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The Wikipedia page says that it would involve the derivative with respect to some normal vector being constant, but I don't quite understand this.

  • Is the value of the normal vector adjusted across the domain, or does it remain the same?
  • Can I say that the standard partial derivative is the derivative with respect to the normal vector if I have a rectangular domain?

Thank you for your help.

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Is the value of the normal vector adjusted across the domain, or does it remain the same?

Consider a domain (nonempty connected open set) $\Omega\subset{\bf R}^n$ with smooth boundary $\partial \Omega$. At each point $x\in\partial \Omega$, we have a normal vector $n(x)$.

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Consider for instance $\Omega$ being a closed unit ball in ${\bf R}^n$ (in the case when $n=2$, you have a closed unit disk). One has $$ n(x)=x. $$

A particular Neumann boundary condition looks like $$ \frac{\partial u}{\partial n}(x)=f(x),\quad x\in\partial \Omega $$ which is understood as $$ \nabla u(x)\cdot n(x)=f(x),\quad x\in\partial\Omega. $$


Can I say that the standard partial derivative is the derivative with respect to the normal vector if I have a rectangular domain?

I would not say that. Note that the directional derivative in the direction of a normal vector only makes sense on the boundary of a domain. But you might have something like this:

enter image description here

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