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{\nopagenumbers
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~
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\centerline{\bf Nonperturbative linearization of dynamical systems}
\footnote{}{{\tt \day }}
\vskip 3truecm
\centerline{G. Gaeta}
\centerline{\it Department of Mathematics,}
\centerline{\it Loughborough University,}
\centerline{\it Loughborough LE11 3TU (G.B.)}
\centerline{\tt G.Gaeta@lut.ac.uk}
\bigskip
\centerline{G. Marmo}
\centerline{\it Dipartimento di Scienze Fisiche,}
\centerline{\it Universit\`a di Napoli,}
\centerline{and {\it INFN, Sezione di Napoli,}}
\centerline{\it 80123 Napoli (Italy)}
\vskip 3 truecm
{\bf Summary.}
{We consider the classical problem of linearizing a vector field $X$
around a fixed point. We adopt a nonperturbative point of view, based
on the symmetry properties of linear vector fields.}
\vfill\eject}
\pageno=1
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\parskip=10pt
{\bf 1. Introduction, and statement of the problem}
The problem of linearizing a nonlinear vector field in
the neighbourhood of a fixed point by means of ${\cal C}^\infty$
transformations is a classical one, its perturbative treatment going
back to Poincar\'e \ref{1-4} in the general case and to Birkhoff for
Hamiltonian systems \ref{5,6}; here we want to consider it from a
non-perturbative point of view; moreover, we will not deal with
general Normal Form transformations \ref{1-6}, but only consider
linearizable systems.
Indeed, although it turns out that the Poincar\'e procedure for
linearizing a linearizable systems is also successfull in reducing a
generic (nonlinearizable) system to its normal form, so that the
problem of formal reduction to normal form is not more difficult in
the general case than in the linearizable one, it is natural to
expect that if we proceed nonperturbatively, the linearizable case
will be much easier to treat. The considerations we will use
in the following are specific to the linearizable case, and cannot be
extended to the general one.
Let us consider a {\it linear} dynamical system in $R^n$ for which
the origin is a fixed point,
$$ \.x^i = A^i_{~j} x^j \eqno(1) $$
(where $i,j=1,...,n$ and $A$ is a $n \times n$ real matrix), and
consider now an invertible (nonlinear)
diffeomorphism\footnote{$^1$}{It will be clear from the following
discussion that we could as well consider a domain $D$ -- containing
the origin -- in $R^n$ rather than the whole $R^n$; similarly we could
as well consider $\Phi$ invertible only locally.} which identifies a
change of coordinates $$ x^i = \Phi^i (y) ~; \eqno(2) $$
we will also denote by $y^i = \Psi^i (x)$ the inverse change of
coordinates. Let us denote by $\La$ the jacobian of this change of
coordinates, and by $\Ga$ its inverse,
$$ \La^i_{~j} = {\pa \Phi^i \over \pa x^j} \equiv {\pa x^i \over \pa
y^j} ~~,~~ \Ga^i_{~j} = {\pa y^i \over \pa x^j} \equiv {\pa \Psi^i
\over \pa x^j} ~~,~~ \La^i_{~j} \Ga^j_{~k} = \delta^i_{~k}
~. \eqno(3) $$
In the new coordinates, (1) is written as
$$ \.y^i = f^i (y) \eqno(4) $$
where the $f$ are now {\it nonlinear} functions given explicitely by
$$ f^i (y) = \Ga^i_{~k} (y) A^k_{~j} \Phi^j (y) ~. \eqno(5) $$
Suppose now that we have to study (4), with $f$ given explicitely, so
that we do not know about $A$ and $\Phi$. How can we find out that (4)
corresponds actually to linear dynamics ``in the wrong coordinates'' ?
The purpose of the present note is indeed to answer this question; it
will turn out that in the process of answering this we also answer the
question of how to concretely linearize the system, i.e. to determine
the linearising change of coordinates $\Phi$.
