Toronto Math Forum
APM3462012 => APM346 Math => Misc Math => Topic started by: Kun Guo on November 24, 2012, 06:23:08 PM

I do not quite understand step b and c(see attached) in the proof of Meanvalue theorem. Did you use Green's identity or some other identity when dragging out terms from the integral?

b. Gauss identity (you may call it Green in 2D). We proved this before.
c. $\Sigma$ is a sphere of radius $r$ with a center $y$ so $G$ so for $x\in \Sigma$ $G(x,y)=c xy^{2n}=cr^{2n}$ is constant and could be moved out of integral.

Hi Professor Ivrii. I have a question on the attachment above. In part b, after dragging out the factor, why the remainder is (partial u)/(partial v) instead of u? Thanks.

Hi Professor Ivrii. I have a question on the attachment above. In part b, after dragging out the factor, why the remainder is (partial u)/(partial v) instead of u? Thanks.
Right, fixed and details added
http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect26.4 (http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect26.4)

thanks professor, now it makes more sense :)

thanks professor, but the formulae are not visible for me.

Try again

Thanks. It works now. :)
Try again

Hi,
what is meant by $\frac{\partial G}{\partial v_x}$ (eq. 7 in Lec. 26)? is $\frac{\partial G}{\partial v_x}=\nabla G \cdot (\nabla \cdot \vec{v})$?
Thanks!

or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?

or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?
No, $\nu_x$ means only that we differentiate with respect to $x$

Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)

Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)
No, what you are writing is wrong. Function $G(x,y)$ depends on both $x$ and $y$ and we need to differentiate with respect to $x$ not $y$. So in fact I mean is
$$
\nabla_x G(x,y) \cdot \nu(x)
$$
where $\nabla_x$ means gradient with respect to $x$. Here $y$ is considered as a parameter

I see, thanks.
Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?

I see, thanks.
Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?
Yes, it is exactly correct except $\Sigma \subset \Omega$. Alternatively we can write $\max_{\bar{\Omega}}u \geq \max_{\Sigma}u$ where $\bar{\Omega}=\Omega \cup \Sigma$.

Here we integrate over V. Is this V the same as omega? And do we know if the coefficient c in G(x,y) is positive or negative? Thanks!

Here we integrate over V. Is this V the same as omega? And do we know if the coefficient c in G(x,y) is positive or negative? Thanks!
Yes, should be $\Omega$. No idea what coefficient your are talking about