Characterizing Orbital-Reversibility Through Normal Forms

Abstract

In this paper, we consider the orbital-reversibility problem for an n-dimensional vector field, which consists in determining if there exists a time-reparametrization that transforms the vector field into a reversible one. We obtain an orbital normal form that brings out the invariants that prevent the orbital-reversibility. Hence, we obtain a necessary condition for a vector field to be orbital-reversible. Namely, the existence of an orbital normal form which is reversible to the change of sign in some of the state variables. The necessary condition provides an algorithm, based on the vanishing of the orbital normal form terms that avoid the orbital-reversibility, that is applied to some families of planar and three-dimensional systems.

Appendices

A Technical Results on Orbital Normal Forms

We present now some technical results about the orbital normal form reduction procedure. Along this appendix, the vector field ${\mathbf{F}}$ is defined in (6). The first result can be found in [11].

Lemma 17

Let us consider $k\in \mathbb {N}$ and a couple of near-identity transformations $\Phi$ and $\Psi$, with generators ${\mathbf{U}}=\sum _{j\ge 1}{\mathbf{U}}_j$ and ${\mathbf{V}}=\sum _{j\ge k}{\mathbf{V}}_j$ (with ${\mathbf{U}}_j,{\mathbf{V}}_j\in \mathcal {Q}_j^{\mathbf{t}}$), respectively. Let also consider the near-identity transformation $\Psi \circ \Phi$, and denote its generator by ${\mathbf{W}}$.

Then, ${\mathbf{U}}$ and ${\mathbf{W}}$ agree up to quasi-homogeneous degree $k-1$ (i.e., $\mathcal {J}^{k-1}\left( {\mathbf{U}} \right) =\mathcal {J}^{k-1}({\mathbf{W}})$), and the k-degree quasi-homogeneous terms are

$$\begin{aligned} {\mathbf{W}}_k={\mathbf{U}}_k+{\mathbf{V}}_k. \end{aligned}$$

Next results show that the orbital normal form reduction procedure can also be carried out by changing the order of the temporal and spatial transformations for each degree, that is, first performing the near-identity transformation and then the time-reparametrization.

Lemma 18

Let us consider $\mu _{k}\in \mathcal {P}_{k}^{\mathbf{t}}$, where $k\in \mathbb {N}$, and ${\mathbf{U}}=\sum _{j\ge 1} {\mathbf{U}}_j$, with ${\mathbf{U}}_j\in \mathcal {Q}_j^{\mathbf{t}}$. Then, there exists $\lambda =\sum _{j\ge k}\lambda _j$, with $\lambda _j \in \mathcal {P}_j^{\mathbf{t}}$, such that

$$\begin{aligned} (1+\mu _{k}) \left( {\mathbf{U}}{\, }_{ **}{\, }{\mathbf{F}} \right) = {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\lambda ){\mathbf{F}} \right) , \end{aligned}$$

where the lowest-degree quasi-homogeneous term of $\lambda$ is $\lambda _k = \mu _{k}$.

Proof

Let us define $T_{\mathbf{U}}^{(0)}\left( {\mathbf{F}} \right) := {\mathbf{F}}$, and

$$\begin{aligned} T_{\mathbf{U}}^{(l)}\left( {\mathbf{F}} \right) := T_{\mathbf{U}}^{(l-1)}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] \right) = {\mathop {\overbrace{\left[ \right. \cdots \left[ \right. }}\limits ^{l\ \text {{times}}}} {\mathbf{F}},{\mathbf{U}}\left. \right] ,\cdots ,{\mathbf{U}}\left. \right] = \left[ T_{\mathbf{U}}^{(l-1)}\left( {\mathbf{F}} \right) , {\mathbf{U}}\right] , \ \text { for } l\ge 1. \end{aligned}$$

In [2] it is shown that the transformed system is given by

$$\begin{aligned} {\mathbf{U}}{\, }_{ **}{\, }{\mathbf{F}}=\sum _{l=0}^{\infty }\frac{1}{l!}T_{\mathbf{U}}^{(l)}\left( {\mathbf{F}} \right) . \end{aligned}$$

Let us denote the successive Lie derivatives along the vector field ${\mathbf{U}}$ of the scalar function $\mu _{k}$ by $\mu ^{[j]}_{k}$, i.e.,

$$\begin{aligned} \mu ^{[0]}_{k} = \mu _{k}, \ \text { and } \mu ^{[j]}_{k}= { \nabla \mu ^{[j-1]}_{k}} \cdot {\mathbf{U}}, \ \text { for } j\ge 1. \end{aligned}$$

Using induction, it is easy to prove that

$$\begin{aligned} \mu _{k} T^{(l)}_{{\mathbf{U}}}\left( {\mathbf{F}} \right) = \sum _{j=0}^l {l \atopwithdelims ()j} (-1)^j T^{(l-j)}_{{\mathbf{U}}} \left( \mu ^{[j]}_{k} {\mathbf{F}} \right) , \ \text { for all } \ l\ge 0. \end{aligned}$$