\bigskip
{\bf 2. Symmetry approach}
Clearly, a possible approach would consist in using the theory of
(Poincar\'e-Dulac) {\it normal forms}; this would amount to a
perturbative construction, order by order, of the inverse change of
coordinates, thus mapping (4) back into (1). This approach is
completely algorithmic and cosntructive, and moreover it is quite
general, in that it works both for linearisable and non-linearisable
systems. However, in the linearisable case this approach has also
several drawbacks, essentially amounting to its perturbative character:
$a)$ If $A$ presents resonances, one would expect nonlinear resonant
terms \ref{1-4} to be present, so that one would realize the inherent
linearity of the system only after checking a series (usually
infinite) of ``miracolous'' cancellations occurr in the normalized
expansion;
$b)$ If $\Phi$ is not analytic -- even if $C^\infty$ -- we cannot hope
to linearize the system by the Poincar\'e procedure, which is
inherently perturbative and polynomial (one could consider
Ecalle's resurgent functions theory \ref{7,8}, but again this means
introducing very complicate tools for a simple problem);
$c)$ In any case, the procedure requires extensive computations,
checks of the convergence of perturbative expansions, and so on;
moreover, we should go to infinite order in perturbation theory to
obtain exact linearization. Even in the most favourable case, in which
one is sure a priori of the linearizability of the problem and of the
convergence of the linearizing transformation (e.g. thanks to Siegel's
theorem \ref{2,6} or to symmetry properties \ref{9-14}), to compute the
explicit linearizing change of coordinates one still has to go at
infinite order in perturbation. In one word, it requires a huge amount
of work to recognize the simple system (1).
Thus, we will look for a different, non-perturbative, approach for
this problem. The natural idea would be to look for properties of the
dynamical system, or equivalently of the vector field
$$ X = A^i_{~j} x^j {\pa ~\over \pa x^i} = f^i (y) {\pa ~\over \pa
y^i} ~, \eqno(6) $$
which are invariant under changes of coordinates -- i.e. they have a
tensorial character -- and which recognize
the linear nature of the system. From this point of view, it is quite
natural to look at the {\it symmetries}\footnote{$^2$}{The symmetry
approach to differential equations -- both ODEs and PDEs -- pioneered
by S. Lie, has received recently diffused attention and is dealt
with, and applied, in several books and many papers, see e.g.
\ref{15-21}.} of (1): indeed, if a vector field $S$ commutes with $X$,
the relation $[X,S]=0$ will hold indipendently of any system of
coordinates.
Symmetries which are related to the linear nature of (1) are those
generated by powers of $A$, i.e. by vector fields of the form (in the
$x$ coordinates)
$$ X_k = [ A^k ]^i_{~j} x^j {\pa ~\over \pa x^i} \eqno(7) $$
for $k$ a non-negative integer. Clearly, as it follows
from $[A^n , A^m ]=0$, these form an abelian algebra (generically, of
dimension $n$). Notice that for $k=0$ (i.e. for $A^0 = I$) we have the
generator of scalings, $X_0 = x^i {\pa / \pa x^i}$, which will be a
symmetry for any linear system.
In the $y$ coordinates, the $X_k$ take the form
$$ X_k = \Ga^i_{~j} (y) [A^k]^j_{~m} \Phi^m (y) {\pa ~\over \pa y^i}
~. \eqno(8') $$
These satisfy therefore, in particular,
$$ X_{k+1} = (\Ga A \Ga^{-1} ) X_k ~. \eqno(8'') $$
Thus, even if we analyze the symmetry algebra of (4) and we detect in
it an abelian algebra, it can be difficult to realize the vector
fields in it are of this form, although of course the relation (8'')
is easier to recognize than the form (8).
The situation is slightly better if we consider $X_0$ alone: indeed,
in this case we have to look for symmetries of the simpler form
$$ X_0 = \Ga^i_{~j} (y) \Phi^j (y) {\pa ~\over \pa y^i} \eqno(9) $$
with $\Ga$ given by (3). Thus, a possible approach would consist in
looking for solutions to the determining equation for symmetries of
dynamical systems\footnote{$^3$}{We recall that if $\phi$ satisfies
(10), then $X_\phi = \phi^i (y,t) (\pa / \pa y^i)$ is a symmetry of
$X$ \ref{15-21}.}
$$ \phi_t + (f \cdot \nabla) \phi - (\phi \cdot \nabla) f =
0 \eqno(10) $$
in the form $ \phi (y,t) = (D \Phi^{-1} ) \Phi$. Recalling that $D
\Phi^{-1} = - \Phi^{-1} (D \Phi ) \Phi^{-1}$, this also reads $\phi
(y,t) = - \Phi^{-1} (D \Phi )$.
