Toronto Math Forum
APM346-2012 => APM346 Math => Misc Math => Topic started by: Kun Guo on November 24, 2012, 06:23:08 PM
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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?
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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.
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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.
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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 (http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4)
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thanks professor, now it makes more sense :)
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thanks professor, but the formulae are not visible for me.
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Try again
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Thanks. It works now. :)
Try again
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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!
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or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?
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or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?
No, $\nu_x$ means only that we differentiate with respect to $x$
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Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)
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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
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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?
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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 $\bar{\Omega}=\Omega \cup \Sigma$.
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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!
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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