**Example 124.** We consider smooth vector fields on $\mathbb{R}^3$. Let $\chi \in C^\infty(\mathbb{R}; \mathbb{R})$ with $\chi \equiv 0$ on $\mathbb{R}_-$ and $\chi(x) > 0$ for $x > 0$. Let $f_0(x) := e_2$ and $f_1(t, x) := f_1(x)$ (independent of time) with

$$f_1(x) := 2\chi^{(1)}(x_2)x_1e_3 + \chi^{(1)}(-x_2)e_1. \qquad (300)$$

For $j \in \mathbb{N}$, one has

$$\mathrm{ad}_{f_0}^j(f_1)(x) = \partial_2^j f_1(x) = 2\chi^{(j+1)}(x_2)x_1e_3 + (-1)^j\chi^{(j+1)}(-x_2)e_1. \qquad (301)$$

Thus, for every $j_1, j_2 \in \mathbb{N}$,

$$[\mathrm{ad}_{f_0}^{j_1}(f_1), \mathrm{ad}_{f_0}^{j_2}(f_1)](x) = 2(-1)^{j_1}\chi^{(j_1+1)}(-x_2)\chi^{(j_2+1)}(x_2)e_3 \\ - 2(-1)^{j_2}\chi^{(j_2+1)}(-x_2)\chi^{(j_1+1)}(x_2)e_3 = 0 \qquad (302)$$

because the supports of $\chi(\cdot)$ and $\chi(-\cdot)$ only touch at $x_2 = 0$, where all derivatives vanish. Hence each bracket of $f_0$ and $f_1$ involving $f_1$ at least twice vanishes identically on $\mathbb{R}^3$. Thus, for every $T > 0$, the family $\mathcal{F} := \{f_1(t, \cdot); t \in [0, T]\} = \{f_1\}$ is semi-nilpotent of index 2 with respect to $f_0$. Let us prove that, despite this property, equality (288) with $M = 1$ fails.

**Computation of the state.** We solve $\dot{x} = f_0(x) + f_1(x)$ for some initial data $p$. Solving the ODE successively for $x_2, x_1$ and $x_3$, we obtain

$$x_1(t) = p_1 + \chi(-p_2) - \chi(-p_2 - t), \qquad (303)$$

$$x_2(t) = p_2 + t, \qquad (304)$$

$$x_3(t) = p_3 + 2(\chi(p_2 + t) - \chi(p_2))(p_1 + \chi(-p_2)). \qquad (305)$$

In particular, with $t = 2\varepsilon$ and $p = -\varepsilon e_2$, $x(2\varepsilon; f_0 + f_1, -\varepsilon e_2) = (\chi(\varepsilon), \varepsilon, 2\chi(\varepsilon)^2)$.

**Computation of the flow.** We compute $e^{\mathcal{Z}_1(t, f_0, f_1)}e^{tf_0}(p)$ for some initial data $p$. One has $\Phi_0(\tau, q) = q + \tau e_2$. Hence, in particular $(\Phi_0(\tau)_{*}f_1)(q) = f_1(q - \tau e_2)$. Moreover $\mathcal{Z}_1(t, f_0, f_1)(q) = \int_0^t g_t(s, q) \mathrm{d}s$ where $g_t(s, q) = (\Phi_0(t - s)_{*}f_1)(q)$. Hence $g_t(s, q) = f_1(q - (t - s)e_2)$ and

$$\begin{aligned} \mathcal{Z}_1(t, f_0, f_1)(q) &= \int_0^t f_1(q + (s-t)e_2) \mathrm{d}s \\ &= 2q_1(\chi(q_2) - \chi(q_2 - t))e_3 + (\chi(-q_2 + t) - \chi(-q_2))e_1. \end{aligned} \qquad (306)$$

Then $e^{\mathcal{Z}_1(t, f_0, f_1)}e^{tf_0}p = e^{\mathcal{Z}_1(t, f_0, f_1)}(p + te_2)$ is $y(1)$ where $y$ is the solution to $y(0) = p + te_2$ and $\dot{y}(s) = \mathcal{Z}_1(t, f_0, f_1)(y(s))$. Solving the ODE successively for $y_2, y_1$ and $y_3$, we obtain

$$y_1(s) = p_1 + s(\chi(-p_2) - \chi(-p_2 - t)), \qquad (307)$$

$$y_2(s) = p_2 + t, \qquad (308)$$

$$y_3(s) = p_3 + (\chi(p_2 + t) - \chi(p_2))(2p_1s + s^2(\chi(-p_2) - \chi(-p_2 - t))). \qquad (309)$$

In particular, with $t = 2\varepsilon$ and $p = -\varepsilon e_2$, $e^{\mathcal{Z}_1(2\varepsilon, f_0, f_1)}e^{2\varepsilon f_0}(-\varepsilon e_2) = (\chi(\varepsilon), \varepsilon, \chi(\varepsilon)^2)$. Thus, for every $\varepsilon > 0$,

$$\left|x(2\varepsilon, f_0 + f_1, -\varepsilon e_2) - e^{\mathcal{Z}_1(2\varepsilon, f_0, f_1)}e^{2\varepsilon f_0}(-\varepsilon e_2)\right| = \chi^2(\varepsilon) > 0. \qquad (310)$$

### 5.3.2 Convergence for linear systems

Let $T > 0$. In this paragraph, we study linear systems of the form

$$\dot{x}(t) = (H_0 + H_1(t))x(t) \quad \text{and} \quad x(0) = p, \qquad (311)$$

where $H_0 \in \mathcal{M}_d(\mathbb{K})$ and $H_1 \in L^1((0, T); \mathcal{M}_d(\mathbb{K}))$. Let $\|\cdot\|$ be a sub-multiplicative norm on $\mathcal{M}_d(\mathbb{K})$.
**Proposition 125.** Let $T > 0$, $H_0 \in \mathcal{M}_d(\mathbb{K})$ and $H_1 \in L^1((0,T); \mathcal{M}_d(\mathbb{K}))$ such that $\|H_1\|_{L^1(0,T)} < \frac{e^{-2T\|H_0\|}}{8}$. Then, for each $t \in [0,T]$ and $p \in \mathbb{K}^d$ the solution to (311) satisfies $x(t) = e^{-Z_\infty(t)}e^{tH_0}p$ where $Z_\infty(t)$ is defined by (264) with

$$A_t(\tau) = e^{(t-\tau)H_0} H_1(\tau) e^{(\tau-t)H_0} = \sum_{k=0}^{+\infty} \frac{(t-\tau)^k}{k!} \mathrm{ad}_{H_0}^k(H_1). \quad (312)$$

*Proof.* The function $y: \tau \in [0,t] \mapsto e^{(t-\tau)H_0}x(\tau)$ satisfies $y'(\tau) = A(\tau)$, $y(0) = e^{tH_0}p$. Thus, by Proposition 111, $y(t) = e^{-Z_\infty(t)}e^{tH_0}p$, which gives the conclusion because $y(t) = x(t)$. □

**Remark 126.** The Magnus expansion in the usual setting (Proposition 111), when applied directly to $A(t) = H_0 + H_1(t)$ requires a smallness assumption on $T\|H_0\|$ (through the condition $\|A\|_{L^1(0,T)} < \frac{1}{8}$), even for small perturbations $H_1$. On the contrary, the Magnus expansion in the interaction picture (Proposition 125) holds even when $T\|H_0\|$ is large, provided that the perturbation $H_1$ is small enough.

**Remark 127.** More generally, in [4, p. 1671], the authors consider the formal power series expressing the chronological logarithm of two flows, associated to two non-autonomous vector fields. They explain that, when the vector fields take values in a Banach algebra, and one of them is small enough, then this series converges. Proposition 125 is an illustration.

### 5.3.3 Divergence for arbitrary small analytic vector fields

Generally speaking, since, as illustrated in Section 5.2.3, the Magnus expansion does not converge for analytic vector fields, one cannot expect that the Magnus expansion in interaction picture converges for analytic vector fields.

For instance, if $f_0 = 0$, or if, for some $a \in [1, d]$, $f_0(x)$ is a linear combination of $e_1, \dots, e_a$ with coefficients depending only on $x_1, \dots, x_a$ and $f_1(t, x)$ is a linear combination of $e_{a+1}, \dots, e_d$, with coefficients depending only on $x_{a+1}, \dots, x_d$, then the vector field $g_t(\tau) = \Phi_0(t - \tau)_{*}f_1(\tau)$ defined in (210) and involved in the Magnus in the interaction picture formula satisfies $g_t(\tau) = f_1(\tau)$.

Hence, each counter-example to the convergence of the usual Magnus expansion also yields counter-examples to the convergence of the Magnus expansion in the interaction picture.

More generally, in [4, p. 1671], the authors consider the formal power series expressing the chronological logarithm of two flows, associated to two non-autonomous vector fields. They claim that, even for analytic vector fields, this series does not converge in general. The counter examples of the present article illustrate this assertion.

## 5.4 Sussmann's infinite product expansion

### 5.4.1 Equality for nilpotent systems

In this section, we study affine systems of the form (226).

**Proposition 128.** Let $\mathcal{B}$ be a Hall basis of $\mathcal{L}(X)$ and $(\xi_b)_{b \in \mathcal{B}}$ be the associated coordinates of the second kind. For every $M \in \mathbb{N}^*$, there exist $\eta_M > 0$ such that the following property holds. Let $T, \delta > 0$, $f_i \in C^\infty(B_{3\delta}; \mathbb{K}^d)$ and $u_i \in L^1((0,T); \mathbb{K})$ for $i \in I$. Assume that the Lie algebra generated by the $f_i$ for $i \in I$ is nilpotent of index at most $M+1$. Then, under the smallness assumption (227), for each $t \in [0,T]$ and $p \in B_\delta$,

$$x(t; f, u, p) = \prod_{b \in \mathcal{B}_{[1,M]}} e^{\xi_b(t,u)f_b} p. \quad (313)$$

*Proof.* The proof strategy is the same as for Proposition 104. We apply the second statement of Lemma 90 instead of the first one, which gives $\varepsilon_j = 0$ for each $j \in [0, k+1]$. The smallness assumption guarantees that all flows are well-defined. □
### 5.4.2 Bilinear systems

Let $T > 0$. In this paragraph, we study the convergence of Sussmann's infinite product expansion for bilinear systems of the form

$$ \dot{x}(t) = \left( \sum_{i \in I} u_i(t) A_i \right) x(t) \quad \text{and} \quad x(0) = p \qquad (314) $$

where $A_i \in \mathcal{M}_d(\mathbb{K})$ are time-invariant and $u_i \in L^1((0, T); \mathbb{K})$. When well-defined, its solution is denoted $x(t; A, u, p)$. Local convergence is proved in Proposition 131 while an example illustrating the lack of global convergence is proposed in Proposition 132.

**Local convergence.** The main goal of this paragraph is to prove Proposition 131 which asserts that Sussmann's infinite product expansion for system (314) converges locally (i.e. for small matrices, small controls or small time).

Before proving this result, we need a definition for an ordered infinite product (given in Definition 129 below) and a sufficient condition for its convergence (given in Lemma 130 below).

Defining the ordered product of a family of matrices indexed by a length-compatible Hall basis is straightforward, because there exists an indexation of the family by $\mathbb{N}$ which is compatible with the order induced by the Hall basis (since it does not involve infinite segments). Hence, one is brought back to the classical case of a sequence of products and usual definitions and convergence criteria can be used.

For arbitrary Hall bases (in the generalized sense of Definition 54), the situation is more intricate, due to the potential infinite segments which can prevent the order of the basis from being compatible with the order of natural integers. For example, in the Lyndon basis of $\mathcal{L}(X)$ for $X = \{X_0, X_1\}$ with $X_0 < X_1$, $\mathrm{ad}_{X_0}^k(X_1) < (X_0, X_1)$ for all $k \ge 2$, so there is an infinite segment before $(X_0, X_1)$. This problem already appears for a product which would be indexed by $\mathbb{N}^2$ with the lexicographic order

$$ (0,0) < (0,1) < (0,2) < \dots < (1,0) < (1,1) < (1,2) < \dots < (2,0) < \dots \qquad (315) $$

We therefore propose a natural definition and a basic sufficient condition for convergence based on absolute convergence. In what follows, $\| \cdot \|$ is a submultiplicative norm on $\mathcal{M}_d(\mathbb{K})$ such that $\|\mathrm{Id}\| = 1$, for instance a subordinated norm.

**Definition 129.** Let $J$ be a totally ordered set and $(A_j)_{j \in J}$ matrices of $\mathcal{M}_d(\mathbb{K})$. We say that the ordered product of the $e^{A_j}$ over $J$ converges when there exists $M \in \mathcal{M}_d(\mathbb{K})$ such that, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $J$ such that, for every finite subset $J_1$ of $J$ containing $J_0$, one has

$$ \left\| M - \left\prod_{j \in J_1} e^{A_j} \right\| \le \varepsilon. \qquad (316) $$

When such an $M$ exists, it is unique and we write

$$ M = \left\prod_{j \in J} e^{A_j}. \qquad (317) $$

The following natural convergence criteria also appears in [68].

**Lemma 130.** Let $J$ be a totally ordered set and $(A_j)_{j \in J}$ matrices of $\mathcal{M}_d(\mathbb{K})$ such that

$$ \sum_{j \in J} \|A_j\| < +\infty. \qquad (318) $$

Then the ordered product of the $e^{A_j}$ over $J$ converges in the sense of Definition 129.
Proof. Let $\alpha$ be the left-hand side of (318).

**Step 1: Basic claims.** We start with straightforward claims. First, for every $j \in J$, one has

$$ \|e^{A_j} - \mathrm{Id}\| \le e^{\|A_j\|} - 1 \le \|A_j\| e^{\|A_j\|} \le \|A_j\| e^{\alpha}. \quad (319) $$

Second, for every finite part $J' \subset J$, one has

$$ \left\| \overleftarrow{\Pi}_{j \in J'} e^{A_j} \right\| \le \overleftarrow{\Pi}_{j \in J'} e^{\|A_j\|} \le e^{\alpha}. \quad (320) $$

Third, for every finite parts $J_0 \subset J_1 \subset J$, one has

$$ \left\| \overleftarrow{\Pi}_{j \in J_1} e^{A_j} - \overleftarrow{\Pi}_{j \in J_0} e^{A_j} \right\| \le e^{3\alpha} \sum_{j \in J_1 \setminus J_0} \|A_j\|. \quad (321) $$

Indeed, writing $J_1 \setminus J_0 = \{j_1 > \dots > j_n\}$, we have the following telescopic decomposition

$$ \overleftarrow{\Pi}_{j \in J_1} e^{A_j} - \overleftarrow{\Pi}_{j \in J_0} e^{A_j} = \sum_{k=1}^{n} \overleftarrow{\Pi}_{\substack{j \in J_0 \\ j > j_k}} e^{A_j} (e^{A_{j_k}} - \mathrm{Id}) \overleftarrow{\Pi}_{\substack{j \in J_1 \\ j < j_k}} e^{A_j}, \quad (322) $$

which, together with the two first claims, proves estimate (321).

**Step 2: Construction of a limit.** We construct a possible limit. For each $n \ge 2$, let

$$ J_n := \left\{ j \in J, \|A_j\| > \frac{1}{n} \right\}. \quad (323) $$

Thanks to assumption (318), the sets $J_n$ are finite and, moreover,

$$ \varepsilon_n := \sum_{j \in J \setminus J_n} \|A_j\| \to 0. \quad (324) $$

Now, for each $n \ge 2$, we define the matrix

$$ M_n := \overleftarrow{\Pi}_{j \in J_n} e^{A_j}. \quad (325) $$

This defines a Cauchy sequence in the complete space $M_d(\mathbb{K})$. Indeed, for every $n < p$, thanks to estimate (321), one has

$$ \|M_n - M_p\| \le e^{3\alpha}\varepsilon_n. \quad (326) $$

Hence, there exists $M \in M_d(\mathbb{K})$ towards which the sequence $(M_n)_{n \ge 2}$ converges. By letting $[p \to \infty]$ in the previous inequality we obtain, for every $n \ge 2$

$$ \|M_n - M\| \le e^{3\alpha}\varepsilon_n. \quad (327) $$

**Step 3: Proof of convergence.** We now prove that the ordered product of the $e^{A_j}$ over $J$ converges to $M$ in the sense of Definition 129. Let $\varepsilon > 0$. Let $n \ge 2$ large enough such that $e^{3\alpha}\varepsilon_n < \varepsilon/2$. For every finite set $J_1$ containing $J_n$, condition (316) holds thanks to (327) and (321). $\square$

**Proposition 131.** Let $\mathcal{B}$ be a Hall basis of $\mathcal{L}(X)$, $(\xi_b)_{b \in \mathcal{B}}$ be the coordinates of the second kind associated to $\mathcal{B}$. There exists $\eta > 0$ such that the following property holds. Let $A_i \in M_d(\mathbb{K})$ for $i \in I$. For $b \in \mathcal{B}$, we define the matrix $A_b := \Lambda(b)$ where $\Lambda: \mathcal{L}(X) \to M_n(\mathbb{K})$ is the homomorphism of Lie algebras such that $\Lambda(X_i) = A_i$ for $i \in I$ (see Lemma 7). Let $T > 0$ and $u_i \in L^1((0, T); \mathbb{K})$ for $i \in I$. Assume that

$$ \|u\|_{L^1(0,T)} \|A\| \le \eta. \quad (328) $$

Then, for each $t \in [0, T]$ and $p \in \mathbb{K}^d$, the ordered product of the $e^{\xi_b(t, u)A_b}$ over $b \in \mathcal{B}$ converges. Moreover, for every $p \in \mathbb{K}^d$,

$$ x(t; A, u, p) = \overrightarrow{\prod}_{b \in \mathcal{B}} e^{\xi_b(t, u)A_b} p. \quad (329) $$
Proof. Let $\eta := 1/(8|I|^2)$. Let $T > 0$. Below, the variable $t$ implicitly belongs to $[0, T]$. To simplify the notations we write $\xi_b(t)$ instead of $\xi_b(t, u)$.

Step 1: Convergence of the ordered product of the $e^{\xi_b(t)A_b}$ over $b \in \mathcal{B}$. One obtains, by induction on $|b|$, that for every $b \in \mathcal{B}$, $\|A_b\| \le (2\|A\|)^{|b|}$. Thus, recalling (107),

$$ \|\xi_b(t)A_b\| \le (2\|A\|\|u\|_{L^1(0,t)})^{|b|}. \qquad (330) $$

Taking into account that $|\mathcal{B}_\ell| \le |I|^{\ell}$, we obtain, using (328),

$$ \sum_{b \in \mathcal{B}} \|\xi_b(t)A_b\| \le \sum_{\ell=1}^{+\infty} (2|I|\|A\|\|u\|_{L^1(0,t)})^{\ell} \le 1 \qquad (331) $$

and Lemma 130 gives the conclusion.

Step 2: Estimates along a Lazard elimination in $\mathcal{B}_{[1,M]}$. Let $M \in \mathbb{N}^*$. We adopt the notations $b_1, \dots, b_{k+1}$ and $Y_0, \dots, Y_{k+1}$ of Definition 53 and we define $x_0(t) := x(t)$ and, for $j \in [1, k+1]$

$$ x_j(t) := e^{-\xi_{b_j}(t, u)A_{b_j}} \dots e^{-\xi_{b_1}(t, u)A_{b_1}} x(t). \qquad (332) $$

We prove by induction on $j \in [0, k+1]$ that

$$ (\mathcal{H}_j): \begin{cases} \dot{x}_j(t) = \left( \sum_{b \in \mathcal{B}_{[1,M]} \cap Y_j} \dot{\xi}_b(t)A_b + \varepsilon_j(t) \right) x_j(t), \\ x_j(0) = p, \end{cases} \qquad (333) $$

where $\varepsilon_0 = 0$ and

$$ \|\varepsilon_j(t)\| \le (M|I|\|A\|\|u(t)\|(4|I|\|A\|\|u\|_{L^1(0,t)})^M + \|\varepsilon_{j-1}(t)\|) e^{2\|\xi_{b_j}(t)A_{b_j}\|}. \qquad (334) $$

First, $(\mathcal{H}_0)$ holds with $\varepsilon_0 = 0$ because $x_0(t) = x(t)$ and $\dot{\xi}_{X_i}(t) = u_i(t)$ for $i \in I$. Let $j \in [1, k+1]$ and assume that $(\mathcal{H}_{j-1})$ holds. We deduce from the definition of $x_j$ that

$$ x_j(t) = e^{-\xi_{b_j}(t)A_{b_j}} x_{j-1}(t) \qquad (335) $$

and from $(\mathcal{H}_{j-1})$ that

$$ \begin{aligned} \dot{x}_j(t) &= -\dot{\xi}_{b_j}(t)A_{b_j}x_j(t) + e^{-\xi_{b_j}(t)A_{b_j}} \left( \sum_{b \in \mathcal{B}_{[1,M]} \cap Y_{j-1}} \dot{\xi}_b(t)A_b + \varepsilon_{j-1}(t) \right) e^{\xi_{b_j}(t)A_{b_j}}x_j(t) \\ &= \left( \sum_{b \in \mathcal{B}_{[1,M]} \cap Y_{j-1} \setminus \{b_j\}} \dot{\xi}_b(t)e^{-\xi_{b_j}(t)A_{b_j}}A_b e^{\xi_{b_j}(t)A_{b_j}} + \tilde{\varepsilon}_{j-1}(t) \right) x_j(t) \end{aligned} \qquad (336) $$

where $\tilde{\varepsilon}_{j-1}(t) := e^{-\xi_{b_j}(t)A_{b_j}}\varepsilon_{j-1}(t)e^{\xi_{b_j}(t)A_{b_j}}$ satisfies,

$$ \|\tilde{\varepsilon}_{j-1}(t)\| \le \|\varepsilon_{j-1}(t)\| e^{2\|\xi_{b_j}(t)A_{b_j}\|}. \qquad (337) $$

For $b \in \mathcal{B}_{[1,M]} \cap Y_{j-1} \setminus \{b_j\}$, let $h(b) \in \mathbb{N}^*$ be the maximal integer such that (241) holds and

$$ \tilde{\varepsilon}_b^j(t) := \dot{\xi}_b(t) e^{-\xi_{b_j}(t)A_{b_j}} A_b e^{\xi_{b_j}(t)A_{b_j}} - \sum_{m=0}^{h(b)-1} \dot{\xi}_b(t) \frac{\xi_{b_j}^m(t)}{m!} A_{\text{ad}_{b_j}^m(b)}. \qquad (338) $$
Then, by definition of $Y_j$, ($\mathcal{H}_j$) holds with $\varepsilon_j$ defined by (245). Using the fourth statement of Lemma 90, (106), (330), we obtain

$$
\begin{align}
\|\bar{\varepsilon}_b^j(t)\| &\le |\dot{\varepsilon}_b(t)| \frac{(2\|\xi_{b_j}(t)A_{b_j}\|)^{h(b)}}{h(b)!} \|A_b\| e^{2\|\xi_{b_j}(t)A_{b_j}\|} \nonumber \\
&\le |b||u(t)||u\|_{L^1}^{|b|-1} \cdot \frac{(2\|A\|\|u\|_{L^1})^{h(b)|b_j|} 2^{h(b)}}{h(b)!} \cdot (2\|A\|)^{|b|} \tag{339} \\
&\le M|u(t)|(4\|A\|\|u\|_{L^1})^M \|A\|, \nonumber
\end{align}
$$

taking into account $M+1 \le |b| + h(b)|b_j| \le 2M$ and $\|A\|\|u\|_{L^1} \le 1$.

