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# Degree 1 Tiniest tensor H(div) on a quadrilateral

◀ Back to Tiniest tensor H(div) definition page In this example:
• $$R$$ is the reference quadrilateral. The following numbering of the subentities of the reference is used:
• • $$\mathcal{V}$$ is spanned by: $$\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle x y\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle x y\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle 0\\\displaystyle \frac{3 y \left(1 - y\right)}{2}\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle \frac{3 x \left(1 - x\right)}{2}\\\displaystyle 0\end{array}\right)$$, $$\left(\begin{array}{c}\displaystyle \frac{3 x \left(2 x y - x - 2 y + 1\right)}{2}\\\displaystyle \frac{3 y \left(2 x y - 2 x - y + 1\right)}{2}\end{array}\right)$$
• $$\mathcal{L}=\{l_0,...,l_{10}\}$$
• Functionals and basis functions: $$\displaystyle l_{0}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{0}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{0}$$
where $$e_{0}$$ is the 0th edge;
and $$\hat{\boldsymbol{n}}_{0}$$ is the normal to facet 0.

$$\displaystyle \boldsymbol{\phi}_{0} = \left(\begin{array}{c}\displaystyle \frac{9 x \left(- 2 x y + x + 2 y - 1\right)}{2}\\\displaystyle - 9 x y^{2} + 15 x y - 6 x + \frac{15 y^{2}}{2} - \frac{23 y}{2} + 4\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{1} = \left(\begin{array}{c}\displaystyle \frac{9 x \left(2 x y - x - 2 y + 1\right)}{2}\\\displaystyle 9 x y^{2} - 15 x y + 6 x - \frac{3 y^{2}}{2} + \frac{7 y}{2} - 2\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{2} = \left(\begin{array}{c}\displaystyle 9 x^{2} y - \frac{15 x^{2}}{2} - 15 x y + \frac{23 x}{2} + 6 y - 4\\\displaystyle \frac{9 y \left(2 x y - 2 x - y + 1\right)}{2}\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{3} = \left(\begin{array}{c}\displaystyle - 9 x^{2} y + \frac{3 x^{2}}{2} + 15 x y - \frac{7 x}{2} - 6 y + 2\\\displaystyle \frac{9 y \left(- 2 x y + 2 x + y - 1\right)}{2}\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{4} = \left(\begin{array}{c}\displaystyle \frac{x \left(18 x y - 15 x - 6 y + 7\right)}{2}\\\displaystyle \frac{9 y \left(2 x y - 2 x - y + 1\right)}{2}\end{array}\right)$$

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

$$\displaystyle \boldsymbol{\phi}_{5} = \left(\begin{array}{c}\displaystyle \frac{x \left(- 18 x y + 3 x + 6 y + 1\right)}{2}\\\displaystyle \frac{9 y \left(- 2 x y + 2 x + y - 1\right)}{2}\end{array}\right)$$

This DOF is associated with edge 2 of the reference element. $$\displaystyle l_{6}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(1 - s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$e_{3}$$ is the 3th edge;
and $$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3.

$$\displaystyle \boldsymbol{\phi}_{6} = \left(\begin{array}{c}\displaystyle \frac{9 x \left(- 2 x y + x + 2 y - 1\right)}{2}\\\displaystyle \frac{y \left(- 18 x y + 6 x + 15 y - 7\right)}{2}\end{array}\right)$$

This DOF is associated with edge 3 of the reference element. $$\displaystyle l_{7}:\boldsymbol{v}\mapsto\displaystyle\int_{e_{3}}\boldsymbol{v}\cdot(s_{0})\hat{\boldsymbol{n}}_{3}$$
where $$e_{3}$$ is the 3th edge;
and $$\hat{\boldsymbol{n}}_{3}$$ is the normal to facet 3.

$$\displaystyle \boldsymbol{\phi}_{7} = \left(\begin{array}{c}\displaystyle \frac{9 x \left(2 x y - x - 2 y + 1\right)}{2}\\\displaystyle \frac{y \left(18 x y - 6 x - 3 y - 1\right)}{2}\end{array}\right)$$

This DOF is associated with edge 3 of the reference element. $$\displaystyle l_{8}:\mathbf{v}\mapsto\displaystyle\int_{R}(\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right))v$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{8} = \left(\begin{array}{c}\displaystyle 9 x \left(2 x y - x - 2 y + 1\right)\\\displaystyle 3 y \left(6 x y - 6 x - 5 y + 5\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element. $$\displaystyle l_{9}:\mathbf{v}\mapsto\displaystyle\int_{R}(\left(\begin{array}{c}\displaystyle 1\\\displaystyle 0\end{array}\right))v$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{9} = \left(\begin{array}{c}\displaystyle 3 x \left(6 x y - 5 x - 6 y + 5\right)\\\displaystyle 9 y \left(2 x y - 2 x - y + 1\right)\end{array}\right)$$

This DOF is associated with face 0 of the reference element. $$\displaystyle l_{10}:\mathbf{v}\mapsto\displaystyle\int_{R}(\left(\begin{array}{c}\displaystyle t_{1}\\\displaystyle s_{0}\end{array}\right))v$$
where $$R$$ is the reference element.

$$\displaystyle \boldsymbol{\phi}_{10} = \left(\begin{array}{c}\displaystyle 18 x \left(- 2 x y + x + 2 y - 1\right)\\\displaystyle 18 y \left(- 2 x y + 2 x + y - 1\right)\end{array}\right)$$

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