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Chapter 3: Problem 13

Construct a differentiable vector field on an open disk of the plane (which is not compact) such that a maximal trajectory \(\alpha(t)\) is not defined for all \(t \in R\) (this shows that the compactness condition of Exercise 12 is essential).

Short Answer

Expert verified
Use the vector field \( \textbf{V}(x, y) = (x, y) \) on an open disk. Trajectories leave the disk in finite time due to exponential growth.

Step by step solution

01

Understand the Problem

The goal is to construct a differentiable vector field on an open disk in the plane where the maximal trajectory is not defined for all time. This is to show that the compactness condition is essential for ensuring the trajectory's existence for all time.
02

Consider the Open Disk

An open disk in the plane is a set of points \( (x, y) \) such that \( x^2 + y^2 < R^2 \) for some radius \( R \). This disk is not compact because it does not include the boundary where \( x^2 + y^2 = R^2 \).
03

Choose a Vector Field

Choose a simple vector field that will cause a trajectory to leave the disk in finite time. One example is the radial vector field \( \textbf{V}(x, y) = (x, y) \). In this field, each point \( (x, y) \) has a vector pointing directly away from the origin with magnitude proportional to its distance from the origin.
04

Define the Trajectory

For a trajectory starting at a point \( \textbf{r}(0) = (x_0, y_0) \), the differential equation governing the trajectory is \( \frac{d\textbf{r}}{dt} = (x(t), y(t)) \). This simplifies to \( \frac{dx}{dt} = x \) and \( \frac{dy}{dt} = y \), with solutions \( x(t) = x_0 e^t \) and \( y(t) = y_0 e^t \).
05

Analyze the Trajectory

The trajectory \( \textbf{r}(t) = (x_0 e^t, y_0 e^t) \) grows exponentially. Thus, for any initial point \( (x_0, y_0) \), there exists a finite time \( t^* \) such that \( x(t^*)^2 + y(t^*)^2 = R^2 \), meaning the trajectory will leave the disk by reaching its boundary in finite time.
06

Conclusion

The chosen vector field \( \textbf{V}(x, y) = (x, y) \) on the open disk causes any trajectory to eventually leave the disk in finite time. Hence, the maximal trajectories are not defined for all \( t \), demonstrating the importance of the compactness condition.

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Key Concepts

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

Open Disk
In this problem, we focus on an open disk in the plane. An open disk consists of all points \( (x, y) \) such that \(x^2 + y^2 < R^2\) for some positive radius \(R\).
Note that this means we are considering all points that lie strictly within the disk and excluding the boundary where \(x^2 + y^2 = R^2\).

This is important because the disk is *not compact*. A set is compact if it is both closed (contains its boundary) and bounded (fits within some finite region).
Since our disk does not include the boundary points, it lacks compactness.
Consequently, it poses interesting behaviors for trajectories in such vector fields.
Maximal Trajectory
The problem asks us to consider a **maximal trajectory** \( \alpha(t) \) which is the path traced by following the vector field over time.
For a given vector field, a trajectory is referred to as maximal if it continues for all time \((t \in \mathbb{R}) \).
This means it doesn't run into any singularities or leave the domain in finite time.

The goal here is to show an example where this isn't the case.
We'll use a vector field in an open disk where the trajectory can leave the disk in a finite time.
This illustrates that without the compactness condition, we can't guarantee the existence of trajectories for all time.
Compactness Condition
The **compactness condition** is crucial for ensuring that maximal trajectories exist for all time in a vector field.
Compactness means the set is closed and bounded, ensuring that trajectories remain within the region as time progresses.

Let's refer to the vector field \( \textbf{V}(x, y) = (x, y) \).
Without compactness, even simple trajectories can eventually leave our set.
Consider starting at point \( (x_0, y_0) \).
The differential equations governing the trajectory are \( \frac{dx}{dt} = x \) and \( \frac{dy}{dt} = y \).
These have exponential solutions \( x(t) = x_0 e^t \) and \( y(t) = y_0 e^t \).
The exponential growth leads to \( r(t) = x_0 e^t, y_0 e^t \) quickly reaching and surpassing the boundary of the disk.
Radial Vector Field
A **radial vector field** is a field where vectors point directly away from (or towards) a central point, usually the origin, with magnitude proportional to the distance from that point.
In our problem, we use \( \textbf{V}(x, y) = (x, y) \).
Here, each point \( (x, y) \) has a vector pointing directly outward from the origin. The length of the vector increases as you move further from the origin.

This is a simple choice of vector field that guarantees our trajectory will eventually grow and leave the open disk.
For example, starting at \( (x_0, y_0) \), the trajectory \( r(t) = (x_0 e^t, y_0 e^t) \) moves outward exponentially.
Eventually, it reaches the boundary where \( x(t)^2 + y(t)^2 = R^2 \).
In this way, the radial vector field ensures that the trajectory cannot remain in the open disk for all time.

