\xi_p$ be, for $p$ close
enough to $p_0$, maximal trajectories of $f_p$ such that
$\xi_p(a)=\bar{x}_p$, and let $x_p$ be the restriction of $\xi_p$ to
$[a,b]$. Then the maps $p\to x_p(b)$ and $p\to J_p(x_p)$ are
differentiable at $p_0$. Let $L_b$, $L_b^0$ be their differentials at
$p_0$.  Then, if $(\lambda^0,\lambda)$ is an adjoint vector for
$[L_{p_0},f_{p_0}]$ along $\xi$, it follows that 
$$
\lambda(b)\cdot L_b(v)+\lambda^0 L_b^0(v)
-\lambda(a)\cdot L_a(v)=\lim_{\ve\downarrow 0}\ve^{-1}\!\int_a^b\!\Big(
H_{p_0+\ve v}(\lambda(t),\lambda_0,\xi(t),t)-
H_{p_0}(\lambda(t),\lambda_0,\xi(t),t)\Big)\,dt\,,\numeqap
$$
where 
$$
H_p(z,z^0,x,t)=\langle z,f_{p}(x,t)\rangle+z^0 L_p(x,t)\numeqap
$$
for $x\in\R^n$, $z\in\R_n$, $z^0\in\R$, $t\in[a,b]$. \proofend

\appen{B}{A lemma}

\xdef\lemincreasing{{\rm \appenlet}.\the\theono}
\lemmap
Let $f$ be a real-valued function on a compact interval $[a,b]$. Assume that
there exists a finite or countable subset $E$ of $[a,b]$ with the following 
properties:

{\parix

{\sl

\item{(\lemincreasing.a)} $\lim\inf_{h\downarrow 0}{f(x+h)-f(x)\over h}\ge 0$
for all $x\in[a,b[\setminus E$, 

\item{(\lemincreasing.b)} $\lim\inf_{h\downarrow 0}f(x+h)\ge f(x)$
for all $x\in[a,b[$,

\item{(\lemincreasing.c)} $\lim\inf_{h\downarrow 0}f(x-h)\le f(x)$
for all $x\in\,]\,a,b]$.

}

}

\medbreak\noindent{\sl Then $f(b)\ge f(a)$.}

\beginproof  Let $\ve>0$ be arbitrary. We will prove that $f(b)\ge f(a)-\ve$. 
Let $\ve'={\ve\over 1+b-a}$ and $E=\{x_k:k\in I\!\!N\}$.

\longversion{
For each $x\in[a,b]$, let $I(x)=\{k:x_k<x\}$. Define $g:[a,b]\to\R$ by 
$$
g(x)=\sum_{k\in I(x)}2^{-k}\ve'\,.
$$
Then $g$ is left-continuous and monotonically nondecreasing. 
}

\shortversion{
For each $x\in[a,b]$, let $I(x)=\{k:x_k<x\}$. Define $g:[a,b]\to\R$ by 
$g(x)=\sum_{k\in I(x)}2^{-k}\ve'$. 
Then $g$ is left-continuous and monotonically nondecreasing. 
}


Let $S$ be the set of all $x\in[a,b]$ such that 
$f(y)\ge f(a)-(y-a)\ve'-g(y)$ for all $y\in[a,x]$.   It is clear that 
$a\in S$, and also that $S$ is an interval. So $S$ is of the form 
$[a,c[$ or $[a,c]$, with $c\in[a,b]$.   

Suppose $S=[a,c\,[\,$.   Then $c>a$.  By  (\lemincreasing.c), given 
$\delta>0$ there is a sequence $\{z_j\}$  of points 
of $[a,b]$ such that $z_j<c$, $z_j\to c$, and 
$f(c)\ge f(z_j)-\delta$.  Then the  $z_j$ belong to $S$, so the inequalities
$f(z_j)\ge f(a)-(z_j-a)\ve'-g(z_j)\ge f(a)-(c-a)\ve'-g(c)$ hold. 
It then follows that
$f(c)\ge f(a)-(c-a)\ve'-g(c)-\delta$. Since $\delta$ is arbitrary,  we have
$f(c)\ge f(a)-(c-a)\ve'-g(c)$, so $c\in S$, which is a contradiction.