Then,

$$\begin{aligned} (1+\mu _{k}) \left( {\mathbf{U}}{\, }_{ **}{\, }{\mathbf{F}} \right)= & {} \sum _{l=0}^{\infty } \frac{1}{l!} T_{\mathbf{U}}^{(l)} \left( {\mathbf{F}} \right) + \sum _{l=0}^{\infty } \frac{1}{l!} \sum _{j=0}^l {l \atopwithdelims ()j} (-1)^j T^{(l-j)}_{{\mathbf{U}}} \left( \mu ^{[j]}_{k} {\mathbf{F}} \right) \\= & {} { \sum _{l=0}^{\infty } \frac{1}{l!} T_{\mathbf{U}}^{(l)} \left( {\mathbf{F}} \right) + \sum _{l=0}^{\infty } \sum _{j=0}^l \frac{1}{j!l!} (-1)^j T^{(l)}_{{\mathbf{U}}} \left( \mu ^{[j]}_{k} {\mathbf{F}} \right) }\\= & {} \sum _{l=0}^{\infty } \frac{1}{l!} T^{(l)}_{{\mathbf{U}}} \left( \left( {\textstyle 1 + \sum _{j=0}^l \frac{1}{j!} (-1)^j \mu ^{[j]}_{k} } \right) {\mathbf{F}} \right) = {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\lambda ) {\mathbf{F}} \right) , \end{aligned}$$

as claimed, where we have introduced $\lambda := \sum _{j=0}^l \frac{1}{j!} (-1)^j \mu ^{[j]}_{k}$. Moreover, it is easy to show that $\lambda _k = \mu _{k}$. $\blacksquare$

Lemma 19

Let us consider $\rho _{k} \in \mathcal {P}_k^{\mathbf{t}}$, where $k\in \mathbb {N}$, and $\mu =\sum _{j\ge 1} \mu _j$, where $\mu _j\in \mathcal {P}_j^{\mathbf{t}}$. Then, there exists $\eta =\sum _{j\ge 1} \eta _j$, with $\eta _j \in \mathcal {P}_j^{\mathbf{t}}$, such that:

$$\begin{aligned} \left( \rho _{k} {\mathbf{F}} \right) {\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) = (1+\eta ) {\mathbf{F}}. \end{aligned}$$

Moreover, $\mathcal {J}^{r+k-1}\left( \eta \right) =\mathcal {J}^{r+k-1}(\mu )$ (i.e., $\eta$ and $\mu$ agree up to quasi-homogeneous degree $r+k-1$), and the $(r+k)$-degree quasi-homogeneous term is $\eta _{r+k} = \mu _{r+k} - \nabla \rho _{k} \cdot {\widetilde{\mathbf{F}}}_r$.

Proof

Firstly, we will show that, for each $l\in \mathbb {N}$, there exists $\eta ^{(l)}=\sum _{ j \ge l(r+k)} \eta ^{(l)}_j$, with $\eta ^{(l)}_j \in \mathcal {P}_j^{\mathbf{t}}$, satisfying:

$$\begin{aligned} T_{\rho _{k} {\mathbf{F}}}^{(l)} \left( (1+\mu ){\mathbf{F}} \right) = \eta ^{(l)}{\mathbf{F}}. \end{aligned}$$

(20)

We proceed by induction on l.

• If $l=1$, using (3) we obtain

  $$\begin{aligned} T_{\rho _{k} {\mathbf{F}}} \left( (1+\mu ){\mathbf{F}} \right)= & {} \left[ (1+\mu ){\mathbf{F}},\rho _{k} {\mathbf{F}} \right] \\= & {} \left( \rho _{k} \left( \nabla \mu \cdot {\mathbf{F}} \right) - (1+\mu ) \left( \nabla \rho _{k} \cdot {\mathbf{F}} \right) \right) {\mathbf{F}}=\eta ^{(1)}{\mathbf{F}}, \end{aligned}$$

  where we have denoted $\eta ^{(1)} := \rho _{k} \left( \nabla \mu \cdot {\mathbf{F}} \right) - (1+\mu ) \left( \nabla \rho _{k} \cdot {\mathbf{F}} \right)$. Observe that the principal part of $\eta ^{(1)}$ is $\eta _{r+k}^{(1)} = - \nabla \rho _{k} \cdot {\widetilde{\mathbf{F}}}_r$.

• Now, suppose that (20) holds for $l-1$, that is, there exists $\eta ^{(l-1)}=\sum _{ j \ge (l-1)(r+k)} \eta ^{(l-1)}_j$, with $\eta ^{(l-1)}_j \in \mathcal {P}_j^{\mathbf{t}}$, such that

  $$\begin{aligned} T_{\rho _{k} {\mathbf{F}}}^{(l-1)} \left( (1+\mu ){\mathbf{F}} \right) = \eta ^{(l-1)}{\mathbf{F}}. \end{aligned}$$

  Using again (3), we obtain

  $$\begin{aligned} T_{\rho _{k} {\mathbf{F}}}^{(l)} \left( (1+\mu ){\mathbf{F}} \right) = \left[ T_{\rho _{k} {\mathbf{F}}}^{(l-1)} \left( (1+\mu ){\mathbf{F}} \right) , \rho _{k} {\mathbf{F}} \right] = \left[ \eta ^{(l-1)}{\mathbf{F}}, \rho _{k} {\mathbf{F}} \right] = \eta ^{(l)}{\mathbf{F}}, \end{aligned}$$

  where we have introduced $\eta ^{(l)} := \rho _{k} \left( \nabla \eta ^{(l-1)}\cdot {\mathbf{F}} \right) - \eta ^{(l-1)} \left( \nabla \rho _{k} \cdot {\mathbf{F}} \right)$. We notice that the principal part of $\eta ^{(l)}$ has quasi-homogeneous degree $l(r+k)$.

