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under context of PDE's

Given a bounded set $\Omega$ and a smooth function $f:\Omega\to \mathbb{R}$ what does it mean to define $\frac{\partial f}{\partial n}=0$ on $\partial \Omega$? the function $f$ is not defined outside $\Omega$ so how can we take derivative for a point on the boundary?

thank you

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It depends on the context. If $\partial\Omega$ has an outward pointing normal $\nu(q)$ almost everywhere for $q\in\partial\Omega$ and $f$ is defined on $\partial\Omega$ as well, this could be interpreted as $$\lim_{h\to 0^+}\frac{f(q-h\nu(q))-f(q)}{h}=0,$$ for almost all $q\in\partial\Omega$.

The normal derivative of the boundary can also be considered in the weak sense, but this depends on the equation $f$ solves. If, for example, $f$ is a harmonic function, we say that $\frac{\partial f}{\partial\nu}=0$, if $$\int_{\Omega}\nabla f\cdot\nabla \phi=0$$ for all $\phi$ which are compactly supported and infinitely differentiable in $\mathbb R^d$.

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  • $\begingroup$ thank you. In the limit you have $f(q-h\nu(q))$. if you do not restrict the sign of $h$ the point $q-h\nu(q)$ might be outside the domain isn't it? $\endgroup$ – Ricky Feb 17 '16 at 0:41
  • $\begingroup$ You are right, I corrected it. $\endgroup$ – detnvvp Feb 17 '16 at 0:42

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