Therefore $S=[a,c]$ for some $c\in[a,b]$.   We now show that $c=b$.
Assume that $c<b$, and consider separately the cases  $c\notin E$,  $c\in E$.

If $c\notin E$, then (\lemincreasing.a) implies that there exists a $c'>c$
such that \hbox{$\,\,f(x)-f(c)\ge -(x-c)\ve'\,\,$} for every $x\in[c,c']$.
But then, 
since $\,\,f(c)\ge f(a)-(c-a)\ve'-g(c)\,\,$, we can conclude that 
\hbox{$f(x)\ge f(a)-(x-a)\ve'-g(c)\ge f(a)-(x-a)\ve'-g(x)$ for
$x\in[c,c']$.}  Therefore 
$c'\in S$, contradicting the fact that $S=[a,c]$.
   
Now suppose $c\in E$. Then $c=x_k$ for some $k$, and $x_k\neq b$. By 
(\lemincreasing.b),  there exists $h>0$ such that $c+h\le b$ and 
$f(y)-f(c)\ge-2^{-k}\ve'$ for all $y\in[c,c+h]$.  If $y\in\,]\,c,c+h]$, then 
$k\notin I(c)$, and $I(c)\cup\{k\}\subseteq I(y)$. Therefore  
$g(y)\ge g(c)+2^{-k}\ve'$.   Since $f(c)\ge f(a)-(c-a)\ve'-g(c)$,
we have $f(y)\ge f(c)-2^{-k}\ve'\ge f(a)-(c-a)\ve'-2^{-k}\ve'-g(c)\ge
f(a)-(y-a)\ve'-g(y)$. So $c+h\in S$, which is a contradiction. 

\longversion{
We have therefore proved that $c=b$. Then 
$$
f(b)\ge f(a)-(b-a)\ve'-g(b)\ge f(a)-(b-a)\ve'-\ve'=f(a)-\ve\,.
$$
Since $\ve$ is arbitrary, we have shown that $f(b)\ge f(a)$, as 
desired.   \proofend  
}

\shortversion{
So we have shown that $c=b$. Then 
$f(b)\ge f(a)-(b-a)\ve'-g(b)\ge f(a)-(b-a)\ve'-\ve'=f(a)-\ve$. 
Since $\ve$ is arbitrary, we have established that $f(b)\ge f(a)$, as 
desired.   \proofend  
}


%\remap
%If $f$ is lower semicontinuous then (\lemincreasing.b) automatically 
%holds. \proofend   

\longversion{
\bigbreak
\appen{C}{A simpler version of Brunovsk\'y's stratification condition} 
}