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Most popular questions from this chapter

Assume that the osculating plane of a line of curvature \(C \subset S\), which is nowhere tangent to an asymptotic direction, makes a constant angle with the tangent plane of \(S\) along \(C\). Prove that \(C\) is a plane curve.

a. Let \(S\) be a regular surface without umbilical points. Prove that \(S\) is a minimal surface if and only if the Gauss map \(N: S \rightarrow S^{2}\) satisfies, for all \(p \in S\) and all \(w_{1}, w_{2} \in T_{p}(S)\), $$ \left\langle d N_{p}\left(w_{1}\right), d N_{p}\left(w_{2}\right)\right\rangle_{N(p)}=\lambda(p)\left\langle w_{1}, w_{2}\right\rangle_{p}, $$ where \(\lambda(p) \neq 0\) is a number which depends only on \(p\). b. Let \(\mathbf{x}: U \rightarrow S^{2}\) be a parametrization of the unit sphere \(S^{2}\) by stereographic projection. Consider a neighborhood \(V\) of a point \(p\) of the minimal surface \(S\) in part a such that \(N: S \rightarrow S^{2}\) restricted to \(V\) is a diffeomorphism (since \(K(p)=\operatorname{det}\left(d N_{p}\right) \neq 0\), such a \(V\) exists by the inverse function theorem). Prove that the parametrization \(y=N^{-1} \circ\) \(\mathbf{x}: U \rightarrow S\) is isothermal (this gives a way of introducing isothermal parametrizations on minimal surfaces without planar points).

(Surfaces of Revolution with Constant Curvature.) \((\varphi(v) \cos u\), \(\varphi(v) \sin u, \psi(v)), \varphi \neq 0\) is given as a surface of revolution with constant Gaussian curvature \(K\). To determine the functions \(\varphi\) and \(\psi\), choose the parameter \(v\) in such a way that \(\left(\varphi^{\prime}\right)^{2}+\left(\psi^{\prime}\right)^{2}=1\) (geometrically, this means that \(v\) is the arc length of the generating curve \((\varphi(v), \psi(v)))\). Show that a. \(\varphi\) satisfies \(\varphi^{\prime \prime}+K \varphi=0\) and \(\psi\) is given by \(\psi=\int \sqrt{1-\left(\varphi^{\prime}\right)^{2}} d v\); thus, \(01, C<1\). Observe that \(C=1\) gives a sphere (Fig. 3-23). c. All surfaces of revolution with constant curvature \(K=-1\) may be given by one of the following types: 1\. \(\varphi(v)=C \cosh v\), \(\psi(v)=\int_{0}^{v} \sqrt{1-C^{2} \sinh ^{2} v} d v .\) 2\. \(\varphi(v)=C \sinh v\), \(\psi(v)=\int_{0}^{v} \sqrt{1-C^{2} \cosh ^{2} v} d v .\) 3\. \(\varphi(v)=e^{v}\), \(\psi(v)=\int_{0}^{v} \sqrt{1-e^{2 v}} d v\) Determine the domain of \(v\) and draw a rough sketch of the profile of the surface in the \(x z\) plane. d. The surface of type 3 in part \(\mathrm{c}\) is the pseudosphere of Exercise \(6 .\) e. The only surfaces of revolution with \(K \equiv 0\) are the right circular cylinder, the right circular cone, and the plane.

Example 4 can be generalized as follows. A one-parameter differentiable family of planes \(\\{\alpha(t), N(t)\\}\) is a correspondence which assigns to each \(t \in I\) a point \(\alpha(t) \in R^{3}\) and a unit vector \(N(t) \in R^{3}\) in such a way that both \(\alpha\), and \(N\) are differentiable maps. A family \(\\{\alpha(t), N(t)\\}, t \in I\), is said to be a family of tangent planes if \(\alpha^{\prime}(t) \neq 0, N^{\prime}(t) \neq 0\), and \(\left\langle\alpha^{\prime}(t), N(t)\right\rangle=0\) for all \(t \in I\). a. Give a proof that a differentiable one-parameter family of tangent planes \(\\{\alpha(t), N(t)\\}\) determines a differentiable one-parameter family of lines \(\left\\{\alpha(t),\left(N \wedge N^{\prime}\right) /\left|N^{\prime}\right|\right\\}\) which generates a developable surface $$ \mathbf{x}(t, v)=\alpha(t)+v \frac{N \wedge N^{\prime}}{\left|N^{\prime}\right|} . $$ The surface (*) is called the envelope of the family \(\\{\alpha(t), N(t)\\}\). b. Prove that if \(\alpha^{\prime}(t) \wedge\left(N(t) \wedge N^{\prime}(t)\right) \neq 0\) for all \(t \in I\), then the envelope (*) is regular in a neighborhood of \(v=0\), and the unit normal vector of \(\mathbf{x}\) at \((t, 0)\) is \(N(t)\). c. Let \(\alpha=\alpha(s)\) be a curve in \(R^{3}\) parametrized by arc length. Assume that the curvature \(k(s)\) and the torsion \(\tau(s)\) of \(\alpha\) are nowhere zero. Prove that the family of osculating planes \(\\{\alpha(s), b\\{s)\\}\) is a oneparameter differentiable family of tangent planes and that the envelope of this family is the tangent surface to \(\alpha(s)\) (cf. Example 5, Sec. 2-3).

Determine the asymptotic curves and the lines of curvature of \(z=x y\).
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