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In my Separation of Variables lecture notes, it is said that we require a rectangular domain for us to apply this method of solving PDEs. What exactly does it mean ?

We have the example where a Circle is not a rectangular domain in Cartesians coordinates but it becomes one in Polar Coordinates.

So the questions are : What exactly is a rectangular domain ?

Cheers !

bsaoptima
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    Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Oct 26 '22 at 17:14
  • Probably rectangular means that in some choice of coordinates it is rectangular. You unfortunately would probably need to know how to transform into those coordinates to use the result. – N8tron Oct 26 '22 at 17:38

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In $\mathbb R^n$, a rectangular domain simply means a domain of the form: $$ \tag{$\ast$}(a_1,b_1)\times \cdots \times(a_n,b_n)$$ with $a_i<b_i$ for each $i=1,\dots,n$. For $n=2$ this is literally a rectangle and for $n=3$ it is a rectangular prism.

It is common to also call any domain in the form of ($\ast$) except with any combination of open intervals, closed interval, or half-open intervals rectangular i.e. $[0,1] \times [0,1)$ is considered rectangular.

Now for separation of variables, I don't entirely agree with how your lecture notes has phrased it - I think it is better to say that separation of variables is possible for domains that are rectangular possibly after a change of coordinates. For example, your lecture notes seems to suggest that a disk is rectangular! This is silly - what it means is that after converting to polar coordinates, a disk is transformed in to a rectangular region.

JackT
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  • This makes way more sense. And for the lecture notes, it was phrased like you so the mistake was more on my side haha. From what I understand having a rectangular domain in the end allows our equations for each variables to act on the same "segments" regardless of how the others behave. I don't know if that is making sense. – bsaoptima Oct 27 '22 at 18:56