an encyclopedia of finite element definitions

◀ Back to Radau 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: $$1$$, $$y$$, $$y^{2}$$, $$x$$, $$x y$$, $$x y^{2}$$, $$x^{2}$$, $$x^{2} y$$, $$x^{2} y^{2}$$
• $$\mathcal{L}=\{l_0,...,l_{8}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:v\mapsto v(0,0)$$

$$\displaystyle \phi_{0} = \frac{4 \sqrt{6} x^{2} y^{2}}{3} + \frac{14 x^{2} y^{2}}{3} - \frac{20 x^{2} y}{3} - \frac{5 \sqrt{6} x^{2} y}{3} + \frac{\sqrt{6} x^{2}}{3} + 2 x^{2} - \frac{20 x y^{2}}{3} - \frac{5 \sqrt{6} x y^{2}}{3} + 2 \sqrt{6} x y + \frac{29 x y}{3} - 3 x - \frac{\sqrt{6} x}{3} + \frac{\sqrt{6} y^{2}}{3} + 2 y^{2} - 3 y - \frac{\sqrt{6} y}{3} + 1$$

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

$$\displaystyle \phi_{1} = - \frac{2 \sqrt{6} x^{2} y^{2}}{3} + 6 x^{2} y^{2} - 10 x^{2} y + \frac{5 \sqrt{6} x^{2} y}{3} - \sqrt{6} x^{2} + 4 x^{2} - 4 x y^{2} + \sqrt{6} x y^{2} - 2 \sqrt{6} x y + 7 x y - 3 x + \sqrt{6} x$$

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

$$\displaystyle \phi_{2} = - \frac{2 \sqrt{6} x^{2} y^{2}}{3} + 6 x^{2} y^{2} - 4 x^{2} y + \sqrt{6} x^{2} y - 10 x y^{2} + \frac{5 \sqrt{6} x y^{2}}{3} - 2 \sqrt{6} x y + 7 x y - \sqrt{6} y^{2} + 4 y^{2} - 3 y + \sqrt{6} y$$

This DOF is associated with vertex 2 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(1,1)$$

$$\displaystyle \phi_{3} = - 8 \sqrt{6} x^{2} y^{2} + 22 x^{2} y^{2} - 18 x^{2} y + 7 \sqrt{6} x^{2} y - 18 x y^{2} + 7 \sqrt{6} x y^{2} - 6 \sqrt{6} x y + 15 x y$$

This DOF is associated with vertex 3 of the reference element.
$$\displaystyle l_{4}:v\mapsto v(\tfrac{3}{5} - \tfrac{\sqrt{6}}{10},0)$$

$$\displaystyle \phi_{4} = - \frac{32 x^{2} y^{2}}{3} - \frac{2 \sqrt{6} x^{2} y^{2}}{3} + \frac{50 x^{2} y}{3} - 6 x^{2} + \frac{2 \sqrt{6} x^{2}}{3} + \frac{2 \sqrt{6} x y^{2}}{3} + \frac{32 x y^{2}}{3} - \frac{50 x y}{3} - \frac{2 \sqrt{6} x}{3} + 6 x$$

This DOF is associated with edge 0 of the reference element.
$$\displaystyle l_{5}:v\mapsto v(0,\tfrac{3}{5} - \tfrac{\sqrt{6}}{10})$$

$$\displaystyle \phi_{5} = - \frac{32 x^{2} y^{2}}{3} - \frac{2 \sqrt{6} x^{2} y^{2}}{3} + \frac{2 \sqrt{6} x^{2} y}{3} + \frac{32 x^{2} y}{3} + \frac{50 x y^{2}}{3} - \frac{50 x y}{3} - 6 y^{2} + \frac{2 \sqrt{6} y^{2}}{3} - \frac{2 \sqrt{6} y}{3} + 6 y$$

This DOF is associated with edge 1 of the reference element.
$$\displaystyle l_{6}:v\mapsto v(1,\tfrac{3}{5} - \tfrac{\sqrt{6}}{10})$$

$$\displaystyle \phi_{6} = - 28 x^{2} y^{2} + \frac{26 \sqrt{6} x^{2} y^{2}}{3} - \frac{26 \sqrt{6} x^{2} y}{3} + 28 x^{2} y - 8 \sqrt{6} x y^{2} + 22 x y^{2} - 22 x y + 8 \sqrt{6} x y$$

This DOF is associated with edge 2 of the reference element.
$$\displaystyle l_{7}:v\mapsto v(\tfrac{3}{5} - \tfrac{\sqrt{6}}{10},1)$$

$$\displaystyle \phi_{7} = - 28 x^{2} y^{2} + \frac{26 \sqrt{6} x^{2} y^{2}}{3} - 8 \sqrt{6} x^{2} y + 22 x^{2} y - \frac{26 \sqrt{6} x y^{2}}{3} + 28 x y^{2} - 22 x y + 8 \sqrt{6} x y$$

This DOF is associated with edge 3 of the reference element.
$$\displaystyle l_{8}:v\mapsto v(\tfrac{3}{5} - \tfrac{\sqrt{6}}{10},\tfrac{3}{5} - \tfrac{\sqrt{6}}{10})$$

$$\displaystyle \phi_{8} = - 8 \sqrt{6} x^{2} y^{2} + \frac{116 x^{2} y^{2}}{3} - \frac{116 x^{2} y}{3} + 8 \sqrt{6} x^{2} y - \frac{116 x y^{2}}{3} + 8 \sqrt{6} x y^{2} - 8 \sqrt{6} x y + \frac{116 x y}{3}$$

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