| Line 18: | Line 18: | ||
<math> 0 = \int\nabla u\ ds </math> | <math> 0 = \int\nabla u\ ds </math> | ||
| + | |||
| + | If we have a domain, <math>\Omega</math> and <math>\gamma</math> a curve in <math>\Omega</math>, where <math>A, B </math> are end points of <math>\gamma</math>, from vector calculus, we have | ||
| + | |||
| + | <math> u(B) - u(A) = \int_\gamma \nabla u \ ds </math>. | ||
| + | |||
| + | but the integral is zero since <math> \nabla u </math> is zero. Hence <math> u(A) = u(B) </math> for any arbitrary point <math>A</math> and <math>B</math> and <math>u</math> is constant in <math>\Omega</math>. | ||
Revision as of 16:15, 30 September 2012
Practice material for Exam 1 collaboration space
You can easily talk about math here, like this:
$ e^{i\theta} = \cos \theta + i \sin \theta. $
Is this the Cauchy Integral Formula?
$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a} \ dz $
This isn't directly related to the practice exam, but is concerning a fact discussed in class.
In one of the first lessons an important fact was provided. Namely, Suppose u is continuously a differentiable function on a connected open set $ \Omega $ and that $ \nabla u \equiv 0 $ Then u must be constant on $ \omega $.
How/Why is
$ 0 = \int\nabla u\ ds $
If we have a domain, $ \Omega $ and $ \gamma $ a curve in $ \Omega $, where $ A, B $ are end points of $ \gamma $, from vector calculus, we have
$ u(B) - u(A) = \int_\gamma \nabla u \ ds $.
but the integral is zero since $ \nabla u $ is zero. Hence $ u(A) = u(B) $ for any arbitrary point $ A $ and $ B $ and $ u $ is constant in $ \Omega $.