Notice that the fact that $x^i (\pa / \pa x^i)$ is a symmetry, without
further assumptions on the $X$ (e.g. analyticity), only ensures that
in the $x$ coordinates we have $X = \~f (x) \pa_x$ with $\~f$
homogeneous of order one; on the other side, as we deal with
nonsingular $f$, and $\Phi$ invertible, we are guaranteed that in this
setting $\~f$ is indeed linear. We have therefore the
{\bf Lemma I.} {\it If the equation (4) admits a symmetry $\phi^i
(y) {\pa / \pa y^i}$ and $\phi$ can be written in the form
$ [D \Phi^{-1} (y)]^i_{~j} \Phi^j (y)$,
then by the change of coordinates $y = \Phi^{-1} (x)$, (4) is reduced
to a system $\.x = \~f (x)$ with $f$ linear.}
We stress that the lemma does {\it not} require to determine the full
symmetry of (4), i.e. the most general solution to (10), but only a
special solution with an appropriate form. Indeed, getting the full
solution to (10) requires to find the most general solution to the
associated homogeneous PDE, namely to solve (4).
Another possibility stems from the obvious observation that (1)
admits a linear superposition principle\footnote{$^4$}{The idea of
using this fact to characterize the linearizability of a nonlinear PDE
belongs to Kumei and Bluman \ref{16,22-24}; here we are actually
specializing their theory to the case of first order ODEs.}; this
means, in particular, that
$$ X_\xi = \xi^i (t) {\pa ~\over \pa x^i} \eqno(11) $$
generates a symmetry of (1), provided $\xi$ obeys (1) itself, i.e.
provided ${\.\xi}^i (t) = A^i_{~j} \xi^j (t)$. Indeed, one can easily
check that this is the case by using equation (10) in the $x$
coordinates, which in this case reduces to $\.\xi = A \xi$.
In the $y$ coordinates we have
$$ X_\xi = \Ga^i_{~j} (y) \xi^j (t) {\pa ~\over \pa y^i} \eqno(12) $$
and therefore we have the
{\bf Lemma II.} {\it If the equation (4) admits a symmetry $\phi^i
(y) {\pa / \pa y^i}$ for $\phi$ of the form $M^i_j (y) \xi^j (t)$
with $\xi$ an arbitrary solution of the linear equation $\.\xi = A
\xi$, then by the change of coordinates $y = \Phi^{-1} (x)$, with $M
= D \Phi^{-1}$, (4) is reduced to the linear system $\.x = Ax$.}
Here again, we stress that it is not required to know the most
general solution of (10).
\bigskip
{\bf 3. Intrinsic approach}
It is possible to look for the linearization of a Dynamical System in
a slightly more general setting, making contact with the general
theory of Nijenhuis operators \ref{25-29}.
We define a {\it separating set} of functions to be a finite
collection $f_1 , f_2 , ... , f_p$ such that $f_a (x) = f_a (y)$ for
any $a=1,...,p$ implies $x=y$; this means that we must have $p \ge
n$.
{\bf Definition.} {\it A separating set of functions is said to be a
{\it linearizing set} for $X$ if it happens that
$$ L_X f_a = A_a^{~b} f_b \ . \eqno(13) $$}
(Here, $L_X$ is the Lie derivative along $X$)
Then, any vector field $X$ admitting a linearizing set $f$ is
$f$-related to a linear system on $R^p$, with a map $f: R^n \to R^p$
which is just given by
$$ f: x \ \to \ \left( f_1 (x) , ... , f_p (x) \right) \ . \eqno(14) $$
If we denote by $z_a$ the coordinates in $R^p$, the image of $R^n$
under $f$ will be given by $z_a = f_a (x)$. So, in the $z$ coordinates
our vector field $X$ is $f$-related to
$$ Y = A_a^{~b} z_b { \pa \over \pa z_a } \ . \eqno(15) $$
When $p=n$, we get a linearization of our system in the usual
sense\footnote{$^5$}{The introduction of linearizing sets in the
general case allows to deal with more general situations \ref{28,29};
e.g. if we have a linear flow in $R^n$ but we consider it on a
nonlinear embedded submanifold $M$ (the simplest case being that of
$M = S^{n-1} \ss R^n$), the flow on $M$ cannot be globally linearized
in the usual sense, but it is recognized as a linear flow by means of
this approach.}.
Therefore, given a vector field $X$ we can look for a linearizing
set for $X$ in the specific case $p=n$.
We consider now a vector field $Z$, and denote by $\zeta$ the
semiflow under $Z$,
$$ { d \over dt} \zeta (t ; y) \big\vert_{t=0} = Z (y) \ ; \eqno(16)
$$
we will also denote by $B_\delta (y_0 )$ the ball of radius $\delta$
centered in $y_0$.
{\bf Definition.} {\it The vector field $Z$ is {\it
dilation-type} if: {\it i)} it exists a unique $y_0$ such that $Z (y_0
) = 0$; {\it ii)}, there exist $n$ functionally independent real
functions $h_i : M \to R$ which are solutions of
$$ Z ( h ) = h \ . \eqno(17) $$}
Notice that the $h_i$'s provide a linearizing set for $Z$, with
matrix $A_{a}^b = \delta_a^b$. We say then that the $h_i$ are a
{\it diagonalizing} set of functions for $Z$.
