/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Suppose that \(\Delta=\Delta_{f_... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: 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 \(x \in \mathcal{O}_{1}\).

Short Answer

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.

Step by step solution

01

Proving Property 1 implies Property 2

Suppose that property 1 holds, which means that \(\Delta\) is invariant under \(X\). This implies that for every \(x \in \mathcal{O}\), the Lie derivative of \(\Delta\) with respect to \(X\) is zero: \(\mathcal{L}_{X} \Delta(x)=0\). Now, let \(\mathcal{O}_{1}\) be an open subset of \(\mathcal{O}\), and let \(t \in \mathbb{R}\) be such that \((t, x) \in \mathcal{D}_{X}\) for all \(x \in \mathcal{O}_{1}\). Define \(\mathcal{O}_{0}:=e^{t X} \mathcal{O}_{1}\). Suppose that \(Y \in \mathrm{V}\left(\mathcal{O}_{0}\right)\) is such that \(Y(z) \in \Delta(z)\) for each \(z \in \mathcal{O}_{0}\). Since \(\mathcal{L}_{X} \Delta(x)=0\) (invariant property) and by the Cartan formula, we have: \[ \mathcal{L}_{X}\Delta(x)=d(\iota_{X}\Delta(x))+\iota_{X}(d\Delta(x)) = 0 \] We know also that \(Y(z) \in \Delta(z)\) for each \(z \in \mathcal{O}_{0}\), which means that \(\iota_{Y} \Delta(z) = 0\) for all \(z \in \mathcal{O}_{0}\). Consider now the adjoint action \(\operatorname{Ad}_{t X} Y(x)\) for each \(x \in \mathcal{O}_{1}\). We have: \[ \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= \frac{d}{dt} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ &= e^{tX}\frac{d}{dt}Y(e^{-t X} x)e^{-tX} \end{aligned} \] Substituting \(z=e^{-t X} x\), we get: \[ \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= e^{tX}\frac{d}{dt}Y(z)e^{-tX} \end{aligned} \] Now, since \(Y(z) \in \Delta(z)\), we have \(\operatorname{Ad}_{tX}Y(x) \in \Delta(x)\) for all \(x \in \mathcal{O}_{1}\). Therefore, property 2 holds.
02

Proving Property 2 implies Property 1

Now, let's assume that property 2 holds, and show that property 1 also holds. To prove that \(\Delta\) is invariant under \(X\), we must show that for each \(x \in \mathcal{O}\) and for all \(Y \in \Delta(x)\), we have \((X \cdot Y)(x) \in \Delta(x)\). Let \(x \in \mathcal{O}_{1}\) and \(Y \in \Delta(x)\). Then, by property 2, we have \(\operatorname{Ad}_{tX}Y(x) \in \Delta(x)\) for each \(x \in \mathcal{O}_{1}\). By definition, the adjoint action is given by: \[ \begin{aligned} \operatorname{Ad}_{t X} Y(x) &= \frac{d}{dt} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ \end{aligned} \] At t=0, we have: \[ \begin{aligned} \operatorname{Ad}_{0} Y(x) &= Y(x) \end{aligned} \] Now, let's differentiate the adjoint action with respect to the flow time t: \[ \begin{aligned} (X \cdot Y)(x) &= \left.\frac{d}{dt}\right|_{t=0}\operatorname{Ad}_{t X} Y(x)\\\ &=\left.\frac{d}{dt}\right|_{t=0} e^{t X} Y(e^{-t X} x) e^{-t X}\\\ \end{aligned} \] Thus, we have \((X \cdot Y)(x) \in \Delta(x)\), which implies that \(\Delta\) is invariant under \(X\). Property 1 holds. Since we have shown that property 1 implies property 2 and property 2 implies property 1, the two properties are equivalent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lie Derivative
The concept of the Lie derivative is crucial in the world of differential geometry and mathematical control theory. It measures how a certain object鈥攐ften a tensor field or a vector field鈥攃hanges as we flow along another vector field on a manifold. Roughly speaking, if you were to flow along the streamlines of a fluid, the Lie derivative tells you how the fields of interest behave relative to this flow.

