Suggested languages for you:

Americas

Europe

Problem 12

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}$.

Expert verified

In summary, the two properties are equivalent as follows:
1. If \(\Delta\) is invariant under \(X\), which means that \(\mathcal{L}_{X} \Delta(x) = 0\), then for any open subset \(\mathcal{O}_{1}\) and vector field \(Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)\) satisfying \(Y(z) \in \Delta(z)\) for all \(z \in \mathcal{O}_{0}\), we can prove that the adjoint action \(\operatorname{Ad}_{t X} Y(x)\) belongs to \(\Delta(x)\) for all \(x \in \mathcal{O}_{1}\).
2. Conversely, if the adjoint action condition holds, we can show that for each \(x \in \mathcal{O}\) and every \(Y \in \Delta(x)\), the Lie derivative of \(Y\) along \(X\), denoted as \((X \cdot Y)(x)\), also belongs to \(\Delta(x)\). This implies that \(\Delta\) is invariant under \(X\).
Therefore, the properties are equivalent.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

- Access over 3 million high quality textbook solutions
- Access our popular flashcard, quiz, mock-exam and notes features
- Access our smart AI features to upgrade your learning

Chapter 4

A distribution on the open subset \(\mathcal{O} \subseteq \mathbb{R}^{n}\) is a map \(\Delta\) which assigns, to each \(x \in \mathcal{O}\), a subspace \(\Delta(x)\) of \(\mathbb{R}^{n}\). A vector field $f \in \mathbb{V}(\mathcal{O})\( is pointwise in \)\Delta\(, denoted \)f \in_{p} \Delta$, if \(f(x) \in \Delta(x)\) for all \(x \in \mathcal{O}\). A distribution is invariant under a vector field \(f \in \mathbb{V}(\mathcal{O})\) if $$ g \in_{p} \Delta \Rightarrow[f, g] \in_{p} \Delta, $$ and it is involutive if it is invariant under all \(f \in_{p} \Delta\), that is, it is pointwise closed under Lie brackets: $$ f \in_{p} \Delta \text { and } g \in_{p} \Delta \quad \Rightarrow \quad[f, g] \in_{p} \Delta . $$ The distribution generated by a set of vector fields $f_{1}, \ldots, f_{r} \in \mathrm{V}(\mathcal{O})\( is defined \)b y$ $$ \Delta_{f_{1}, \ldots, f_{r}}(x):=\operatorname{span}\left\\{f_{1}(x), \ldots, f_{r}(x)\right\\} $$ for each \(x \in \mathcal{O}\). A distribution has constant rank \(r\) if \(\operatorname{dim} \Delta(x)=r\) for all \(x \in \mathcal{O}\).

Chapter 4

If the vector fields \(Y_{1}, \ldots, Y_{\ell}\) and \(X\) are analytic, then $\operatorname{span}\left\\{\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, t \in \mathcal{I}_{X, x^{0}}\right\\}$ equals $$ \operatorname{span}\left\\{\operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, k \geq 0\right\\} $$ for each \(x^{0} \in \mathcal{O}\). Proof. Fix \(x^{0} \in \mathcal{O}\). Let \(S_{0}\) and \(S_{1}\) be, respectively, the sets of vectors \(\nu \in \mathbb{R}^{n}\) and \(\mu \in \mathbb{R}^{n}\) such that $$ \nu^{\prime} \operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right)=0, \quad j=1, \ldots \ell, k \geq 0 $$ and $$ \mu^{\prime} \operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right)=0, \quad j=1, \ldots \ell, t \in \mathcal{I}_{X, x^{0}} $$ Take any \(\nu \in S_{0}\). For each \(j=1, \ldots \ell\), by Lemma \(4.4 .3\), $$ \left.\nu^{\prime} \frac{\partial^{k} \mathrm{Ad}_{t X} Y_{j}\left(x^{0}\right)}{\partial t^{k}}\right|_{t=0}=0, \forall k \geq 0 . $$ Since, by Lemma 4.4.4, \(\operatorname{Ad}_{t} X Y_{j}\left(x^{0}\right)\) is analytic as a function of \(t\), this means that $\nu^{\prime} \operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right) \equiv 0\(, so \)\nu \in S_{1}$. Conversely, if \(\nu \in S_{1}\), then $\nu^{\prime} \operatorname{Ad}_{\ell X} Y_{j}\left(x^{0}\right) \equiv 0$ implies that all derivatives at zero vanish, so \(\nu \in S_{0}\) (analyticity is not needed here). Thus \(S_{0}=S_{1}\), and the result is proved.

Chapter 4

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.

Chapter 4

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}$.

Chapter 4

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is a distribution of constant rank \(r\). Then, 1\. The following two properties are equivalent, for any $f \in \mathbb{V}(\mathcal{O})$ : (a) \(f \in_{p} \Delta\) (b) For each \(x^{0} \in \mathcal{O}\), there are a neighborhood \(\mathcal{O}_{0}\) of \(x^{0}\) and \(r\) smooth functions $\alpha_{i}: \mathcal{O}_{0} \rightarrow \mathbb{R}_{1} i=1, \ldots, r$, so that $$ f(x)=\sum_{i=1}^{r} \alpha_{i}(x) f_{i}(x) \text { for all } x \in \mathcal{O}_{0} $$ 2\. The following two properties are equivalent, for any $f \in \mathbb{V}(\mathcal{O})$ : (a) \(\Delta\) is invariant under \(f\). (b) \(\left[f, f_{j}\right] \in_{p} \Delta\) for each \(j \in\\{1, \ldots, r\\}\). 3\. Finally, the following two properties are equivalent: (a) \(\Delta\) is involutive. (b) \(\left[f_{i}, f_{j}\right] \in_{p} \Delta\) for all $i, j \in\\{1, \ldots, r\\}$.

The first learning app that truly has everything you need to ace your exams in one place.

- Flashcards & Quizzes
- AI Study Assistant
- Smart Note-Taking
- Mock-Exams
- Study Planner