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# Degree 2 Nédélec (first 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: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle - x^{2}\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y^{2}\\\displaystyle - x y\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{7}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1)\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle y \left(- 4 x - 4 y + 3\right)\\\displaystyle x \left(4 x + 4 y - 3\right)\end{array}\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{1}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(\sqrt{3} \cdot \left(2 s_{0} - 1\right))\hat{\boldsymbol{t}}_{0}$$
where $$e_{0}$$ is the 0th edge;
$$\hat{\boldsymbol{t}}_{0}$$ is the tangent to edge 0;
and $$s_{0},s_{1}$$ is a parametrisation of $$e_{0}$$.

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y \left(4 x - 4 y + 1\right)}{3}\\\displaystyle \frac{\sqrt{3} x \left(- 4 x + 4 y + 1\right)}{3}\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle y \left(1 - 4 x\right)\\\displaystyle 4 x^{2} - 5 x + 1\end{array}\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{3}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{1}}\boldsymbol{v}\cdot(\sqrt{3} \cdot \left(2 s_{0} - 1\right))\hat{\boldsymbol{t}}_{1}$$
where $$e_{1}$$ is the 1st edge;
$$\hat{\boldsymbol{t}}_{1}$$ is the tangent to edge 1;
and $$s_{0},s_{1}$$ is a parametrisation of $$e_{1}$$.

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y \left(4 x + 8 y - 5\right)}{3}\\\displaystyle \frac{\sqrt{3} \left(- 4 x^{2} - 8 x y + 7 x + 6 y - 3\right)}{3}\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle 4 y^{2} - 5 y + 1\\\displaystyle x \left(1 - 4 y\right)\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{5}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{2}}\boldsymbol{v}\cdot(\sqrt{3} \cdot \left(2 s_{0} - 1\right))\hat{\boldsymbol{t}}_{2}$$
where $$e_{2}$$ is the 2nd edge;
$$\hat{\boldsymbol{t}}_{2}$$ is the tangent to edge 2;
and $$s_{0},s_{1}$$ is a parametrisation of $$e_{2}$$.

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} \left(- 8 x y + 6 x - 4 y^{2} + 7 y - 3\right)}{3}\\\displaystyle \frac{\sqrt{3} x \left(8 x + 4 y - 5\right)}{3}\end{array}\right)$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}\sqrt{2}\\0\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle 4 \sqrt{2} y \left(- x - 2 y + 2\right)\\\displaystyle 4 \sqrt{2} x \left(x + 2 y - 1\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.
$$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{R}\boldsymbol{v}\cdot\left(\begin{array}{c}0\\\sqrt{2}\end{array}\right)$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle 4 \sqrt{2} y \left(2 x + y - 1\right)\\\displaystyle 4 \sqrt{2} x \left(- 2 x - y + 2\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element.