In mathematical terms, if you have a vector field \( X \) and another field \( Y \) for which you want to compute the Lie derivative, we denote the Lie derivative of \( Y \) with respect to \( X \) as \( \mathcal{L}_{X} Y \). One can use the Cartan's magic formula to compute the Lie derivative, especially when dealing with differential forms: \( \mathcal{L}_{X} = d\iota_{X} + \iota_{X}d \), where \( d \) is the exterior derivative, and \( \iota_{X} \) is the interior product by \( X \).

How does this relate to control theory? When working with dynamical systems, knowing how perturbations or controls affect the system is vital. The Lie derivative gives insight into the invariance of a system's behavior under continuous symmetries, which is a cornerstone for stability analysis and control design.
Invariant Property
In the realm of control systems and physics, invariance refers to properties or quantities that remain unchanged under certain transformations or flow of a dynamical system. An invariant property under the action of a vector field \( X \) on a manifold signifies that the value of the property is constant along the trajectories generated by \( X \).

This concept is manifested mathematically by stating that the Lie derivative of the quantity or property with respect to the vector field is zero (\( \mathcal{L}_{X} \Delta = 0 \)). The invariant property plays a pivotal role in control theory when analyzing the stability of equilibria, and in constrained motion where certain quantities, like the conservation of energy in mechanics, must not change over time.
Adjoint Action
In control theory, the notion of adjoint action has a well-defined place, particularly in the study of Lie groups and Lie algebras. The adjoint action is an operation where an element of a Lie group 'acts' on another element of the associated Lie algebra.

To comprehend the adjoint action, let's denote it as \( \operatorname{Ad} \), where for a given time-dependent flow \( e^{tX} \) generated by a vector field \( X \) and vector field \( Y \) in the Lie algebra, the adjoint action of \( X \) on \( Y \) is given as \( \operatorname{Ad}_{tX}Y \). Intuitively, it tells us how the vector field \( Y \) transforms as we flow along \( X \) for a time \( t \) and then back. This is particularly useful when considering symmetries and conserved quantities in control systems, crucial for understanding system behavior and designing controls that respect these invariances.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let the vector fields \(Y_{1}, \ldots, Y_{\ell}\), and \(X\) be analytic, and pick any \(\left(t, x^{0}\right) \in \mathcal{D}_{X}\). Let $$ d:=\operatorname{dim} \operatorname{span}\left\\{Y_{j}\left(e^{t X} x^{0}\right), j=1, \ldots \ell\right\\} . $$ Then, $$ d \leq \operatorname{dim} \operatorname{span}\left\\{\operatorname{ad}_{X}^{k} Y_{j}\left(x^{0}\right), j=1, \ldots \ell, k \geq 0\right\\} . $$ Proof. Since \(\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right)=Q \cdot Y_{j}\left(e^{t X} x^{0}\right)\), where \(Q:=\left(e^{-t X}\right)_{*}\left(e^{t X} x^{0}\right)\) is a nonsingular matrix, \(d=\operatorname{dim} \operatorname{span}\left\\{\operatorname{Ad}_{t X} Y_{j}\left(x^{0}\right), j=1, \ldots \ell\right\\}\). The result follows then from Proposition 4.4.5.

Let \(G: X \rightarrow \mathbb{R}^{n \times m}\) be a matrix function of class \(C^{\infty}\), defined on an open subset \(X \subseteq \mathbb{R}^{n}\). Consider the following properties: (C) For each \(x^{0} \in X\), there is some neighborhood \(X_{0}\) of \(x^{0}\), and there are two matrix functions \(C: x_{0} \rightarrow \mathbb{R}^{n \times n}\) and \(D: x_{0} \rightarrow \mathbb{R}^{m \times m}\), both of class \(C^{\infty}\) and nonsingular for each \(x \in X_{0}\), such that $$ C(x) G(x) D(x)=\left(\begin{array}{l} I \\ 0 \end{array}\right) \quad \text { for all } x \in X_{0}, $$ where \(I\) is the \(m \times m\) identity matrix. (F) For each \(x^{0} \in X\) there are \(X_{0}, C\), and \(D\) as above, and there is some diffeomorphism II : \(x_{0} \rightarrow V_{0}\) into an open subset of \(\mathbb{R}^{n}\) so that \(C(x)=\) \(\Pi_{*}(x)\) for all \(x \in X_{0}\). (a) Show that (C) holds if and only if rank \(G(x)=m\) for all \(x \in X\). (b) Show that (F) holds if and only if the columns \(g_{1}, \ldots, g_{m}\) generate a distribution \(\Delta_{g_{1}, \ldots, g_{m}}\) which is involutive and has constant rank \(m\). (That is, rank \(G(x)=m\) for all \(x \in X\) and the \(n^{2}\) Lie brackets \(\left[g_{i}, g_{j}\right]\) are pointwise linear combinations of the \(g_{i}\) 's.) 180 (c) Give an example where (C) holds but (F) fails. (d) Interpret (b), for a system without drift \(\dot{x}=G(x) u\), in terms of a change of variables (a feedback equivalence) \((x, u) \mapsto(z, v):=\left(\Pi(x), D(x)^{-1} u\right)\). What are the equations for \(\dot{z}\) in terms of \(v\), seen as a new input? (Hint: (a) and (b) are both easy, from results already given.)

