Chapter 15: Problem 21

On the manifold \(\mathbb{R}^{2}\) with coordinates \((r, y)\), let \(X\) be the vector field \(X=-y \partial_{x}+\) \(x \partial_{y}\). Determine the integral curve through any point \((x, y)\), and the one-parameter group generated by \(X\). Find coordinates \(\left(x^{\prime}, y^{\prime}\right)\) such that \(X=\partial_{x}\),

Short Answer

Expert verified

The integral curve is \(x(t)=r\cos(t+\phi)\) and \(y(t)= r\sin(t+\phi)\), and the one-parameter group is \(x(t)=x\cos(t)-y\sin(t)\) and \(y(t)=x\sin(t)+y\cos(t)\). The transformed coordinates are \(x^{\prime}=x\cos(t)-y\sin(t)\) and \(y^{\prime}=x\sin(t)+y\cos(t)\).

Step by step solution

Firstly, solving the integral curve

The integral curve is described by the ordinary differential equation derived from the vector field \(X\). Set up the differential equations \( \frac{dx}{dt} = -y, \frac{dy}{dt} = x \). These are the equations one needs to solve for the integral curve.

Secondly, solving the set of Ordinary Differential Equations (ODEs)

Every ODE solution requires a unique method. In this case, the coupled ODEs can be solved using trigonometric function solution. The solutions are \(x(t)=r\cos(t+\phi)\) and \(y(t)= r\sin(t+\phi)\), where \(r=\sqrt{x^2+y^2}\) is the initial radius to the origin and \(\phi \) is the initial angle.

Thirdly, Generating the one-parameter group

This step is worked by evaluating the integral curves at various points. Computing X at \(x(0)\) and \(y(0)\) gives the one-parameter group of transformations as \(x(t)=x\cos(t)-y\sin(t)\) and \(y(t)=x\sin(t)+y\cos(t)\)

Finally, Transforming the coordinates

Simply letting \(x'= r\cos(t+\phi)\) and \(y'= r\sin(t+\phi)\) makes \( X=\partial_{x'}\). Hence, a set of coordinates \(x^{\prime}\) and \(y^{\prime}\) that transforms the vector field into a simpler form is given as \(x^{\prime}=x\cos(t)-y\sin(t)\) and \(y^{\prime}=x\sin(t)+y\cos(t)\)