By the principle of induction, equality (20) holds for all $l\in \mathbb {N}$. Hence, we have:

$$\begin{aligned} \left( \rho _{k} {\mathbf{F}} \right) {\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right)= & {} {\textstyle \sum _{l=0}^{\infty } {\textstyle {\frac{1}{l!}}} T_{\rho _{k} {\mathbf{F}}}^{(l)}\left( (1+\mu ){\mathbf{F}} \right) }=(1+\mu ){\mathbf{F}}+{\textstyle \sum _{l=0}^{\infty }{\textstyle {\frac{1}{l!}}}\eta ^{(l)}{\mathbf{F}}}\\= & {} \left( 1+\mu +{\textstyle \sum _{l=1}^{\infty } {\textstyle {\frac{1}{l!}}}\eta ^{(l)}} \right) {\mathbf{F}}= (1+\eta ) {\mathbf{F}}, \end{aligned}$$

where we have denoted $\eta :=\mu +\sum _{l\ge 1}\frac{1}{l!}\eta ^{(l)}$. To show that the principal part of $\eta$ is $\eta _{r+k} = \mu _{r+k} - \nabla \rho _{k} \cdot {\widetilde{\mathbf{F}}}_r$, it is enough to recall that the principal part of $\eta ^{(1)}$ is $\eta _{r+k}^{(1)} = - \nabla \rho _{k} \cdot {\widetilde{\mathbf{F}}}_r$. $\blacksquare$

The following two lemmas deal with vector fields having a Lie symmetry up to order $r+m$.

Lemma 20

Let us consider $m\in \mathbb {N}$ and assume that there exists ${\mathbf{U}}=\sum _{j=1}^{m}{\mathbf{U}}_j$, with ${\mathbf{U}}_j\in \mathcal {Q}_j^{\mathbf{t}}$, such that

$$\begin{aligned} \mathcal {J}^{r+m}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] \right) = \mathcal {J}^{r+m}\left( \nu {\mathbf{F}} \right) , \ \text { for some }\nu =\sum _{j=1}^{m}\nu _j, \text { with } \nu _j\in \mathcal {P}_j^{\mathbf{t}}. \end{aligned}$$

Then, for any $\mu =\sum _{j=1}^{m}\mu _j$, with $\mu _j\in \mathcal {P}_j^{\mathbf{t}}$, $l\ge 0$, and $k=l+1,\dots ,l+m$, we have:

$$\begin{aligned} \left( T_{\mathbf{U}}^{(l)}\left( \mu {\mathbf{F}} \right) \right) _{r+k}= & {} \sum _{j=l+1}^{k}\mu _{j}^{(l)}{\mathbf{F}}_{r+k-j}, \end{aligned}$$

for some $\mu _{j}^{(l)} \in \mathcal {P}_{j}^{{\mathbf{t}}}$. Moreover, $\mu _{{j}}^{(l)}$ depends univocally on ${\mathbf{U}}_1,\dots ,{\mathbf{U}}_{{j}-l},\nu _1,\dots ,\nu _{{j}-l},\mu _1,\dots ,\mu _{{j}-l}$.

Proof

We use induction on l.

• For $l=0$, the result is trivial because $\left( T_{\mathbf{U}}^{(0)}\left( \mu {\mathbf{F}} \right) \right) _{r+k}=\left( \mu {\mathbf{F}} \right) _{r+k}=\sum _{j=1}^{k}\mu _{j}{\mathbf{F}}_{r+k-j}$.

• Assume that the result is true for $l-1$, being $l>0$. Then:

  $$\begin{aligned} \left( T_{\mathbf{U}}^{(l)}\left( \mu {\mathbf{F}} \right) \right) _{r+k}= & {} \sum _{{j}=1}^{k-l} \left[ \left( T_{\mathbf{U}}^{(l-1)}\left( \mu {\mathbf{F}} \right) \right) _{r+k-{j}},{\mathbf{U}}_{j} \right] \\= & {} \sum _{{i}=l}^{k-1} \sum _{{j}=1}^{k-{i}} \left[ \mu _{{i}}^{(l-1)} {\mathbf{F}}_{r+k-{i}-{j}},{\mathbf{U}}_{j} \right] \\= & {} \sum _{{i}=l}^{k-1} \left( \sum _{{j}=1}^{k-{i}} \left( \nabla \mu _{{i}}^{(l-1)} \cdot {\mathbf{U}}_{j} \right) {\mathbf{F}}_{r+k-{i}-{j}} + \mu _{{i}}^{(l-1)} \sum _{{j}=1}^{k-{i}} \left[ {\mathbf{F}}_{r+k-{i}-{j}},{\mathbf{U}}_{j} \right] \right) \\= & {} \sum _{{i}=l}^{k-1} \left( \sum _{{j}=1}^{k-{i}} \left( \nabla \mu _{{i}}^{(l-1)} \cdot {\mathbf{U}}_{j} \right) {\mathbf{F}}_{r+k-{i}-{j}} \right) + \sum _{{i}=l}^{k-1}\mu _{{i}}^{(l-1)} \left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k-{i}} . \end{aligned}$$