A distribution is said to be smooth if it is locally generated by sets (possibly infinite) of vector fields, that is, for each \(x^{0} \in \mathcal{O}\) there is a subset \(\mathcal{F} \subseteq V(\mathcal{O})\), and there is an open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\), such that, for each \(x \in \mathcal{O}_{0}, \Delta(x)\) is the span of the vectors \(\\{f(x), f \in \mathcal{F}\\}\). Show that, if \(\Delta\) is a smooth distribution of constant rank \(r\), then for each \(x^{0} \in \mathcal{O}\) there is some open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\), and a set of \(r\) vector fields \(f_{1}, \ldots, f_{r}\), such that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) on \(\mathcal{O}_{0}\), that is, \(\Delta(x)=\operatorname{span}\left\\{f_{1}(x), \ldots, f_{r}(x)\right\\}\) for each \(x \in \mathcal{O}_{0}\). Invariance of a distribution under \(f\) is equivalent to invariance under the linear operators \(\mathrm{Ad}_{t f}\), in the sense that \(\Delta\) is invariant under \(f\) if and only if \(\operatorname{Ad}_{t f} g(x) \in \Delta(x)\) for each \((t, x) \in \mathcal{D}_{f}\) and each \(g \in_{p} \Delta\). This fact, proved next, is perhaps the most important property of the Lie bracket operation. (We need a slightly stronger version, in which \(g\) is not necessarily defined globally.)

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is a distribution of constant rank \(r\) on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\). Show that the following two properties are equivalent: \- \(\Delta\) is involutive. 176 4\. Nonlinear Controllability \- For each \(x^{0} \in \mathcal{O}\) there is an open subset \(\mathcal{O}_{0} \subseteq \mathcal{O}\) which contains \(x^{0}\) and a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow \mathcal{V}\) into some open subset \(\mathcal{V} \subseteq \mathbb{R}^{n}\) such that \(\left(\Pi_{*} \Delta\right)(z)=\operatorname{span}\left\\{e_{1}, \ldots, e_{r}\right\\}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (A distribution of the form \(\Delta_{e_{1}, \ldots, e_{r}}\) is said to be a flat distribution. So, the claim is that a distribution is involutive if and only if it is locally flat in appropriate coordinates.)

Suppose that \(\Delta=\Delta_{f_{1}, \ldots, f_{r}}\) is involutive of constant rank \(r\). Then, for each \(x^{0} \in \mathcal{O}\) there exists some open subset \(\mathcal{O}_{0}\) containing \(x^{0}\), an \(\varepsilon>0\), and a diffeomorphism \(\Pi: \mathcal{O}_{0} \rightarrow(-\varepsilon, \varepsilon)^{n}\), so that the following property holds. If we partition $$ \Pi=\left(\begin{array}{l} \Pi_{1} \\ \Pi_{2} \end{array}\right), \quad \Pi_{2}: \mathcal{O}_{0} \rightarrow(-\varepsilon, \varepsilon)^{n-r}, $$ then \(\Delta(x)=\operatorname{ker}\left(\Pi_{2}\right)_{*}(x)\) for all \(x \in \mathcal{O}_{0}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.