Author Topic: Mean-value theorem  (Read 9501 times)

Kun Guo

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Mean-value theorem
« 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?

Victor Ivrii

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Re: Mean-value theorem
« Reply #1 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.

Ziting Zhou

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Re: Mean-value theorem
« Reply #2 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.

Victor Ivrii

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Re: Mean-value theorem
« Reply #3 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.

Right, fixed and details added
http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4

Kun Guo

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Re: Mean-value theorem
« Reply #4 on: November 25, 2012, 05:46:40 PM »
thanks professor, now it makes more sense  :)

Ziting Zhou

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Re: Mean-value theorem
« Reply #5 on: November 25, 2012, 05:54:05 PM »
thanks professor, but the formulae are not visible for me.

Victor Ivrii

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Re: Mean-value theorem
« Reply #6 on: November 25, 2012, 06:19:55 PM »
Try again

Ziting Zhou

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Re: Mean-value theorem
« Reply #7 on: November 25, 2012, 06:23:30 PM »

Thomas Nutz

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Re: Mean-value theorem
« Reply #8 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!

Thomas Nutz

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Re: Mean-value theorem
« Reply #9 on: November 26, 2012, 01:51:41 PM »
or rather $\nabla G \cdot  \frac{\partial}{\partial x}(\vec{v})$?

Victor Ivrii

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Re: Mean-value theorem
« Reply #10 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 $x$

Tse Chiang Chen

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Re: Mean-value theorem
« Reply #11 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)
« Last Edit: November 26, 2012, 03:51:53 PM by Tse Chiang Chen »

Victor Ivrii

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Re: Mean-value theorem
« Reply #12 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

Thomas Nutz

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Re: Mean-value theorem
« Reply #13 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?

Victor Ivrii

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Re: Mean-value theorem
« Reply #14 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 $\bar{\Omega}=\Omega \cup \Sigma$.