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Let's say I have the quarter plane domain on the x-y plane that is the region $x>0$ and $y>0$, then I transform this region into the $(\omega,\tau)$ plane defiend by the following equations:

$$ x = \omega \tau$$

$$y= \omega - \omega \tau$$

And by rearranging: $$ x+y = \omega \tag{1}$$

and,

$$ \frac{x}{x+y} = \tau \tag{2}$$

How do I fugre out how to sketch the region on the $(\omega,\tau)$ plane? / The corresponding bounds? So far, I've got that the lines parallel $x+y=C$ where $C$ is some constant in the $(x,y)$ plane are mapped to lines of constant $\omega$ and hence $\omega$ runs from $0 \to \infty$, and, curves of the form $$ \frac{x}{x+y} =C$$ are mapped to lines of constant $\tau$ but I'm having a bit of a hard time visualizing the way $\tau$ is plotted on the target plane and hence the bounds.

Clemens Bartholdy
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  • These are lines through the origin of the form $y=\left(\frac{1}{C}-1\right)x$, which means $\tau$ sweeps out the domain like an angle, from the $x$ axis to the $y$ axis, which are the lines $\tau = 1$ and $0$, respectively. Thus $0\leq \tau \leq 1$ – Ninad Munshi Apr 30 '21 at 19:45
  • Could you explain how you got that $\tau$ sweeps the domain like an angle? I got the manipulation to the line form but I don't get it after that @Ninad Munshi – Clemens Bartholdy Apr 30 '21 at 20:50

1 Answers1

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Keep in mind your original bounds over the first quadrant $x>0$ and $y>0$. Taking your first equation, we see immediately that since $x,y\in (0,\infty)$ then the sum $ x+y = \omega \in (0,\infty)$.

Using your second equation, it is clear that $$\tau= \frac{x}{x+y} $$ can never be negative nor can it ever be $0$ since $x$ cannot be $0$. In fact, you can quickly convince yourself that no combination of $x,y\in (0,\infty)$ can yield a value for $\tau$ outside of $\tau \in (0,1)$. Note that this is an open interval equivalent to $0<\tau<1$ and NOT the closed interval $0 \leq \tau \leq 1$.

If you need more convincing, start looking at the single variable limits as $x$ or $y$ goes to either $0$ or $\infty$.

vb628
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  • Perhaps a stupid question, but how would we rule out the idea that the boundary region on the $(\omega,\tau)$ plane is not a curve ? – Clemens Bartholdy May 01 '21 at 07:05
  • @Buraian look at the Jacobian for the transformation. It is easy to see that it has full rank. In fact, the Jacobian determinant is $$\bigg|\dfrac{\partial (x,y)}{\partial(\omega,\tau)}\bigg| = -\omega$$ which cannot be zero. – vb628 May 01 '21 at 15:50
  • How does the jacobian determinant ratio relate to the nature of the boundary? This is my first time encountering this idea, I think this may also be the point I was looking for. Could you give me the keywords to search up to learn more of the idea you have mentioned? – Clemens Bartholdy May 01 '21 at 16:08
  • @Buraian, I think I misread your first comment (my fault). The region of the first quadrant as you have described above does not contain its boundary in $(x,y)$ space or in $(\omega, \tau)$ space. The region is an open set. When we integrate, we oftentimes needn't worry about when our region doesn't contain its boundaries. I thought you were asking how we knew that the transformation spanned the entire first quadrant. This is when we look at the Jacobian to see if it has full rank (thus we look at its determinant). – vb628 May 01 '21 at 17:27
  • Open set.. hmm isn't open and closed described in reference to a topology? From my understanding, you say a set is open if it is in a topology and closed if it is not. Other than that, I am still interested in how the rank of the jacobian relates to how the area is on the new plane. Do you have any reference for studying that? – Clemens Bartholdy May 01 '21 at 19:07
  • @Buraian, open really just means that a space doesn't contain the boundary. We clearly can't include the boundary as we move off to $+\infty$ and because we started with conditions $x,y>0$ and not $x,y\geq 0$ we clearly don't have a minimum value for $x$ or $y$ but only an infimum value of zero. You can find the technical definition explained in the answer to this question. – vb628 May 01 '21 at 19:29
  • @Buraian The Jacobian determinant itself is what gives you the "volume form" you're looking for. Literally we have that

    $$ \text{d}x \text{d}y = \bigg|\dfrac{\partial (x,y)}{\partial(\omega,\tau)}\bigg| \text{d}\omega \text{d}\tau \text{ .} $$

    The book that I like most on this matter is The Geometry of Physics by Frankel. Any good vector calculus book will include discussion about this including a lot of analysis books if you prefer that format.

     – vb628 May 01 '21 at 19:30
  • Well yes, I know the equation you have written right now, that's just how the area element changes but I am asking for the point relating the rank and span. I'll check the q youve linked – Clemens Bartholdy May 01 '21 at 19:44
  • After two weeks I finally understood. Though I must ask, this really required a 'nice set up' to be there to work. Isn't there like an algorithmic approach to these things? – Clemens Bartholdy May 17 '21 at 18:42
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    @Buraian it seems like looking at the limits of $\tau$ and $\omega$ as $x$ and $y$ approach their boundaries is a good place to start. – vb628 May 17 '21 at 23:42
  • Ah, right I'll keep that in mind. See what happens at the boundaries – Clemens Bartholdy May 18 '21 at 07:45