We deduce from (245), (337), (339) and the relation $|\mathcal{B}_{\ll 1, M}]| \le |I|^{M+1}$ that (334) holds.

Step 4: *Proof of an estimate on the ordered product of the $e^{\xi_b(t)A_b}$ over $\mathcal{B}_{\ll 1, M}]$. We deduce from (334), (331) and the relation $k+1 = |\mathcal{B}_{\ll 1, M}]| \le |I|^{M+1}$ that*

$$
\|\varepsilon_{k+1}(t)\| \le eM \|A\| |I|^2 |u(t)| (4|I|^2 \|A\| \|u\|_{L^1(0,t)})^M. \quad (340)
$$

Hence, using (328),

$$
\|\varepsilon_{k+1}\|_{L^1(0,t)} \le \frac{e}{4}(4|I|^2 \|A\| \|u\|_{L^1(0,t)})^{M+1} \le 2^{-M}. \quad (341)
$$

We deduce from ($\mathcal{H}_{k+1}$), (99) and Grönwall's lemma that

$$
\left| \left. \prod_{b \in \mathcal{B}_{\ll 1, M}] e^{-\xi_b(t, u)A_b} x(t) - p \right| \right| = |x_{k+1}(t) - p| \le \int_0^t |\varepsilon_{k+1}(\tau)x_{k+1}(\tau)| d\tau \le 2^{-M} e|p| \quad (342)
$$

Multiplying both sides by the finite product $\prod_{b \in \mathcal{B}_{\ll 1, M}] e^{\xi_b(t, u)A_b}$ gives

$$
\left| x(t) - \prod_{b \in \mathcal{B}_{\ll 1, M}] e^{\xi_b(t, u)A_b} p \right| \le e^2 2^{-M} |p| \quad (343)
$$

Passing to the limit $[M \to \infty]$ in the previous estimate gives (329). $\square$

<figure><img src="image_8.png" /></figure>

**Lack of global convergence.** The goal of this paragraph is to illustrate that the smallness assumption (328) in Proposition 131 is necessary because the equality does not hold globally.

**Proposition 132.** Consider the constant control $\bar{u} : t \in \mathbb{R}_+ \mapsto (1, 1) \in \mathbb{R}^2$.

1. There exist a Hall basis $\mathcal{B}$ of $\mathcal{L}(\{X_1, X_2\})$ and a subsequence $(b_k)_{k \in \mathbb{N}}$ of $\mathcal{B}$ such that

$$
\exists \gamma > 0, \forall k \in \mathbb{N}, t \ge 0, \quad \xi_{b_k}(t, \bar{u}) \ge \left(\frac{t}{\gamma}\right)^{|b_k|} \quad (344)
$$

2. There exists $A_1, A_2 \in \mathcal{M}_3(\mathbb{C})$ and $t > \gamma$ such that $(e^{\xi_{b_k}(t, \bar{u})A_{b_k}})_{k \in \mathbb{N}}$ does not converge to $\text{Id}_3$ in $\mathcal{M}_3(\mathbb{C})$. Thus, the ordered product of the $e^{\xi_b(t, \bar{u})A_b}$ over $\mathcal{B}$ does not converge in $\mathcal{M}_3(\mathbb{C})$.

*Proof.* For the first point we adapt an argument due to Sussmann in [100, pages 333-335]. We define by induction two sequences $(b_k^1)_{k \in \mathbb{N}}$ and $(b_k^2)_{k \in \mathbb{N}}$ of $\text{Br}(\{X_1, X_2\})$ by

$$
b_0^1 = X_1, \quad b_0^2 = X_2, \quad b_{k+1}^1 := [b_k^2, [b_k^1, b_k^2]], \quad b_{k+1}^2 := [b_k^1, [b_k^1, b_k^2]]. \quad (345)
$$

There exists a Hall basis of $\mathcal{L}(\{X_1, X_2\})$, whose order, denoted $<$, is compatible with length and such that, for every $k \in \mathbb{N}$, $b_k^1, b_k^2 \in \mathcal{B}$ and $b_k^1 < b_k^2$. It suffices to choose, on the brackets with length $3^k$, some order such that $b_k^1 < b_k^2$. Then, automatically, $[b_k^1, b_k^2] \in \mathcal{B}$ and thus $b_{k+1}^1, b_{k+1}^2 \in \mathcal{B}$. Such
a process indeed allows to construct a Hall basis (see Remark 55), provided that one chooses an arbitrary length-compatible order on all other brackets.

To lighten the notations, we write $\xi_b(t)$, instead of $\xi_b(t, \bar{u})$. We have $\xi_{X_1}(t) = \xi_{X_2}(t) = t$. An easy induction shows that, for every $b \in B$, $\dot{\xi}_b(t) = t^{|b|-1}/\alpha_b$, where $\alpha_b \in \mathbb{N}^*$. The constants $\alpha_b$ can be computed recursively: $\alpha_{X_1} = \alpha_{X_2} = 1$ and, if $b = \text{ad}_{b_1}^m(b_2)$ with $m \in \mathbb{N}^*$, $b_1 < b_2$ and $\lambda(b_2) < b_1$ then $\alpha_b = \alpha_{b_1}^m|b_1|^m m! \alpha_{b_2}$. In particular, for every $k \in \mathbb{N}$,

$$ \alpha_{b_{k+1}^1} = \alpha_{b_k^1} \alpha_{b_k^2}^2 |b_k^2| |b_k^1| = \alpha_{b_k^1} \alpha_{b_k^2}^2 3^{2k}, \quad \alpha_{b_{k+1}^2} = 2\alpha_{b_k^1}^2 \alpha_{b_k^2} |b_k^1|^2 = 2\alpha_{b_k^1}^2 \alpha_{b_k^2}^2 3^{2k}. \quad (346) $$

Let $\beta_k = \max\{\alpha_{b_k^1}, \alpha_{b_k^2}\}$. Then, $\beta_0 = 1$ and, by the previous relations,

$$ \beta_{k+1} \le 3^{2k+1} \beta_k^3. \quad (347) $$

Thus $\theta_k := 3^{-k} \ln(\beta_k)$ satisfies $\theta_0 = 0$ and

$$ \theta_{k+1} \le \theta_k + (2k+1)3^{-(k+1)} \ln(3), \quad (348) $$

which leads to $\theta_k \le \eta := \sum_{j=1}^{+\infty} (2j+1)3^{-(j+1)} \ln(3)$ i.e. $\beta_k \le (\gamma')^{3k}$ where $\gamma' = e^\eta$. Therefore, for every $k \in \mathbb{N}$ and $j \in \{1, 2\}$ we have

$$ \left| \xi_{b_k^j}(t) \right| \ge \frac{1}{|b_k^j|} \left( \frac{t}{\gamma'} \right)^{|b_k^j|} . \quad (349) $$

Let $\gamma > \gamma'$ be such that, for every $k \in \mathbb{N}$, $\frac{1}{3^k} \left(\frac{\gamma}{\gamma'}\right)^{3k} \ge 1$. Then (344) holds, for instance with $b_k = b_k^1$.

For the second point, let, for $j \in \{1, 2, 3\}$, $F_j \in \mathcal{M}_3(\mathbb{R})$ be the matrix of the linear map $x \in \mathbb{R}^3 \mapsto e_j \wedge x$. Then $[F_1, F_2] = F_3$, $[F_2, F_3] = F_1$ and $[F_3, F_1] = F_2$. In particular

$$ [F_2, [F_1, F_2]] = F_1, \quad [F_1, [F_1, F_2]] = -F_2. \quad (350) $$

We consider $A_1 = e^{i \frac{\pi}{6}} F_1$ and $A_2 = e^{i \frac{\pi}{6}} F_2$ in $\mathcal{M}_3(\mathbb{C})$. One easily proves by induction on $k \in \mathbb{N}^*$ that $A_{b_k^1} = (-1)^{k+1} i F_1$ and $A_{b_k^2} = -i F_2$. We have, for every $k \in \mathbb{N}$ and $t \in \mathbb{R}$

$$ e^{\xi_{b_k}(t)A_{b_k}} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cosh(\xi_{b_k}(t)) & i(-1)^k \sinh(\xi_{b_k}(t)) \\ 0 & i(-1)^{k+1} \sinh(\xi_{b_k}(t)) & \cosh(\xi_{b_k}(t)) \end{pmatrix} \quad (351) $$

By (344), this sequence of matrices diverges for every $t > \gamma$. $\square$

### 5.4.3 Investigation for analytic vector fields

In this paragraph, we study affine systems (226). Our goal is to explain the difficulty of the convergence question for Sussmann’s infinite product for arbitrary analytic vector fields. First, we state a definition (Definition 133) and a sufficient condition for the convergence (Lemma 134), in the same spirit as for matrices. Then we show that they do not provide convergence for general analytic vector fields and we formulate an open problem.

**Definition 133.** Let $J$ be a totally ordered set, $\delta > 0$ and $(f_j)_{j \in J}$ a family of $\mathbb{C}^1(B_{2\delta}; \mathbb{K}^d)$. We say that the ordered product of the $e^{f_j}$ over $J$ converges uniformly on $B_\delta$ if there exists $g \in \mathbb{C}^0(B_\delta; \mathbb{K}^d)$
such that, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $J$ such that, for every finite subset $J_1$ of $J$ containing $J_0$, and for every $p \in B_\delta$ one has

$$ \left\| g(p) - \left. \prod_{j \in J_1} e^{f_j} p \right| \right\| \le \varepsilon. \qquad (352) $$

When such a $g$ exists, it is unique and we write

$$ g = \left. \prod_{j \in J} e^{f_j} \right\}. \qquad (353) $$

**Lemma 134.** Let $J$ be a totally ordered set, $\delta > 0$ and $(f_j)_{j \in J}$ a family of $C^1(B_{2\delta}; \mathbb{K}^d)$ such that

$$ \sum_{j \in J} \|f_j\|_{C^0} < \delta \quad \text{and} \quad \alpha := \sum_{j \in J} \|f_j\|_{C^1} < \infty. \qquad (354) $$

Then the ordered product of the $e^{f_j}$ over $J$ converges uniformly on $B_\delta$ and is $e^\alpha$-Lipschitz.

*Proof.* We proceed as in the proof of Lemma 130.

**Step 1: Basic claims.** First, for every finite subset $J' \subset J$ and $p \in \mathbb{K}^d$ with $|p| \le 2\delta - \sum_{j \in J'} \|f_j\|_{C^0}$, then

$$ \left. \prod_{j \in J'} e^{f_j} p \right| \in B_{2\delta} \quad \text{and} \quad \left\| \partial_p \left[ \left. \prod_{j \in J'} e^{f_j} p \right| \right] \right\| \le \prod_{j \in J'} e^{\|f_j\|_{C^0}} \le e^\alpha \qquad (355) $$

because of Lemma 85 and the chain rule.

Second, for every finite parts $J_0 \subset J_1 \subset J$ and $p \in \mathbb{K}^d$ with $|p| \le 2\delta - \sum_{j \in J_1} \|f_j\|_{C^0}$ one has

$$ \left\| \left. \prod_{j \in J_1} e^{f_j} p \right| - \left. \prod_{j \in J_0} e^{f_j} p \right| \right\| \le e^{\alpha} \sum_{j \in J_1 \setminus J_0} \|f_j\|_{C^0}. \qquad (356) $$

Indeed, writing $J_1 \setminus J_0 = \{j_1 > \dots > j_n\}$, we have the following telescopic decomposition

$$ \left. \prod_{j \in J_1} e^{f_j} p - \prod_{j \in J_0} e^{f_j} p \right| = \sum_{k=1}^{n} \left\{ \left( \left. \prod_{\substack{j \in J_0 \\ j > j_k}} e^{f_j} \right| e^{f_{j_k}} \left( \left. \prod_{\substack{j \in J_1 \\ j < j_k}} e^{f_j} \right| p - \left( \left. \prod_{\substack{j \in J_0 \\ j > j_k}} e^{f_j} \right| \left( \left. \prod_{\substack{j \in J_1 \\ j < j_k}} e^{f_j} \right| p \right) \right) \right| \right\}. \qquad (357) $$

For $k \in [1, n]$, let $x_k := \left. \prod_{\substack{j \in J_1 \\ j < j_k}} e^{f_j} p \right|$ which is a point in $B_{2\delta - \|f_{j_k}\|_{C^0}}$. By (355) and (149), the term with index $k$ in the previous sum is bounded by

$$ \left| \left( \left. \prod_{\substack{j \in J_0 \\ j > j_k}} e^{f_j} \right| e^{f_{j_k}} x_k - \left( \left. \prod_{\substack{j \in J_0 \\ j > j_k}} e^{f_j} \right| x_k \right) \right| \le e^{\alpha} \left| e^{f_{j_k}} x_k - x_k \right| \le e^{\alpha} \|f_{j_k}\|_{C^0}. \qquad (358) $$

which, together with (357) proves (356).

**Step 2: Construction of a limit.** We construct a possible limit. For each $n \ge 2$, let

$$ J_n := \left\{ j \in J, \|f_j\|_{C^1} > \frac{1}{n} \right\}. \qquad (359) $$

Thanks to assumption (354), the sets $J_n$ are finite and, moreover,

$$ \varepsilon_n := \sum_{j \in J \setminus J_n} \|f_j\|_{C^1} \to 0. \qquad (360) $$
Now, for each $n \ge 2$, we define $g_n \in C^0(B_\delta; \mathbb{K}^d)$ by

$$g_n(p) := \left. \prod_{j \in J_n} e^{f_j} p. \right\} \qquad (361)$$

This defines a Cauchy sequence in the complete space $C^0(B_\delta; \mathbb{K}^d)$. Indeed, for every $n < n'$ and $p \in B_\delta$, thanks to estimate (356), one has

$$|g_n(p) - g_{n'}(p)| \le e^{\alpha} \varepsilon_n. \qquad (362)$$

Hence, there exists $g \in C^0(B_\delta; \mathbb{K}^d)$ towards which the sequence $(g_n)_{n \ge 2}$ uniformly converges on $B_\delta$. By (355), $g_n$ is $e^\alpha$-Lipschitz on $B_\delta$ for every $n \in \mathbb{N}$, thus so is $g$. By letting $[n' \to \infty]$ in the previous inequality we obtain, for every $n \ge 2$ and $p \in B_\delta$

$$|g_n(p) - g(p)| \le e^{\alpha} \varepsilon_n. \qquad (363)$$

**Step 3: Proof of convergence.** We now prove that the ordered product of the $e^{f_j}$ over $J$ converges uniformly to $g$ on $B_\delta$ in the sense of Definition 133. Let $\varepsilon > 0$. Let $n \ge 2$ large enough such that $e^\alpha \varepsilon_n < \varepsilon/2$. For every finite set $J_1$ containing $J_n$, condition (352) holds thanks to (363) and (356). $\square$

Now, let us emphasize that, by using estimates on $\xi_b(t, u)$ and $f_b$ depending only on the length of the Lie bracket $b$, it is not possible to prove the convergence of $\sum |\xi_b(t, u)||f_b||_{C^1}$, where the sum ranges over $b \in \mathcal{B}$, an arbitrary Hall basis of $\mathcal{L}(X)$.

On the one hand, one easily proves by induction on $|b|$ that, for every $b \in \mathcal{B}$ and $u \in L^\infty$ with $\|u\|_{L^\infty} \le 1$, there holds $|\xi_b(t, u)| \le t^{|b|}$. However, by the first statement of Proposition 132, when $X$ contains at least two indeterminates, there are Hall bases (even compatible with length) for which one may not expect an upper bound, function of $|b|$ alone, that behaves better than geometrically. Hence, we should consider the $t^{|b|}$ bound to be sharp, when one restricts to bounds depending only on $|b|$.

On the other hand, if the vector fields are locally analytic, there exists $r, \delta > 0$ such that $f_i \in C^{\omega,r}(B_\delta; \mathbb{K}^d)$ for $i \in I$. By (140) with $r_1 \leftarrow r$ and $r_2 \leftarrow r/e$ for every $b \in \mathcal{B}$,

$$\|f_b\|_{C^1} \le \left(1 + \frac{e}{r}\right) (|b| - 1)! \left(\frac{9}{r}\right)^{|b|-1} F^{|b|}, \qquad (364)$$

where $F := \max_{i \in I} |||f_i|||_r$. However, by Remark 81, the dependence in $(|b| - 1)!$ is optimal (again, if one restricts to bounds involving only $|b|$).

We deduce from the previous estimates that there exists $C > 0$ such that

$$|\xi_b(t, u)||f_b||_{C^1} \le (Ct)^{|b|} |b|! \qquad (365)$$

This bound does not provide the convergence of the considered series. Indeed, for every $t > 0$, $(Ct)^{|b|} |b|! \to +\infty$ as $|b| \to +\infty$, so an argument depending on $|b|$ alone doesn't even prove that the general term tends to zero.

To prove the convergence of Sussmann's infinite product expansion, one therefore either needs a better sufficient condition than Lemma 134 or one needs to prove estimates on $\xi_b$ and $f_b$ that take into account the structure of the bracket $b$, and not only its length.

**Open problem 135.** Does Sussmann's infinite product converge for analytic vector fields?

In Section 6.4, we prove the convergence (for analytic vector fields) of some infinite subproducts, by applying Lemma 134 with estimates on $\xi_b$ that depend on the structure of $b$.
# 6 Error estimates for control systems

In this section, we consider control-affine systems with drift, i.e. of the form

$$ \dot{x}(t) = f_0(x(t)) + \sum_{i=1}^{q} u_i(t) f_i(x(t)) \quad \text{and} \quad x(0) = p, \qquad (366) $$

where $f_0, \dots, f_q$ are vector fields and $u = (u_1, \dots, u_q) \in L^1(\mathbb{R}; \mathbb{K}^q)$. When well-defined, the solution is denoted $x(t; f, u, p)$ where $f = (f_0, \dots, f_q)$ and $u = (u_1, \dots, u_q)$.

We prove error formulas at every order in $\|u\|_{L^1}$ for the Chen-Fliess expansion, the Magnus expansion in the interaction picture and for Sussmann's infinite product expansion. In each case, the error formula involves an infinite sum or an infinite product which turns out to be well-defined. We also propose a counter-example for the validity of such error estimates for the usual Magnus expansion, for which the infinite sum involved is not well-defined.

## 6.1 Chen-Fliess expansion

The convergence of the Chen-Fliess series, for control affine systems (366) with analytic vector fields, under a smallness assumption on $t$ and a uniform bound on $u$, is a classical result, see for instance [36, Proposition 3.37] or [99, Proposition 4.3]. In this section we prove the convergence of the Chen-Fliess expansion, (Proposition 136) under a smallness assumption on $\|u\|_{L^1}$. We also generalize the Chen-Fliess expansion to nonlinear systems (not necessarily affine) with scalar input (Proposition 137), because this fact will be used in Section 7.2.