Notice also that it would be natural to require that for a dilation
field there is $\delta > 0$ such that
$$ \lim_{t \to - \infty} \zeta (t ; y ) = y_0 \qquad \forall y \in
B_\delta (y_0 ) ~; \eqno(18) $$
however, this is automatically satisfied when a linearizing set
exists.
{\bf Lemma III.} {\it If $Z$ is a dilation-type vector field, with
$\{ h_1 , ... , h_n \}$ a diagonalizing set of functions for $Z$, then
any $f$ solution of $Z(f)=f$ can be uniquely written as
$$ f(y) = \sum_{i=1}^n c^i h_i (y) \ . \eqno(19) $$}
Indeed, since the $h_i$'s are functionally independent (which is a
condition stronger than the separating condition) we have for any
function $f$
$$ {\rm d} f = g^i {\rm d} h_i \qquad g^i \in {\cal F} \eqno(20) $$
By requiring $f$ to be a solution to (18) we get
$$ {\rm d} f (Z ) = f = g^i {\rm d} h_i (Z ) = g^i h_i = L_Z f
\eqno(21) $$
and therefore $( L_Z g^i ) h_i = 0 $ implies $L_Z g^i = 0$, as
the $h_i$ are functionally independent. Due to the regularity
requirement on the $g^i$ in the neighbourhood of $y_0$, we have $g^i
\in R$. Thus, we conclude that any solution to $Z(f)=f$ can be written
in the form (44), with the $c^i$ real constants; the lemma is proved.
Clearly, if $Z$ is just $X_0$, then $y_0 = \Phi (0)$ and the $ h_i$
are nothing else than the $x_i$ as functions of the $y$, i.e. $h_i =
\Phi^i (y) $.
Using lemma III, we have immediately:
{\bf Lemma IV.} {\it Let $Z$ be a dilation-type vector field, with
$\{ h_1 , ... , h_n \}$ a diagonalizing set of functions for $Z$.
If $[X,Z ]=0$, then $\{ h_1 , ... , h_n \}$ is a linearizing set for
$X$.}
Indeed, if $[X,Z ] =0$, we have
$$ L_X ( L_Z h_i ) = L_Z L_X h_i = L_X h_i ~, \eqno(22) $$
which also means
$$ L_X h_i = A_i^{~j} h_j \eqno(23) $$
because of the properties of solutions to (18). As the $h_i$'s define
a change of coordinates $x^i = h_i (y)$, the linearized vector field
will be
$$ Y = A_i^{~j} h_j {\pa \over \pa h_i} \ . \eqno(24) $$
We finally notice that generically (i.e. under suitable nondegeneracy
conditions, satisfied by generic vector fields) if $h_1$ is a solution
to $L_Z h = h$, we may get new functionally independent solutions by
applying repeatedly $L_X$ to $h_1$. This simple fact can be of help in
constructing the diagonalizing set for $Z$.
\bigskip
{\bf 4. Symmetry and recursion operators}
In this section, we would like to point out how the approach delined
in the previous section is related to recursion operators and the Lax
formalism for integrable systems. Notice, indeed, that a system which
is linearizable by a change of coordinate (C-linearizable in the
Calogero terminology) is by this also integrable.
When $X$ is a linear vector field,
$$ X = A^i_{~j} x^j {\pa \over \pa x^i} \eqno(25) $$
we can associate to the matrix $A$ a (1,1) tensor field
$$ R \equiv T_A = A_i^{~j} \ \left[ d x^i \otimes {\pa ~ \over \pa x^j
} \right] \ . \eqno(26) $$
This tensor satisfies automatically the two equations
$$ L_X (R ) = 0 \eqno(27) $$
$$ ~~~ N_R = 0 \eqno(28) $$
where $L_X$ is the Lie derivative along $X$, and $N_R$ is the Nijenhuis
tensor \ref{25-29} associated with $R$, i.e.
$$ N_R [X,Y] \ = \ \[ RX , RY \] + R^2 \[ X,Y \] - R \[ \[ RX , Y \]
+ \[ X , RY \] \) ~. \eqno(29) $$
By applying $R$ to $X$ we get the vector fields $X_k = (R )^k X$,
which have the property that
$$ [ X_k , X_m ] = 0 \eqno(30) $$
i.e. we can generate pairwise commuting symmetries.
