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Let $\Gamma$ be a smooth boundary of a domain $\Omega$. Let $u = g$ on $\Gamma$. How can I compute the tangential derivative of the function $u$ using the information that $u = g$ on $\Gamma$? Please provide a reference.

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  • $\begingroup$ You mean mathematically, or in actual practice? If the latter, how do you represent $g$? $\endgroup$ Commented May 12 at 17:02

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As @WolfgangBangerth already mentioned, the practical implementation depends on the representation of $\mathbf g$. I will give a short answer for a polynomial curve.


If $\mathbf{g}$ are point values on a smooth curve $\boldsymbol\Gamma$ you can calculate the tangential derivative

$$ \begin{align} \frac{ \partial \mathbf g}{ \partial \Gamma} = \frac{1}{\|\boldsymbol{\dot\gamma}(\xi)\|}\circ \frac{ \partial \mathbf g}{ \partial \xi}, \\ \end{align} $$

with

$$ \begin{align} \|\boldsymbol{\dot\gamma}(\xi)\| = \sqrt{\left(\frac{ \partial \mathbf{x}}{ \partial \xi}\right)^2 + \left(\frac{ \partial \mathbf y}{ \partial \xi}\right)^2}, \\ \end{align} $$

and

$$ \begin{align} \frac{ \partial \mathbf x}{ \partial \xi} = \mathbf{\mathcal D}\mathbf x, \quad \frac{ \partial \mathbf y}{ \partial \xi} = \mathbf{\mathcal D}\mathbf y, \quad \frac{ \partial \mathbf g}{ \partial \xi} = \mathbf{\mathcal D}\mathbf g. \\ \end{align} $$

Here $\mathcal D$ is defined as derivative matrix, $\circ$ is the elementwise multiplication and $\mathbf x, \mathbf y$ are coordinates. The curve may be parametrized using $\xi \in [0,1]$. Note that $\mathbf x,\mathbf y, \mathbf g$ are nodal values

$$ \begin{align} \mathbf x = \left(x_1, x_2, \dots, x_n\right), \quad \mathbf y = \left(y_1, y_2, \dots, y_n\right), \quad \mathbf g = \left(g_1, g_2, \dots, g_n\right). \\ \end{align} $$


Example: ($n=2$)

If $\Gamma$ consists of many linear segments $\gamma_i$, you can calculate the tangential derivative on each segment using

$$ \begin{align} \frac{ \partial \mathbf g}{ \partial \gamma_i} = \frac{ g_2 - g_1}{ \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2}} . \\ \end{align} $$

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