\longversion{
\noindent  We prove that {\sl (Br.3.A) holds if and only if ${\cal P}$ is a 
stratification of $S$.} It is clear that if ${\cal P}$ is a stratification
of $S$ then (Br.3.A) holds.  Let us prove the converse. Suppose that
(Br.3.A) holds. Then ${\cal P}$ is a locally finite partition of $S$ into
nonempty connected embedded ${\cal C}^1$ submanifolds of $S$.  Let us prove by
contradiction that the Frontier Axiom (FA) holds.  Suppose the axiom was
violated for some pair $(P_1,P_2)$ of members of ${\cal P}$.  Of all these
``bad'' pairs, choose one for which $P_1$ has the largest possible
dimension.  Then $P_1,P_2$ are in ${\cal P}$, $P_1\neq P_2$ (so $P_1\cap
P_2=\emptyset$), and $P_1\cap clos(P_2)\neq\emptyset$. Let $\tilde
P_1=P_1\cap clos(P_2)$, so $\tilde P_1$ is a nonempty relatively closed
subset of $P_1$. We will show that $\tilde P_1=P_1$, contradicting the fact
that $(P_1,P_2)$ is a bad pair.  To begin with, we cannot have $P_2=\{0\}$,
for if $P_2=\{0\}$ then $P_2$ would be closed, so $P_1\cap P_2=P_1\cap
clos(P_2)=\tilde P_1\neq\emptyset$, which is a contradiction.  Also, ${\rm
dim}(P_1)<n$, for if ${\rm dim}(P_1)=n$ then $P_1$ would be open, so the
fact that $P_1\cap clos(P_2)\neq\emptyset$ would imply that $P_1\cap
P_2\neq\emptyset$, which is a contradiction.  Finally, we may assume that
$P_1\neq\{0\}$, because if $P_1=\{0\}$ then $\tilde P_1=\{0\}=P_1$, because
$\tilde P_1\subseteq P_1$ and $\tilde P_1\neq\emptyset$. So from on we take
it for granted that $P_1\neq\{0\}$, $P_2\neq\{0\}$, and ${\rm dim}(P_1)<n$,
so in particular $P_1\in{\cal P}'$.  Let us first assume that ${\rm
dim}(P_2)<n$. Then $P_2\in{\cal P}'$.  So both $P_2$ and $P_1$ belong to
${\cal P}'$, and in addition $P_1\cap clos(P_2)\neq\emptyset$. Since ${\cal
P}'$ is a stratification of $S'$ by (Br.3.A), it follows that $P_1\subseteq
clos(P_2)$, contradicting the fact that the pair $(P_1,P_2)$ was bad.  This
excludes the possibility that ${\rm dim}(P_2)<n$, and we are left with the
case when ${\rm dim}(P_2)=n$, so $P_2$ is open in $S$.  Let $\hat
P_1\equaldef P_1\backslash\tilde P_1$.  The fact that $(P_1,P_2)$ is a bad
pair implies that $\hat P_1\neq\emptyset$. Since $\tilde P_1\neq\emptyset$,
$\tilde P_1$ is relatively closed in $P_1$, and $P_1$ is connected, it
follows that $\hat P_1$ is not relatively closed in $P_1$. So $\tilde
P_1\cap clos_{P_1}(\hat P_1)\neq\emptyset$.  Pick a point $\xb\in\tilde
P_1\cap clos_{P_1}(\hat P_1)$. Let $m={\rm dim}(P_1)$. We already know that
$m<n$. Choose a coordinate chart ${\bf x}=(x^1,\ldots,x^n):N\to\R^n$ that
maps an open neighborhood $N$ of $\xb$ diffeomorphically onto the open cube
$\,]\!-\!1,1\,[\,^n$ in $\R^n$, in such a way that ${\bf x}(\xb)=0$ and
${\bf x}(P_1\cap N)=\,]\!-\!1,1\,[\,^m\times\{0\}^{n-m}$. For $\delta>0$,
let $N(\delta)={\bf x}^{-1}\Big(\,]\!-\!\delta,\delta\,[\,^n\Big)$.  Since
${\cal P}$ is locally finite, if we let $Q(\delta)$ be the set of all
$P\in{\cal P}$ that intersect $N(\delta)$, then there must exist a
$\bar\delta$ such that the set $Q=Q(\delta)$ is finite and independent of
$\delta$ for $\delta\in\,]\,0,\bar\delta]$.  Clearly, $P_1\in Q$ and