Thus for a given $X$, the existence of a (1,1) tensor field $R$ such
that (27),(28) are satisfied is a necessary condition for $X$ to be
linearizable.
It would be possible \ref{28,29} to look for a separating set of
functions by searching for invariant subspaces of exact 1-forms under
the endomorphism associated to $R$ on 1-forms, i.e.
$$ R ( {\rm d} f_a ) = B_a^{~b} {\rm d} f_b \eqno(31) $$
with $B$ a real matrix. Clearly,
$$ (R )^k ( {\rm d} f_a ) = (B^k)_a^{~b} {\rm d}f_b \eqno(32) $$
Generalized eigenspaces of $R$ are invariant subspaces for $X$.
It should be stressed, finally, that this $R$ has all the
properties of a {\it recursion operator} (in the sense
encountered in the theory of integrable systems \ref{15,28})
for our finite dimensional evolution equation; thus, it permits to
also obtain a Lax representation, as discussed e.g. in \ref{28}.
Rather than discussing this point here, we refer to \ref{28,29} for a
general discussion, and more specifically to \ref{28-35} for the
geometry of Lax systems, to \ref{25-29} for the geometry of Nijenhuis
operators, to \ref{25-29,36-39} for how the Nijenhuis tensor describes
the geometry of the tangent bundle; and to \ref{36-40} for the
geometry of the Nijenhuis tensor in relation with a distinguished
vector field $X$ on $M$ describing dynamical evolution. Finally, for
the Hamiltonian setting see \ref{31-35,41,42}.
\bigskip \vfill\eject
{\bf 5. Examples: linearizable vector fields}
We will now consider some examples of applications of our results. We
will for each example consider, in the order, the application of th
emethods based on lemma I, lemma II and on lemma IV.
In the following we write all indices as lower ones, to avoid any
confusion between indices and exponents.
{\tt Example 1.}
As a first, although trivial, test, we consider the case $n=1$. Now
we have for (1) $\.x = a x$, and
$$ f(y) = a {\Phi \over \Phi '} \eqno(33) $$
Looking first for $\phi = \phi(y)$ as solution to (10), we get
$$ {\phi_y \over \phi} = g(y) \equiv {f_y \over f} \eqno(34) $$
and in this case we actually have
$$ g(y) = {\Phi ' \over \Phi} - {\Phi '' \over \Phi '} \eqno(35) $$
so that indeed
$$ \phi (y) = c \Phi (y) / \Phi ' (y) = \~c f(y) \eqno(36) $$
Thus, applying Lemma I is just equivalent to determining directly
if, given $f(y)$, there exists a $\Phi$ such that (33) is verified;
obviously this just yields
$$ \Phi (y) = c_1 \exp \left[ \int_{y_0}^y [a/ f(y)] dy \right]
\eqno(37) $$
To make a concrete example, in this way we immediately get that
$$ \.y = {1 + y^2 \over 1 + 3 y^2} a y \eqno(38) $$
is transformed into $\.x = a x$ with
$$ x = \Phi (y) = y + y^3 \eqno(39) $$
Let us now look for
$$ \phi = \phi (y,t) = \alpha (y) \xi (t) \eqno(40) $$
so that (10) reads now
$$ \a \.\xi + f \xi \a_y = \a \xi f_y \eqno(41) $$
which for $\a$ yields
$$ {d \a \over \a } = \left[ {f_y \over f } - {\.\xi \over \xi} {1
\over f} \right] d y \eqno(42) $$
If $\.\xi = k \xi$, we get
$$ {d \a \over \a } = \left[ \left( 1 - {k\over a} \right) {\Phi_y
\over \Phi } - {\Phi_{yy} \over \Phi} \right] \eqno(43) $$
and choosing $k=a$ we get
$$ \a = {c_2 \over \Phi_y} \eqno(44) $$
For a concrete example, one could use this approach to obtain $\Ga$
which, for $f$ as in (38), yields the same $\Phi$ as in (39).
In the approach based on lemma III (and with the notation of section
3), $Z$ is just given by $Z = g(y) d / dy$, see equation (34). With
this, (18) just yields $f(y) = c \Phi (y)$, which corresponds to Lemma
III.