In the following statement $q \in \mathbb{N}^*$, $I = [0, q]$. For a word $\sigma = \sigma_1 \cdots \sigma_\ell \in I^*$, with $\ell \in \mathbb{N}^*$, $\sigma_1, \dots, \sigma_\ell \in I$, and vector fields $f_0, f_1, \dots, f_q$, we denote by $f_\sigma$ the differential operator $f_{\sigma_1} \cdots f_{\sigma_\ell}$ (with the notations of Remark 74). For $t > 0$ and $u = (u_1, \dots, u_q) \in L^1(0, t)$, the quantity $\int_0^t u_\sigma$ is defined in (19), with $u_0 = 1$.

**Proposition 136.** Let $\delta, r > 0$ and $f_0, f_1, \dots, f_q \in C^{\omega,r}(B_{2\delta}; \mathbb{K}^d)$. There exists $\eta > 0$ such that, for every $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, $t \in [0, \eta]$ and $u \in L^1((0, t); \mathbb{K}^q)$ such that $\|u\|_{L^1} \le \eta$ and $p \in B_\delta$, then

$$ \varphi(x(t; f, u, p)) = \sum_{\sigma \in I^*} \left( \int_0^t u_\sigma \right) (f_\sigma \varphi)(p) \qquad (367) $$

where the sum converges absolutely, uniformly with respect to $(t, u, p)$. Moreover, for every $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, there exists $C > 0$ such that, for every $M \in \mathbb{N}$, $p \in B_\delta$, $t \in [0, \eta]$ and $u \in L^1((0, t); \mathbb{K}^q)$ such that $\|u\|_{L^1} \le \eta$, then

$$ \left| \varphi(x(t; f, u, p)) - \sum_{n(\sigma) \le M} \left( \int_0^t u_\sigma \right) (f_\sigma \varphi)(p) \right| \le (C \|u\|_{L^1})^{M+1}, \qquad (368) $$

where the sum ranges over words $\sigma \in I^*$ such that the number of non-zero letters is at most $M$.

*Proof.* For $\sigma = \sigma_1 \cdots \sigma_\ell \in I^*$, let $n(\sigma)$ be the number of non-zero letters in $\sigma$, i.e. $n(\sigma) = |\{i \in [1, \ell]; \sigma_i \neq 0\}|$ and $n_0(\sigma)$ be the number of occurrences of the letter zero in $\sigma$, i.e. $n_0(\sigma) = |\{i \in [1, \ell]; \sigma_i = 0\}|$. Then $\ell = n(\sigma) + n_0(\sigma)$. One proves by induction on the length $\ell$ of $\sigma \in I^*$ the following estimate, for every $t > 0$ and $u \in L^1((0, t); \mathbb{K}^q)$,

$$ \left| \left( \int_0^t u_\sigma \right) \right| \le \frac{\|u\|_{L^1(0,t)}^{n(\sigma)}}{n(\sigma)!} \frac{t^{n_0(\sigma)}}{n_0(\sigma)!}. \qquad (369) $$
Let $\|f\| = \sum_{i=0}^{q} \|f_i\|_r$, $\eta = r/(10\|f\|)$, $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, $t \in [0, \eta]$ and $u \in L^1((0, t); \mathbb{K}^q)$ such that $\|u\|_{L^1(0,t)} = \sum_{i=1}^{q} \|u_i\|_{L^1(0,t)} \le \eta$ and $p \in B_\delta$. Using (369) and (135), we get

$$ \left| \left( \int_0^t u_\sigma \right) (f_\sigma \varphi) (p) \right| \le \|u\|_{L^1(0,t)}^{n(\sigma)} t^{n_0(\sigma)} \left( \frac{10}{r} \|f\| \right)^\ell \|\varphi\|_r \quad (370) $$

which proves the absolute convergence of the sum in (367), uniformly with respect to $(t, u, p)$

The proof of the equality in (367) consists in applying (255) to $f(t, x) = f_0(x) + \sum_{i=1}^q u_i(t)f_i(x)$. In particular the sum involved in (368) is the Taylor expansion of order $M$ of $u \mapsto \varphi(x(t; f, u, p))$ at $u=0$. By adapting Lemma 83 to affine systems with $L^1$ controls, we get the real-analyticity of the map $u \mapsto \varphi(x(t; f, u, p))$ on $B_{L^1(0,t)}(0, \eta)$ uniformly with respect to $(t, p) \in [0, \eta] \times B_\delta$ which ends the proof of (368). $\square$

The last statement of this section focuses on nonlinear control systems with scalar input

$$ \dot{x} = f(x, u) \quad (371) $$

where $f: \mathbb{K}^d \times \mathbb{K} \to \mathbb{K}^d$, When well-defined, the solution of this ODE, with initial condition $x(0) = p$ is denoted $x(t; f, u, p)$. We introduce the notation

$$ \int_0^t u^k := \int_{\Delta^n(t)} u(\tau_1)^{k_1} \cdots u(\tau_n)^{k_n} d\tau \quad (372) $$

for every $t > 0$, $u \in L^1((0, t); \mathbb{K})$, and every multi-index $k = (k_1, \dots, k_n) \in \mathbb{N}^n$ with $n \in \mathbb{N}^*$.

**Proposition 137.** Let $r, \delta, \delta_u > 0$, $f \in C^{\omega,r}(B_{2\delta} \times [-\delta_u, \delta_u]; \mathbb{K}^d)$ and $f_k := \frac{1}{k!} \partial_u^k f(\cdot, 0)$ for every $k \in \mathbb{N}$. There exists $T^*, \eta > 0$ such that, for every $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, $t \in [0, T^*]$, $u \in L^\infty((0, t); \mathbb{K})$ with $\|u\|_{L^\infty} \le \eta$ and $p \in B_\delta$, with the notations of Remark 74,

$$ \varphi(x(t; f, u, p)) = \sum_{\substack{n \in \mathbb{N} \\ k \in \mathbb{N}^n}} \left( \int_0^t u^k \right) (f_{k_1} \cdots f_{k_n}) (\varphi)(p), \quad (373) $$

where the sum converges absolutely, uniformly with respect to $(t, u, p)$. Moreover, for every $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, there exists $C > 0$ such that, for every $M \in \mathbb{N}$, $t \in [0, T^*]$, $u \in L^\infty((0, t); \mathbb{K})$ with $\|u\|_{L^\infty} \le \eta$ and $p \in B_\delta$

$$ \left| \varphi(x(t; f, u, p)) - \sum_{\substack{n \in \mathbb{N} \\ k \in \mathbb{N}^n, |k| \le M}} \left( \int_0^t u^k \right) (f_{k_1} \cdots f_{k_n}) (\varphi)(p) \right| \le (C \|u\|_{L^\infty})^{M+1} \quad (374) $$

where the sum is taken over $n \in \mathbb{N}$ and $k = (k_1, \dots, k_n) \in \mathbb{N}^n$ such that $k_1 + \dots + k_n \le M$.

*Proof.* We define $r' = r/e$,

$$ T^* := \min \left\{ \frac{r'}{10 \|f\|_r}, \frac{\delta}{\|f\|_{C^0}} \right\}, \quad \eta := \min \left\{ \delta_u, \frac{r}{10} \right\}. \quad (375) $$

Let $\varphi \in C^{\omega,r}(B_{2\delta}; \mathbb{K})$, $t \in [0, T^*]$, $u \in L^\infty((0, t); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le \eta$ and $p \in B_\delta$. Then $x(t; f, u, p) \in B_{2\delta}$.

**Step 1: Uniform absolute convergence of the sum in (373).** Using the iterated version of (127) and (116), we get, for every $k \in \mathbb{N}$,

$$ \|f_k\|_{r'} \le \frac{1}{k!} \left( \frac{k}{r-r'} \right)^k \|f\|_r \le \left( \frac{e}{r-r'} \right)^k \|f\|_r \le \left( \frac{5}{r} \right)^k \|f\|_r. \quad (376) $$
For every $n \in \mathbb{N}^*$ and $k_1, \dots, k_n \in \mathbb{N}$, we have, using (135) and (376)

$$ \begin{aligned} |(f_{k_1} \cdots f_{k_n})(\varphi)(p)| &\le n! \left(\frac{5}{r'}\right)^n \|f_{k_n}\|_{r'} \cdots \|f_{k_1}\|_{r'} \|\varphi\|_{r'} \\ &\le n! \left(\frac{5}{r'}\right)^n \left(\frac{5}{r}\right)^{k_1+\cdots+k_n} \|f\|_r^n \|\varphi\|_{r'} \end{aligned} \quad (377) $$

and

$$ \left| \int_0^t u^k \right| = \left| \int_{\Delta^n(t)} u(\tau_1)^{k_1} \cdots u(\tau_n)^{k_n} d\tau \right| \le \frac{t^n}{n!} \|u\|_{L^\infty}^{k_1+\cdots+k_n}. \quad (378) $$

By definition of $T^*$ and $\eta$ we have $\frac{5t}{r'} \|f\|_r \le \frac{1}{2}$ and $\frac{5}{r} \|u\|_{L^\infty} \le \frac{1}{2}$, which gives the conclusion.

**Step 2: Equality in (373) and error formula (374).** We have $f(\cdot, u) = \sum_{j=0}^{+\infty} u^j f_j$ with convergence in $C^{\omega,r'}(B_{2\delta}; \mathbb{K}^d)$ uniformly with respect to $u \in B_{\mathbb{K}^q}(0, \eta)$. Thus, the equality (373) is a consequence of Fubini theorem and (255) applied to $(t, x) \mapsto f(x, u(t))$. In particular the finite sum involved in (374) is the Talyor expansion of order $M$ of $u \mapsto \varphi(x(t; f, u, p))$ at $u = 0$. By Lemma 83 $u \mapsto \varphi(x(t; f, u, p))$ is analytic on $B_{L^\infty(0,T^*)}(0, \eta)$ uniformly with respect to $(t, p) \in [0, T^*] \times B_\delta$, which ends the proof of (374). $\square$

## 6.2 Magnus expansion in the usual setting: a counter-example

Contrary to other expansions, the usual Magnus expansion does not yield, in general, error estimates involving the size of the control. Indeed, the infinite segments which would need to be summed do not converge, even for analytic vector fields, arbitrarily small times and even when the drift vector field vanishes at the origin. The following statement illustrates that even the series defining the terms which are linear with respect to the control does not converge.

**Proposition 138.** Let $d := 2$. There exists $T, \delta > 0$, $f_0, f_1 \in C^{\omega, \delta}(B_\delta; \mathbb{K}^d)$ with $f_0(0) = 0$ and a control $u \in C^\infty([0, T])$, such that, if one defines, for $t \in (0, T)$, the sequence of vector fields

$$ F_n(t) := \sum_{k=0}^{n} \zeta_{\mathrm{ad}_{X_0}^k}(X_1)(t, u) \mathrm{ad}_{f_0}^k(f_1), \quad (379) $$

then, for each $\delta^* \in (0, \delta)$ and $t \in (0, T)$, $F_n(t) \in C^\infty(B_\delta; \mathbb{K}^d)$ does not converge in $C^0(B_{\delta^*}; \mathbb{K}^d)$.

*Proof.* We define the following vector fields for $x \in \mathbb{R}^2$ with $|x| < 1$,

$$ f_0(x) := x_2 e_1 \quad \text{and} \quad f_1(x) := \frac{1}{1-x_1} e_2. \quad (380) $$

Then,

$$ \mathrm{ad}_{f_0}^k(f_1)(x) = x_2^k \partial_1^k \left( \frac{1}{1-x_1} \right) e_2 = \frac{k! x_2^k}{(1-x_1)^{k+1}} e_2. \quad (381) $$

We now choose the particular control $u(t) := t$ for $t \in (0, T)$ with $T = 1$ (the simpler choice, $u(t) := 1$, would not produce a diverging counter-example). Using the expression (71) from Example 40 for the coordinates of the first kind along the brackets $\mathrm{ad}_{X_0}^k(X_1)$ for this particular control, we obtain, for $t \in (0, T)$,

$$ F_n(t)(x) = \sum_{k=0}^{n} (-1)^{k+1} t^{k+2} \frac{B_{k+1}}{k+1} \frac{x_2^k}{(1-x_1)^{k+1}}. \quad (382) $$

Thus, for each $t, \delta^* > 0$, the sequence of vector fields $F_n(t)$ does not converge in $C^0(B_{\delta^*}; \mathbb{K}^d)$, since for every $x_2 \ne 0$, the general term of the series does not tend to zero because of the asymptotic (531) for Bernoulli numbers. $\square$
## 6.3 Magnus expansion in the interaction picture

The following statement is an immediate consequence of Proposition 99. It illustrates that, contrary to the classical Magnus expansion, our “Magnus in the interaction picture” expansion allows to obtain error estimates involving the size of the control, at any order.

**Proposition 139.** Let $M \in \mathbb{N}$, $\delta > 0$, $T > 0$, $f_0 \in C^{M^2+1}(B_{5\delta}; \mathbb{K}^d)$ with $T \|f_0\|_{C^0} \le \delta$ and $f_1, \dots, f_q \in C^{M^2}(B_{5\delta}; \mathbb{K}^d)$. There exists $\gamma, C > 0$ such that, for every $u = (u_1, \dots, u_q) \in L^1((0,T); \mathbb{K})$ with

$$ \|u\|_{L^1} \le \gamma \qquad (383) $$

$p \in B_\delta$ and $t \in [0, \gamma]$ then

$$ \left| x(t; f, u, p) - e^{\mathcal{Z}_M(t, f, u)} e^{tf_0} p \right| \le C \|u\|_{L^1(0,t)}^{M+1}. \qquad (384) $$

In (384), $\mathcal{Z}_M(t, f, u)$ (where implicitly $f = (f_0, f_1, \dots, f_q)$ and $u = (u_1, \dots, u_q)$) is a notation for the vector field $\mathcal{Z}_M(t, f_0, \sum_{i=1}^q u_i f_i)$, defined in Definition 98 for the affine perturbation $f_\sharp(t, x) = \sum_{i=1}^q u_i(t) f_i(x)$. This notation is chosen by analogy with Theorem 41.

## 6.4 Sussmann’s infinite product expansion

The goal of this section is to prove Proposition 143 which states that, despite the difficulties mentioned in Section 5.4.3 concerning the full convergence of Sussmann’s infinite product expansion, some (infinite) subproducts of it do converge and yield error estimates at every order in the size of the control for control-affine systems with drift of the form (366).

We start with an elementary remark (Lemma 140) on the structure of brackets of a Hall set which allows to prove nice asymmetric estimates on the associated coordinates of the second kind (see Lemma 141). The following result proves that, when one tries to factorize the lateral $X_0$ factors outside of a bracket of a Hall set $B \subset \text{Br}(X)$ with $X_0 \in X$, these $X_0$ factors cannot alternate sides more than once.

**Lemma 140.** Let $q \in \mathbb{N}^*$, $X = \{X_0, X_1, \dots, X_q\}$ and $B \subset \text{Br}(X)$ be a Hall set. For each $b \in B$, there exist $m, \bar{m} \in \mathbb{N}$ such that

$$ b = \text{ad}_{X_0}^m \text{ad}_{X_0}^{\bar{m}}(b^*), \qquad (385) $$

where $\text{ad}_{X_0}^{\bar{m}}$ denotes the iterated right bracketing $\bar{m}$ times by $X_0$ and $b^* \in B$ is such that either $b^* \in X$ or $b = (b_1, b_2)$ with $b_1 \ne X_0$ and $b_2 \ne X_0$.

*Proof.* The key point is that, by the third condition in Definition 54, for each $b \in B \setminus X$, $\lambda(b) < b$. Let $b \in B$. We disjunct cases.

* If $b \in X$ or $(\lambda(b) \ne X_0$ and $\mu(b) \ne X_0)$, then (385) holds with $m = \bar{m} = 0$ and $b^* = b$.

* If $\lambda(b) = X_0$, there exists a unique $m \in \mathbb{N}^*$ and $\tilde{b} \in B$ such that $b = \text{ad}_{X_0}^m(\tilde{b})$ where $\tilde{b} \in X$ or $\lambda(\tilde{b}) \ne X_0$.

- If $\tilde{b} \in X$ or $\mu(\tilde{b}) \ne X_0$, (385) holds with $\bar{m} = 0$ and $b^* = \tilde{b}$.

- Otherwise, there exists a unique $\bar{m} \in \mathbb{N}^*$ and $b^* \in B$ such that $\tilde{b} = \text{ad}_{X_0}^{\bar{m}}(b^*)$ where $b^* \in X$ or $\mu(b^*) \ne X_0$.

* If $b^* \in X$, (385) holds.

* Else $\mu(b^*) \ne X_0$. one has $\lambda(b^*) < b^*$ as recalled. Moreover, since $\bar{m} \ge 1$, $(b^*, X_0) \in B$ so $b^* < X_0$ (by the second point of Definition 54). Hence $\lambda(b^*) < X_0$. So we also have $\lambda(b^*) \ne X_0$ and (385) holds.
• If $\mu(b) = X_0$, there exists a unique $\bar{m} \in \mathbb{N}^*$ and $\tilde{b} \in B$ such that $b = \text{ad}_{X_0}^{\bar{m}}(\tilde{b})$ where $\tilde{b} \in X$ or $\mu(\tilde{b}) \neq X_0$.

- If $\tilde{b} \in X$, (385) holds with $m = 0$ and $b^* = \tilde{b}$.

- Else $\mu(\tilde{b}) \neq X_0$. Since $\bar{m} \ge 1$, $(\tilde{b}, X_0) \in B$, so $\tilde{b} < X_0$. Since $\lambda(\tilde{b}) < \tilde{b}$, this proves $\lambda(\tilde{b}) \neq X_0$. So (385) holds with $m = 0$ and $b^* = \tilde{b}$.

Hence, the decomposition (385) always holds. $\square$

We now turn to asymmetric estimates for the coordinates of the second kind, which, contrary to Lemma 63, isolate the role of $X_0$ associated with the implicit control $u_0 = 1$.

**Lemma 141.** Let $q \in \mathbb{N}^*$, $X = \{X_0, X_1, \dots, X_q\}$, $B \subset \text{Br}(X)$ a Hall set and $(\xi_b)_{b \in B}$ the associated coordinates of the second kind. For every $k \in \mathbb{N}^*$, there exists $c_k \ge 1$ such that, for each $b \in B$ with $n(b) = k$, $T > 0$, $u \in L^1((0, T); \mathbb{K}^q)$ and $t \in [0, T]$,

$$|\xi_b(t, 1, u)| \le \|u\|_{L^1(0,t)}^k \frac{(c_k t)^{n_0(b)}}{n_0(b)!} \quad (386)$$

and

$$|\dot{\xi}_b(t; 1, u)| \le \begin{cases} k|u(t)||u|_{L^1(0,t)}^{k-1} & \text{when } n_0(b) = 0, \\ \|u|_{L^1(0,t)}^{k-1} (kt|u(t)| + n_0(b)\|u\|_{L^1(0,t)}) \frac{c_k(c_k t)^{n_0(b)-1}}{n_0(b)!} & \text{when } n_0(b) > 0. \end{cases} \quad (387)$$

*Proof.* In this proof, we write $\xi_b(t)$ instead of $\xi_b(t, 1, u)$ by concision for the value at time $t \in [0, T]$ of the coordinate of the second kind associated with the control $u_0 = 1$ and $u_i$ for $i \in [1, q]$. First, when (387) holds on $[0, T]$, then so does (386) by time-integration (with the same constant). Hence, we only need to prove the bound on the time derivative of the coordinates.

**Step 1:** Persistence of the estimates by right bracketing with $X_0$. Let $k \in \mathbb{N}^*$ and $b \in B$ such that $n(b) = k$. We assume that (386) holds and we prove that $\tilde{b} := (b, X_0)$ satisfies both estimates with the same constant. Since $\dot{\xi}_{X_0}(t) = 1$, we have

$$|\dot{\xi}_b(t)| = |\xi_b(t)\dot{\xi}_{X_0}(t)| \le \|u\|_{L^1(0,t)}^k \frac{(c_k t)^{n_0(b)}}{n_0(b)!}. \quad (388)$$

Hence $\tilde{b}$ satisfies (387) (and (386) by integration) because $c_k \ge 1$ and $n_0(\tilde{b}) > 0$.

**Step 2:** Persistence of the estimates by arbitrary long left bracketing with $X_0$, up to $c_k \leftarrow 2c_k$. Let $k \in \mathbb{N}^*$ and $b \in B$ with $n(b) = k$. We assume that (387) holds and we prove that, for every $m \in \mathbb{N}^*$, $\tilde{b} := \text{ad}_{X_0}^m(b)$ satisfies both estimates with a constant $c_k \leftarrow 2c_k$. If $n_0(b) = 0$, it is straightforward to check that $\tilde{b}$ satisfies (387) with $c_k \leftarrow 1$. If $n_0(b) = 1$, we have

$$\begin{align} |\dot{\xi}_b(t)| &= \frac{1}{m!} |\xi_{X_0}^m(t) \dot{\xi}_b(t)| \nonumber \\ &\le \frac{t^m}{m!} \|u\|_{L^1}^{k-1} (kt|u(t)| + n_0(b)\|u\|_{L^1}) \frac{c_k(c_k t)^{n_0(b)-1}}{n_0(b)!} \tag{389} \\ &\le \|u\|_{L^1}^{k-1} (kt|u(t)| + (m+n_0(b))\|u\|_{L^1}) 2^{m+n_0(b)} c_k^{n_0(b)} \frac{t^{m+n_0(b)-1}}{(m+n_0(b))!} \nonumber \\ &\le \|u\|_{L^1}^{k-1} (kt|u(t)| + n_0(\tilde{b})\|u\|_{L^1}) (2c_k)^{n_0(\tilde{b})} \frac{t^{n_0(\tilde{b})-1}}{n_0(\tilde{b})!} \nonumber \end{align}$$

because $n_0(\tilde{b}) = m + n_0(b)$ and $c_k \ge 1$. So $\tilde{b}$ satisfies (387) with a constant $c_k \leftarrow 2c_k$.
Step 3: Proof of the estimates by induction on $k \in \mathbb{N}^*$.

