Solving Vector Field with Poincare's Lemma

[Ted123]

Homework Statement

[PLAIN]http://img130.imageshack.us/img130/8540/vecx.jpg

The Attempt at a Solution

I've done (i).

First of all Poincare's Lemma says that if the domain $U$ of ${\bf F}$ is simply connected then:

${\bf F}$ is irrotational $\iff {\bf F}$ is conservative.

So for (ii)(a), does $V$ being simply connected (is it or not?) mean Poincare's Lemma implies there is a potential function for $\bf F$ (since it would mean $\bf F$ is conservative and hence is a gradient)
DESPITE $U$ not being simply connected - it has a hole in the middle at (0,0)?

 

[ystael]

Yes. $$V$$ is, in fact, simply connected, and the restriction of $$\nabla\times\mathbf{F}$$ to $$V$$ is still zero. (The simplest way to prove that $$V$$ is simply connected is to see that it's star-shaped with respect to any point on the positive $$x$$-axis.)

 

[Ted123]

Yes. $$V$$ is, in fact, simply connected, and the restriction of $$\nabla\times\mathbf{F}$$ to $$V$$ is still zero. (The simplest way to prove that $$V$$ is simply connected is to see that it's star-shaped with respect to any point on the positive $$x$$-axis.)

So for (ii)(b), how do I find $\phi$ in $S_1$ ?

Presumably then I can solve the simultaneous equations $x=r\cos\,\phi$ and $y=r\sin\,\phi$ in terms of $x$ and $y$ (e.g. using $\tan^{-1}$) and then verify by explicit differentiation that $\nabla \phi (x,y) = \mathbf{F} (x,y)$ . And then do $S_2$ and $S_3$ in a similar way.

 

[ystael]

$$(r, \phi)$$ are just polar coordinates with a particular choice of range for the angular coordinate; what does that tell you the angular coordinate should be on the positive $$x$$-axis?

 

[Ted123]

$$(r, \phi)$$ are just polar coordinates with a particular choice of range for the angular coordinate; what does that tell you the angular coordinate should be on the positive $$x$$-axis?

Well for $S_1$ the angle would be 0 but what is the range of $r$, is it $\pi$ ?

For $S_2$ the angle is $\frac{\pi}{2}$ but again what is $r$, again is it $\pi$ ?

For $S_3$ the angle is $\frac{3\pi}{2}$, is $r, -\pi$ ?

So are the angular coordinates of $S_1, S_2,S_3, (\pi, 0), (\pi, \pi/2), (-\pi,3\pi/2)$ respectively?

 

[Ted123]

Well for $S_1$ the angle would be 0 but what is the range of $r$, is it $\pi$ ?

For $S_2$ the angle is $\frac{\pi}{2}$ but again what is $r$, again is it $\pi$ ?

For $S_3$ the angle is $\frac{3\pi}{2}$, is $r, -\pi$ ?

So are the angular coordinates of $S_1, S_2,S_3, (\pi, 0), (\pi, \pi/2), (-\pi,3\pi/2)$ respectively?

$\frac{y}{x} = \tan(\phi)$

so $\phi (x,y) = \tan^{-1} \left(\frac{y}{x}\right)$

and $\nabla \phi(x,y) = \mathbf{F}$

But how do I find $\phi$ in $S_2$ and $S_3$? $x$ can be 0 in both of these.

 

[ystael]

Sounds like you need to go back to a trigonometry text and review how polar coordinates work. $$r$$ is not an angular coordinate; it's a distance (from the origin). And if $$x = 0$$, then you already know what the value of $$\phi$$ should be, depending on whether $$y > 0$$ or $$y < 0$$.

 

[Ted123]

Sounds like you need to go back to a trigonometry text and review how polar coordinates work. $$r$$ is not an angular coordinate; it's a distance (from the origin). And if $$x = 0$$, then you already know what the value of $$\phi$$ should be, depending on whether $$y > 0$$ or $$y < 0$$.

I know r is a distance but I thought seen as $\phi\in(-\pi, \pi)$ that that distance would be those values.

 

[Ted123]

I know r is a distance but I thought seen as $\phi\in(-\pi, \pi)$ that that distance would be those values.

Can anyone see how I do the last part of (ii)(c)? Why can't $\phi$ be extended to x<0 ?

 

[ystael]

Let $$\varepsilon$$ be small. What is $$\phi$$ close to at $$(-1, \varepsilon)$$? What is $$\phi$$ close to at $$(-1, -\varepsilon)$$? How does this make it difficult to find a value for $$\phi$$ at $$(-1, 0)$$?

 

[Ted123]

I know r is a distance but I thought seen as $\phi\in(-\pi, \pi)$ that that distance would be those values.

$\displaystyle \oint = 0 \neq 2\pi$