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Kreyszig's introduction of the normal derivative

$u_n = \partial u / \partial n$

leaves me a bit puzzled as to what variable $n$ is if $u$ is a function of $x$ and $y$.

The trailing example uses a condition for one of the boundaries specified as $u_n = 6x$ and the solution steps deduces $\dfrac {\partial u_{ij}} {\partial n} = \dfrac {\partial u_{ij}} {\partial y} = 6x$ out of thin air.

So what exactly is the normal derivative? Is it simply an interpretation of a normal vector standing up from $u$ if it's treated as a surface, i.e. $\dfrac {\partial u_{ij}} {\partial n} = \dfrac {\partial u_{ij}} {\partial x} + \dfrac {\partial u_{ij}} {\partial y}$ ?

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1 Answer 1

up vote 3 down vote accepted

The normal derivative is a directional derivative in a direction that is outwardly normal (perpendicular) to some curve, surface or hypersurface (that is assumed from context) at a specific point on the aforementioned curve, surface or hypersurface. If $\mathbf{N}$ is the normal vector then $\partial u / \partial n$ stands for $\vec{\nabla} u \cdot \mathbf{N}$. From the example you give it seems that $\mathbf{N}=(0,1)$ on whatever boundary is being considered.

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