\bigskip
{\tt Example 2.}
We consider now a two-dimensional system $\.y = f(y)$, with only one
nonlinear term:
$$ \eqalign{
{\.y}_1 =& y_1 - \eps y_2 - y_2^2 \cr
{\.y}_2 =& y_2 \cr} \eqno(45) $$
Let us first try to apply Lemma I. From (10), we obtain that the vector
field $X_0$ identified by
$$ \phi_1 = y_1 - y_2^2 ~~,~~ \phi_2 = y_2 \eqno(46) $$
is a symmetry of our system; this is of the form $ \phi_i = \Ga_{ij}
\Phi_j$ with
$$ \Phi = \pmatrix{y_1 + y_2^2 \cr y_2} ~~,~~ \Ga =
\pmatrix{1&2 y_2 \cr 0&1} \eqno(47) $$
which indeed satisfies $[\Ga^{-1}]_{ij} = \pa \Phi_i / \pa y_j$;
correspondingly we get the linearizing transformation $\Phi^{-1}$ as
$$ y_1 = x_1 - x_2^2 ~~,~~ y_2 = x_2 \eqno(48) $$
and in this coordinates we get, as expected,
$$ X_0 = x_i {\pa \over \pa x_i } ~. \eqno(49) $$
Let us now come to the procedure based on Lemma II, i.e. let us look
for $\phi$ in the form $\phi (y,t) = \Ga_{ij} (y) \xi_j (t)$. The
equations (10) are now\footnote{$^6$}{Notice that $A$ and $\Ga$ should
be seen as the unknowns of the problem.}, assuming that ${\.\xi}_i =
A_{ij} \xi_j$ for some $A$ and eliminating the common factor $\xi_k$,
$$ f_j {\pa \Ga_{ik} \over \pa y_j} = {\pa f_i \over \pa y_j}
\Ga_{jk} - \Ga_{ij} A_{jk} \eqno(50) $$
With the explicit expression of $f$, and therefore $\pa f / \pa y$,
given above, we have that indeed for $\Ga$ as in (47) above and
$$ A = \pmatrix{1&-\eps\cr0&1} \eqno(51) $$
the equation is satisfied. This leads to the same linearizing
transformation (48) as above, and the linear equation is indeed just
$\.x = A x$.
Let us now consider, for the same problem, the approach of section 3;
we take
$$ Z = (y_1 - y_2^2 ) {\pa \over \pa y_1} + y_2 {\pa \over \pa
y_2} \eqno(52) $$
and the diagonalizing $h_i$'s are given by
$$ h_1 = y_1 + y_2^2 \quad , \quad h_2 = y_2 \ . \eqno(53) $$
One can check that in this case, solving (18) -- e.g. by the method of
characteristics -- yields
$$ f(y) = c_1 h_1 (y) + c_2 h_2 (y) \eqno(54) $$
with arbitrary constants $c_1 , c_2$; and moreover that
$$ \cases{ X (h_1 ) = h_1 - \eps h_2 & \cr X (h_2 ) = h_2 &
\cr} \eqno(55) $$
so that $x_i = h_i (y)$ takes the system into the form ${\dot x} = Ax$
with the same $A$ as in $X(h_i ) = A_{ij} h_j$.
\bigskip
{\tt Example 3.}
We will consider again a system in $R^2$, given now by
$$ f = \pmatrix{
2 \left( y_1 + y_2 + e^{y_1} \right) \cr
(y_2 - y_1) \left( 1 + 2 e^{y_1} \right) +
e^{y_1} - 2 e^{2 y_1} \cr} \eqno(56) $$
If we look for solutions of (10) in the form $\phi = \phi (y)$, we
are confronted with a system of two PDEs, i.e.
