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# Degree 3 Wu–Xu 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: $$1$$, $$x$$, $$x^{2}$$, $$x^{3}$$, $$y$$, $$x y$$, $$x^{2} y$$, $$y^{2}$$, $$x y^{2}$$, $$y^{3}$$, $$x^{2} y \left(- x - y + 1\right)$$, $$x y^{2} \left(- x - y + 1\right)$$
• $$\mathcal{L}=\{l_0,...,l_{11}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:v\mapsto v(0,0)$$

$$\displaystyle \phi_{0} = 12 x^{3} y + 2 x^{3} + 24 x^{2} y^{2} - 18 x^{2} y - 3 x^{2} + 12 x y^{3} - 18 x y^{2} + 6 x y + 2 y^{3} - 3 y^{2} + 1$$

This DOF is associated with vertex 0 of the reference element.
$$\displaystyle l_{1}:v\mapsto\frac{\partial}{\partial x}v(0,0)$$

$$\displaystyle \phi_{1} = x \left(- 4 x^{2} y + x^{2} + 6 x y - 2 x + 4 y^{3} - 3 y^{2} - 2 y + 1\right)$$

This DOF is associated with vertex 0 of the reference element.
$$\displaystyle l_{2}:v\mapsto\frac{\partial}{\partial y}v(0,0)$$

$$\displaystyle \phi_{2} = y \left(4 x^{3} - 3 x^{2} - 4 x y^{2} + 6 x y - 2 x + y^{2} - 2 y + 1\right)$$

This DOF is associated with vertex 0 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(1,0)$$

$$\displaystyle \phi_{3} = x \left(- 6 x^{2} y - 2 x^{2} - 12 x y^{2} + 9 x y + 3 x - 6 y^{3} + 9 y^{2} - 3 y\right)$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{4}:v\mapsto\frac{\partial}{\partial x}v(1,0)$$

$$\displaystyle \phi_{4} = x^{2} \left(x - 1\right)$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{5}:v\mapsto\frac{\partial}{\partial y}v(1,0)$$

$$\displaystyle \phi_{5} = x y \left(- 4 x^{2} - 12 x y + 9 x - 8 y^{2} + 12 y - 4\right)$$

This DOF is associated with vertex 1 of the reference element.
$$\displaystyle l_{6}:v\mapsto v(0,1)$$

$$\displaystyle \phi_{6} = y \left(- 6 x^{3} - 12 x^{2} y + 9 x^{2} - 6 x y^{2} + 9 x y - 3 x - 2 y^{2} + 3 y\right)$$

This DOF is associated with vertex 2 of the reference element.
$$\displaystyle l_{7}:v\mapsto\frac{\partial}{\partial x}v(0,1)$$

$$\displaystyle \phi_{7} = x y \left(- 8 x^{2} - 12 x y + 12 x - 4 y^{2} + 9 y - 4\right)$$

This DOF is associated with vertex 2 of the reference element.
$$\displaystyle l_{8}:v\mapsto\frac{\partial}{\partial y}v(0,1)$$

$$\displaystyle \phi_{8} = y^{2} \left(y - 1\right)$$

This DOF is associated with vertex 2 of the reference element.
$$\displaystyle l_{9}:\mathbf{V}\mapsto\displaystyle\frac{\sqrt{2}}{2}\int_{e_{0}}\frac{\partial}{\partial\left(\begin{array}{c}\displaystyle - \frac{\sqrt{2}}{2}\\\displaystyle - \frac{\sqrt{2}}{2}\end{array}\right)}v$$
where $$e_{0}$$ is the 0th edge.

$$\displaystyle \phi_{9} = 3 \sqrt{2} x y \left(- 2 x^{2} - 4 x y + 3 x - 2 y^{2} + 3 y - 1\right)$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{10}:\mathbf{V}\mapsto\displaystyle\int_{e_{1}}\frac{\partial}{\partial\left(\begin{array}{c}\displaystyle -1\\\displaystyle 0\end{array}\right)}v$$
where $$e_{1}$$ is the 1st edge.

$$\displaystyle \phi_{10} = 6 x y \left(- 2 x^{2} - 2 x y + 3 x + y - 1\right)$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{11}:\mathbf{V}\mapsto\displaystyle\int_{e_{2}}\frac{\partial}{\partial\left(\begin{array}{c}\displaystyle 0\\\displaystyle 1\end{array}\right)}v$$
where $$e_{2}$$ is the 2nd edge.

$$\displaystyle \phi_{11} = 6 x y \left(2 x y - x + 2 y^{2} - 3 y + 1\right)$$

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