Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) has constant rank \(r\) and is invariant under each of \(X_{1}, \ldots, X_{k} \in \mathrm{V}(\mathcal{O})\). Let \(\mathcal{O}_{k}\) be an open subset of \(\mathcal{O}\) and let $s_{1}, \ldots, s_{k}$ be real numbers with the following property: $$ \left(s_{i}, x\right) \in \mathcal{D}_{X_{i}}, \quad \forall x \in \mathcal{O}_{i}, i=1, \ldots, k, $$ where we define $$ \mathcal{O}_{k-1}:=e^{s_{k} X_{k}} \mathcal{O}_{k}, \ldots, \mathcal{O}_{1}:=e^{s_{2} X_{2}} \mathcal{O}_{2}, \mathcal{O}_{0}:=e^{s_{1} X_{1}} \mathcal{O}_{1} . $$ Assume that \(Y \in \mathbb{V}\left(\mathcal{O}_{0}\right)\) is such that $Y(z) \in \Delta(z)\( for each \)z \in \mathcal{O}_{0}$. Then, $\operatorname{Ad}_{s_{k} X_{k}} \ldots \operatorname{Ad}_{s_{1} X_{1}} Y(x) \in \Delta(x)\( for all \)x \in \mathcal{O}_{k}$.

(a) Show that for systems of dimension one: \(\dot{x}=f(x, u), x=\) \(\mathbb{R}\), the accessibility rank condition holds at a state \(x^{0}\) if and only if \(\mathcal{R}\left(x^{0}\right)\) has a nonempty interior. (b) Consider the following system with \(\mathcal{X}=\mathbb{R}^{2}\) and \(\mathcal{U}=\mathbb{R}^{2}\) : $$ \begin{aligned} &\dot{x}_{1}=u_{1} \\ &\dot{x}_{2}=\varphi\left(x_{1}\right) u_{2} \end{aligned} $$ where \(\varphi: \mathbb{R} \rightarrow \mathbb{R}\) is a \(C^{\infty}\) function with the property that \(\varphi(z) \neq 0\) for each \(z \neq 0\) but \(\frac{d^{j} \varphi}{d z^{j}}(0)=0\) for all \(j=0,1,2, \ldots\) (for instance, \(\varphi(z):=e^{-1 / z^{2}}\) for \(z \neq 0, \varphi(0):=0)\). Show that, for all \(x^{0}, x^{0} \in\) int $\mathcal{R}_{\mathcal{V}}^{\leq T}\left(x^{0}\right)$, but, on the other hand, the accessibility rank condition does not hold at the points of the form \(\left(0, x_{2}\right)\).

Suppose \(X\) and \(Y\) are analytic vector fields defined on \(\mathcal{O}\). For any \(x^{0} \in \mathcal{O}\), let $\mathcal{I}=\mathcal{I}_{X, x^{0}}:=\left\\{t \in \mathbb{R} \mid\left(t, x^{0}\right) \in \mathcal{D}_{X}\right\\}\(. Then, the function \)\gamma: \mathcal{I} \rightarrow \mathbb{R}^{n}: t \mapsto \operatorname{Ad}_{t X} Y\left(x^{0}\right)$ is analytic. Proof. Let $\alpha(t):=\left(e^{-t X}\right),\left(e^{t X} x^{0}\right)\(, seen as a function \)\mathcal{I} \rightarrow \mathbb{R}^{n \times n}\(. Note that \)\alpha(0)=I$, and that, by Equation (4.28), the vector \(\left(e^{t X} x^{0}, \alpha(t)\right)\) is the solution of the differential equation $$ \begin{array}{lll} \dot{x}(t) & =X(x(t)) & x(0) & =x^{0} \\ \dot{\alpha}(t) & =-\alpha(t) \cdot X_{*}(x(t)) & & \alpha(0)=1 . \end{array} $$ This is a differential equation with analytic right-hand side, so \(x(\cdot)\) and \(\alpha(\cdot)\) are both analytic (see, for instance, Proposition C.3.12). Then, \(\gamma(t)=\alpha(t) Y(x(t))\) is also analytic.

