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I have $\Omega$ the following domain $$ \Omega = \left\{\left(x_1,x_2\right) \in \mathbb{R}^2, \ 1 \leq \sqrt{x_1^2+x_2^2} \leq 2\right\} \text{ and }u\left(x_1,x_2\right)=\ln\left(\sqrt{x_1^2+x_2^2}\right) $$

I'm asked to calculate $\displaystyle \frac{ \partial u }{\partial n}$ on $\Gamma$.

I guess $\Gamma$ is the boundary of $\Omega$ but I dont know what is $n$ and i've no idea how to calculate this. I know how to calculate derivative relatively to $x_1$ or $x_2$

Atmos
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2 Answers2

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Recall that for a differentiable function the directional derivative is given by

$$\frac{ \partial u }{\partial n}=\nabla u \cdot n$$

where, in that case if we refer to the boundary, $n$ is the unit normal vectors to the circle centered at the origin that is for $P=(x,y)$

$$n=\pm \left( \frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} \right)$$

and

$$\nabla u=(u_x,u_y)=\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)$$

user
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  • I've to admit that I dont understand the formula ( I understand nabla etc but how can from what it stems ? ) – Atmos Sep 24 '18 at 10:02
  • @Atmos It is just a generalization of partial derivative. For a differentiable function, in a generic direction $n$ the directional derivative is given by the dot product of the gradient vector and the vector $n$. – user Sep 24 '18 at 10:04
  • @Atmos Refer also to the Related OP – user Sep 24 '18 at 10:06
  • Thanks a lot @gimusi – Atmos Sep 24 '18 at 10:09
  • @Atmos I need to fix something on n! – user Sep 24 '18 at 10:12
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    @Atmos Now it should be ok, indeed of course n depends upon x and y! Let me know for any other doubts on that! Bye – user Sep 24 '18 at 10:15
  • is this $f$ or $\nabla u$ ? :p – Atmos Sep 24 '18 at 10:23
  • @Atmos Yes of course, I change the synmol according to your one! – user Sep 24 '18 at 10:25
  • Last question : if i find that such a derivative is null what can i conclude ? – Atmos Sep 24 '18 at 10:28
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    @Atmos Nothing special, but for that particular function it is never true. Indeed the function in any radial direction form the origin is $u(r)= \ln r$ with $1\le r \le \sqrt 2$ and $u'(r)=\frac1r$ which indeed is the same result we obtain by the directional derivative. – user Sep 24 '18 at 10:32
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As you are dealing with an annulus and the function is radially symmetric, switch to polar coordinates.

The normal to your boundary is then just $r/|r|$ as you want to move orthogonally to the circle.

b00n heT
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