  As $\mathcal {J}^{r+m}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] \right) =\mathcal {J}^{r+m}\left( \nu {\mathbf{F}} \right)$, then $\left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k-{i}} = \left( \nu {\mathbf{F}} \right) _{r+k-{i}} = \sum _{{j}=1}^{k-{i}} \nu _{j} {\mathbf{F}}_{r+k-{i}-{j}}$ for $k=l+1,\dots ,l+m$. Hence:

  $$\begin{aligned} \left( T_{\mathbf{U}}^{(l)}\left( \mu {\mathbf{F}} \right) \right) _{r+k}= & {} \sum _{{i}=l}^{k-1}\sum _{{j}=1}^{k-{i}} \left( \left( \nabla \mu _{{i}}^{(l-1)}\cdot {\mathbf{U}}_{j} \right) + \mu _{i}^{(l-1)}\nu _{j} \right) {\mathbf{F}}_{r+k-{i}-{j}}\\= & {} \sum _{{j}=l+1}^{k}\left( \sum _{{i}=l}^{{j}-1} \left( \left( \nabla \mu _{{i}}^{(l-1)}\cdot {\mathbf{U}}_{{j}-{i}} \right) + \mu _{i}^{(l-1)}\nu _{{j}-{i}} \right) \right) {\mathbf{F}}_{r+k-{j}}, \end{aligned}$$

  and the result holds by defining $\mu _{{j}}^{(l)}:=\sum _{{i}=l}^{{j}-1} \left( \left( \nabla \mu _{{i}}^{(l-1)}\cdot {\mathbf{U}}_{{j}-{i}} \right) + \mu _{i}^{(l-1)}\nu _{{j}-{i}} \right) \in \mathcal {P}_{j}^{{\mathbf{t}}}$, that depends univocally on ${\mathbf{U}}_1,\dots ,{\mathbf{U}}_{{j}-l},\nu _1,\dots ,\nu _{{j}-l},\mu _1,\dots ,\mu _{{j}-l}$.

$\blacksquare$

Lemma 21

Let us consider $m\in \mathbb {N}$ and assume that there exists ${\mathbf{U}}=\sum _{j=1}^{m+1}{\mathbf{U}}_j$, with ${\mathbf{U}}_j\in \mathcal {Q}_j^{\mathbf{t}}$, such that

$$\begin{aligned} \mathcal {J}^{r+m}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] \right) = \mathcal {J}^{r+m}\left( \nu {\mathbf{F}} \right) , \ \text { for some } \ \nu = \sum _{j=1}^{m}\nu _j, \text { with } \nu _j\in \mathcal {P}_j^{\mathbf{t}}. \end{aligned}$$

Then:

• (a) For any $\mu = \sum _{j=1}^{m+1}\mu _j$, with $\mu _j\in \mathcal {P}_j^{\mathbf{t}}$, there exists $\lambda = \sum _{j=1}^{m+1}\lambda _j$, with $\lambda _j\in \mathcal {P}_j^{\mathbf{t}}$, such that

  $$\begin{aligned} \mathcal {J}^{r+m+1} \left( {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) \right) = \mathcal {J}^{r+m+1} \left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +(1+\lambda ){\mathbf{F}} \right) . \end{aligned}$$

  (21)

• (b) For any $\lambda = \sum _{j=1}^{m+1}\lambda _j$, with $\lambda _j\in \mathcal {P}_j^{\mathbf{t}}$, there exists $\mu = \sum _{j=1}^{m+1}\mu _j$, with $\mu _j\in \mathcal {P}_j^{\mathbf{t}}$, such that (21) holds.

Proof

As $[{\mathbf{F}},{\mathbf{U}}]_{r+k}=\left( \nu {\mathbf{F}} \right) _{r+k}$ for $k=1,\dots , m$, it can be shown that

$$\begin{aligned} \left( T_{\mathbf{U}}^{(l)}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] \right) \right) _{r+k} = \left( T_{\mathbf{U}}^{(l)}\left( \nu {\mathbf{F}} \right) \right) _{r+k}, \end{aligned}$$

for all $l\ge 0$, $k=l+1,\dots ,l+m$. Using that

$$\begin{aligned} {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) = {\mathbf{F}}+ \left[ {\mathbf{F}},{\mathbf{U}} \right] + \mu {\mathbf{F}}+ \sum _{l=1}^{\infty }{\textstyle {\frac{1}{(l+1)!}}} T_{\mathbf{U}}^{(l)}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] + (l+1) \mu {\mathbf{F}} \right) , \end{aligned}$$

we obtain

$$\begin{aligned} \left( {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) \right) _{r+k}= & {} {\mathbf{F}}_{r+k} + \left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +\mu {\mathbf{F}} \right) _{r+k}\\&+ \sum _{l=1}^{k-1} \frac{1}{(l+1)!} \left( T_{\mathbf{U}}^{(l)}\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +(l+1)\mu {\mathbf{F}} \right) \right) _{r+k}\\= & {} {\mathbf{F}}_{r+k}+\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +\mu {\mathbf{F}} \right) _{r+k}\\&+\sum _{l=1}^{k-1} \frac{1}{(l+1)!} \left( T_{\mathbf{U}}^{(l)}\left( (\nu +(l+1)\mu ){\mathbf{F}} \right) \right) _{r+k}. \end{aligned}$$

From Lemma 20, we get:

$$\begin{aligned} \left( {\mathbf{U}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) \right) _{r+k}= & {} {\mathbf{F}}_{r+k}+\left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +\mu {\mathbf{F}} \right) _{r+k} +\sum _{l=1}^{k-1}\sum _{j=l+1}^{k}\mu _{j}^{(l)}{\mathbf{F}}_{r+k-j}\\= & {} {\mathbf{F}}_{r+k}+\left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k} +\sum _{j=1}^{k}\mu _{j}{\mathbf{F}}_{r+k-j} +\sum _{j=2}^{k}\sum _{l=1}^{j-1}\mu _{j}^{(l)}{\mathbf{F}}_{r+k-j}\\= & {} {\mathbf{F}}_{r+k}+\left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k}+ \sum _{j=1}^{k}\left( \mu _{j}+ \sum _{l=1}^{j-1}\mu _{j}^{(l)} \right) {\mathbf{F}}_{r+k-j}. \end{aligned}$$