$$ \eqalign{
2 \left[ y_1 + y_2 + e^{y_1} \right] {\pa \phi_1 \over \pa y_1} + &
\left[ (y_2 - y_1) \left( 1 + 2 e^{y_1} \right) +
e^{y_1} - 2 e^{2 y_1} \right] {\pa \phi_1 \over \pa y_2} = \cr
& = 2 \left( 1 + e^{y_1} \right) \phi_1 + 2 \phi_2 \cr
2 \left[ y_1 + y_2 + e^{y_1} \right] {\pa \phi_2 \over \pa y_1} + &
\left[ (y_2 - y_1) \left( 1 + 2 e^{y_1} \right) +
e^{y_1} - 2 e^{2 y_1} \right] {\pa \phi_2 \over \pa y_2} = \cr
& = - \left[ \left( 1 + e^{y_1} \right) + 2 (y_1 + y_2 ) e^{y_1} + 4
e^{2 y_1} \right] \phi_1 + \left( 1 - 2 e^{y_1} \right) \phi_2 \cr }
\eqno(57) $$
It is quite clear that this is not an easy equation to solve;
however, one can check that
$$ \phi = \pmatrix{
y_1 \cr
y_2 + (1 - y_1 ) e^{y_1} \cr} \eqno(58) $$
provides a special solution to (57). This is indeed of the form (9),
as required to apply Lemma I, with
$$ \Phi = \pmatrix{ y_1 + y_2 + e^{y_1} \cr y_2 + e^{y_1} \cr}
\eqno(59) $$
$$ \Ga = \pmatrix{ 1 & -1 \cr - e^{y_1} & 1 + e^{y_1}
\cr} ~=~ \left[ {\pa \Phi \over \pa y } \right]^{-1} \eqno(60) $$
In order to apply Lemma II we would instead look for solutions to
(10) in the form $\phi = \Ga_{ij} (y) \xi_i (t)$. The equations
satisfied by the $\Ga_{ij}$ are now even more complicate, but we can
take advantage of the freedom given by the $A$. Indeed, the $\Ga$
given above is also a solution to the set of equations one obtains in
this way, provided one chooses
$$ A = \pmatrix{1&2\cr-1&2} \eqno(61) $$
Indeed, by (59) we have that, with the inverse change of coordinates
$$ \eqalign{
y_1 =& x_1 - x_2 \cr
y_2 =& x_2 - e^{(x_1 - x_2 )} \cr} \eqno(62) $$
we reduce $\.y = f(y)$, with $f$ given by (56), to $\.x = Ax$ with
$A$ given by (61).
If we apply the approach of section 3, based on the existence of a
dilation-type field $Z$ commuting with $X$, the $Z$ is given by $Z =
\phi_i \pa / \pa y_i$ with $\phi$ as in (58); the $h_i$ are the
$\Phi_i$ given in (59), and equation (18) gives $f$ as in (54);
again, it is easily verified that
$$ X (h_i ) = A_{ij} h_j \eqno(63) $$
with $A$ given explicitely by (61).
\bigskip\vfill\eject
{\bf 6. Examples: nonlinearizable vector fields}
We will now give examples in which our results are used to show
that a given vector field (dynamical system) can {\it not} be
linearized; we consider systems in $R^2$ for simplicity.
\bigskip
{\tt Example 4.}
We give first an example of a system which is not linearizable
because it does not admit enough symmetries. We consider $R^2$ with
coordinates $(x,y)$ and the vector field
$$ \Ga = \phi (x^2 + y^2 ) \ \( x {\pa \over \pa y} - y {\pa \over
\pa x} \) \ . \eqno(64) $$
If this vector field is linearizable, it has to admit at least two
symmetries.
We notice that $X_0 = x \pa_y - y \pa_x$ and $\D = x \pa_x + y \pa_y
$ are a basis for the module of vector fields which have the origin
as a fixed point. Therefore our symmetries should have the form
$ Y = a X_0 + b \D$, with $a,b \in {\cal F} \( R^2 \)$ smooth
functions\footnote{$^7$}{We recall that symmetries of a vector field
with isolated fixed points should have the same points as fixed
points.}.
We then look for $a,b$ such that $[Y , \Ga ] = 0$; we have to require
that (with $L_X$ the Lie derivative along $X$)
$$ \[ a X_0 + b \D , \phi X_0 \] = \( b L_\D \phi - \phi L_{X_0} a
\) X_0 - \phi \( L_{X_0} b \) \D = 0 \ . \eqno(65) $$
Therefore -- for the vanishing of the term along $\D$ -- we need
$L_{X_0} b = 0$, which implies $ b = b(x^2 + y^2 )$.
For the other term, i.e. from $b L_\D \phi = \phi L_{X_0} a$, we
integrate both sides along a circle centered in the origin:
$$ \int_{S^1} \( b L_\D \phi \) \ d \theta \ = \ \phi \ \int_{S^1}
\( L_{X_0} a \) \ d \theta \ , \eqno(66) $$
where $\phi$ has been taken out of the integral because it depends
only on $(x^2 + y^2 )$, i.e. is a constant on $S^1$. By using this
same argument we arrive at
$$ \( b L_\D \phi \) (2 \pi ) \ = \ \phi \( a (2 \pi ) - a (0) \) \
; \eqno(67) $$
if the function $a$ is regular, $a (2 \pi ) = a (0)$ and we get $b
L_\D \phi = 0$. Thus, either $b = 0$ or $L_\D = 0$ (or both). Now,
$\D$ does not have any smooth constant of motion, and thus $L_\D
\phi = 0 $ implies that $\phi$ is a constant, or otherways it has to
be $b=0$.
It follows from this that $L_\D b \not= 0$ requires $b=0$, i.e. there
is only one family of symmetries (depending on a constant) for our
system, which therefore cannot be linearized.