Initialization for $k = 1$. For $i \in [1, q]$, $\dot{\xi}_{X_i}(t) = u_i(t)$ so both estimates are satisfied with constant 1 when $b \in \{X_1, \dots, X_q\}$. By Lemma 140, Step 1 and Step 2, we deduce that (386) and (387) hold for $k = 1$ with $c_1 = 2$.

Induction ($k-1 \to k$). Let $k \ge 2$ and let us assume that the estimates are proved for every $b \in \mathcal{B}$ with $n(b) \le (k-1)$. Let $b \in \mathcal{B}$ with $n(b) = k$. By Lemma 140, Step 1 and Step 2, we can assume that $b = \text{ad}_{b_1}^m(b_2)$ with $b_1, b_2 \in \mathcal{B}$, $b_1 \ne X_0$ and $(b_2 \in X$ or $\lambda(b_2) < b_1$) and $(b_2 \ne X_0$ or $m > 1)$. Assume that $b_2 \ne X_0$. Then the induction assumption applies to both $b_1$ and $b_2$. Let $k_1 := n(b_1)$ and $k_2 := n(b_2)$. Then $k = mk_1 + k_2$, $n_0(b) = mn_0(b_1) + n_0(b_2) \ge n_0(b_2)$. Using the induction assumption and (117) with $a \leftarrow (m+1)$, we obtain, when $n_0(b_2) > 0$,

$$ 
\begin{aligned}
|\dot{\xi}_b(t)| &= \left| \frac{1}{m!} \xi_{b_1}^m(t) \dot{\xi}_{b_2}(t) \right| \\
&\le \frac{1}{m!} \left( \|u\|_{L^1}^{k_1} \frac{(c_{k_1} t)^{n_0(b_1)}}{n_0(b_1)!} \right)^m \|u\|_{L^1}^{k_2-1} (k_2 t|u(t)| + n_0(b_2)\|u\|_{L^1}) \frac{c_{k_2} (c_{k_2} t)^{n_0(b_2)-1}}{n_0(b_2)!} \\
&\le \|u\|_{L^1}^{k_1-1} (kt|u(t)| + n_0(b)\|u\|_{L^1}) 2^{mn_0(b)} c_{k_1}^{mn_0(b_1)} c_{k_2}^{n_0(b_2)} \frac{t^{n_0(b)-1}}{n_0(b)!}.
\end{aligned} 
\quad (390) $$

Since $m \le k$, we have the two desired estimates with $c_k := 2 \cdot 2^k \max\{c_j; j \in [1, k-1]\}$, where the first factor 2 comes from Step 2. When $n_0(b_2) = 0$, the proof is similar and easier. When $b_2 = X_0$, the induction hypothesis applies because $m > 1$ so $n(b_1) < n(b)$ and the proof is straightforward. $\square$

**Remark 142.** The “persistence” of the estimates with respect to left or right bracketing by $X_0$, as mentioned and derived in Steps 1 and 2 of the proof of Lemma 141 might be linked with sufficient conditions for small-time local controllability which “ignore” the number of leading (or trailing) $X_0$ factors (see [17], [63, Theorem 6] or [73, Theorem 3.7]).

These estimates allow to prove the main result of this section.

**Proposition 143.** Let $q \in \mathbb{N}^*$, $X = \{X_0, X_1, \dots, X_q\}$, $\mathcal{B}$ a Hall basis of $\mathcal{L}(X)$ and $(\xi_b)_{b \in \mathcal{B}}$ the associated coordinates of the second kind. Let $M \in \mathbb{N}$, $r, \delta > 0$, $f_0, \dots, f_q \in C^{\omega,r}(B_{4\delta}; \mathbb{K}^d)$. There exists $\eta, C_M > 0$ such that, for every $u \in L^1((0, T); \mathbb{K}^q)$ with $T \le \eta$ and $\|u\|_{L^1(0,T)} \le \eta$, the ordered product of the $e^{\xi_b(t,1,u)f_b}$ over the infinite set $\mathcal{B} \cap S_M = \{b \in \mathcal{B}; n(b) \le M\}$ (using Definition 10) converges uniformly on $B_\delta$ and, for each $t \in [0, T]$ and $p \in B_\delta$,

$$ \left| x(t; f, u, p) - \prod_{b \in \mathcal{B} \cap S_M} e^{\xi_b(t, 1, u) f_b} p \right| \le C_M \|u\|_{L^1(0,t)}^{M+1}. \quad (391) $$

*Proof.* In this proof, to simplify the notations, we write $x(t)$, $\xi_b(t)$ and $\|u\|$ instead of $x(t; f, u, p)$, $\xi_b(t, 1, u)$ and $\|u\|_{L^1(0,t)}$. Let $(c_k)_{k \in \mathbb{N}^*}$ be the increasing sequence of constants of Lemma 141. We define

$$ C_* := \frac{18 \|f\|_r}{r} \max_{k \in [1, 2M]} c_k, \quad (392) $$

$$ \eta := \min \left\{ \frac{\delta}{2\|f\|_{C^1}}, \frac{\min\{1, \delta\}}{2C_*(q+1)M!(1+r)} \right\} \quad (393) $$

$$ C_M := e^{2\delta}(1+r)(2M)!(q+1)^{M+1}C_*^{M+1}. \quad (394) $$

For $t \in [0, T]$ and $u \in L^1((0, T); \mathbb{K}^q)$ with $T \le \eta$ and $\|u\| \le \eta$, using (393),

$$ t \|f_0\|_{C^0} + \sum_{i=1}^{q} \|u_i\|_{L^1(0,t)} \|f_i\|_{C^0} \le \eta \|f\|_{C^0} \le \delta. \quad (395) $$
Hence, for each $p \in B_\delta$, $x(t; f, u, p) \in B_{2\delta}$.

**Strategy.** Since the product involved in (391) is indexed by the infinite set $B \cap S_M$, the proof strategy consists in considering the sequence of finite products $B_{[1,L]} \cap S_M$ for $L \in \mathbb{N}^*$ and let $L \to +\infty$. The error between the true solution and the finite product contains both a term scaling like $\|u\|^{M+1}$ which will persist in the limit and a transitory error term which vanishes as $L \to +\infty$. Each bracket in $b \in B$ is either, not involved at all in the process, involved in the final error, involved in the transitory error term, or involved in the finite product, depending on $L, M, n(b)$ and $n_0(b)$ as pictured in Fig. 1. In Steps 2, 3 and 4, $L \ge M+1$ is fixed. In Step 5, we take the limit $L \to +\infty$.

<figure><img src="image_9.png" /><figcaption>Figure 1: Decomposition of $B$ along the Lazard elimination process for the product on $B \cap S_M$.</figcaption></figure>

**Step 0: Preliminary estimates.** First, using estimate (387) from Lemma 141, for each $b \in B$ with $n(b) = k$, one has in particular

$$ \|\dot{\xi}_b\|_{L^1} \le \|u\|^k \frac{(c_k t)^{n_0(b)}}{n_0(b)!}. \qquad (396) $$

Taking into account that for every $m \in \mathbb{N}^*$, $|B_m| \le (q+1)^m$ and using the analytic estimate (140), we obtain the following estimate for the terms which can be part of the final error

$$ \begin{aligned} & \sum_{b \in B \cap (S_{2M} \setminus S_M)} \|\dot{\xi}_b\|_{L^1} \|f_b\|_{C^1} \\ & \le \sum_{k=M+1}^{2M} \sum_{n_0=0}^{+\infty} |B_{k+n_0}| \|u\|^k \frac{(c_k t)^{n_0}}{n_0!} (1+r) \left(\frac{9 \|f\|_r}{r}\right)^{k+n_0} (k+n_0-1)! \quad (397) \\ & \le (1+r)(2M-1)! \sum_{k=M+1}^{2M} ((q+1)C_* \|u\|)^k \sum_{n_0=0}^{+\infty} ((q+1)C_* T)^{n_0} \\ & \le (1+r)(2M)!(q+1)^{M+1} C_*^{M+1} \|u\|^{M+1}, \end{aligned} $$

because $\|u\| \le \eta$, $T \le \eta$ and $(q+1)C_*\eta \le \frac{1}{2}$. For the terms which can be part of the finite product
or of the transitory error, there holds similarly

$$
\begin{align}
& \sum_{b \in \mathcal{B} \cap S_M} \|\dot{\xi}_b\|_{L^1} \|f_b\|_{C^1} \nonumber \\
& \le T \|f_0\|_{C^1} + \sum_{k=1}^{M} \sum_{n_0=0}^{+\infty} |\mathcal{B}_{k+n_0}| \|u\|^k \frac{(c_k t)^{n_0}}{n_0!} (1+r) \left(\frac{9 \|f\|_r}{r}\right)^{k+n_0} (k+n_0-1)! \tag{398} \\
& \le T \|f_0\|_{C^1} + (1+r)(M-1)! \sum_{k=1}^{M} ((q+1)C_* \|u\|)^k \sum_{n_0=0}^{+\infty} ((q+1)C_* T)^{n_0} \nonumber \\
& \le T \|f_0\|_{C^1} + (1+r)M!(q+1)C_* \|u\| \le \delta. \nonumber
\end{align}
$$

Step 1: Convergence of the ordered product of the $e^{\xi_b(t)f_b}$ over $\mathcal{B} \cap S_M$, uniformly on $B_\delta$, towards a Lipschitz map. Thanks to (398), we have

$$
\sum_{b \in \mathcal{B} \cap S_M} |\xi_b(t)| \|f_b\|_{C^1} \le \sum_{b \in \mathcal{B} \cap S_M} \|\dot{\xi}_b\|_{L^1} \|f_b\|_{C^1} \le \delta \quad (399)
$$

and Lemma 134 gives the conclusion of Step 1.

Step 2: Lazard structure on $\mathcal{B}_{[1,L]} \cap S_M$. We use the notations of Definition 53 to describe $\mathcal{B}_{[1,L]}$.
There exists $m \in \mathbb{N}$ and an extraction $\phi$ such that

$$
\mathcal{B}_{[1,L]} \cap S_M = \{b_{\phi(1)} < \dots < b_{\phi(m+1)}\}. \quad (400)
$$

Let $i \in [1, m+1]$ and $n = \phi(i)$. By Definition 53, there exists a unique factorization

$$
b_{\phi(i)} = b_n = \mathrm{ad}_{b_{n-1}}^{j_{n-1}} \cdots \mathrm{ad}_{b_1}^{j_1}(b_0) \quad (401)
$$

where $b_0 \in X$, $j_1, \dots, j_{n-1} \in \mathbb{N}$ (one just identifies left and right factors in $\mathrm{Br}(X)$). For every $j \in [1, n-1] \setminus \phi([1, i-1])$, $b_j$ contains at least $(L+1)$ occurrences of the variables $X_1, \dots, X_q$, thus it cannot be involved in the factorization of $b_n$. This proves that

$$
\left\{
\begin{array}{l}
b_{\phi(1)} \in \tilde{Y}_0 := X, \\
b_{\phi(2)} \in \tilde{Y}_1 := \{ \mathrm{ad}_{b_{\phi(1)}}^j (v); j \in \mathbb{N}, v \in \tilde{Y}_0 \setminus \{b_{\phi(1)}\} \}, \\
\dots \\
b_{\phi(m+1)} \in \tilde{Y}_m := \{ \mathrm{ad}_{b_{\phi(m)}}^j (v); j \in \mathbb{N}, v \in \tilde{Y}_{m-1} \setminus \{b_{\phi(m)}\} \},
\end{array}
\right.
\qquad (402)
$$

$$
\mathcal{B}_{[1,L]} \cap S_M \cap \tilde{Y}_{m+1} = \emptyset, \quad (403)
$$

$$
\text{where } \tilde{Y}_{m+1} := \{ \mathrm{ad}_{b_{\phi(m+1)}}^j (v); j \in \mathbb{N}, v \in \tilde{Y}_m \setminus \{b_{\phi(m+1)}\} \}.
$$

Step 3: *Proof of estimates along the Lazard elimination on* $\mathcal{B}_{[1,L]} \cap S_M$. To simplify the notations, from now on, we write $\mathcal{B}_{[1,L]} \cap S_M = \{b_1 < \dots < b_{m+1}\}$ and we use (402) and (403) with $\phi = \mathrm{Id}$. Let $x_0(t) := x(t)$. By (399), for every $j \in [1, m+1]$,

$$
x_j(t) := e^{-\xi_{b_j}(t)} f_{b_j} \dots e^{-\xi_{b_1}(t)} f_{b_1} x(t) \quad (404)
$$

is well-defined and belongs to $B_{3\delta}$. The goal of Step 3 is to prove by induction on $j \in [0, m+1]$ that

$$
(\mathcal{H}_j) : \begin{cases} \dot{x}_j(t) = \sum_{b \in \mathcal{B}_{[1,L]} \cap S_M \cap \bar{Y}_j} \dot{\xi}_b(t) f_b(x_j(t)) + \varepsilon_j(t), \\ x_j(0) = p, \end{cases} \quad (405)
$$
where

$$
\|\epsilon_j\|_{L^1} \le e^{\|\xi_{b_j}(t)\|} \|f_{b_j}\|_{C^1} \|\epsilon_{j-1}\|_{L^1} + \sum_{\tilde{b} \in Z_j} \|\dot{\xi}_{\tilde{b}}\|_{L^1} \|f_{\tilde{b}}\|_{C^0}, \quad (406)
$$

where $Z_j \subset (\mathcal{B} \cap S_{2M}) \setminus (\mathcal{B}_{[1,L]} \cap S_M)$ is defined in (413).

First ($\mathcal{H}_0$) holds with $\varepsilon_0 = 0$ because $\dot{\xi}_{X_0}(t) = 1$ and $\dot{\xi}_{X_i}(t) = u_i(t)$ for $i \in [1, q]$. Now, let $j \in [1, m+1]$ and assume that $(\mathcal{H}_{j-1})$ holds. We deduce from the definition of $x_j$ that

$$
x_j(t) = e^{-\xi_{b_j}(t)} f_{b_j}(x_{j-1}(t)) = \Phi_j(-\xi_{b_j}(t), x_{j-1}(t)) \quad (407)
$$

and thus that

$$
\dot{x}_j(t) = \sum_{b \in \mathcal{B}_{[1,L]} \cap S_M \cap \tilde{Y}_{j-1} \setminus \{b_j\}} \dot{\xi}_b(t) (\Phi_j(-\xi_{b_j}(t))_* f_b)(x_j(t)) + \tilde{\varepsilon}_{j-1}(t), \quad (408)
$$

where $\tilde{\varepsilon}_{j-1}(t) = \partial_p \Phi_j(-\xi_{b_j}(t), x_{j-1}(t)) \varepsilon_{j-1}(t)$. We get $(\mathcal{H}_j)$ with

$$
\varepsilon_j(t) := \sum_{b \in \mathcal{B}_{[1,L]} \cap S_M \cap \tilde{Y}_{j-1} \setminus \{b_j\}} \bar{\varepsilon}_b^j(t) + \tilde{\varepsilon}_{j-1}(t) \quad (409)
$$

where, for every $b \in \mathcal{B}_{[1,L]} \cap S_M \cap \tilde{Y}_{j-1} \setminus \{b_j\}$,

$$
\bar{\varepsilon}_b^j(t) := \dot{\xi}_b(t) (\Phi_j(-\xi_{b_j}(t))_* f_b)(x_j(t)) - \sum_{k=0}^{h(b)-1} \dot{\xi}_b(t) \frac{\xi_{b_j}^k(t)}{k!} f_{\text{ad}_{b_j}^k(b)}(x_j(t)) \quad (410)
$$

where $h(b) \in \mathbb{N}^*$ is the maximal integer such that

$$
n(b) + (h(b) - 1)n(b_j) \le M \quad \text{and} \quad |b| + (h(b) - 1)|b_j| \le L. \quad (411)
$$

By (162),

$$
|\bar{\varepsilon}_b^j(t)| \le |\dot{\xi}_b(t)| \frac{|\xi_{b_j}(t)|^{h(b)}}{h(b)!} \|f_{\text{ad}_{b_j}^{h(b)}(b)}\|_{C^0} = |\dot{\xi}_b(t)| \|f_b\|_{C^0}, \quad (412)
$$

for $\tilde{b} := \text{ad}_{b_j}^{h(b)}(b)$. Hence, (406) holds with

$$
Z_j := \{ \text{ad}_{b_j}^{h(b)}(b) ; b \in \mathcal{B}_{[1,L]} \cap S_M \cap \tilde{Y}_{j-1} \setminus \{b_j\} \}. \quad (413)
$$

This yields $Z_j \subset (\mathcal{B} \cap S_{2M}) \setminus (\mathcal{B}_{[1,L]} \cap S_M)$ thanks to (411).

Step 4: Proof of an estimate on the finite product over $\mathcal{B}_{[1,L]} \cap S_M$. By (406), (399) and (397), we have

$$
\begin{align}
\|\epsilon_{m+1}\|_{L^1} &\le e^{\delta} \sum_{b \in \mathcal{B} \cap (S_{2M} \setminus S_M)} \|\dot{\xi}_b\|_{L^1} \|f_b\|_{C^0} + e^{\delta} \sum_{b \in (\mathcal{B} \cap S_M) \setminus \mathcal{B}_{[1,L]}} \|\dot{\xi}_b\|_{L^1} \|f_b\|_{C^0} \tag{414} \\
&\le e^{-\delta} C_M \|u\|^{M+1} + o_{L \to +\infty}(1), \nonumber
\end{align}
$$

because the series in (398) converges. We deduce from (405) and (403) that

$$
\left| \left. \leftarrow \prod_{b \in \mathcal{B}_{[1,L]} \cap S_M} e^{-\xi_b(t,1,u)} f_b x(t) - p \right| \right| = \left| x_{m+1}(t) - p \right| \le e^{-\delta} C_M \|u\|^{M+1} + o_{L \to +\infty}(1) \quad (415)
$$

By (399), the map $\underset{b \in \mathcal{B}_{[1,L]} \cap S_M}{\overset{\rightarrow}{\Pi}} e^{-\xi_b(t,1,u)} f_b$ is $e^\delta$ Lipschitz on $B_{3\delta}$. Then, by (415),

$$
\left| x(t) - \underset{b \in \mathcal{B}_{[1,L]} \cap S_M}{\overset{\rightarrow}{\Pi}} e^{-\xi_b(t,1,u)} f_b p \right| \le C_M \|u\|^{M+1} + o_{L \to +\infty}(1) \quad (416)
$$

Step 5: Infinite subproduct limit. By Step 1, the infinite product over $\mathcal{B} \cap S_M$ is well-defined. By letting $L \to +\infty$ in estimate (416), we obtain the conclusion of Proposition 143. $\square$
# 7 Refined error estimates for scalar-input affine systems

In this section, we consider scalar-input affine systems with drift, i.e. of the form

$$ \dot{x}(t) = f_0(x(t)) + u(t)f_1(x(t)) \quad \text{and} \quad x(0) = p, \qquad (417) $$

where $f_0, f_1$ are vector fields on $K^d$ and $u \in L^1((0, T); K)$. When well-defined, its solution is denoted $x(t; f, u, p)$. Such systems have been extensively studied in control theory, as toy models for more complex situations.

The goal of this section is to improve, in this particular framework, the error estimates of the previous section: the new bound is not expressed in terms of $\|u\|_{L^1}$ but in terms of the $L^\infty$ norm of the time-primitive of the input, which heuristically corresponds to the $W^{-1,\infty}$ norm of $u$.

This refined estimate is somehow optimal in the scale of Sobolev spaces (as shown by the one dimensional system $\dot{x}(t) = u(t)$) and specific to the scalar-input case (see Section 7.5).

Lowering the Sobolev regularity required on the input is of paramount interest for applications in control theory (see e.g. [14]) and might also be useful for applications to stochastic differential equations where the input is a noise with low regularity (see e.g. [15]).

**Definition 144** (Integrated input). Let $T > 0$ and $u \in L^1((0, T); K)$. In this section, $U$ always denotes the time-primitive of $u$ vanishing at zero, i.e. defined by $U(t) := \int_0^t u(s) \, ds$ for $t \in [0, T]$.

## 7.1 Auxiliary system trick

Enhancing the estimates relies on the following trick which factorizes the dependence of the input and introduces an auxiliary system involving the time-primitive $U$ of the input (and not $u$ itself).