Notice that $\mu _{j}^{(l)}$ depends univocally on ${\mathbf{U}}_1,\dots ,{\mathbf{U}}_{{j}-1},\nu _1,\dots ,\nu _{{j}-1},\mu _1,\dots ,\mu _{{j}-1}$. On the other hand, we have:

$$\begin{aligned} \left( \left[ {\mathbf{F}},{\mathbf{U}} \right] +(1+\lambda ){\mathbf{F}} \right) _{r+k}= & {} {\mathbf{F}}_{r+k}+\left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k}+\left( \lambda {\mathbf{F}} \right) _{r+k}={\mathbf{F}}_{r+k}+\left[ {\mathbf{F}},{\mathbf{U}} \right] _{r+k}+\sum _{j=1}^{k}\lambda _j{\mathbf{F}}_{r+k-j}. \end{aligned}$$

To prove item (a), it is enough to define $\lambda _j=\mu _{j}+ \sum _{l=1}^{j-1}\mu _{j}^{(l)}$, for $j=1,\dots , k$.

To prove item (b), it is enough to solve the equation $\lambda _j=\mu _j+ \sum _{l=1}^{j-1}\mu _{j}^{(l)}$ (this can be done because $\mu _{j}^{(l)}$ depends univocally on ${\mathbf{U}}_1,\dots ,{\mathbf{U}}_{{j}-1},\nu _1,\dots ,\nu _{{j}-1},\mu _1,\dots ,\mu _{{j}-1}$) and define $\lambda _j=\mu _{j}+ \sum _{l=1}^{j-1}\mu _{j}^{(l)}$, for $j=1,\dots , k$. $\blacksquare$

B The Range of $\overline{\mathcal {NL}}^{(N)}$

In this appendix, we show that ${\mathrm {Range}}(\overline{\mathcal {NL}}^{(N)})={\mathrm {Range}}(\overline{\mathcal {L}}^{(N)})$. This is a consequence of Propositions 24 and 25.

The following propositions are necessary to prove Propositions 24 and 25.

Proposition 22

Let us consider the vector field ${\mathbf{F}}$ given in (6). Then, for each ${\widetilde{\mathbf{U}}}\in \bigoplus _{j=1}^m\mathcal {R}_j^{\mathbf{t}}$ and ${\widetilde{\lambda }}\in \bigoplus _{j=1}^{m}\mathcal {O}_j^{\mathbf{t}}$ ($m\in \mathbb {N}$), there exist ${\widetilde{\mathbf{V}}}\in \bigoplus _{j=1}^{m}\widehat{\mathcal {R}}_j^{\mathbf{t}}$ and ${\widetilde{\nu }}\in \bigoplus _{j=1}^{m}\widehat{\mathcal {O}}_j^{\mathbf{t}}$ such that

$$\begin{aligned} \mathcal {J}^{r+m}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\mathbf{F}}} \right) = \mathcal {J}^{r+m}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right) . \end{aligned}$$

Proof

Let us denote

$$\begin{aligned} \kappa :=\min \left\{ j\in \mathbb {N}:{\mathrm {Proj}}_{{\mathrm {Range}}\left( {\ell }_{j-r} \right) } \left( {\widetilde{\lambda }}_j \right) \ne 0 \ \text { or } \ {\mathrm {Proj}}_{{\mathrm {Ker}}\left( {\ell }_{j-r} \right) {\widetilde{\mathbf{F}}}_r} \left( {\widetilde{\mathbf{U}}}_j \right) \ne {\mathbf{0}} \right\} . \end{aligned}$$

Observe that $\kappa \le m$, otherwise ${\widetilde{\mathbf{V}}}={\widetilde{\mathbf{U}}}$ and ${\widetilde{\nu }}={\widetilde{\lambda }}$.

From (11), we can write ${\widetilde{\lambda }}_{\kappa } = {\widetilde{\lambda }}_{\kappa }^{(1)} + {\widetilde{\lambda }}_{\kappa }^{(2)}$, where ${\widetilde{\lambda }}_{\kappa }^{(1)} = \nabla {\overline{\rho }}\cdot {\widetilde{\mathbf{F}}}_r \in {\mathrm {Range}}\left( \ell ^{(\texttt {e})}_{\kappa -r} \right)$ (${\overline{\rho }}\in \mathcal {E}_{\kappa -r}^{\mathbf{t}}$) and ${\widetilde{\lambda }}_{\kappa }^{(2)} \in \widehat{\mathcal {O}}_{\kappa }^{\mathbf{t}}$. We observe that ${\overline{\rho }}=0$ if ${\widetilde{\lambda }}_{\kappa }^{(1)} = 0$, and ${\overline{\rho }}\in \mathcal {E}_{\kappa -r}^{\mathbf{t}}{\setminus } {\mathrm {Ker}}\left( {\ell }_{\kappa -r} \right)$ otherwise.

Moreover, by (13), we can write ${\widetilde{\mathbf{U}}}_{\kappa } = {\widetilde{\mathbf{U}}}_{\kappa }^{(1)} + {\widetilde{\mathbf{U}}}_{\kappa }^{(2)}$, where ${\widetilde{\mathbf{U}}}_{\kappa }^{(1)} = {\overline{\zeta }}{\widetilde{\mathbf{F}}}_r\in {\mathrm {Ker}}\left( \ell ^{(\texttt {e})}_{\kappa -r} \right) {\widetilde{\mathbf{F}}}_r$ (${\overline{\zeta }}\in {\mathrm {Ker}}\left( \ell ^{(\texttt {e})}_{\kappa -r} \right)$) and ${\widetilde{\mathbf{U}}}_{\kappa }^{(2)} \in \widehat{\mathcal {R}}_{\kappa }^{\mathbf{t}}$.