$P_2\in Q$. Let $Z(\delta)=clos_{N(\delta)}(P_2\cap N(\delta))$.  Then
$(P_2\cup\tilde P_1)\cap N(\delta)\subseteq Z(\delta)$.  Let us exclude the
possibility that $(P_2\cup\tilde P_1)\cap N(\delta)=Z(\delta)$.  If
$(P_2\cup\tilde P_1)\cap N(\delta)=Z(\delta)$, then
$P_2\cap(N(\delta)\backslash P_1)$ is relatively closed in
$N(\delta)\backslash P_1$. Since $P_2\cap(N(\delta)\backslash P_1)$ is also
relatively open in $N(\delta)\backslash P_1$, and nonempty, it follows that
$P_2\cap(N(\delta)\backslash P_1)$ must be a union of connected components
of $N(\delta)\backslash P_1$. If $m<n-1$, then $N(\delta)\backslash P_1$ is
connected, so $P_2\cap(N(\delta)\backslash P_1)=N(\delta)\backslash P_1$.
If $m=n-1$, then $N(\delta)\backslash P_1$ has two connected components,
and $P_1\cap N(\delta)$ is entirely contained in the closure of each one of
them. So in both cases it follows that $P_1\cap N(\delta)\subseteq
clos(P_2)$. But then $P_1\cap N(\delta)\subseteq \tilde P_1$, showing that
$\tilde P_1$ is a relative neighborhood of $\xb$ in $P_1$, which
contradicts our choice of $\xb$ as a limit point in $P_1$ of
$P_1\backslash\tilde P_1$.  We now know that $(P_2\cup\tilde P_1)\cap
N(\delta)$ is a proper subset of $Z(\delta)$, for each
$\delta\in\,]\,0,\bar\delta]$.  Fix a sequence $\{\delta_j\}$ such that
$0<\delta_j\le\bar\delta$ for all $j$ and $\delta_j\downarrow 0$. Pick
$y_j\in Z(\delta_j)\backslash\Big((P_2\cup\tilde P_1)\cap
N(\delta_j)\Big)$.  Then every $y_j$ belongs to a $P^j\in Q$, but
$y_j\notin P_1\cup P_2$.  (It is obvious that $y_j\notin P_2$ and
$y_j\notin \tilde P_1$, since $y_j\in N(\delta_j)$.  On the other hand,
$y_j\in clos(P_2)$, so $y_j$ cannot be in $P_1$, because if it was in $P_1$
it would belong to $\tilde P_1$.)  So $P_1\neq P^j\neq P_2$ for every
$j$. Since $Q$ is finite, we may assume, after passing to a subsequence,
that all the $P_j$'s are equal to one and the same $P\in Q$. By
construction, $P\cap clos(P_2)\neq\emptyset$, because all the $y_j$ are in
$clos(P_2)$. Since $P\neq P_2$, $P$ cannot be open, because $P\in{\cal P}$,
so $P\cap P_2=\emptyset$, thanks to the fact that ${\cal P}$ is a
partition, so if $P$ was open then $P\cap clos(P_2)$ would be empty.  So
${\rm dim}(P)<n$. Moreover, $P\neq\{0\}$ because $y_j\neq\xb$ but
$y_j\to\xb$ and $y_j\in P$. So $P\in{\cal P}'$.  Since $\xb\in clos(P)\cap
P_1$, and both $P$ and $P_1$ are in ${\cal P}'$, it follows that
$P_1\subseteq clos(P)$ and ${\rm dim}(P_1)<{\rm dim}(P)$.  Since we have
chosen $(P_1,P_2)$ to be a bad pair for which ${\rm dim}(P_1)$ is
maximized, we can infer that $(P,P_2)$ is not a bad pair. Since $P$ and
$P_2$ are in ${\cal P}$, and $P\cap clos(P_2)\neq\emptyset$, the inclusion
$P\subseteq clos(P_2)$ must hold. But then $clos(P)\subseteq clos(P_2)$ as
well, and this implies that $P_1\subseteq clos(P_2)$, since we know that
$P_1\subseteq clos(P)$. So $(P_1,P_2)$ is not a bad pair after all, and our
proof is complete. \proofend}







\bigbreak
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\sussh

\sussi

\sussl

\sussn

\longversion{\sussenva}

\sussenvb

\sussm

\zelikin

\bye