Consider a model for the "shopping cart" shown in Figure 4.2 ("knife-edge" or "unicycle" are other names for this example). The state is given by the orientation \(\theta\), together with the coordinates \(x_{1}, x_{2}\) of the midpoint between the back wheels. Figure 4.2: Shopping cart. The front wheel is a castor, free to rotate. There is a non-slipping constraint on movement: the velocity $\left(\dot{x}_{1}, \dot{x}_{2}\right)^{\prime}\( must be parallel to the vector \)(\cos \theta, \sin \theta)^{\prime} .$ This leads to the following equations: $$ \begin{aligned} \dot{x}_{1} &=u_{1} \cos \theta \\ \dot{x}_{2} &=u_{1} \sin \theta \\ \dot{\theta} &=u_{2} \end{aligned} $$ where we may view \(u_{1}\) as a "drive" command and \(u_{2}\) as a steering control (in practice, we implement these controls by means of differential forces on the two back corners of the cart). We view the system as having state space \(\mathbb{R}^{3}\) (a more accurate state space would be the manifold \(\mathbb{R}^{2} \times \mathbb{S}^{1}\) ). (a) Show that the system is completely controllable. (b) Consider these new variables: $z_{1}:=\theta, z_{2}:=x_{1} \cos \theta+x_{2} \sin \theta, z_{3}:=\( \)x_{1} \sin \theta-x_{2} \cos \theta, v_{1}:=u_{2}\(, and \)v_{2}:=u_{1}-u_{2} z_{3}$. (Such a change of variables is called a "feedback transformation".) Write the system in these variables, as \(\dot{z}=\widetilde{f}(z, v) .\) Note that this is one of the systems \(\Sigma_{i}\) in Exercise 4.3.14. Explain why controllability can then be deduced from what you already concluded in that previous exercise.

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) has constant rank \(r\), and let \(X \in\) \(\mathbb{V}(\mathcal{O})\). Then, the following two properties are equivalent: 1\. \(\Delta\) is invariant under \(X\). 2\. Let \(\mathcal{O}_{1}\) be an open subset of \(\mathcal{O}\) and let $t \in \mathbb{R}\( be so that \)(t, x) \in \mathcal{D}_{X}\( for all \)x \in \mathcal{O}_{1}\(. Define \)\mathcal{O}_{0}:=e^{t X} \mathcal{O}_{1}$. Assume that \(Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)\) is such that $Y(z) \in \Delta(z)\( for each \)z \in \mathcal{O}_{0}\(. Then, \)\operatorname{Ad}_{t X} Y(x) \in \Delta(x)\( for each \)x \in \mathcal{O}_{1}$.

Let \(X \in \mathbb{V}(\mathcal{O})\) and $\left(t^{0}, x^{0}\right) \in \mathcal{D}_{X}\(. Pick any \)Y \in \mathbb{V}\left(\mathcal{O}_{Y}\right)$ such that \(e^{t^{0} X} x^{0} \in \mathcal{O}_{Y}\). Then, $$ \left.\frac{\partial \operatorname{Ad}_{t X} Y\left(x^{0}\right)}{\partial t}\right|_{t=t^{0}}=\operatorname{Ad}_{t^{0} X}[X, Y]\left(x^{0}\right) . $$ Proof. We have, for all \((t, x) \in \mathcal{D}_{X}, e^{-t X} e^{t X} x=x\), so taking \(\partial / \partial x\) there results $$ \left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot\left(e^{t X}\right)_{*}(x)=I $$ and, taking \(\partial / \partial t\), $\frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right)\right) \cdot\left(e^{t X}\right)_{*}(x)+\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot \frac{\partial}{\partial t}\left(\left(e^{t X}\right)_{*}(x)\right)=0 .$ On the other hand, $$ \begin{aligned} \frac{\partial}{\partial t}\left(\left(e^{t X}\right)_{*}(x)\right) &=\frac{\partial}{\partial t} \frac{\partial}{\partial x}\left(e^{t X} x\right)=\frac{\partial}{\partial x} \frac{\partial}{\partial t}\left(e^{t X} x\right) \\ &=\frac{\partial}{\partial x} X\left(e^{t X} x\right)=X_{*}\left(e^{t X} x\right) \cdot\left(e^{t X}\right)_{*}(x) \end{aligned} $$ so substituting in (4.27) and postmultiplying by $\left(\left(e^{t X}\right)_{*}(x)\right)^{-1}$ there results the identity: $$ \frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right)\right)=-\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot X_{*}\left(e^{t X} x\right) . $$ Thus, $$ \begin{gathered} \frac{\partial}{\partial t}\left(\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot Y\left(e^{t X} x\right)\right) \\ =-\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot X_{*}\left(e^{t X} x\right) \cdot Y\left(e^{t X} x\right) \\ +\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot Y_{*}\left(e^{t X} x\right) \cdot X\left(e^{t X} x\right) \\ =\left(e^{-t X}\right)_{*}\left(e^{t X} x\right) \cdot[X, Y]\left(e^{t X} x\right) \end{gathered} $$ as required. A generalization to higher-order derivatives is as follows; it can be interpreted, in a formal sense, as saying that $\operatorname{Ad}_{t X}=e^{t a d x}\(. For each \)k=0,1, \ldots\(, and each vector fields \)X, Y$, we denote \(\operatorname{ad}_{X}^{0} Y=Y\) and $\operatorname{ad}_{X}^{k+1} Y:=\operatorname{ad}_{X}\left(\operatorname{ad}_{X}^{k} Y\right)$.