\magnification=1200 \def\day{30/1/96} \parindent=0pt \parskip=10pt \def\La{\Lambda} \def\Ga{\Gamma} \def\phi{\varphi} \def\pa{\partial} \def\a{\alpha} \def\b{\beta} \def\th{\theta} \def\eps{\epsilon} \def\D{\Delta} \def\({\left(} \def\){\right)} \def\[{\left[} \def\]{\right]} \def\.#1{{\dot #1}} \def\~#1{{\widetilde #1}} \def\ref#1{[#1]} {\nopagenumbers \parindent=0pt \parskip=0pt ~ \vskip 3truecm \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 \parindent=0pt \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. 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