Let us denote ${\widetilde{\mathbf{F}}}$ the reversible part of ${{\mathbf{F}}}$ and define ${\widetilde{\mathbf{V}}}:= {\widetilde{\mathbf{U}}}- ({\overline{\rho }}+{\overline{\zeta }}){\widetilde{\mathbf{F}}}$, where

$$\begin{aligned} {\widetilde{\nu }}:= {\widetilde{\lambda }}- \nabla {\overline{\rho }}\cdot {\widetilde{\mathbf{F}}}- \nabla {\overline{\zeta }}\cdot {\widetilde{\mathbf{F}}}. \end{aligned}$$

From Lemma 5, we obtain:

$$\begin{aligned} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\mathbf{F}}} \right) _{r+\kappa }= & {} \left( \left[ {\widetilde{\mathbf{F}}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\widetilde{\mathbf{F}}} \right) _{r+\kappa } = \left( \left[ {\widetilde{\mathbf{F}}},{\widetilde{\mathbf{V}}}+({\overline{\rho }}+{\overline{\zeta }}){\widetilde{\mathbf{F}}} \right] +{\widetilde{\lambda }}{\widetilde{\mathbf{F}}} \right) _{r+\kappa }\\= & {} \left( \left[ {\widetilde{\mathbf{F}}},{\widetilde{\mathbf{V}}} \right] +\left[ {\widetilde{\mathbf{F}}},({\overline{\rho }}+{\overline{\lambda }}){\widetilde{\mathbf{F}}} \right] +{\widetilde{\lambda }}{\widetilde{\mathbf{F}}} \right) _{r+\kappa } \\= & {} \left( \left[ {\widetilde{\mathbf{F}}},{\widetilde{\mathbf{V}}} \right] + \left( {\widetilde{\lambda }}-\nabla {\overline{\rho }}\cdot {\widetilde{\mathbf{F}}}- \nabla {\overline{\zeta }}\cdot {\widetilde{\mathbf{F}}} \right) {\widetilde{\mathbf{F}}} \right) _{r+\kappa } \\= & {} \left( \left[ {\widetilde{\mathbf{F}}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\widetilde{\mathbf{F}}} \right) _{r+\kappa } \\= & {} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right) _{r+\kappa }. \end{aligned}$$

To complete the proof, it is enough to observe that:

$$\begin{aligned} {\widetilde{\mathbf{V}}}_{\kappa }= & {} {\widetilde{\mathbf{U}}}_{\kappa }^{(1)} + {\widetilde{\mathbf{U}}}_{\kappa }^{(2)} - ({\overline{\rho }}+{\overline{\zeta }}){\widetilde{\mathbf{F}}}_r = {\widetilde{\mathbf{U}}}_{\kappa }^{(2)} - {\overline{\rho }}{\widetilde{\mathbf{F}}}_r \in \widehat{\mathcal {R}}_{\kappa }^{\mathbf{t}},\\ {\widetilde{\nu }}_{\kappa }= & {} {\widetilde{\lambda }}_{\kappa }^{(1)} + {\widetilde{\lambda }}_{\kappa }^{(2)} - \nabla {\overline{\rho }}\cdot {\widetilde{\mathbf{F}}}_r - \nabla {\overline{\zeta }}\cdot {\widetilde{\mathbf{F}}}_r = {\widetilde{\lambda }}_{\kappa }^{(2)} \in \widehat{\mathcal {O}}_{\kappa -r}^{\mathbf{t}}. \end{aligned}$$

$\blacksquare$

Proposition 23

Let us assume that the vector field ${\mathbf{F}}$ given in (6) is $R_x$-reversible up to order $r+N-1$, with $N\in \mathbb {N}$. Then, for all ${\widetilde{\mathbf{U}}}\in \bigoplus _{j=1}^N \widehat{\mathcal {R}}_j^{\mathbf{t}}$ and ${\widetilde{\mu }}\in \bigoplus _{j=1}^N \widehat{\mathcal {O}}_j^{\mathbf{t}}$ such that $({\widetilde{\mathbf{U}}},{\widetilde{\mu }})$ belongs to the domain of $\overline{\mathcal {NL}}^{(N)}$, there exists $\lambda \in \bigoplus _{j=1}^{N}\mathcal {P}_j^{\mathbf{t}}$ such that

$$\begin{aligned} \mathcal {J}^{r+N} \left( {\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) \right) = \mathcal {J}^{r+N} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}} \right) . \end{aligned}$$

Proof

We use induction on N. If $N=1$, the result is trivial. Moreover

$$\begin{aligned} \left( {\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) \right) _{r+1} = {\mathbf{F}}_{r+1} + \left( [{\mathbf{F}},{\widetilde{\mathbf{U}}}]+{\widetilde{\mu }}{\mathbf{F}} \right) _{r+1}, \end{aligned}$$

and it is enough to take $\lambda _1={\widetilde{\mu }}_1$.