**Proposition 145.** Let $\delta > 0$, $f_0, f_1 \in C^\omega(B_{3\delta}; K^d)$ and $\eta^* > 0$ small enough so that the two following maps are well defined and (globally) analytic

$$ \Phi_1 : \begin{cases} [-\eta^*, \eta^*] \times B_{2\delta} & \to B_{3\delta} \\ (\tau, q) & \mapsto e^{\tau f_1}(q) \end{cases} \quad \text{and} \quad F : \begin{cases} B_{2\delta} \times [-\eta^*, \eta^*] & \to K^d \\ (q, \tau) & \mapsto (\Phi_1(-\tau)_* f_0)(q). \end{cases} \qquad (418) $$

Let $T > 0$ be such that $T \|F\|_{C^0} \le \delta$.

1. For every $p \in B_\delta$ and $U \in C^0([0, T]; K)$ with $\|U\|_{L^\infty} \le \eta^*$, there exists a unique solution $x_1 \in C^1([0, T]; K^d)$ to

$$ \begin{cases} \dot{x}_1(t) = F(x_1(t), U(t)), \\ x_1(0) = p, \end{cases} \qquad (419) $$

denoted $x_1(t; F, U, p)$. It takes values in $B_{2\delta}$. Moreover, the map $(p, U) \mapsto x_1(\cdot; F, U, p)$ is analytic from $B_\delta \times B_{C^0[0,T]}(0, \eta^*)$ to $C^1([0, T]; K^d)$.

2. For every $p \in B_\delta$, $t \in [0, T]$ and $u \in L^1((0, T); K)$ such that $\|U\|_{L^\infty} \le \eta^*$,

$$ x(t; f, u, p) = \Phi_1(U(t); x_1(t; F, U, p)). \qquad (420) $$

*Proof.* The existence of $\eta^*$ such that $\Phi_1$ and $F$ are well defined and globally analytic results from the third statement of Lemma 90. The analytic dependence of $x_1$ with respect to $(p, U)$ is given by Lemma 83. By definition of $x_1$, the right-hand side of (420) satisfies the same Cauchy problem as $x$ thus the two functions are equal. □
## 7.2 A new formulation of the Chen-Fliess expansion

The goal of this section is to derive a new formulation of the Chen-Fliess expansion for scalar-input affine systems (417).

**Proposition 146.** Let $\delta, r > 0$ and $f_0, f_1 \in C^{\omega,r}(B_{3\delta}; \mathbb{K}^d)$. There exists $\eta > 0$ such that for every $\varphi \in C^{\omega,r}(B_{3\delta}; \mathbb{K})$, $t \in [0, \eta]$, $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta$ and $p \in B_\delta$, with the notations of Remark 74,

$$ \varphi(x(t; f, u, p)) = \sum_{\substack{\ell \in \mathbb{N}, n \in \mathbb{N} \\ k \in \mathbb{N}^n}} \frac{U(t)^\ell}{\ell! k!} \left( \int_0^t U^k \right) \left( f_1^\ell (\mathrm{ad}_{f_1}^{k_1}(f_0)) \cdots (\mathrm{ad}_{f_1}^{k_n}(f_0)) \right) (\varphi)(p) \quad (421) $$

with the notation (372), where the sum converges absolutely, uniformly with respect to $(t, u, p)$. Moreover, for every $\varphi \in C^{\omega,r}(B_{3\delta}; \mathbb{K})$, there exists $C > 0$ such that, for every $M \in \mathbb{N}^*$, $t \in [0, \eta]$, $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta$ and $p \in B_\delta$,

$$ \left| \varphi(x(t; f, u, p)) - \sum_{\substack{\ell \in \mathbb{N}, n \in \mathbb{N} \\ \ell+|k| \le M}} \frac{U(t)^\ell}{\ell! k!} \left( \int_0^t U^k \right) \left( f_1^\ell (\mathrm{ad}_{f_1}^{k_1}(f_0)) \cdots (\mathrm{ad}_{f_1}^{k_n}(f_0)) \right) (\varphi)(p) \right| \le C^{M+1} \left( |U(t)|^{M+1} + \int_0^t |U|^{M+1} \right) \quad (422) $$

where the sum is taken over $\ell \in \mathbb{N}$, $n \in \mathbb{N}$ and $k = (k_1, \dots, k_n) \in \mathbb{N}^n$ such that $\ell + k_1 + \dots + k_n \le M$. Proof. Let $\eta^*, T, x_1$ be as in Proposition 145, $\|f\| := \|f_0\|_r + \|f_1\|_r$ and

$$ \eta := \min \left\{ T, \eta^*, \frac{\delta}{\|f\|}, \frac{r}{28 \|f\|} \right\}. \quad (423) $$

Let $\varphi \in C^{\omega,r}(B_{3\delta}; \mathbb{K})$, $t \in [0, \eta]$, $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta$ and $p \in B_\delta$. Then $x_1(t; F, U, p) \in B_{3\delta}$ and, by (420) and (423), $x(t; f, u, p) \in B_{3\delta}$.

**Step 1: Proof of the absolute convergence in (421) uniformly with respect to $p \in B_\delta$.** Let $r' := r/e$. Then, by Lemma 80, for every $k \in \mathbb{N}$, $\mathrm{ad}_{f_1}^k(f_0) \in C^{\omega,r'}(B_{3\delta}; \mathbb{K}^d)$ and

$$ \left\| \mathrm{ad}_{f_1}^k(f_0) \right\|_{r'} \le \frac{k!}{e} \left(\frac{9}{r}\right)^k \|f\|^{k+1}. \quad (424) $$

Thus, by (135),

$$ \begin{aligned} & \left| (f_1^\ell (\mathrm{ad}_{f_1}^{k_1}(f_0)) \cdots (\mathrm{ad}_{f_1}^{k_n}(f_0)))(\varphi)(p) \right| \\ & \le (n + \ell)! \left(\frac{5}{r'}\right)^{n+\ell} \left\| f_1 \right\|_{r'}^\ell \left\| \mathrm{ad}_{f_1}^{k_1}(f_0) \right\|_{r'} \cdots \left\| \mathrm{ad}_{f_1}^{k_n}(f_0) \right\|_{r'} \\ & \le (n + \ell)! \left(\frac{14}{r}\right)^{n+\ell} \left\| f \right\|^{\ell} \frac{k_1!}{e} \left(\frac{9}{r}\right)^{k_1} \left\| f \right\|^{k_1+1} \cdots \frac{k_n!}{e} \left(\frac{9}{r}\right)^{k_n} \left\| f \right\|^{k_n+1} \\ & \le e^{-n} (n + \ell)! k_1! \cdots k_n! \left(\frac{14 \|f\|}{r}\right)^{n+\ell+k_1+\cdots+k_n}. \end{aligned} \quad (425) $$

Moreover, recalling notation (372),

$$ \left| \frac{U(t)^\ell}{\ell! k!} \int_0^t U^k \right| = \left| \frac{U(t)^\ell}{\ell!} \int_{\Delta^n(t)} \frac{U(\tau_1)^{k_1} \cdots U(\tau_n)^{k_n}}{k_1! \cdots k_n!} d\tau \right| \le \|U\|_{L^\infty}^{\ell+k_1+\cdots+k_n} \frac{t^n}{n! \ell! k_1! \cdots k_n!}. \quad (426) $$
Thus it is sufficient to prove the summability over $\ell \in \mathbb{N}, n \in \mathbb{N}^*, k_1, \dots, k_n \in \mathbb{N}$ of the following quantity

$$ \left(\frac{t}{e}\right)^n \frac{(n+\ell)!}{n!\ell!} \left(\frac{14\|f\|}{r}\right)^{n+\ell+k_1+\cdots+k_n} \|U\|_{L^\infty}^{\ell+k_1+\cdots+k_n} \\ \le \left(\frac{t}{e}\right)^n 2^{n+\ell} \left(\frac{14\|f\|}{r}\right)^{n+\ell+k_1+\cdots+k_n} \|U\|_{L^\infty}^{\ell+k_1+\cdots+k_n} \\ \le \left(\frac{28t\|f\|}{er}\right)^n \left(\frac{28\|f\|}{r}\|U\|_{L^\infty}\right)^{\ell+k_1+\cdots+k_n} \quad (427) $$

which is ensured by (423).

**Step 2: Proof of (421) and (422).** Applying Lemma 86 and Proposition 137 we get

$$ \begin{aligned} \varphi(x(t; f, u, p)) &= \varphi \left( e^{U(t)f_1} x_1(t; F, U, p) \right) \\ &= \sum_{\ell=0}^{+\infty} \frac{U(t)^\ell}{\ell!} f_1^\ell \varphi(x_1(t; F, U, p)) \\ &= \sum_{\ell=0}^{+\infty} \frac{U(t)^\ell}{\ell!} f_1^\ell \sum_{\substack{n \in \mathbb{N} \\ k \in \mathbb{N}^n}} \frac{1}{k!} \left( \int_0^t U^k \right) \left( (\mathrm{ad}_{f_1}^{k_1}(f_0)) \cdots (\mathrm{ad}_{f_1}^{k_n}(f_0)) \right) (\varphi)(p) \end{aligned} \quad (428) $$

The bound proved in Step 1 allows to exchange the differential operator $f_1^\ell$ and the second sum, which proves (421). To prove (422), one bounds the queue of the series thanks to (425) and the following consequence of Hölder's inequality, valid when $\ell + |k| \ge (M+1)$

$$ \left| U(t)^\ell \int_{\Delta^n(t)} U(\tau_1)^{k_1} \cdots U(\tau_n)^{k_n} d\tau \right| \le C(\eta) \left( |U(t)|^{M+1} + \int_0^t |U|^{M+1} \right). \quad (429) $$

□

One of the ingredients of the above proof is the Chen-Fliess expansion of the auxiliary system $x_1(t; F, U, p)$, which appears in [3, Section 3] under the denomination “representation of the perturbation flow”.

**Remark 147.** The bound (422) between the exact solution and the truncated Chen-Fliess series (in its' original formulation) is used by Stefani in [96, Lemma 3.1 and Corollary 3.1]. Our proof is both different and shorter.

**Remark 148.** Equality (421) where the sum converges absolutely proves that appropriate packages of the Chen-Fliess expansion are absolutely summable under a smallness assumption on $\|U\|_{L^\infty}$, which is weaker than the smallness assumption on $\|u\|_{L^1}$ which is used in Proposition 136 for multi-input systems.

## 7.3 Magnus expansion in the interaction picture

In this section, we prove the following enhanced error estimate for the magnus expansion in the interaction picture with scalar input. Our proof relies on an appropriate approximation for the auxiliary system $x_1$ introduced in Section 7.1.

**Proposition 149.** Let $\delta > 0$ and $f_0, f_1 \in C^\omega(B_{3\delta}; \mathbb{K}^d)$. For every $M \in \mathbb{N}$, there exist $\eta_M, C_M > 0$ such that, for every $T \in [0, \eta_M]$, $u \in L^1((0, T); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta_M$, $t \in [0, T]$ and $p \in B_\delta$,

$$ \left| x(t; f, u, p) - e^{\mathcal{Z}_M(t, f, u)} e^{tf_0} p \right| \le C_M \left( |U(t)|^{M+1} + \int_0^t |U|^{M+1} \right). \quad (430) $$
Proof. In Section 7.3.1, we introduce a vector field $\mathcal{Y}_M(t, f, U)$ such that $e^{\mathcal{Y}_M(t,f,U)}e^{tf_0}(p)$ is a good approximation of the auxiliary state $x_1$ defined in (419). Since, by (420), $x(t) = e^{U(t)f_1}(x_1(t))$, the desired estimate then relies on the following decomposition

$$x(t; f, u, p) - e^{\mathcal{Z}_M(t,f,u)}e^{tf_0}p = x(t; f, u, p) - e^{U(t)f_1}e^{\mathcal{Y}_M(t,f,U)}e^{tf_0}p + e^{U(t)f_1}e^{\mathcal{Y}_M(t,f,U)}e^{tf_0}p - e^{\mathcal{Z}_M(t,f,u)}e^{tf_0}p. \quad (431)$$

Using Proposition 150 and Proposition 154 (see further) for the first and second lines, we get

$$\left|x(t; f, u, p) - e^{\mathcal{Z}_M(t,f,u)}e^{tf_0}p\right| \le C_M \left(\|U\|_{L^1}^{M+1} + |U(t)|^{M+1} + \|U\|_{L^{M+1}}^{M+1}\right) \quad (432)$$

which gives the conclusion since $\|U\|_{L^1(0,t)} \le t^{\frac{M}{M+1}} \|U\|_{L^{M+1}(0,t)}$. $\square$

In Section 7.3.1, we define $\mathcal{Y}_M(t, f, U)$ and prove in Proposition 150 that it indeed provides a good approximation of the auxiliary state. In Section 7.3.2, we explain the link between $e^{U(t)X_1}e^{\mathcal{Y}_M(t,X,U)}$ and $e^{\mathcal{Z}_M(t,X,u)}$ at the formal level. In Section 7.3.3, we show in Proposition 154 that this formal link entails that $e^{U(t)f_1}e^{\mathcal{Y}_M(t,f,U)}$ is close to $e^{\mathcal{Z}_M(t,f,u)}$.

### 7.3.1 An approximation of the auxiliary state

We use the error formula of Proposition 99 for the Magnus expansion in the interaction picture to obtain an approximation of the auxiliary state.

**Proposition 150.** Let $\delta, \rho > 0$, $f_0, f_1 \in C^{\omega, \rho}(B_{3\delta}; \mathbb{K}^d)$. For every $M \in \mathbb{N}$, there exist $\eta_M, C_M > 0$ such that, for every $p \in B_\delta$, $t \in [0, \eta_M]$, $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta_M$,

$$\left|x(t; f, u, p) - e^{U(t)f_1}e^{\mathcal{Y}_M(t,f,U)}e^{tf_0}p\right| \le C_M \|U\|_{L^1(0,t)}^{M+1} \quad (433)$$

where $\mathcal{Y}_M(t, f, U) := \text{Log}_M\{G_t\}(t)$, and $G_t : [0, t] \times B_{3\delta} \to \mathbb{K}^d$ is defined by

$$G_t(s, y) := \sum_{\substack{k \in \mathbb{N}^* \\ \ell \in \mathbb{N}}} \frac{(s-t)^\ell}{\ell!} \frac{U(s)^k}{k!} \text{ad}_{f_0}^\ell \text{ad}_{f_1}^k(f_0)(y) \quad (434)$$

and this sum converges absolutely in $C^{\omega, \rho'}(B_{3\delta}; \mathbb{K}^d)$ with $\rho' = \rho/e$. Moreover,

$$\mathcal{Y}_M(t, f, U) = \sum \frac{(-1)^{m-1}}{rm} \int_{\Delta^r(t)} \frac{(\tau_1 - t)^{\ell_1}}{\ell_1!} \frac{U(\tau_1)^{k_1}}{k_1!} \dots \frac{(\tau_r - t)^{\ell_r}}{\ell_r!} \frac{U(\tau_r)^{k_r}}{k_r!} d\tau \quad (435) \\ \left[ \dots \left[ \text{ad}_{f_0}^{\ell_1}(\text{ad}_{f_1}^{k_1}(f_0)), \text{ad}_{f_0}^{\ell_2}(\text{ad}_{f_1}^{k_2}(f_0)) \right], \dots, \text{ad}_{f_0}^{\ell_r}(\text{ad}_{f_1}^{k_r}(f_0)) \right],$$

where the sum is taken over $r \in [1, M]$, $m \in [1, r]$, $\mathbf{r} \in \mathbb{N}_r^m$, $\ell_1, \dots, \ell_r \in \mathbb{N}$, $k_1, \dots, k_r \in \mathbb{N}^*$ and the sum converges absolutely in $C^{\omega, \rho'}(B_{3\delta}; \mathbb{K}^d)$.

**Proof.** Step 1: Convergence in (434) and (435). By (138), for every $s \in [0, t]$,

$$\left\| \frac{(s-t)^\ell}{\ell!} \frac{U(s)^k}{k!} \text{ad}_{f_0}^\ell \text{ad}_{f_1}^k(f_0) \right\|_{\rho'} \le t^\ell \|U\|_{L^\infty}^k \frac{(k+\ell)!}{\ell!k!} \left(\frac{9}{\rho}\right)^{k+\ell} \|f\|_{\rho}^{k+\ell+1} \quad (436)$$

thus the sum in (434) converges absolutely in $C^{\omega, \rho'}(B_{3\delta}; \mathbb{K}^d)$ when $t$ and $\|U\|_{L^\infty}$ are smaller than $\frac{\rho}{18\|f\|_{\rho}}$.
Proof. Let $\eta_M > 0$ be as in Proposition 150, $t \in [0, \eta_M]$ and $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} < \eta_M$. The sum involved in $\mathcal{Y}_M^j(t, f, U)$ converges absolutely in $C^{\omega, \rho'}(B_{3\delta}; \mathbb{K}^d)$ because it is a sub-family of the one considered in Proposition 150. By (437), there exists $C_M > 0$ (independent of $t$ and $U$) such that, for every $j \in \mathbb{N}^*$, (443) holds. The non-decreasing of $q \in [1, \infty] \mapsto \|\cdot\|_{L^q(0,t)}$ (since $t \le 1$) gives the last conclusion. $\square$

### 7.3.2 Identification procedure at the formal level

In this paragraph, we highlight at the formal level the link between $e^{U(t)X_1}e^{\mathcal{Y}_M(t,X,U)}$ and $e^{\mathcal{Z}_M(t,X,u)}$ in $\hat{\mathcal{L}}(X)$. We start with a new formal factorization, well adapted to estimates with respect to the primitive of the scalar input.

**Proposition 152.** Let $X = \{X_0, X_1\}$ and $u \in L^1(\mathbb{R}_+; \mathbb{K})$. For every $x^* \in \hat{\mathcal{A}}(X)$, the solution $x$ to the formal differential equation

$$ \begin{cases} \dot{x}(t) = x(t)(X_0 + u(t)X_1), \\ x(0) = x^*, \end{cases} \qquad (444) $$

satisfies, for every $t \in \mathbb{R}_+$,

$$ x(t) = x^* \exp(tX_0) \exp(\mathcal{Y}_\infty(t, X, U)) \exp(U(t)X_1) \qquad (445) $$

where $\mathcal{Y}_\infty(t, X, U) \in \hat{\mathcal{L}}(X)$ is defined by $\mathcal{Y}_\infty(t, X, U) = \text{Log}_\infty\{\beta_t\}(t)$ and $\beta_t: [0, t] \to \hat{\mathcal{L}}(X)$ is defined by

$$ \beta_t(s) = e^{-(t-s)X_0} \left( e^{U(s)X_1} X_0 e^{-U(s)X_1} - X_0 \right) e^{(t-s)X_0} = \sum_{\substack{k \in \mathbb{N}^* \\ \ell \in \mathbb{N}}} \frac{(s-t)^\ell}{\ell!} \frac{U(s)^k}{k!} \text{ad}_{X_0}^\ell \text{ad}_{X_1}^k (X_0) \qquad (446) $$

i.e.

$$ \mathcal{Y}_{\infty}(t, X, U) = \sum \frac{(-1)^{m-1}}{rm} \int_{\Delta^r(t)} \frac{(\tau_1 - t)^{\ell_1}}{\ell_1!} \frac{U(\tau_1)^{k_1}}{k_1!} \cdots \frac{(\tau_r - t)^{\ell_r}}{\ell_r!} \frac{U(\tau_r)^{k_r}}{k_r!} d\tau \left[ \cdots \left[ \text{ad}_{X_0}^{\ell_1} \left( \text{ad}_{X_1}^{k_1}(X_0) \right) \right] \cdots \left( \text{ad}_{X_0}^{\ell_2} \left( \text{ad}_{X_1}^{k_2}(X_0) \right) \right) \cdots \cdots \left( \text{ad}_{X_0}^{\ell_r} \left( \text{ad}_{X_1}^{k_r}(X_0) \right) \right) \right] \qquad (447) $$

where the sum is taken over $r \in \mathbb{N}^*$, $m \in [1, r]$, $\mathbf{r} \in \mathbb{N}_r^m$, $\ell_1, \dots, \ell_r \in \mathbb{N}$, $k_1, \dots, k_r \in \mathbb{N}^*$.

**Proof.** First, in the same way as Theorem 27 has been generalized to an infinite alphabet in the proof of Theorem 41, it is possible to generalize Theorem 41 to an infinite alphabet.

The function $x_1: [0, T] \to \hat{\mathcal{A}}(X)$ defined by $x_1(t) := x(t)e^{-U(t)X_1}$ satisfies $x_1(0) = x^*$ and

$$ \dot{x}_1 = x_1(t)e^{U(t)X_1} X_0 e^{-U(t)X_1} = x_1(t) \left( X_0 + \sum_{k \in \mathbb{N}^*} \frac{U(t)^k}{k!} \text{ad}_{X_1}^k (X_0) \right) \qquad (448) $$

This equation is of the form $\dot{x}_1(t) = x_1(t)(X_0 + \sum_{k \in \mathbb{N}^*} a_k(t)Y_k)$ for some indeterminates $Y_k$. Thus, Theorem 41 (adapted to an infinite alphabet) and the homomorphism of algebras sending $Y_k$ to $\text{ad}_{X_1}^k(X_0)$ prove that

$$ x_1(t) = x^* \exp(tX_0) \exp(\mathcal{Y}_\infty(t, X, U)). \qquad (449) $$

which gives the conclusion. $\square$
We now use the formal expansion (445) to obtain an alternative formula for $\mathcal{Z}_{\infty}(t, X, u)$ defined by Theorem 41, in terms of the primitive of the scalar input. For $r, \nu \in \mathbb{N}$, we introduce the finite dimensional subspace of $\mathcal{L}(X)$

$$ \mathcal{L}_{r,\nu}(X) := \operatorname{span}\{\mathbb{E}(b); b \in \operatorname{Br}(X), n_0(b) = \nu, n_1(b) = r\} \quad (450) $$

and $P_{r,\nu}: \hat{\mathcal{L}}(X) \to \mathcal{L}_{r,\nu}(X)$ the associated canonical projection.

