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futurebird
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Complex analysis: having partials is the same as being "well defined?"
My professor proved this theorem in class and I don't know if I even wrote it down correctly in my notes. I don't have access to the book so I need to know if this makes sense. Here is the theorem:
Under these conditions:
R is a simply connected region in \mathbb{C} and
p,q : R \rightarrow \mathbb{R} continuous.
\sigma - rect is a rectangle in \mathcal{R}.
The existence of a function U on R such that:
\frac{\partial U}{\partial x}=p, \frac{\partial U}{\partial y}=q
\Longleftrightarrow \displaystyle\oint_{\sigma - rect} pdx + qdy =0
---
QUESTIONS:
1. My professor said that having the partials for p and q is the same thing as saying that U is well defined. Why is this the case?
2. Then he said U would be well defined if it was path independent, well, if we know it's path independent then, of course \displaystyle\oint_{\sigma - rect} pdx + qdy =0. So, am I right to think of path independence as a consequence of this proof, not a condition? (I know this seems obvious, but I keep losing track of what we "know" when we start the proof and what we are trying to prove so, I want to get this absolutely clear!)
My professor proved this theorem in class and I don't know if I even wrote it down correctly in my notes. I don't have access to the book so I need to know if this makes sense. Here is the theorem:
Under these conditions:
R is a simply connected region in \mathbb{C} and
p,q : R \rightarrow \mathbb{R} continuous.
\sigma - rect is a rectangle in \mathcal{R}.
The existence of a function U on R such that:
\frac{\partial U}{\partial x}=p, \frac{\partial U}{\partial y}=q
\Longleftrightarrow \displaystyle\oint_{\sigma - rect} pdx + qdy =0
---
QUESTIONS:
1. My professor said that having the partials for p and q is the same thing as saying that U is well defined. Why is this the case?
2. Then he said U would be well defined if it was path independent, well, if we know it's path independent then, of course \displaystyle\oint_{\sigma - rect} pdx + qdy =0. So, am I right to think of path independence as a consequence of this proof, not a condition? (I know this seems obvious, but I keep losing track of what we "know" when we start the proof and what we are trying to prove so, I want to get this absolutely clear!)
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