Now, let us consider $N>1$ and assume that there exists $\lambda = \sum _{j=1}^{N-1} \lambda _j$, such that

$$\begin{aligned} \mathcal {J}^{r+N-1} \left( {\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) \right) = \mathcal {J}^{r+N-1} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}} \right) . \end{aligned}$$

The hypothesis $\mathcal {J}^{r+N-1} \left( \overline{{\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) } \right) ={\mathbf{0}}$ yields. If we take $\lambda =\bar{\lambda }+\tilde{\lambda }$, where $\bar{\lambda }=\sum _{j=1}^{N-1}\bar{\lambda }_j$, $\bar{\lambda }_j\in \mathcal {E}_j^{\mathbf{t}}$ and $\tilde{\lambda }=\sum _{j=1}^{N-1}\tilde{\lambda }_j$, $\tilde{\lambda }_j\in \mathcal {O}_j^{\mathbf{t}}$ we obtain

$$\begin{aligned} \mathcal {J}^{r+N-1} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}}} \right) =\mathcal {J}^{r+N-1} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+{\widetilde{\lambda }}){\mathbf{F}} \right) ={\mathbf{0}}. \end{aligned}$$

Hence $\mathcal {J}^{r+N-1} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\mathbf{F}} \right) = {\mathbf{0}}$ and from Lemma 21 item (a), the statement holds for the value N. Consequently, we have

$$\begin{aligned} \mathcal {J}^{r+N} \left( {\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) \right)= & {} \mathcal {J}^{r+N} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}} \right) . \end{aligned}$$

$\blacksquare$

Next result shows that ${\mathrm {Range}}\left( \overline{\mathcal {NL}}^{(N)} \right) \subset {\mathrm {Range}}\left( \overline{\mathcal {L}}^{(N)} \right)$.

Proposition 24

Let us assume that the vector field ${\mathbf{F}}$ given in (6) is $R_x$-reversible up to order $r+N-1$, with $N\in \mathbb {N}$. Then for all generators $({\widetilde{\mathbf{U}}},{\widetilde{\mu }})$, belonging to the domain of the nonlinear operator $\overline{\mathcal {NL}}^{(N)}$, there exists $({\widetilde{\mathbf{V}}},{\widetilde{\nu }})$, belonging to the domain of the linear operator $\overline{\mathcal {L}}^{(N)}$, such that $\overline{\mathcal {NL}}^{(N)}\left( {\widetilde{\mathbf{U}}},{\widetilde{\mu }} \right) =\overline{\mathcal {L}}^{(N)} \left( {\widetilde{\mathbf{V}}},{\widetilde{\nu }} \right)$.

Proof

By Proposition 23 there exists $\lambda \in \bigoplus _{j=1}^{N}\mathcal {P}_j^{\mathbf{t}}$ such that

$$\begin{aligned} \mathcal {J}^{r+N} \left( {\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) \right) = \mathcal {J}^{r+N} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}} \right) . \end{aligned}$$

If we take $\lambda =\bar{\lambda }+\tilde{\lambda }$, where $\bar{\lambda }=\sum _{j=1}^{N-1}\bar{\lambda }_j$, $\bar{\lambda }_j\in \mathcal {E}_j^{\mathbf{t}}$ and $\tilde{\lambda }=\sum _{j=1}^{N-1}\tilde{\lambda }_j$, $\tilde{\lambda }_j\in \mathcal {O}_j^{\mathbf{t}}$ we obtain

$$\begin{aligned} \mathcal {J}^{r+N} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+\lambda ){\mathbf{F}}} \right) =\mathcal {J}^{r+N} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\mathbf{F}}} \right) +\mathcal {J}^{r+N} \left( \overline{(1+\bar{\lambda }){\mathbf{F}}} \right) . \end{aligned}$$

As ${\mathbf{F}}$ is $R_x$-reversible up to order $r+N-1$, from Lemma 5(a) we obtain $\mathcal {J}^{r+N} \left( \overline{(1+\bar{\lambda }){\mathbf{F}}} \right) ={\mathbf{0}}$ and by Proposition 22 there exist ${\widetilde{\mathbf{V}}}\in \bigoplus _{j=1}^{N}\widehat{\mathcal {R}}_j^{\mathbf{t}}$ and ${\widetilde{\nu }}\in \bigoplus _{j=1}^{N}\widehat{\mathcal {O}}_j^{\mathbf{t}}$ such that

$$\begin{aligned} \mathcal {J}^{r+N}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +{\widetilde{\lambda }}{\mathbf{F}}} \right) = \mathcal {J}^{r+N}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right) . \end{aligned}$$

Hence $\mathcal {J}^{r+N} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{U}}} \right] +(1+{\widetilde{\mu }}){\mathbf{F}}} \right) =\mathcal {J}^{r+N} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right)$.

The hypothesis $\mathcal {J}^{r+N-1} \left( \overline{{\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) } \right) ={\mathbf{0}}$ yields, since $(\tilde{{\mathbf{U}}},\tilde{\mu })$ belongs to the domain of $\overline{\mathcal {NL}}^{(N)}$. Therefore $\mathcal {J}^{r+N-1}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right) ={\mathbf{0}}$ and we prove that $({\widetilde{\mathbf{V}}},{\widetilde{\nu }})$ belongs to the domain of ${\overline{\mathcal {L}}^{(N)}_{\scriptstyle {\left\{ {\widetilde{\mathbf{F}}}_r,\dots ,{\widetilde{\mathbf{F}}}_{r+N-1} \right\} }}}$, and $\overline{\mathcal {NL}}^{(N)}\left( {\widetilde{\mathbf{U}}},{\widetilde{\mu }} \right) =\overline{\mathcal {L}}^{(N)}\left( {\widetilde{\mathbf{V}}},{\widetilde{\nu }} \right)$.

$\blacksquare$

In the following result, we prove that ${\mathrm {Range}}\left( \overline{\mathcal {L}}^{(N)} \right) \subset {\mathrm {Range}}\left( \overline{\mathcal {NL}}^{(N)} \right)$.

