Toronto Math Forum

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

Title: Mean-value theorem
Post 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 Mean-value theorem. Did you use Green's identity or some other identity when dragging out terms from the integral?
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 24, 2012, 07:07:57 PM
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 $x\in \Sigma$  $G(x,y)=c |x-y|^{2-n}=cr^{2-n}$ is constant and could be moved out of integral.
Title: Re: Mean-value theorem
Post by: Ziting Zhou on November 25, 2012, 04:56:23 PM
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.
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 25, 2012, 05:26:24 PM
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.

http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4 (http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4)
Title: Re: Mean-value theorem
Post by: Kun Guo on November 25, 2012, 05:46:40 PM
thanks professor, now it makes more sense  :)
Title: Re: Mean-value theorem
Post by: Ziting Zhou on November 25, 2012, 05:54:05 PM
thanks professor, but the formulae are not visible for me.
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 25, 2012, 06:19:55 PM
Try again
Title: Re: Mean-value theorem
Post by: Ziting Zhou on November 25, 2012, 06:23:30 PM
Thanks. It works now.  :)
Try again
Title: Re: Mean-value theorem
Post by: Thomas Nutz on November 26, 2012, 01:48:06 PM
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!
Title: Re: Mean-value theorem
Post by: Thomas Nutz on November 26, 2012, 01:51:41 PM
or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 26, 2012, 03:15:53 PM
or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?

No, $\nu_x$ means only that we differentiate with respect to $x$
Title: Re: Mean-value theorem
Post by: Tse Chiang Chen on November 26, 2012, 03:48:43 PM
Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 26, 2012, 04:00:59 PM
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
Title: Re: Mean-value theorem
Post by: Thomas Nutz on November 26, 2012, 06:34:06 PM
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?
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 27, 2012, 03:46:53 AM
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 $\bar{\Omega}=\Omega \cup \Sigma$.
Title: Re: Mean-value theorem
Post by: Qitan Cui on November 27, 2012, 08:37:08 AM
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!
Title: Re: Mean-value theorem
Post by: Victor Ivrii on November 27, 2012, 09:37:58 AM
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