ordinary differential equations
Consider the system
$$\begin {cases}
\frac{dx}{dt} = 2x^2-1\\
\frac{dy}{dt} = 2xy
\end{cases}$$
Compute the index of the unit circle with respect to the vector field.
My attempt:
We can parametrize the curve using polar coordinates: $x = \cos \theta, y = \sin \theta$, then $\frac{dx}{dt} = 2\cos^2\theta – 1 = \cos2\theta$, $\frac{dy}{dt} = 2\sin\theta \cos\theta = \sin2\theta.$ How should I proceed?
The vector field:
$\hat{e}_R$ is the unit radial vector. This is simply $(x,y,z)$ divided by its length, $r$: $\hat{e}_R=\frac{(x,y,z)}{r}$.
$\hat{e}_\theta$ is the unit vector tangent to the sphere, thus perpendicular to $(x,y,z)$, which is also perpendicular to $(0,0,1)$ since changing $\theta$ does not change $z$. Therefore, we simply need to take the cross product of $(x,y,z)$ and $(0,0,1)$ and normalize to get $\hat{e}_\theta=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}$. The sign is chosen so that the vector points counter-clockwise.
$\hat{e}_\phi$ is perpendicular to the other two, so we take the cross product and normalize: $\hat{e}_\phi=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}}$. Again, the sign is chosen to have positive $z$ component.
Thus, we get $$ \begin{align} \hat{e}_R&=\frac{(x,y,z)}{r}\\ \hat{e}_\theta&=\frac{(-y,x,0)}{\sqrt{r^2-z^2}}\\ \hat{e}_\phi&=\frac{(-xz,-yz,r^2-z^2)}{r\sqrt{r^2-z^2}} \end{align} $$
References
Solving (*)
We can solve $(\ast)$ without changing to polar coordinates $$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2xy}{x^2-y^2}\tag{1} $$ which can be manipulated to $$ \frac{\mathrm{d}}{\mathrm{d}y}\frac{x^2}{y}=\frac{2x}{y}\frac{\mathrm{d}x}{\mathrm{d}y}-\frac{x^2}{y^2}=-1\tag{2} $$ Integrating $(2)$ yields $$ \frac{x^2}{y}=2c-y\tag{3} $$ For some $c$. Therefore, $$ x^2+y^2=2cy\tag{4} $$
Answer to the Question
A change of coordinate may make the solution more apparent, but the equation should be solvable in either coordinate system.
Best Answer
As you go once around the unit circle ($\theta$ runs from $0$ to $2\pi$), the vector $(\cos2\theta, \sin2\theta)$ makes two full turns counterclockwise, therefore the index is $2$.