\bigskip
{\tt Example 5.}
We will now consider examples in which we have the required number of
symmetries, but none of them is a dilation-type vector field.
Let us write $r^2 = x^2 + y^2$, and consider the (Van der Pol- like)
system
$$ \eqalign{
{\dot x} = & - (r^2 -1) x + (r^2 - 2) y \cr
{\dot y} = & - (r^2 -2) x - (r^2 - 1) y \ ; \cr } \eqno(68) $$
we denote the corresponding vector field as $X$.
When we look for symmetries, i.e. for vector fields
$$ Y = f(x,y) \pa_x + g (x,y) \pa_y \eqno(69) $$
such that $[X,Y]=0$, it turns out that the only solutions are of the
form (with $c_1 , c_2$ real constants)
$$ Y_1 = c_1 X \quad , \quad Y_2 = c_2 \( x \pa_y - y \pa_x \) \ .
\eqno(70) $$
It is clear that $Y_2$ is not a dilation-type vector field (it is
just a homogeneous rotation), and $Y_1$ is just proportional to $X$
(which is, by the way, not dilation-type as well). Thus, we can
conclude -- using any of the proposed approaches -- that $X$ is not
linearizable.
Indeed, as for the first proposal, $X$ does not admit symmetries
depending on an arbitrary solution of a linear equation; for the
second one, it does not admit a dilation-type symmetry.
\bigskip
{\tt Example 6.}
We consider now the following generalization (again in $R^2$) of the
situation encountered in example 5:
$$ \eqalign{
{\dot x} = & \a (r) x - \b (r) y \cr
{\dot y} = & \b (r) x + \a (r) y \cr} \eqno(71) $$
(where both $\a (r)$ and $\b(r)$ are not identically zero) so that we
deal with the vector field
$$ X = \( \a (r) x - \b (r) y \) \pa_x + \( \b (r) x + \a (r) y \)
\pa_y \eqno(72) $$
or, as it is convenient to use polar coordinates $(r,\th )$,
$$ X = \a (r) \pa_r + \b (r) \pa_\th \ . \eqno(73) $$
We write in full generality
$$ Y = f (r,\th ) \pa_r + g (r, \th ) \pa_\th \eqno(74) $$
and now the condition $[X,Y]=0$ gives two PDEs, i.e.
$$ \eqalignno{
\a (r) {\pa f \over \pa r} + \b (r) {\pa f \over \pa \th } = & f
\cdot {\a}' (r) \ , & (75) \cr
\a (r) {\pa g \over \pa r} + \b (r) {\pa g \over \pa \th } = & f
\cdot {\b}' (r) \ . & (76) \cr} $$
These can be solved using the method of characteristics; for the
first one we get
$$ {d r \over \a (r) } = { d \th \over \b (r) } = {df \over f {\a}'
(r) } \ ; \eqno(77) $$
equating the first and the third term we get
$$ f(r, \th ) = \xi (\th ) \ \a (r) \ , \eqno(78) $$
and by using the other term -- or going back to (75) -- we get $\xi '
(\th ) = 0$, i.e.
$$ f (r , \th ) = c_1 \a (r) \ . \eqno(79) $$
Let us look at (76); this yields
$$ {d r \over \a (r) } = { d \th \over \b (r) } = {dg \over c_1 \a
(r) {\b}' (r) } \ ; \eqno(80) $$
equating, here again, the first and the third term we get
$$ g (r , \th ) = c_1 \b (r) + \xi (\th ) \eqno(81) $$
and again, from the other term or going back to (76) we get $\xi '
(\th ) = 0$, i.e.
$$ g (r , \th ) = c_1 \b (r) + c_2 \ . \eqno(82) $$
Thus, we have only two symmetries,
$$ \eqalignno{
Y_1 = & c_1 X & (83) \cr Y_2 = & c_2 \pa_\th \ , & (84) \cr} $$
and we are in the same situation as in example 5, and we can derive
the same conclusions, i.e. that $X$ is not linearizable. Clearly, the
present discussion does not apply if the condition $\a (r) \not\equiv
0 \not\equiv \b ' (r)$ is not satisfied.
\vfill
{\bf Acknowledgement}
This paper originated from discussions we had in the I.A.S.S. (Vietri,
Italy) in occasion of the workshop on ``Completely Integrable Systems
and Separability'' in September 1995; we would like to thank the
I.A.S.S. and the organizers for their hospitality on that occasion.
\vfill\eject
\parskip=5pt
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\bye