**Proposition 153.** Let $X = \{X_0, X_1\}$, $T > 0$, $u \in L^1((0, T); \mathbb{K})$, $t \in [0, T]$, $\mathcal{Y}_{\infty}(t, X, U)$ defined by Proposition 152 and $\mathcal{Z}_{\infty}(t, X, u)$ defined by Theorem 41. Then, in $\hat{\mathcal{L}}(X)$,

$$ \mathcal{Z}_{\infty}(t, X, u) = \mathrm{CBHD}_{\infty}(\mathcal{Y}_{\infty}(t, X, U), U(t)X_1). \quad (451) $$

In particular, for every $M \in \mathbb{N}^*$, $r \in [1, M]$ and $\nu \in \mathbb{N}$,

$$ P_{r,\nu}\mathcal{Z}_M(t, X, u) = P_{r,\nu}\mathrm{CBHD}_M(\mathcal{Y}_M(t, X, U), U(t)X_1). \quad (452) $$

In this statement, $\mathrm{CBHD}_{\infty}$ is defined in Corollary 33, $\mathrm{CBHD}_M$ is its truncation used in Corollary 94 and $\mathcal{Z}_M(t, X, u)$ is defined in Theorem 41 and used in Proposition 99.

*Proof.* We deduce from Proposition 152 and Theorem 41 that

$$ \exp(\mathcal{Z}_{\infty}(t, X, u)) = \exp(\mathcal{Y}_{\infty}(t, X, U)) \exp(U(t)X_1). \quad (453) $$

Thus Corollary 33 proves (451). Let $M \in \mathbb{N}^*$, $r \in [1, M]$, $\nu \in \mathbb{N}$. We deduce from (451) that

$$ P_{r,\nu}\mathcal{Z}_{\infty}(t, X, u) = P_{r,\nu}\mathrm{CBHD}_{\infty}(\mathcal{Y}_{\infty}(t, X, U), U(t)X_1). \quad (454) $$

By definition, $\mathcal{Z}_{\infty}(t, X, u) - \mathcal{Z}_M(t, X, u)$ is a linear combination of brackets all involving at least $(M+1)$ occurrences of $X_1$, thus $P_{r,\nu}\mathcal{Z}_{\infty}(t, X, u) = P_{r,\nu}\mathcal{Z}_M(t, X, u)$. By definition, $\mathcal{Y}_{\infty}(t, X, U)$ is a sum of brackets involving all at least one occurrence of $X_1$, thus

$$ P_{r,\nu}\mathrm{CBHD}_{\infty}(\mathcal{Y}_{\infty}(t, X, U), U(t)X_1) = P_{r,\nu}\mathrm{CBHD}_M(\mathcal{Y}_{\infty}(t, X, U), U(t)X_1). \quad (455) $$

Moreover $\mathcal{Y}_{\infty}(t, X, U) - \mathcal{Y}_M(t, X, U)$ is a linear combination of brackets involving all at least $(M+1)$ occurrences of $X_1$ thus

$$ P_{r,\nu}\mathrm{CBHD}_M(\mathcal{Y}_{\infty}(t, X, U), U(t)X_1) = P_{r,\nu}\mathrm{CBHD}_M(\mathcal{Y}_M(t, X, U), U(t)X_1), \quad (456) $$

which ends the proof of (452).

### 7.3.3 Error formula for analytic vector fields

We prove in Proposition 154 an error bound between $e^{U(t)f_1}e^{\mathcal{Y}_M(t,f,U)}$ and $e^{\mathcal{Z}_M(t,f,u)}$.

**Proposition 154.** Let $\delta, \rho > 0$, $f_0, f_1 \in C^{\omega,\rho}(B_{3\delta}; \mathbb{K}^d)$. For every $M \in \mathbb{N}$, there exist $\eta_M, C_M > 0$ such that, for every $t \in [0, \eta_M]$, $p \in B_\delta$ and $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta_M$,

$$ \left| e^{U(t)f_1} e^{\mathcal{Y}_M(t,f,U)} e^{tf_0} p - e^{\mathcal{Z}_M(t,f,u)} e^{tf_0} p \right| \le C_M \left( |U(t)|^{M+1} + \int_0^t |U|^{M+1} \right). \quad (457) $$

*Proof.* We split the difference as

$$ \begin{aligned} & e^{U(t)f_1} e^{\mathcal{Y}_M(t,f,U)} e^{tf_0} p - e^{\mathrm{CBHD}_M(\mathcal{Y}_M(t,f,U), U(t)f_1)} e^{tf_0} p \\ & \qquad + e^{\mathrm{CBHD}_M(\mathcal{Y}_M(t,f,U), U(t)f_1)} e^{tf_0} p - e^{\mathcal{Z}_M(t,f,u)} e^{tf_0} p. \end{aligned} \quad (458) $$

Taking into account that $\|\mathcal{Y}_M(t, f, U)\|_{C^2} \le C\|U\|_{L^1(0,t)}$, the first line is bounded by Corollary 94. Using Grönwall's lemma and Proposition 155 bounds the second line. □
**Proposition 155.** Let $\delta, \rho > 0$, $f_0, f_1 \in C^{\omega, \rho}(B_{3\delta}; \mathbb{K}^d)$ and $\rho' := \rho/e$. For every $\rho'' \in (0, \rho')$, $M \in \mathbb{N}$, there exist $\eta_M, C_M > 0$ such that, for every $t \in [0, \eta_M]$, $u \in L^1((0, t); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta_M$,

$$ \|\mathcal{Z}_M(t, f, u) - \text{CBHD}_M(\mathcal{Y}_M(t, f, U), U(t)f_1)\|_{\rho''} \le C_M \left( |U(t)|^{M+1} + \int_0^t |U|^{M+1} \right). \quad (459) $$

In particular, $\mathcal{Z}_M(t, f, u)$ is the sum of the terms homogeneous with degree at most $M$ with respect to $U$ in $\text{CBHD}_M(\mathcal{Y}_M(t, f, U), U(t)f_1)$.

*Proof. Step 1: Finite approximation of $\mathcal{Y}_M(t, f, U)$.* First, by Lemma 151, one can write

$$ \mathcal{Y}_M(t, f, U) = \sum_{j=1}^{M} \mathcal{Y}_M^j(t, f, U) + \sum_{j>M} \mathcal{Y}_M^j(t, f, U) =: \bar{\mathcal{Y}}_M(t, f, U) + \mathcal{R}_M(t, f, U), \quad (460) $$

where the remainder satisfies $\|\mathcal{R}_M(t, f, U)\|_{\rho'} \le C \|U\|_{L^{M+1}(0,t)}^{M+1}$. By the triangular and Young inequalities, it is therefore sufficient to prove (459) with $\mathcal{Y}_M$ replaced by the finite truncation $\bar{\mathcal{Y}}_M(t, f, U)$.

*Step 2: Identification at the free level.* Let $\Lambda: \mathcal{L}(X) \to C^\omega(B_{3\delta}; \mathbb{K}^d)$ be the homomorphism of Lie algebras such that $\Lambda(X_i) = f_i$. The relation (452) is made of finite linear combinations of brackets of $X_0$ and $X_1$. Let $M \in \mathbb{N}$. By applying $\Lambda$ to this equality, we get, for every $r \in [1, M]$, $\nu \in \mathbb{N}$

$$ P_{r,\nu}\mathcal{Z}_M(t, f, u) = P_{r,\nu}\text{CBHD}_M(\bar{\mathcal{Y}}_M(t, f, U), U(t)f_1). \quad (461) $$

By definition

$$ \mathcal{Z}_M(t, f, u) = \sum_{\nu \in \mathbb{N}} \sum_{r=1}^{M} P_{r,\nu} \mathcal{Z}_M(t, f, u) \quad (462) $$

where the sum converges in $C^{\omega, \rho'}(B_{3\delta}; \mathbb{K}^d)$ for appropriate $\rho' \in (0, \rho)$, by Proposition 100. Thus, with the notations of (47),

$$ \mathcal{Z}_M(t, f, u) - \text{CBHD}_M(\bar{\mathcal{Y}}_M(t, f, U), U(t)f_1) = \sum_{jh_1+h_2>M} F_{2,h}(\mathcal{Y}_M^j(t, f, U), U(t)f_1), \quad (463) $$

where the sum is taken over $j, h_1, h_2 \in [1, M]$.

*Step 3: Proof of (459).* From now on, $\eta_M > 0$ is given by Proposition 150 and Lemma 151, $t \in [0, \eta_M]$, $u \in L^1((0, t); \mathbb{K})$ is such that $\|U\|_{L^\infty} < \eta_M$ and $\rho'' \in (0, \rho')$. For each term in the finite sum (463), one has, thanks to Lemma 151,

$$ \begin{aligned} \left\| F_{2,h}(\mathcal{Y}_M^j(t, f, U), U(t)f_1) \right\|_{\rho''} &\le C \left\| \mathcal{Y}_M^j(t, f, U) \right\|_{\rho'}^{h_1} \left\| U(t)f_1 \right\|_{\rho'}^{h_2} \\ &\le C' \|U\|_{L^j(0,t)}^{jh_1} |U(t)|^{h_2} \le C' \|U\|_{L^{M+1}(0,t)}^{jh_1} t^{h_1 - \frac{jh_1}{M+1}} |U(t)|^{h_2} \end{aligned} \quad (464) $$

which concludes the proof thanks to Young's inequality since $jh_1 + h_2 \ge M + 1$. $\square$

## 7.4 Sussmann's infinite product expansion

When the input is scalar, the estimates of the coordinates obtained in Lemma 141 can be enhanced to involve only the primitive of the input, at least for Hall bases where $X_1$ is minimal, which in turn improves the estimate of Proposition 143 (see Proposition 157 below). The hypothesis that $X_1$ is the minimal element can be seen as the formal counterpart of the auxiliary system trick of Section 7.1.
**Lemma 156.** Let $X = \{X_0, X_1\}$, $\mathcal{B}$ be a Hall basis of $\mathcal{L}(X)$ for which $X_1$ is the minimal element and $(\xi_b)_{b \in \mathcal{B}}$ the associated coordinates of the second kind. For every $k \ge 1$, there exists $c_k \ge 1$ such that, for each $b \in \mathcal{B} \setminus X$ with $n(b) = k$, $T > 0$, $u \in L^1((0, T); \mathbb{K})$ and $t \in [0, T]$,

$$ |\xi_b(t, 1, u)| \le c_k \|U\|_{L^k(0,t)}^k \frac{(c_k t)^{n_0(b)-1}}{(n_0(b)-1)!} \quad (465) $$

and

$$ |\dot{\xi}_b(t; 1, u)| \le \begin{cases} c_k |U(t)|^k & \text{when } n_0(b) = 1, \\ c_k |U(t)|^k \frac{(c_k t)^{n_0(b)-1}}{(n_0(b)-1)!} + c_k^2 \|U\|_{L^k(0,t)}^k \frac{(c_k t)^{n_0(b)-2}}{(n_0(b)-2)!} & \text{when } n_0(b) \ge 2. \end{cases} \quad (466) $$

*Proof.* As for Lemma 141, estimate (465) is obtained, for each $b$, by time integration of (466). Moreover, still as in Lemma 141, both estimates are invariant by right-bracketing with $X_0$, and also by arbitrary long left-bracketing with $X_0$, up to $c_k \leftarrow 2c_k$. Let us prove (465) and (466) by induction on $k$.

*Initialization for $k=1$.* We have $\xi_{X_1}(t) = U(t)$ and $\dot{\xi}_{[X_1, X_0]}(t) = U(t)$. Hence $[X_1, X_0] \in \mathcal{B}$ (because $X_1 < X_0$) satisfies both estimates. By Lemma 140, when $n(b) = 1$, there exist $m, \bar{m} \in \mathbb{N}$ such that $b = \text{ad}_{X_0}^m \text{ad}_{X_0}^{\bar{m}}(X_1)$. Since $X_1$ is minimal, if $b \ne X_1$, $\bar{m} > 0$. Thus, by the previous invariant properties, we get the conclusion with $c_1 := 2$.

*Induction ($k-1 \to k$).* Let $k \ge 2$ and let us assume that the two estimates are proved for every $b \in \mathcal{B} \setminus X$ with $n(b) \le (k-1)$. Let $b \in \mathcal{B}$ with $n(b) = k$. By Lemma 140 and the previous invariant properties, we may assume that $b = \text{ad}_{b_1}^m(b_2)$ with $m \in \mathbb{N}^*, b_1 < b_2 \in \mathcal{B}$, $b_1 \ne X_0$, $(b_2 \in X$ or $\lambda(b_2) < b_1$) and $(b_2 \ne X_0$ or $m > 1)$.

* If $b_1 = X_1$, then $b_2 = X_0$ (otherwise, if $b_2 \notin X$, $\lambda(b_2) < X_1$, which is impossible since $X_1$ is minimal). Thus

$$ |\dot{\xi}_b(t)| = \frac{|U(t)|^m}{m!} \quad (467) $$

so (466) with $c_k = 1$ holds since $n_0(b) = 1$ and $k = m$.

* If $b_1 \ne X_1$, then $b_1$ satisfies (466) for some $k_1 \in [1, k-1]$. Moreover, either $b_2 = X_0$ or $b_2 \notin X$ (because it cannot be $X_1$). The case ($b_2 = X_0$ and $m > 1$) is easier and left to the reader. Thus we are left with the case where $b_2$ satisfies (466) for some $k_2 \in [1, k-1]$. One has $k = mk_1 + k_2$ and $\nu := n_0(b) = mn_0(b_1) + n_0(b_2) =: m\nu_1 + \nu_2$. Thus,

$$ |\dot{\xi}_b(t)| \le \frac{c_{k_1}^m \|U\|_{L^{k_1}}^{mk_1} (c_{k_1} t)^{m\nu_1-m}}{m! (\nu_1-1)!^m} \left( c_{k_2} |U(t)|^{k_2} \frac{(c_{k_2} t)^{\nu_2-1}}{(\nu_2-1)!} + c_{k_2}^2 \|U\|_{L^{k_2}}^{k_2} \frac{(c_{k_2} t)^{\nu_2-2}}{(\nu_2-2)!} \mathbf{1}_{\nu_2 \ge 2} \right). \quad (468) $$

Thanks to Hölders' inequality,

$$ \|U\|_{L^{k_1}}^{mk_1} \|U\|_{L^{k_2}}^{k_2} \le \|U\|_{L^k}^k t^m. \quad (469) $$

Thanks to Hölder's inequality and Young's inequality,

$$ \|U\|_{L^{k_1}}^{mk_1} |U(t)|^{k_2} \le t^m (\|U\|_{L^k}^k t^{-1} + |U(t)|^k). \quad (470) $$

Moreover, thanks to (117), for $i \in [1, 2]$,

$$ \frac{1}{m!} \frac{1}{(\nu_1 - 1)!^m} \frac{1}{(\nu_2 - i)!} \le 2^{(m+1)(\nu - i)} \frac{1}{(\nu - i)!}. \quad (471) $$

Combining these inequalities proves (466) with $c_k := 2^{k+2} \max\{c_j; j \in [1, k-1]\}$.
□

These enhanced estimates yield the following result.

**Proposition 157.** Let $X = \{X_0, X_1\}$, $\mathcal{B}$ a Hall basis of $\mathcal{L}(X)$ for which $X_1$ is the minimal element and $(\xi_b)_{b \in \mathcal{B}}$ the associated coordinates of the second kind. Let $r, \delta > 0$, $f_0, f_1 \in C^{\omega,r}(B_{4\delta}; \mathbb{K}^d)$. For each $M \in \mathbb{N}^*$, there exist $\eta_M, C_M > 0$ such that, for every $u \in L^1((0,T); \mathbb{K})$ with $T \le \eta_M$ and $\|U\|_{L^{M+1}(0,T)} \le \eta_M$, the ordered product of the $e^{\xi_b(t,1,u)f_b}$ over the infinite set $\mathcal{B} \cap S_M = \{b \in \mathcal{B}; n(b) \le M\}$ (using Definition 10) converges uniformly on $B_\delta$ and, for each $t \in [0,T]$ and $p \in B_\delta$,

$$ \left| x(t; f, u, p) - \prod_{b \in \mathcal{B} \cap S_M} e^{\xi_b(t,1,u)f_b} p \right| \le C_M \|U\|_{L^{M+1}(0,t)}^{M+1}. \quad (472) $$

*Proof.* The proof is the same as the proof of Proposition 143. The only difference is that we use estimates of Lemma 156 instead of those of Lemma 141. The fact that these enhanced estimates are not valid for $b \in X$ doesn't come into play. Indeed, neither $X_0$ nor $X_1$ are involved in the final error term (397). □

## 7.5 Failure of the primitive estimate for multiple inputs

Proposition 149 relying only on the primitive of the input is specific to the scalar-input case and fails for multiple inputs. As an illustration, for $\delta > 0$ and $f_0, f_1 \in C^\infty(B_\delta; \mathbb{K}^d)$, in the degenerate case $M = 0$ and the particular case $f_0(0) = 0, p = 0$, estimate (430) implies that, for every $T > 0$, there exists $C_T > 0$ such that, for $t \in [0, T]$ and $u \in L^1(0, T)$ with $\|U\|_{L^\infty} \le 1$,

$$ |x(t; u, 0)| \le C_T \|U\|_{L^\infty}. \quad (473) $$

As illustrated by the following example, even this very crude estimate fails for multiple inputs, because the $W^{-1,\infty}$ norms are not sufficient to bound the nonlinear terms arising in the dynamic.

**Example 158.** Let $T > 0$ and consider the following system on $\mathbb{R}^2$:

$$ \begin{cases} \dot{x}_1 = u, \\ \dot{x}_2 = vx_1, \end{cases} \quad (474) $$

where $u$ and $v$ are two scalar inputs. There exists $u_n, v_n \in L^1(0, T)$ such that

$$ \|U_n\|_{L^\infty} + \|V_n\|_{L^\infty} \to 0 \quad \text{and} \quad |x(t; (u_n, v_n), 0)| \not\to 0, \quad (475) $$

where $U_n$ is the primitive of $u_n$ and $V_n$ the primitive of $V_n$. Indeed, let $n \in \mathbb{N}^*$ and define $u_n(t) := n \cos n^2 t$ and $v_n(t) := n \sin n^2 t$. Then one has

$$ \|U_n\|_{L^\infty} + \|V_n\|_{L^\infty} \le \frac{2}{n}. \quad (476) $$

Moreover, $x_1(t) = U_n(t) = (\sin n^2 t)/n$ and

$$ x_2(T) = \int_0^T v_n(t) U_n(t) dt = \int_0^T \sin^2(n^2 t) dt \to \frac{T}{2}, \quad (477) $$

as $n \to +\infty$. This proves (475).

**Remark 159.** Although Proposition 149 does not hold for multiple inputs, we expect that the proof method can be adapted to obtain asymmetric estimates, involving for example $\|U\|_{L^\infty} + \|v\|_{L^\infty}$ in the two-inputs case (or the converse). Such asymmetric estimates have been used successfully to obtain sharp results for particular control systems in [51].
# 8 On direct intrinsic representations of the state

The expansions studied above in this article unfortunately don't provide a direct intrinsic representation of the state. The Magnus and Sussmann expansions are given with intrinsic quantities (Lie brackets of the vector fields) but they require to compute one or multiple flows in order to recover the state. The Chen-Fliess expansion gives directly a formula for the state, but it depends on non-intrinsic quantities (see Remark 17 and Remark 167). In this section, we investigate the possibility of finding a direct intrinsic formula for the state. We discuss this possibility in the context of affine systems.

## 8.1 Approximate direct intrinsic representations

We prove in this section approximate direct intrinsic representations which achieve the desired goal up to a small error. We believe that the formulas we derive can be of interest for applications to control theory as they give approximate expressions for the state in terms of the inputs and Lie brackets of the vector fields evaluated at the origin.

We start with an elementary result, which bounds the error when replacing a flow by the value of the vector field.

**Lemma 160.** Let $\delta > 0$ and $z \in C^1(B_\delta; \mathbb{K}^d)$ such that $\|z\|_{C^0} \le \delta$. Then

$$|e^z(0) - z(0)| \le |z(0)||Dz\|_{C^0}e^{|Dz|_{C^0}}. \quad (478)$$

*Proof.* Let $x(t) := e^{tz}(0)$ for $t \in [0, 1]$. Then, for every $t \in [0, 1]$,

$$|x(t) - tz(0)| \le \int_{0}^{t} |z(x(\tau)) - z(0)| d\tau \le \frac{t^2}{2} \|Dz\|_{C^0} |z(0)| + \int_{0}^{t} \|Dz\|_{C^0} |x(\tau) - \tau z(0)| d\tau \quad (479)$$

and by Grönwall's lemma, $|x(t) - tz(0)| \le \frac{t^2}{2} \|Dz\|_{C^0} |z(0)| e^{t \|Dz\|_{C^0}}$.

