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Chapter 4: Problem 4

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.

Short Answer

Expert verified
The function \(\gamma: \mathcal{I} \rightarrow \mathbb{R}^{n}: t \mapsto \operatorname{Ad}_{t X} Y\left(x^{0}\right)\) is analytic because both 伪(t) and x(t) are analytic functions, and Y is an analytic vector field. This is demonstrated through the definition of 伪(t) and the given differential equation, which confirms the analytic properties of both x(t) and 伪(t). Therefore, the product 伪(t) Y(x(t)) is also analytic, concluding that 纬(t) is indeed an analytic function.

Step by step solution

01

Define the function 伪(t) and find the differential equation it satisfies

Define 伪(t) as \(\alpha(t):=\left(e^{-t X}\right),\left(e^{t X} x^{0}\right)\) and note that \(\alpha(0)=I\). The given differential equation is \[ \begin{array}{lll} \dot{x}(t) & =X(x(t)) & x(0) & =x^{0} \\ \dot{\alpha}(t) & =-\alpha(t) \cdot X_{*}(x(t)) & & \alpha(0)=1 . \end{array} \]
02

Determine the analytic property of x(t) and 伪(t)

Since the right-hand side of the differential equation is analytic, and according to Proposition C.3.12, both x(t) and 伪(t) are analytic functions.
03

Define the function 纬(t) and find its analytic property

The function 纬(t) is defined as \(\gamma(t)=\alpha(t) Y(x(t))\). Since both 伪(t) and x(t) are analytic, and Y is an analytic vector field, the product 伪(t) Y(x(t)) is also analytic. Hence, the function 纬(t) is analytic, as desired.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. They are profoundly impactful in modeling real-world phenomena where change is observed. Think of them as stories written in the language of calculus, describing how things evolve.

In the exercise above, the differential equation shows how the vector field evolves over time with respect to each state. It's given by two parts: one for a position variable \(x(t)\) and the other for a function \(\alpha(t)\), which both depend on time \(t\):
  • \( \dot{x}(t) = X(x(t)) \) with \( x(0) = x^0 \)
  • \( \dot{\alpha}(t) = -\alpha(t) \cdot X_{*}(x(t)) \) with \( \alpha(0) = 1 \)
These equations provide instructions on how to compute the instants by knowing the rate of change provided by \(X\) and \(X_*\), respectively.

Solving these types of differential equations, particularly in the context of vector fields, frequently involves deriving expressions that describe how a point in space evolves over time.
Analytic Functions
An analytic function is one that is infinitely differentiable and equal to its Taylor series. This means the function behaves very predictably, allowing us to determine values and derivatives simply by knowing its power series. Analytic functions are powerful because they provide a definite structure for understanding complex mathematical behaviours.

In our context, both the solution \(x(t)\) to the position differential equation and \(\alpha(t)\), which arises in our analysis, are analytic. The exercise leverages the nature of these functions so that they remain analytic throughout the transformation processes operating on them. This means they can be represented as a power series about any point in their domain, making computations more manageable.

Analytic functions essentially guarantee smoothness and continuity, ensuring accurate behavior when observing these systems. This is why such good properties are desired in math, helping us ensure our solutions work without surprises.
Proposition on Analyticity
The proposition referred to in the exercise likely builds on the significance of analytic functions when solving differential equations. This proposition ensures that if the differential equations have analytic right-hand sides, then the solutions to these equations are also analytic.

This is what guarantees that our solutions, \(x(t)\) and \(\alpha(t)\), maintain their analytic form throughout the computations. Essentially, if you start with an analytic framework (as in the example, with \(X\), \(x(t)\), and \(\alpha(t)\) being analytic), then the proposition confirms that the resulting function \(\gamma(t) = \alpha(t)Y(x(t))\) will also be analytic, as desired.

Thus, analytic functions propagate through operations like differentiation and multiplication, maintaining their beneficial properties. In the exercise, this means all connected entities - the vector fields, transformations, and the final derived function \(\gamma(t)\) - adhere to analytic principles. This creates a coherent, reliable system for tackling complex mathematical models.

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Most popular questions from this chapter

Consider a rigid body which is being controlled by means of one or more applied torques (for example, a satellite in space, under the action of one or more pairs of opposing thruster jets). We only study here the effect of controls on angular momenta; more complete models incorporating orientations and even linear displacements are of course also possible. With the components of \(x=\left(x_{1}, x_{2}, x_{3}\right)^{\prime}\) denoting the angular velocity coordinates with respect to the principal axes, and the positive numbers \(I_{1}, I_{2}, I_{3}\) denoting the respective principal moments of inertia, this is a system with \(X=\mathbb{R}^{3}, \mathcal{U}=\mathbb{R}^{m}\), where \(m\) is the number of torques; the evolution is represented by (the Euler equations for rotational movement): $$ I \dot{x}=S(x) I x+T u, $$ where \(I\) is the diagonal matrix with entries \(I_{1}, I_{2}, I_{3}\) and where \(T\) is a matrix whose columns describe the axes along which the torques act. The matrix \(S(x)\) is the rotation matrix $$ S(x)=\left(\begin{array}{ccc} 0 & x_{3} & -x_{2} \\ -x_{3} & 0 & x_{1} \\ x_{2} & -x_{1} & 0 \end{array}\right) $$ (Equivalently, the equations can be written as \(I \dot{x}=I \times x+T u\), where " \(\times "\) is the vector product in \(\mathbb{R}^{3}\).) (a) Consider the case in which there are two torques \((m=2)\), which act about the first two principal axes, that is, \(T \in \mathbb{R}^{3 \times 2}\) has columns \((1,0,0)^{\prime}\) and \((0,1,0)^{\prime}\). The equations can be written as follows: $$ \begin{aligned} &\dot{x}_{1}=a_{1} x_{2} x_{3}+b_{1} u_{1} \\ &\dot{x}_{2}=a_{2} x_{1} x_{3}+b_{2} u_{2} \\ &\dot{x}_{3}=a_{3} x_{1} x_{2} \end{aligned} $$ where \(a_{1}=\left(I_{2}-I_{3}\right) / I_{1}, a_{2}=\left(I_{3}-I_{1}\right) / I_{2}\), and \(a_{3}=\left(I_{1}-I_{2}\right) / I_{3}\), and \(b_{1}, b_{2}\) are both nonzero. You may assume that \(b_{1}=b_{2}=1\). Show that the accessibility rank condition holds at every point, if and only if \(I_{1} \neq I_{2}\). (b) Now consider the case in which there is only one torque, acting about a mixed axis. Taking for simplicity the case in which there is rotational symmetry, \(I_{1}=I_{2}\), the equations can be written as follows: $$ \begin{aligned} \dot{x}_{1} &=a x_{2} x_{3}+b_{1} u \\ \dot{x}_{2} &=-a x_{1} x_{3}+b_{2} u \\ \dot{x}_{3} &=b_{3} u \end{aligned} $$ where we assume \(a \neq 0\), and the \(b_{i}\) 's are real numbers. Show that the accessibility rank condition holds at every point of the state space if and only if \(b_{3} \neq 0\) and \(b_{1}^{2}+b_{2}^{2} \neq 0\).

