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I know that when a function is zero on a boundary, say $u=0$ on $\Gamma$, then $\nabla u = (\nabla u \cdot \nu)\nu$ where $\nu$ is the outward normal vector to $\Gamma$. My question is why is $\nabla u \cdot \tau$ ($\tau$ is the unit tangent vector) zero in this case? What if $u = const.$ on $\Gamma$. Is $\nabla u \cdot \tau$ also zero in this case?

Julienne Franz
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    we know that on the level surface, $\nabla u$ is parallel to the normal vector. It means that $\nabla u = c \nu$ for some $c$. Moreover, normal vector $\nu$ is perpendicular to the tangent vector, say $\tau$. Thus, $\nabla u \cdot \tau = c (\nu \cdot \tau ) = 0.$ – Shanna May 21 '21 at 07:14
  • So, in general, if $u = const.$, $\nabla u \cdot \tau$ is also zero, right? Thanks! – Julienne Franz May 22 '21 at 02:31

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