This elementary estimate allows to obtain approximate direct intrinsic representations from the various Magnus expansions described above.

**Proposition 161.** Let $M \in \mathbb{N}^*$, $\delta > 0$ and $q \in \mathbb{N}^*$.

1. Let $I = [0, q]$ or $I = [1, q]$. Let $f_i \in C^{M^2}(B_\delta; \mathbb{K}^d)$ for $i \in I$. For $T > 0$ and $u \in L^\infty((0, T); \mathbb{K}^q)$, if $x(t; f, u, 0)$ denotes the solution to (226) with $p = 0$ and $Z_M(t, f, u)$ denotes the vector field defined in Proposition 93 (called $Z_M(t, \sum_{i \in I} u_i f_i)$ in this statement), then, as $T \to 0$,

$$x(t; f, u, 0) = Z_M(t, f, u)(0) + O\left(t^{M+1} + t|x(t; f, u, 0)|\right). \quad (480)$$

in the following sense: there exist $C, \eta > 0$ such that, for every $T \in (0, \eta]$ and $u \in L^\infty((0, T); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$, for each $t \in [0, T]$,

$$|x(t; f, u, 0) - Z_M(t, f, u)(0)| \le C \left(t^{M+1} + t|x(t; f, u, 0)|\right). \quad (481)$$

2. Let $T > 0$, $f_0, \dots, f_q \in C^{M^2+1}(B_{2\delta}; \mathbb{K}^d)$ with $f_0(0) = 0$ and $T \|f_0\|_{C^0} \le \delta$. For $u \in L^1((0, T); \mathbb{K}^q)$, if $x(t; f, u, 0)$ denotes the solution to (366) with $p = 0$ and $Z(t; f, u)$ denotes the vector field defined in Proposition 139, then, as $\|u\|_{L^1} \to 0$,

$$x(t; f, u, 0) = Z_M(t, f, u)(0) + O\left(\|u\|_{L^1(0,t)}^{M+1} + |x(t; f, u, 0)|^{1+\frac{1}{M}}\right). \quad (482)$$
3. Let $f_0, f_1 \in C^\omega(B_{3\delta}; \mathbb{K}^d)$ with $f_0(0) = 0$. Let $T > 0$ as in Proposition 145. For $u \in L^1((0,T); \mathbb{K})$, if $x(t; f, u, 0)$ denotes the solution to (417) with $p = 0$ and $\mathcal{Z}(t, f, u)$ denotes the vector field defined in Proposition 139 (with $q = 1$), then, as $(T, \|U\|_{L^\infty}) \to 0$,

$$x(t; f, u, 0) = \mathcal{Z}_M(t, f, u)(0) + O\left(\|U\|_{L^{M+1}(0,t)}^{M+1} + |x(t; f, u, 0)|^{1+\frac{1}{M}}\right), \quad (483)$$

*Proof. Proof of the first statement.* By Proposition 93, there exists $C_1 > 0$ and $T^* > 0$ such that for every $u \in L^\infty((0, T^*); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$ and $t \in [0, T^*]$,

$$|x(t; f, u, 0) - e^{Z_M(t, f, u)}(0)| \le C_1 t^{M+1}. \quad (484)$$

By the explicit expression of $Z_M(t, f, u)$, there exists $C_2 > 0$ such that for every $u \in L^\infty((0, T^*); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$ and $t \in [0, T^*]$,

$$\|Z_M(t, f, u)\|_{C^1} \le C_2 t. \quad (485)$$

Thus, by Lemma 160, there exists $C_3 > 0$ such that, for every $u \in L^\infty((0, T^*); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$ and $t \in [0, T^*]$,

$$|e^{Z_M(t, f, u)}(0) - Z_M(t, f, u)(0)| \le C_3 t |Z_M(t, f, u)(0)|. \quad (486)$$

Then, by triangular inequality, for every $u \in L^\infty((0, T^*); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$ and $t \in [0, T^*]$

$$|x(t; f, u, 0) - Z_M(t, f, u)(0)| \le C_1 t^{M+1} + C_3 t |Z_M(t, f, u)(0)| \quad (487)$$

and in particular, for $t \le T \le 1/(2C_3)$

$$|Z_M(t, f, u)(0)| \le 2|x(t; f, u, 0)| + 2C_1 t^{M+1}. \quad (488)$$

This gives (481) with $C = \max\{2C_1; 2C_3\}$ and $\eta := \min\{T^*, 1/(2C_3)\}$.

*Proof of the second statement.* The strategy is the same: one starts from the estimate in Proposition 139, then applies Lemma 160 to $\mathcal{Z}_M(t, f, u)$ and concludes thanks to the following estimate, implied by the explicit expressions of the vector field

$$\|\mathcal{Z}_M(t, f, u)\|_{C^1} = \mathcal{O}\left(\|u\|_{L^1(0,t)}\right), \quad (489)$$

and Young's inequality.

*Proof of the third statement.* First, one can assume that $f_1(0) \ne 0$. Indeed, otherwise, both $x$ and $\mathcal{Z}_M$ vanish identically, so the desired estimate is void. Using Proposition 155 and the explicit expression of the vector field $\text{CBHD}_M(\mathcal{Y}_M(t, f, U), U(t)f_1)$, we obtain in the asymptotics $(t, \|U\|_{L^\infty}) \to 0$

$$\|\mathcal{Z}_M(t, f, u)\|_{C^1} = \mathcal{O}\left(|U(t)| + \|U\|_{L^1(0,t)}\right). \quad (490)$$

Thus, using $f_0(0) = 0$, Proposition 149 and the same strategy as above, we obtain in the asymptotics $(t, \|U\|_{L^\infty}) \to 0$

$$x(t; f, u, 0) = \mathcal{Z}_M(t, f, u)(0) + \mathcal{O}\left(|U(t)|^{M+1} + \int_0^t |U|^{M+1} + (|U(t)| + \|U\|_{L^1}) |x(t; f, u, 0)|\right). \quad (491)$$

The following proposition and Young's inequality give the conclusion. $\square$

**Proposition 162.** Let $\delta > 0$, $f_0, f_1 \in C^\omega(B_{3\delta}; \mathbb{K}^d)$ with $f_0(0) = 0$ and $f_1(0) \ne 0$. There exists $T, \eta, C > 0$ such that, for every $u \in L^1((0, T), \mathbb{K})$ with $\|U\|_{L^\infty} < \eta$ and $t \in [0, T]$,

$$|U(t)| \le C (|x(t; f, u, 0)| + \|U\|_{L^1(0,t)}). \quad (492)$$
Proof. With the notations of Proposition 145, $x(t; f, u, 0) = e^{U(t)f_1}x_1(t; F, U, 0)$ tends to zero when $\|U\|_{L^\infty} \to 0$. A Taylor expansion of order 2 in $x(t; f, u, 0) = e^{U(t)f_1}x_1(t; F, U, 0)$ provides $C_1 > 0$ such that, for every $t \in [0, T]$ and $u \in L^1((0, T); \mathbb{K})$ such that $\|U\|_{L^\infty} \le \eta^*$,

$$|x(t; f, u, 0) - x_1(t; F, U, 0) - U(t)f_1(0)| \le C_1|U(t)|^2 + C_1|U(t)||x_1(t; F, U, 0)|. \quad (493)$$

Moreover, by Grönwall's lemma, there exists $C_2 > 0$ such that

$$|x_1(t; F, U, 0)| \le C_2 \|U\|_{L^1(0,t)}. \quad (494)$$

Let $P: \mathbb{K}^d \to \mathbb{K}^d$ defined by $P(y) = \langle y, f_1(0) \rangle / |f_1(0)|^2$. Applying $P$ to the vector in the left-hand side of (493) and using (494), we get the conclusion, when $\|U\|_{L^\infty}$ is small enough. $\square$

Under additional nilpotency assumptions, one can omit the truncation errors in the representation formulas of Proposition 161.

**Corollary 163.** Under the same assumptions as in Proposition 161.

1. Assume moreover that $\mathcal{L}(\{f_i; i \in I\})$ is nilpotent of index at most $M+1$. Then, as $T \to 0$,

$$x(t; f, u, 0) = Z_M(t, f, u)(0) + O \left( \|x(t; f, u, 0)\| \right). \quad (495)$$

2. Assume moreover that $f_i \in \mathcal{C}^\omega(B_{4\delta}; \mathbb{K}^d)$ for $i \in I := [0, q]$ and that $\{f_i; i \in [1, q]\}$ is semi-nilpotent of index at most $M+1$ with respect to $f_0$. Then, as $\|u\|_{L^1} \to 0$,

$$x(t; f, u, 0) = \mathcal{Z}_M(t, f, u)(0) + O \left( \|u\|_{L^1(0,t)} |x(t; f, u, 0)| \right). \quad (496)$$

3. Assume moreover that $f_0, f_1 \in \mathcal{C}^\omega(B_{4\delta}; \mathbb{K}^d)$ and that $\{f_1\}$ is semi-nilpotent of index at most $M+1$ with respect to $f_0$. Then, as $(T, \|U\|_{L^\infty}) \to 0$,

$$x(t; f, u, 0) = \mathcal{Z}_M(t, f, u)(0) + O \left( (\|U\|_{L^1(0,t)} + |U(t)|) |x(t; f, u, 0)| \right). \quad (497)$$

Proof. These are straightforward consequences of Proposition 110 (for the first item) and Corollary 122 (for the second and third item, thanks to the analyticity assumption), using the same approach as in the proof of Proposition 161. $\square$

**Remark 164.** Estimate (483) proves that, for a situation in which $\int_0^t |U|^{M+1}$ is negligible, the state is well approximated by $\mathcal{Z}_M(t, f, u)(0)$, which is a convergent series of iterated Lie brackets of $f_0$ and $f_1$ evaluated at 0. We expect that this representation can be useful for applications to control theory, where one tries to relate controllability of the system with geometric relations on the Lie brackets evaluated at zero.

## 8.2 Diffeomorphisms and Lie brackets

Lie brackets behave very nicely with respect to local changes of coordinates. Let $f_i$ be smooth vector fields for $i \in I$, $p \in \mathbb{K}^d$ and $\theta$ be a smooth local diffeomorphism near $p$. If $x(t)$ denotes the solution to (226), we define $y(t) := \theta(x(t))$. Then, one checks that $y$ is the solution to

$$\dot{y}(t) = \sum_{i \in I} u_i(t) g_i(y(t)) \quad \text{and} \quad y(0) = p', \quad (498)$$

where $g_i := \theta_*f_i$ and $p' := \theta(p)$. By iterating Lemma 89, Lie brackets of the vector fields defining the dynamics for $y$ can be computed explicitly from those of $x$. More precisely, for every $b \in \text{Br}(X)$,

$$g_b = \theta_* f_b \quad (499)$$
with the notation of Definition 78. In particular, there exists a linear invertible map $L_p : \mathbb{K}^d \to \mathbb{K}^d$, $L_p := D\theta(p)$, such that, for every $b \in \text{Br}(X)$,

$$g_b(p') = L f_b(p). \qquad (500)$$

Conversely, if the $f_i$ and $g_i$ for $i \in I$ are analytic vector fields, the existence of points $p$ and $p'$ and a linear invertible map $L_p$ such that (500) holds is a sufficient condition for the existence of a local smooth diffeomorphism $\theta$ with $\theta(p) = p'$ and such that, for all controls $u_i$, there holds $y(t) = \theta(x(t))$ where $x$ and $y$ denote the solutions to (226) and (498) for the same set of controls. This nice property is proved in [75, Theorem 1] and was then extended with a more general geometric viewpoint in [98] (see also [5, Theorem 5.5] for a modern presentation).

When (500) only holds for brackets up to some length $M \in \mathbb{N}$ and the controls are uniformly bounded in $L^\infty$, one can prove (see [76]) the existence of a local smooth diffeomorphism $\theta$ and a constant $C$ such that

$$|y(t) - \theta(x(t))| \leq Ct^{M+1}. \qquad (501)$$

Up to our knowledge, the converse, which is conjectured to be true in [76], is a nice open problem.

**Open problem 165.** Let $I = [1, q]$ and $X = \{X_1, \dots, X_q\}$. Let $p, p' \in \mathbb{K}^d$. Assume that there exists a smooth diffeomorphism $\theta$ from a neighborhood of $p$ to a neighborhood of $p'$ and $M \in \mathbb{N}$ such that, for all controls $u_1, \dots, u_q \in L^\infty(0, T)$ with $\|u_i\| \leq 1$, estimate (501) holds for the trajectories $x$ and $y$ corresponding to the same controls. Does this imply that there exists a linear invertible map such that, for each $b \in \text{Br}(X)$ with $|b| \leq M$, (500) holds?

**Open problem 166.** Same question in the context of affine systems with drift, i.e. when $I = [0, q]$, $X = \{X_0, X_1, \dots, X_q\}$ and the first control $u_0$ is constrained to be identically equal to 1. This question might be harder because one gets less information from (501) as it is valid for less choices of controls since $u_0$ is heavily constrained.

**Remark 167.** Property (500) is specific to Lie brackets and does not hold for products of differential operators. As an illustration, consider the case $K = \mathbb{R}$, $d = 2$, $X = \{X_0, X_1\}$, $p = p' = 0$ with $L_0 := D\theta(0) = \text{Id}_2$. Then, for every $\phi \in C^\infty(\mathbb{R}^2; \mathbb{R})$, $(g_b\phi)(0) = (f_b\phi)(0)$, but this relation does not extend to a similar relation between products of $f_0$ and $f_1$ and those of $g_0$ and $g_1$. For example, with the vector fields $f_0(x) := (0, x_1)$ and $f_1(x) := (1, 0)$ and the smooth diffeomorphism $\theta(x) := (x_1, x_2+x_1^2)$, one has $g_0(y) = (0, y_1)$ and $g_1(y) = (1, 2y_1)$. In particular, $(f_1^2\phi)(0) = \partial_{11}\phi(0)$ but $(g_1^2\phi)(0) = \partial_{11}\phi(0) + 2\partial_{12}\phi(0)$. This explains why we consider that the Chen-Fliess expansion is not an intrinsic representation of the state, as it depends on quantities which are not invariant through local changes of coordinates.

## 8.3 Replacing the Magnus flow by a diffeomorphism

Let $f_i$ for $i \in I$ be smooth vector fields. We consider the solution $x(t; u)$ to (226) with $p = 0$. Let $Z_M(t, u)$ be the vector field defined in Proposition 93 (and called $Z_M(t, \sum_{i \in I} u_i f_i)$ in this statement). By Proposition 93, for each $M \in \mathbb{N}$, $x(t; u)$ is given by the time-one flow of the autonomous vector field $Z_M(t, u)$, up to an error scaling like $t^{M+1}$ when the controls $u_i$ are uniformly bounded in $L^\infty$.

In this paragraph, inspired by the nice properties of Lie brackets with respect to diffeomorphisms recalled above, we attempt to replace the computation of the time-one flow by a diffeomorphism. This can be seen as being related with the converse of the classical question of whether a given diffeomorphism can be represented as the time-one flow of an autonomous vector field (see e.g. [7, 8] for positive answers in particular cases, [7, Section 2] for an elementary necessary condition, and [52] or [88] for statements highlighting that the answer is only rarely positive).

This also corresponds to replacing the terms $x(t; u)+o(|x(t; u)|)$ in Proposition 161 by $\theta(x(t; u))$, where $\theta$ is a smooth local diffeomorphism of $K^d$.

We start with a definition.
**Definition 168.** Let $T > 0$ and $n \in \mathbb{N}$. We say that a functional $\beta : [0, T] \times L^\infty((0, T); \mathbb{K}^q) \to \mathbb{K}$ is homogeneous of degree $n$ with respect to time when, for every $u \in L^\infty((0, T); \mathbb{K}^q)$, $\lambda \in (0, 1]$ and $t \in [0, T]$,

$$ \beta(\lambda t, u^{\lambda}) = \lambda^{n} \beta(t, u) \qquad (502) $$

where $u^{\lambda}$ is defined by $u^{\lambda}(\lambda t) := u(t)$ for $t \in [0, T]$ and $u^{\lambda}(\lambda t) := 0$ for $t > T$.

In particular, the product of two homogeneous functionals of degree $n$ and $m$ with respect to time is an homogeneous functional of degree $n + m$. The coordinates of the first kind $\zeta_b(t, u)$, pseudo-first kind $\eta_b(t, u)$ and second kind $\xi_b(t, u)$ are all homogeneous of degree $|b|$ with respect to time. An interesting property of homogeneous functionals is given by the following statement.

**Lemma 169.** Let $T > 0$, $n \in \mathbb{N}$ and $\beta : [0, T] \times L^\infty((0, T); \mathbb{K}^q) \to \mathbb{K}$, homogeneous of degree $n$ with respect to time. Assume that there exists $C > 0$ such that, for every $u \in L^\infty((0, T); \mathbb{K}^q)$ with $\|u\|_{L^\infty((0,T))} \le 1$ and each $t \in [0, T]$,

$$ |\beta(t, u)| \le C t^{n+1}. \qquad (503) $$

Then $\beta = 0$.

*Proof.* Let $t \in [0, T]$ and $u \in L^\infty((0, T); \mathbb{K}^q)$ such that $\|u\|_{L^\infty((0,T))} \le 1$. On the one hand, for each $\lambda \in (0, 1]$, $\beta(\lambda t, u^{\lambda}) = \lambda^n \beta(t, u)$. On the other hand, $|\beta(\lambda t, u^{\lambda})| \le C\lambda^{n+1}t^{n+1}$ because $\|u^{\lambda}\|_{L^\infty} = \|u\|_{L^\infty} \le 1$. Hence $|\beta(t, u)| \le C\lambda t^{n+1}$ for each $\lambda \in (0, 1]$ so $\beta(t, u) = 0$. $\square$

One could wonder if the following proposition holds.

**False proposition 170.** Let $X = \{X_i; i \in I\}$, $\mathcal{B}$ be a monomial basis of $\mathcal{L}(X)$. Let $T > 0$. There exists a family $(\beta_b)_{b \in \mathcal{B}}$ of functionals from $[0, T] \times L^\infty((0, T); \mathbb{K}^q)$ to $\mathbb{K}$, with $\beta_b$ homogeneous of degree $|b|$ with respect to time, such that the following statement holds. Let $\delta > 0$ and $f_i \in C^\infty(B_\delta; \mathbb{K}^d)$ for $i \in I$. There exists a smooth diffeomorphism $\theta$ of $\mathbb{K}^d$ near $p = 0$ such that, for each $M > 0$, there exists $C_M, T_M > 0$ such that, for every $u \in L^\infty((0, T); \mathbb{K}^q)$ with $\|u\|_{L^\infty} \le 1$, for each $t \in [0, T_M]$,

$$ |\theta(x(t; u)) - y_M(t; u)| \le C_M t^{M+1}, \qquad (504) $$

and

$$ y_M(t; u) = \theta(0) + \sum_{|b| \le M} \beta_b(t, u)g_b(\theta(0)), \qquad (505) $$

where $g_b = \theta_* f_b$ and $x(t; u)$ is the solution to (226) starting from $p = 0$.

The functionals $\beta_b$ would be the analog of the coordinates of the first and second kind described earlier. A formula such as (505) would be ideal for applications to control theory for example, since it is expressed on intrinsic quantities (Lie brackets) and allows to compute $x(t; u)$ directly without solving for flows (one recovers $x(t; u) \approx \theta^{-1}(y(t; u))$). In some sense, it corresponds to asking if there exists a local change of coordinates for which the Chen-Fliess expansion only involves Lie bracket terms (and all the non-Lie bracket terms vanish).

Unfortunately, it is impossible in general, as illustrated by the following counter-example.

**Proposition 171.** Let $X = \{X_0, X_1\}$. Let $T > 0$ and consider, in $\mathbb{R}^3$, $f_0(x) := (0, x_1 + x_1^2, x_1x_2)$ and $f_1(x) := (1, 0, 0)$, i.e. the following affine system with drift

$$ \left\{ \begin{array}{l} \dot{x}_1 = u, \\ \dot{x}_2 = x_1 + x_1^2, \\ \dot{x}_3 = x_1x_2, \end{array} \right. \qquad (506) $$

together with the initial data $x(0) = 0$. There exists a monomial basis $\mathcal{B}$ of $\mathcal{L}(X)$, such that, for all functionals $\beta_b : [0, T] \times L^\infty((0, T); \mathbb{R}) \to \mathbb{R}$ for $b \in \mathcal{B}$, homogeneous of degree $|b|$ with respect to time and for every local $C^6$ diffeomorphism $\theta$ of $\mathbb{R}^3$, there exists $M \in [1, 6]$ and a control $u \in L^\infty((0, T); \mathbb{R})$ with $\|u\|_{L^\infty} \le 1$ such that (504) does not hold, even for small times.
Proof. Let $\mathcal{B}$ be a length-compatible Hall basis of $\mathcal{L}(X)$ with $X_0 < X_1$.

