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# Degree 1 Nédélec (second kind) on a triangle

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In this example:
• $$R$$ is the reference triangle. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{5}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle 2 y\\\displaystyle 4 x\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle - 4 y\\\displaystyle - 2 x\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{2}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{t}}_{1}$$ is the tangent to edge 1.

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle - 2 y\\\displaystyle - 4 x - 6 y + 4\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{1}$$
where $$e_{1}$$ is the 1st edge;
and $$\hat{\boldsymbol{t}}_{1}$$ is the tangent to edge 1.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle 4 y\\\displaystyle 2 x + 6 y - 2\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{4}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{t}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{t}}_{2}$$ is the tangent to edge 2.

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle - 6 x - 4 y + 4\\\displaystyle - 2 x\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{t}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
and $$\hat{\boldsymbol{t}}_{2}$$ is the tangent to edge 2.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle 6 x + 2 y - 2\\\displaystyle 4 x\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.