Proposition 25

Let us assume that the vector field ${\mathbf{F}}$ given in (15) is $R_x$-reversible up to order $r+N-1$, with $N\in \mathbb {N}$. Then for all generators $\left( {\widetilde{\mathbf{V}}},{\widetilde{\nu }} \right)$ belonging to the domain of $\overline{\mathcal {L}}^{(N)}$, there exists $\left( {\widetilde{\mathbf{U}}},{\widetilde{\mu }} \right)$ belonging to the domain of the nonlinear operator $\overline{\mathcal {NL}}^{(N)}$ such that $\overline{\mathcal {L}}^{(N)}(\tilde{{\mathbf{V}}},\tilde{\nu })=\overline{\mathcal {NL}}^{(N)}({\widetilde{\mathbf{U}}},{\widetilde{\mu }})$.

Proof

We will prove that

$$\begin{aligned}&\mathcal {J}^{r+N-1}\left( \overline{{\widetilde{\mathbf{V}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) } \right) ={\mathbf{0}}\ \ \text { and }\\&\overline{\mathcal {L}}^{(N)}\left( {\widetilde{\mathbf{V}}},{\widetilde{\nu }} \right) ={\overline{\mathbf{F}}}_{r+N} -{\overline{{\widetilde{\mathbf{V}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) }}_{r+N}. \end{aligned}$$

As $\left( {\widetilde{\mathbf{V}}},{\widetilde{\nu }} \right)$ belongs to the domain of the operator $\overline{\mathcal {L}}^{(N)}_{\scriptstyle {\left\{ {\widetilde{\mathbf{F}}}_r,\dots ,{\widetilde{\mathbf{F}}}_{r+N-1} \right\} }}$, then we obtain $\mathcal {J}^{r+N-1}\left( \overline{[{\mathbf{F}},{\widetilde{\mathbf{V}}}]+{\widetilde{\nu }}{\mathbf{F}}} \right) ={\mathbf{0}}$. Also, as ${\mathbf{F}}$ is $R_x$-reversible up to order $r+N-1$, from Lemma 5 items (b) and (c), we get $\mathcal {J}^{r+N-1}\left( [{\mathbf{F}},{\widetilde{\mathbf{V}}}]+{\widetilde{\nu }}{\mathbf{F}} \right) ={\mathbf{0}}$.

From Lemma 21 item (b), for $m=N$, $\lambda ={\widetilde{\nu }}$ there exists $\mu =\sum _{j=1}^{N}\mu _j$, with $\mu _j\in \mathcal {P}_j^{\mathbf{t}}$, such that

$$\begin{aligned} \mathcal {J}^{r+N} \left( {\widetilde{\mathbf{V}}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) \right) = \mathcal {J}^{r+N} \left( \left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +(1+{\widetilde{\nu }}){\mathbf{F}} \right) . \end{aligned}$$

In particular, as ${\mathbf{F}}$ is $(N-1)$-$R_x$-reversible, we obtain $\mathcal {J}^{r+N-1}\left( {\overline{\mathbf{F}}} \right) ={\mathbf{0}}$ and

$$\begin{aligned} {\mathbf{0}}= & {} \mathcal {J}^{r+N-1}\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +{\widetilde{\nu }}{\mathbf{F}}} \right) +\mathcal {J}^{r+N-1}\left( {\overline{\mathbf{F}}} \right) \\= & {} \mathcal {J}^{r+N-1} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +(1+{\widetilde{\nu }}){\mathbf{F}}} \right) = \mathcal {J}^{r+N-1} \left( \overline{{\widetilde{\mathbf{V}}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) } \right) . \end{aligned}$$

By Proposition 8 there exist ${\widetilde{\mathbf{U}}}\in \bigoplus _{j=1}^N\widehat{\mathcal {R}}_j^t$ and ${\widetilde{\mu }}\in \bigoplus _{j=1}^N\widehat{\mathcal {O}}_{j}^{\mathbf{t}}$ such that

$$\begin{aligned} \mathcal {J}^{r+N} \left( \overline{{\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) } \right) =\mathcal {J}^{r+N} \left( \overline{{\widetilde{\mathbf{V}}}{\, }_{ **}{\, }\left( (1+\mu ){\mathbf{F}} \right) } \right) = \mathcal {J}^{r+N} \left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +(1+{\widetilde{\nu }}){\mathbf{F}}} \right) . \end{aligned}$$

Therefore $({\widetilde{\mathbf{U}}},{\widetilde{\mu }})$ belongs to the domain of nonlinear operator $\overline{\mathcal {NL}}^{(N)}$. On the other hand

$$\begin{aligned} \overline{\mathcal {NL}}^{(N)}\left( {\widetilde{\mathbf{U}}},{\widetilde{\mu }} \right)= & {} -\left( \overline{{\widetilde{\mathbf{U}}}{\, }_{ **}{\, }\left( (1+{\widetilde{\mu }}){\mathbf{F}} \right) } \right) _{r+N}+\overline{{\mathbf{F}}}_{r+N}\\= & {} -\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +(1+{\widetilde{\nu }}){\mathbf{F}}} \right) _{r+N}+\overline{{\mathbf{F}}}_{r+N}\\= & {} -\left( \overline{\left[ {\mathbf{F}},{\widetilde{\mathbf{V}}} \right] +(1+{\widetilde{\nu }}){\mathbf{F}}} \right) _{r+N}=\mathcal {L}^{(N)}({\widetilde{\mathbf{V}}},{\widetilde{\nu }}). \end{aligned}$$

$\blacksquare$