I have a question about a directional derivative. I put the complete proposition but my question is only in the last line, that is:

$\boxed{\text{why}\;\,-\dfrac{\partial u}{\partial\mathbf{n}}(x_*)=\displaystyle\lim_{x\to x_*}\dfrac{u(x)-u(x_*)}{R-|x|}}$ ?

This is a screenshot of the book:


Thank respond in detail!!

P.D.: part of the book: Partial Differential Equations, Emmanuele DiBenedetto

  • 2
    $\begingroup$ Please update your question to give proper credit to the textbook you used / its author. It is poor form to scan other peoples' copyrighted works, especially without attribution! $\endgroup$ – Nick Peterson Sep 21 '13 at 0:49
  • $\begingroup$ What is $(5.1)$? $\endgroup$ – robjohn Sep 21 '13 at 0:59
  • $\begingroup$ Nicholas: apology, now I've written the author credits and book. robjohn. (5.1) is a previous result, whose direct application gives that inequality. It is not important. My question is only about the last line: the limit. $\endgroup$ – yemino Sep 21 '13 at 1:05

It seems the only thing missing is that as you take $x$ to $x^\ast$ via an orthogonal direction , you can express the distance $|x-x^\ast|$ as $R-|x-x_0|$. I think there is a typo here, probably an artifact from a previous version which said something like we can take $x_0=0$ without loss of generality or something like that.

  • $\begingroup$ Sorry evan, but I don't know how to relacionate $\dfrac{\partial u}{\partial\mathbf{n}}(x_*)$ with $\displaystyle\lim_{x\to x_*}\frac{u(x)-u(x_*)}{R-|x|}$ The definition that I know is: $\dfrac{\partial u}{\partial\mathbf{n}}(x_*)=\displaystyle\lim_{h\to0}\frac{u(x_*+h\mathbf{n})}{h}$ $\endgroup$ – yemino Sep 22 '13 at 13:48
  • $\begingroup$ In this case your $h$ is the same as my $x-x_\ast$, evaluate at two points and divide by the distance then take limit. $\endgroup$ – Evan Sep 22 '13 at 16:31

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