Step 1: Computation of $y_6(t)$. We define $\dot{\mathcal{B}}_\ell = \{b \in \mathcal{B}; n_1(b) = \ell\}$ for every $\ell \in \mathbb{N}$. Then $\dot{\mathcal{B}}_1 = \{\mathrm{ad}_{X_0}^k(X_1); k \in \mathbb{N}\}$. The computation shows that the only elements $b \in \dot{\mathcal{B}}_1$ such that $f_b \neq 0$ are

$$b_1 = X_1, \qquad b_2 = [X_0, X_1], \qquad c_1 = [X_0, [X_0, X_1]], \qquad (507)$$
$$f_{b_1}(x) = e_1, \qquad f_{b_2}(x) = -(1+2x_1)e_2 - x_2e_3, \qquad f_{c_1}(x) = x_1^2 e_3. \qquad (508)$$

Thus, the only elements $b \in \dot{\mathcal{B}}_2$ that could satisfy $f_b \neq 0$ are $[b_1, b_2]$, $[b_1, c_1]$, $[b_2, c_1]$. The computation shows that, among them, only the two first ones do satisfy the condition:

$$b_3 = [X_1, [X_0, X_1]], \qquad c_2 = [X_1, \mathrm{ad}_{X_0}^2(X_1)], \qquad (509)$$
$$f_{b_3}(x) = -2e_2, \qquad f_{c_2}(x) = 2x_1 e_3. \qquad (510)$$

Thus, the only elements $b \in \dot{\mathcal{B}}_3$ with length at most 6 that could satisfy $f_b \neq 0$ are $[b_1, b_3]$, $[b_1, c_2]$, $[b_2, b_3]$, $[b_2, c_2]$, $[c_1, b_3]$. The computation shows that, among them, only the second and the third ones do satisfy the condition:

$$b_4 = \mathrm{ad}_{X_1}^2 \mathrm{ad}_{X_0}^2(X_1), \qquad b_5 = [[X_0, X_1], [X_1, [X_0, X_1]]], \qquad (511)$$
$$f_{b_4}(x) = 2e_3, \qquad f_{b_5}(x) = -2e_3. \qquad (512)$$

Thus the only elements $b \in \dot{\mathcal{B}}_4$ with length at most 6 that could satisfy $f_b \neq 0$ are $[b_1, b_4]$ and $[b_1, b_5]$, but the computation shows that they satisfy $f_b = 0$. Therefore, for every $b \in \dot{\mathcal{B}}_4 \cup \dot{\mathcal{B}}_5 \cup \dot{\mathcal{B}}_6$, $f_b = 0$. In conclusion, $b_1, \dots, b_5$ are the only elements $b \in \mathcal{B}$ such that $f_b(0) \neq 0$. In particular, none of them have length 4 or 6, thus

$$y_6(t) = \theta(0) + D\theta(0) \left( \beta_1(t, u)e_1 - \beta_2(t, u)e_2 - 2\beta_3(t, u)e_2 + 2(\beta_4(t, u) - \beta_5(t, u))e_3 \right) \qquad (513)$$

is the sum of 4 homogeneous functionals of degree 1, 2, 3 and 5. Here and below we write $\beta_j$ instead of $\beta_{b_j}$ for brevity.

Step 2: Computation of homogeneous terms with degree 4 and 6 in $\theta(x(t))$. In this step, we consider a local $\mathcal{C}^6$ diffeomorphism $\theta$ of $\mathbb{R}^3$ defined on a neighborhood of $p = 0$. For $u \in L^\infty((0, T); \mathbb{R})$, we denote by $U$ the primitive of $u$ such that $U(0) = 0$ and $V$ the primitive of $U$ such that $V(0) = 0$. Straightforward explicit integration of (506) yields

$$x(t; u) = U(t)e_1 + V(t)e_2 + \int_0^t U^2(s) \, ds e_2 + \frac{1}{2} V^2(t) e_3 + \int_0^t U(s) \int_0^s U^2(s') \, ds' \, ds e_3, \qquad (514)$$

where the five terms are respectively functionals homogeneous of degree 1 through 5 with respect to time in the sense of Definition 168. Using a Taylor expansion of $\theta$ at 0, one obtains (vector-valued) functionals $\gamma_k$ for $k \in [1, 6]$, homogeneous of degree $k$ with respect to time such that for every $M \in [1, 6]$

$$\theta(x(t)) = \theta(0) + \sum_{k=1}^{M} \gamma_k(t, u) + \mathcal{O}(t^{M+1}). \qquad (515)$$

In particular

$$\gamma_4(t, u) = \frac{1}{2} V^2(t) \partial_3 \theta(0) + U(t) \int_0^t U^2 \partial_{12} \theta(0) + \frac{1}{2} V^2(t) \partial_{22} \theta(0) \qquad (516)$$
$$+ \frac{1}{2} U^2(t) V(t) \partial_{112} \theta(0) + \frac{1}{4!} U^4(t) \partial_1^4 \theta(0)$$
and

$$
\begin{align}
\gamma_6(t, u) = U(t) \int_0^t U(s) \int_0^s U(s')^2 ds'ds \partial_{13}\theta(0) &+ \frac{1}{2} V^3(t) \partial_{23}\theta(0) + \frac{1}{2} \left(\int_0^t U^2\right)^2 \partial_{22}\theta(0) \nonumber \\
&+ \frac{1}{4} U^2(t) V^2(t) \partial_{113}\theta(0) + \frac{1}{2} U(t) V(t) \int_0^t U^2 \partial_{122}\theta(0) + \frac{1}{6} V^3(t) \partial_{222}\theta(0) \tag{517} \\
&+ \frac{1}{6} U^3(t) \int_0^t U^2 \partial_{1112}\theta(0) + \frac{1}{4} U^2(t) V^2(t) \partial_{1122}\theta(0) \nonumber \\
&+ \frac{1}{4!} V(t) U^4(t) \partial_1^4 \partial_2 \theta(0) + \frac{1}{6!} U^6(t) \partial_1^6 \theta(0). \nonumber
\end{align}
$$

Step 3: Denying (504). We proceed by contradiction, assuming that there exists a local $C^6$ diffeomorphism $\theta$ of $\mathbb{R}^3$ such that, for each $M \in [1, 6]$, there exists $C_M, T_M > 0$ such that (504) holds for every $t \in [0, T_M]$ and $u \in L^\infty((0, T_M); \mathbb{R})$ with $\|u\|_{L^\infty} \le 1$.

By induction on $M$, estimate (504), Lemma 169 and (513) imply that $\gamma_1 = \beta_1 \partial_1 \theta(0)$, $\gamma_2 = -\beta_2 \partial_2 \theta(0)$, $\gamma_3 = -2\beta_3 \partial_2 \theta(0)$, $\gamma_4 = 0$, $\gamma_5 = 2(\beta_4 - \beta_5) \partial_3 \theta(0)$ and $\gamma_6 = 0$.

On the one hand, by choosing $u$ such that $U(t) = 0$ but $V(t) \neq 0$, the relation $\gamma_4(t, u) = 0$ implies that $\partial_{22}\theta(0) = -\partial_3\theta(0) \neq 0$ because $\theta$ is a local diffeomorphism. On the other hand, by choosing $u$ such that $U(t) = V(t) = 0$ but $\int_0^t U^2 \neq 0$, the relation $\gamma_6(t, u) = 0$ implies that $\partial_{22}\theta(0) = 0$. This concludes the proof, since we have found incompatible conditions on $\partial_{22}\theta(0)$. $\square$

**Remark 172.** This section is written with a focus on time-based estimates. However, a similar "false proposition" could be stated for control-based estimates. The same counter-example also negates this possibility.

## Acknowledgements

The authors would like to thank Matthias Kawski for the time taken to read carefully this long article, many relevant bibliographical and notational suggestions, and his enthusiasm concerning some of the results contained in this work.

The authors also thank Joackim Bernier for discussions about the numerical literature and the importance of the convergence issues discussed in this article for numerical splitting schemes.

The authors benefit from the support of ANR project LEBESGUE, grant ANR-11-LABX-0020. Karine Beauchard and Frédéric Marbach benefit from the support of ANR project TRECOS, grant ANR-20-CE40-0009-01.

# A Proving that the logarithm of the flow is a Lie series

## A.1 Using shuffle relations and Ree's theorem

In this paragraph, we describe a proof of Theorem 20 relying on Ree's theorem and shuffle relations satisfied by the coefficients of the Chen series. This approach is notably used in [71, 74]. We start with some definitions.

**Definition 173 (Shuffle product).** The shuffle product is the map from $I^* \times I^*$ to the free vector space over $I^*$ defined by induction on the length of the words by $\emptyset \sqcup \sigma = \sigma \sqcup \emptyset := \sigma$ for every $\sigma \in I^*$ (where $\emptyset$ denotes the empty word) and, for every $\sigma, \sigma' \in I^*$ and $\ell, \ell' \in I$,

$$ (\sigma\ell) \sqcup (\sigma'\ell') := (\sigma \sqcup (\sigma'\ell'))\ell + ((\sigma\ell) \sqcup \sigma')\ell' \quad (518) $$

Intuitively, the shuffle product of two words is the sum of all the ways of riffling these two words together, interleaving their letters (exactly as one would riffling shuffle two packets of a
card deck). For example, the shuffle product of the words ab and cd (over the Latin alphabet) is abcd + acbd + acdb + cabd + cadb + cdab.

The following result was introduced in [89, Theorem 2.5] to prove Theorem 20.

**Lemma 174** (Ree's theorem). Let $\gamma : I^* \to \mathbb{K}$ with $\gamma_0 = 1$. We still denote by $\gamma$ its linear extension to the free vector space over $I^*$. Consider the formal series $x := \sum_{\sigma \in I^*} \gamma_{\sigma} X_{\sigma}$. Then $\log x \in \hat{\mathcal{L}}(X)$ iff (the linear extension of) $\gamma$ satisfies the so-called "shuffle relations", i.e. iff for every $\sigma, \sigma' \in I^*$,

$$ \gamma_{\sigma\omega\sigma'} = \gamma_{\sigma\gamma\sigma'}. \qquad (519) $$

*Proof.* This statement is item (iii) in [90, Theorem 3.2]. □

Therefore, to show that $\log x(t)$ is a Lie series, it suffices to check that the coefficients $\int_0^t a_\sigma$ (defined in (19)) of the Chen series satisfy these shuffle relations. We proceed as in [89, Section 2], by induction on $|\sigma| + |\sigma'|$ in (519). Definition (19) can also be written as, for every $\sigma \in I^*$ and $\ell \in I$,

$$ \int_0^t a_{\sigma\ell} = \int_0^t a_{\sigma}(s)a_{\ell}(s) \, ds. \qquad (520) $$

Since we set $\int_0^t a_\emptyset = 1$ by definition for the empty word $\emptyset$, (519) holds for every $\sigma, \sigma' \in I^*$ when $|\sigma| + |\sigma'| = 1$. Assume now that it holds for $|\sigma| + |\sigma'| \le n$ for some $n \in \mathbb{N}^*$. Let $\sigma, \sigma' \in I^*$ and $\ell, \ell' \in I$ such that $|\sigma\ell| + |\sigma'\ell'| = n + 1$. Applying successively, (518) and the linearity of the extension of $\gamma$, (520), the induction hypothesis and eventually (520) again, we obtain, for every $t \ge 0$,

$$ \begin{aligned} \gamma_{(\sigma\ell)\omega(\sigma'\ell')} (t) &= \gamma_{(\sigma\omega(\sigma'\ell'))\ell} (t) + \gamma_{((\sigma\ell)\omega\sigma')\ell'} (t) \\ &= \int_0^t \gamma_{\sigma\omega(\sigma'\ell')} (s) a_\ell (s) \, ds + \int_0^t \gamma_{(\sigma\ell)\omega\sigma'} (s) a_{\ell'} (s) \, ds \\ &= \int_0^t \gamma_\sigma (s) \gamma_{\sigma'\ell'} (s) a_\ell (s) \, ds + \int_0^t \gamma_{\sigma\ell} (s) \gamma_{\sigma'} (s) a_{\ell'} (s) \, ds \\ &= \int_0^t \dot{\gamma}_{\sigma\ell} (s) \gamma_{\sigma'\ell'} (s) + \gamma_{\sigma\ell} (s) \dot{\gamma}_{\sigma'\ell'} (s) \, ds, \end{aligned} \qquad (521) $$

which proves that $\gamma_{(\sigma\ell)\omega(\sigma'\ell')} = \gamma_{\sigma\ell}\gamma_{\sigma'\ell'}$.

## A.2 Using Friedrich's criterion

In this paragraph, we describe a proof of Theorem 20 relying on Friedrich's criterion. This approach is notably used in [97, Section 3]. We start with some definitions.

Let $\mathcal{A}(X) \otimes \mathcal{A}(X)$ be the tensor product of the algebra $\mathcal{A}(X)$ with itself (i.e. the tensor product of $\mathcal{A}(X)$ and $\mathcal{A}(X)$, endowed with the product rule $(a \otimes b)(a' \otimes b') := (aa') \otimes (bb')$, see [26, Chapter 3, Section 4.1, Definition 1] for a precise construction). The algebra $\mathcal{A}(X)$ is the universal enveloping algebra of the Lie algebra $\mathcal{L}(X)$, and as such is a Hopf algebra (see [1]). The coproduct homomorphism $\Delta : \mathcal{A}(X) \to \mathcal{A}(X) \otimes \mathcal{A}(X)$ is defined by setting the values $\Delta(1) := 1 \otimes 1$ and $\Delta(X_i) := X_i \otimes 1 + 1 \otimes X_i$ for $1 \le i \le q$. This defines a unique homomorphism because $\mathcal{A}(X)$ is freely generated by $X$ as an algebra (see [90, Proposition 1.2] for more detail). The coproduct $\Delta$ can then be used to characterize Lie elements, as in the following result, which was proposed by Friedrichs in [48], then proved by multiple authors in the same period [35, 44, 78, 80].

**Lemma 175** (Friedrichs' criterion). For $a \in \mathcal{A}(X)$, $a \in \mathcal{L}(X)$ if and only if the condition $\Delta(a) = a \otimes 1 + 1 \otimes a$ holds.

*Proof.* This statement is the equivalence between (i) and (iii) in [90, Theorem 1.4]. □
**Example 176.** The element $X_1X_2$ does not belong to $\mathcal{L}$. And indeed,

$$ \begin{aligned} \Delta(X_1X_2) &= \Delta(X_1)\Delta(X_2) = (X_1 \otimes 1 + 1 \otimes X_1)(X_2 \otimes 1 + 1 \otimes X_2) \\ &= X_1X_2 \otimes 1 + X_1 \otimes X_2 + X_2 \otimes X_1 + 1 \otimes X_1X_2 \tag{522} \\ &\neq X_1X_2 \otimes 1 + 1 \otimes X_1X_2. \end{aligned} $$

On the contrary, the element $[X_1, X_2] = X_1X_2 - X_2X_1$ belongs to $\mathcal{L}$. And indeed,

$$ \begin{aligned} \Delta([X_1, X_2]) &= \Delta(X_1X_2) - \Delta(X_2X_1) \\ &= (X_1X_2 \otimes 1 + X_1 \otimes X_2 + X_2 \otimes X_1 + 1 \otimes X_1X_2) \\ &\quad - (X_2X_1 \otimes 1 + X_2 \otimes X_1 + X_1 \otimes X_2 + 1 \otimes X_2X_1) \tag{523} \\ &= [X_1, X_2] \otimes 1 + 1 \otimes [X_1, X_2]. \end{aligned} $$

The tensor product $\mathcal{A}(X) \otimes \mathcal{A}(X)$ also has a graded structure, with $(\mathcal{A}(X) \otimes \mathcal{A}(X))_n = \bigoplus_{i=0}^n \mathcal{A}_i(X) \otimes \mathcal{A}_{n-i}(X)$. Since the homomorphism $\Delta$ is linear and degree preserving, it can be extended as an homomorphism from $\hat{\mathcal{A}}(X)$ to $\mathcal{A}(X) \widehat{\otimes} \mathcal{A}(X)$, the formal power series over $\mathcal{A}(X) \otimes \mathcal{A}(X)$. For such series with zero constant term, one can define, as in (11), an exponential, say $\exp_\otimes$, which also verifies a uniqueness property such as Lemma 13. One can then derive a criterion to determine whether the logarithm of a formal power series is a Lie element.

**Corollary 177.** Let $a \in \hat{\mathcal{A}}(X)$ with $a_0 = 1$. Then $\log(a) \in \hat{\mathcal{L}}(X)$ if and only if $\Delta(a) = a \otimes a$.

*Proof.* We follow [90, Theorem 3.2]. By linearity and degree preservation, Lemma 175 implies that, for $a \in \hat{\mathcal{A}}(X)$, $a \in \hat{\mathcal{L}}(X)$ if and only if $\Delta(a) = a \otimes 1 + 1 \otimes a$. For $a \in \hat{\mathcal{A}}(X)$ with constant term 1,

$$ \begin{aligned} \log a \in \hat{\mathcal{A}}(X) &\iff \Delta(\log(a)) = \log(a) \otimes 1 + 1 \otimes \log(a) \\ &\iff \exp_\otimes (\Delta(\log(a))) = \exp_\otimes (\log(a) \otimes 1 + 1 \otimes \log(a)) \tag{524} \\ &\iff \Delta(\exp(\log(a))) = \exp_\otimes (\log(a) \otimes 1) \exp_\otimes (1 \otimes \log(a)) \\ &\iff \Delta(a) = ((\exp \log a) \otimes 1)(1 \otimes (\exp \log a)) = a \otimes a, \end{aligned} $$

where we used the equality $\Delta(\exp(\cdot)) = \exp_\otimes(\Delta(\cdot))$, because $\Delta$ is an homomorphism, and the fact that $\exp_\otimes(b \otimes 1 + 1 \otimes c) = \exp_\otimes(b \otimes 1) \exp_\otimes(1 \otimes c)$, because $b \otimes 1$ and $1 \otimes c$ commute. $\square$

Therefore, to show that $\log(x(t))$ is a Lie series, it suffices to check that $\Delta(x(t)) = x(t) \otimes x(t)$. This can be checked as in [97, Section 3] using the following argument. At the initial time $\Delta(x(0)) = \Delta(1) = 1 \otimes 1 = x(0) \otimes x(0)$. On the one hand

$$ \frac{d}{dt} \Delta(x) = \Delta(\dot{x}) = \Delta(xa) = \Delta(x) \Delta(a) = \Delta(x)(a \otimes 1 + 1 \otimes a). \tag{525} $$

On the other hand,

$$ \frac{d}{dt} (x \otimes x) = \dot{x} \otimes x + x \otimes \dot{x} = (xa) \otimes x + x \otimes (xa) = (x \otimes x)(a \otimes 1 + 1 \otimes a). \tag{526} $$

Hence, both quantities satisfy the same formal differential equation with the same initial condition, so they are equal for every $t \ge 0$.

# B Elementary numerical identities

## B.1 Bernoulli numbers

We use the notation $(B_n)_{n \in \mathbb{N}}$ to denote the Bernoulli numbers, which are defined (using the modern NIST sign and indexing convention) by the identity

$$ \forall z \in \mathbb{C}, |z| < 2\pi, \quad \frac{z}{e^z - 1} = \sum_{n=0}^{+\infty} B_n \frac{z^n}{n!} = 1 - \frac{z}{2} + \sum_{n=1}^{+\infty} B_{2n} \frac{z^{2n}}{(2n)!}. \tag{527} $$
**Lemma 178.** The Bernoulli numbers satisfy, for every $n \ge 2$

$$ \sum_{k=0}^{n-1} \binom{n}{k} B_k = 0, \qquad (528) $$

$$ \sum_{k=0}^{n} \binom{n}{k} \frac{B_k}{n+1-k} = 0, \qquad (529) $$

$$ \sum_{k=0}^{n} \frac{B_{n-k}}{(n-k)!(k+2)!} = - \frac{B_{n+1}}{(n+1)!}. \qquad (530) $$

Moreover, the odd Bernoulli numbers except $B_1$ vanish and, for every $n \ge 1$,

$$ B_{2n} = (-1)^{n+1} \frac{2(2n)!}{(2\pi)^{2n}} \zeta(2n) \sim (-1)^{n+1} 2\sqrt{2\pi n} \left(\frac{n}{\pi}\right)^{2n}, \qquad (531) $$

where $\zeta$ is the Riemann zeta function.

*Proof.* The first two identities are classical and can be proved using the generating series of the Bernoulli numbers of (527), respectively by identification in $z = (e^z - 1) \times (z/(e^z - 1))$ for (528) and in $1 = ((e^z - 1)/z) \times (z/(e^z - 1))$ for (529).

The third identity (530) follows from (529) and the computation

$$ \begin{aligned} \sum_{k=0}^{n} \frac{B_{n-k}}{(n-k)!(k+2)!} &= \frac{1}{(n+1)!} \sum_{\ell=0}^{n} \binom{n}{n-\ell} \frac{n+1}{n+1-\ell} \frac{B_{\ell}}{n-\ell+2} \\ &= \frac{1}{(n+1)!} \sum_{\ell=0}^{n} \binom{n+1}{\ell} \frac{B_{\ell}}{(n+1)-\ell+1} = -\frac{B_{n+1}}{(n+1)!}. \end{aligned} \qquad (532) $$

Eventually, the relationship with the Riemann zeta function is proved in [9, equation (12.38)]. The asymptotic is a consequence of the Stirling's approximation and $\zeta(s) \to 1$ as $s \ge 1$ tends to $+\infty$ (which is a direct consequence of the formula $\zeta(s) = \sum n^{-s}$). $\square$

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