Let \(X \in \mathrm{V}(\mathcal{O}),\left(t^{0}, x^{0}\right) \in \mathcal{D}_{X}, \mathcal{O}_{Y} \subseteq \mathcal{O}\), and \(Y \in \mathrm{V}\left(\mathcal{O}_{Y}\right)\) be so that \(e^{t^{0} X} x^{0} \in \mathcal{O}_{Y}\). Then, for each \(k=0,1,2, \ldots\), $$ \left.\frac{\partial^{k} \mathrm{Ad}_{t X} Y\left(x^{0}\right)}{\partial t^{k}}\right|_{t=t^{0}}=\operatorname{Ad}_{t^{0} X} \operatorname{ad}_{X}^{k} Y\left(x^{0}\right) . $$ In particular, for any \(X, Y \in \mathrm{V}(\mathcal{O})\) and each \(x^{0} \in \mathcal{O}\), $$ \left.\frac{\partial^{k} \operatorname{Ad}_{t X} Y\left(x^{0}\right)}{\partial t^{k}}\right|_{t=0}=\operatorname{ad}_{X}^{k} Y\left(x^{0}\right) $$ for each \(k\). Proof. Fix the state \(x^{0}\) and consider the function \(\gamma(t):=\operatorname{Ad}_{t X} Y\left(x^{0}\right)\); we want to prove that \(\gamma^{(k)}(t)=\operatorname{Ad}_{t X} \operatorname{ad}_{X}^{k} Y\left(x^{0}\right)\) for each \(k\) and for all \(t\) such that \(\left(t, x^{0}\right) \in \mathcal{D}_{X}\) and \(e^{t X} x^{0} \in \mathcal{O}_{Y}\). We proceed by induction on \(k\). The case \(k=0\) is trivial, since, by definition, \(\operatorname{ad}_{X}^{0} Y=Y\). Now, assume that the conclusion has been shown for \(k\). Then, $$ \gamma^{(k+1)}(t)=\frac{d}{d t} \gamma^{(k)}(t)=\frac{d}{d t} \operatorname{Ad}_{\ell X Z}\left(x^{0}\right), $$ where \(Z\) is \(\operatorname{ad}_{X}^{k} Y\), seen as a vector field defined on \(\mathcal{O}_{Y}\). By Lemma 4.4.2 (the case \(k=1), \gamma^{(k+1)}(t)=\operatorname{Ad}_{t X} \operatorname{ad}_{X} Z\left(x^{0}\right)=\operatorname{Ad}_{t X} \operatorname{ad}_{X}^{k+1} Y\left(x^{0}\right)\), as desired. In particular, consider any \(X, Y \in \mathbb{V}(\mathcal{O})\) and \(x^{0} \in \mathcal{O} .\) As \(\left(0, x^{0}\right) \in \mathcal{D}_{X}\), the conclusions hold for \(t=0\). The next few technical results will be needed later, when we specialize our study to analytic systems.

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 \(x \in \mathcal{O}\).

A distribution \(\Delta\) of constant rank \(r\) is completely integrable if, for each \(x^{0} \in \mathcal{O}\) there exists some neighborhood \(\mathcal{O}_{0}\) of \(x^{0}\) and a smooth function $$ \Psi: \mathcal{O}_{0} \rightarrow \mathbb{R}^{n-r} $$ such that $$ \operatorname{ker} \Psi_{\star}(x)=\Delta(x) $$ for all \(x \in \mathcal{O}_{0}\).

Suppose that \(f_{1}, \ldots, f_{r}\) are smooth vector fields on an open set \(\mathcal{O} \subseteq \mathbb{R}^{n}\), and that \(f_{1}(x), \ldots, f_{r}(x)\) are linearly independent for each \(x \in \mathcal{O}\). Show that the following two properties are equivalent: \- \(\left[f_{i}, f_{j}\right]=0\) for each \(i, j \in\\{1, \ldots, r\\} .\) \- 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 \(\left(\Pi_{*} f_{i}\right)(z)=e_{i}\) for each \(z \in \mathcal{V}\), where \(e_{i}\) is the \(i\) th canonical basis vector. (That is, the vector fields commute if and only if there is a local change of variables where they all become \(\left.f_{i} \equiv e_{i} \cdot\right)\)
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