an encyclopedia of finite element definitions

Rotated Buffa–Christiansen

 Abbreviated names RBC Orders $$k=1$$ Reference elements dual polygon Number of DOFs dual polygon(n): $$n$$ Notes These elements are defined on the barycentric dual grid. These elements are defined as a linear combination of Nédélec first kind basis functions on the fine grid. Categories Vector-valued elements, H(curl) conforming elements

Implementations

 Bempp "RBC"↓ Show Bempp examples ↓ Symfem "RBC"↓ Show Symfem examples ↓

Examples

dual polygon(4)
order 1
dual polygon(5)
order 1
dual polygon(6)
order 1
• $$R$$ is the reference dual polygon. The following numbering of the subentities of the reference is used:
• Basis functions:
$$\displaystyle \boldsymbol{\phi}_{0} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{1} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{2} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{3} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{1}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle \frac{1}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle \frac{3}{8} - \frac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} - \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \frac{y}{4} + \frac{3}{8}\\\displaystyle - \frac{x}{4} - \frac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
• $$R$$ is the reference dual polygon. The following numbering of the subentities of the reference is used:
• Basis functions:
$$\displaystyle \boldsymbol{\phi}_{0} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} + \frac{2}{5}\\\displaystyle \frac{- 2 \sqrt{2} x - 3 \sqrt{2} - \sqrt{10}}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle - \frac{3 \sqrt{5} y \sqrt{2 \sqrt{5} + 10}}{25} - \frac{\sqrt{5} y \sqrt{10 \sqrt{5} + 50}}{25} + \frac{y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{1}{5}\\\displaystyle \frac{- 6 \sqrt{2} x - 2 \sqrt{10} x + 3 \sqrt{10} + 11 \sqrt{2}}{5 \left(\sqrt{5} + 3\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{- \frac{y \sqrt{10 \sqrt{5} + 50}}{10} - \frac{y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{\sqrt{5}}{2} + \frac{1}{2}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{3 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{x \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{9 \sqrt{10 - 2 \sqrt{5}}}{40} - \frac{\sqrt{2 \sqrt{5} + 10}}{40} + \frac{3 \sqrt{10 \sqrt{5} + 50}}{40} + \frac{\sqrt{50 - 10 \sqrt{5}}}{8}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8)), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 4 \sqrt{\sqrt{5} + 5} - \frac{8 \sqrt{5 \sqrt{5} + 25}}{5}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x + 10 \sqrt{2} + 6 \sqrt{10}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y - \sqrt{5} \sqrt{5 - \sqrt{5}} - \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{\sqrt{2} \left(- 4 x - \sqrt{5} - 1\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8)), (-sqrt(5)/4 - 1/4, 0)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y + \sqrt{5 - \sqrt{5}} + \sqrt{5} \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} - \sqrt{2}}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{\frac{3 y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{7 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{7 y \sqrt{2 \sqrt{5} + 10}}{10} + 4 + \frac{9 \sqrt{5}}{5}}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{7 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{7 x \sqrt{10 - 2 \sqrt{5}}}{10} - \frac{3 x \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{\sqrt{10 - 2 \sqrt{5}}}{10} + \frac{\sqrt{50 - 10 \sqrt{5}}}{20} + \frac{11 \sqrt{2 \sqrt{5} + 10}}{40} + \frac{\sqrt{10 \sqrt{5} + 50}}{8}}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8)), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 4 \sqrt{5 - \sqrt{5}} - \frac{8 \sqrt{25 - 5 \sqrt{5}}}{5} + \frac{16 \sqrt{5 \sqrt{5} + 25}}{5} + 8 \sqrt{\sqrt{5} + 5}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x + 50 \sqrt{2} + 26 \sqrt{10}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle - \frac{y \sqrt{10 \sqrt{5} + 50}}{50} + \frac{y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{1}{5}\\\displaystyle \frac{- 2 \sqrt{10} x + 2 \sqrt{2} x - 7 \sqrt{2} + 5 \sqrt{10}}{5 \left(-1 + \sqrt{5}\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8)), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} - \frac{2}{5}\\\displaystyle \frac{- 2 \sqrt{2} x - 3 \sqrt{2} - \sqrt{10}}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{1} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} - \frac{2}{5}\\\displaystyle \frac{- 2 \sqrt{2} x - 3 \sqrt{2} + \sqrt{10}}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle - \frac{3 \sqrt{5} y \sqrt{2 \sqrt{5} + 10}}{25} - \frac{\sqrt{5} y \sqrt{10 \sqrt{5} + 50}}{25} + \frac{y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{2}{5}\\\displaystyle \frac{2 \left(- 3 \sqrt{2} x - \sqrt{10} x - 7 \sqrt{2} - 3 \sqrt{10}\right)}{5 \left(\sqrt{5} + 3\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{- 4 \sqrt{10} y + 4 \sqrt{2} y - 4 \sqrt{5} \sqrt{\sqrt{5} + 5} - 4 \sqrt{5} \sqrt{5 - \sqrt{5}} + 4 \sqrt{5 - \sqrt{5}} + 4 \sqrt{\sqrt{5} + 5}}{- 85 \sqrt{\sqrt{5} + 5} - 65 \sqrt{5 - \sqrt{5}} + 25 \sqrt{5} \sqrt{5 - \sqrt{5}} + 35 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 20 \sqrt{2} x + 4 \sqrt{10} x - 4 \sqrt{10} + 20 \sqrt{2}}{- 500 \sqrt{\sqrt{5} + 5} - 325 \sqrt{5 - \sqrt{5}} + 155 \sqrt{5} \sqrt{5 - \sqrt{5}} + 230 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8)), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 8 \sqrt{\sqrt{5} + 5} - \frac{16 \sqrt{5 \sqrt{5} + 25}}{5} + \frac{8 \sqrt{25 - 5 \sqrt{5}}}{5} + 4 \sqrt{5 - \sqrt{5}}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x + 50 \sqrt{2} + 26 \sqrt{10}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{2 \cdot \left(2 \sqrt{2} y \left(1 + \sqrt{5}\right) - 3 \sqrt{5} \sqrt{5 - \sqrt{5}} - 3 \sqrt{5 - \sqrt{5}}\right)}{5 \left(1 + \sqrt{5}\right)^{2} \sqrt{5 - \sqrt{5}}}\\\displaystyle \frac{\sqrt{2} \left(- 4 x - \sqrt{5} - 1\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8)), (-sqrt(5)/4 - 1/4, 0)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y - \sqrt{5} \sqrt{5 - \sqrt{5}} - \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} - \sqrt{2}}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{12 \sqrt{10} y + 28 \sqrt{2} y + 6 \sqrt{5} \sqrt{5 - \sqrt{5}} + 14 \sqrt{5 - \sqrt{5}}}{45 \sqrt{5 - \sqrt{5}} + 30 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5} \sqrt{5 - \sqrt{5}} + 20 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- \frac{6 \sqrt{10} x}{5} - 2 \sqrt{2} x - 2 \sqrt{2} - \frac{4 \sqrt{10}}{5}}{\sqrt{5} \sqrt{5 - \sqrt{5}} + \sqrt{5} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8)), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y + \frac{8 \sqrt{5 \sqrt{5} + 25}}{5} + 4 \sqrt{\sqrt{5} + 5}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 12 \sqrt{2} x - 4 \sqrt{10} x + 2 \sqrt{2} + 2 \sqrt{10}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 15 \sqrt{5} \sqrt{5 - \sqrt{5}} + 15 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{2 \left(- \sqrt{10} x + \sqrt{2} x - \sqrt{2} + \sqrt{10}\right)}{5 \left(-1 + \sqrt{5}\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8)), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} - \frac{1}{5}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} + 9 \sqrt{2}}{10 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{2} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} - \frac{1}{5}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{2} + \sqrt{10}}{10 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle - \frac{3 \sqrt{5} y \sqrt{2 \sqrt{5} + 10}}{25} - \frac{\sqrt{5} y \sqrt{10 \sqrt{5} + 50}}{25} + \frac{y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{2}{5}\\\displaystyle \frac{2 \left(- 3 \sqrt{2} x - \sqrt{10} x - 2 \sqrt{2}\right)}{5 \left(\sqrt{5} + 3\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{- \frac{y \sqrt{10 \sqrt{5} + 50}}{10} - \frac{y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} - 1 + \frac{2 \sqrt{5}}{5}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{3 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{x \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{3 \sqrt{10 \sqrt{5} + 50}}{40} - \frac{\sqrt{50 - 10 \sqrt{5}}}{10} - \frac{\sqrt{2 \sqrt{5} + 10}}{40} + \frac{3 \sqrt{10 - 2 \sqrt{5}}}{20}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8)), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{44 \sqrt{10} y}{5} + 20 \sqrt{2} y + \frac{44 \sqrt{25 - 5 \sqrt{5}}}{5} + 20 \sqrt{5 - \sqrt{5}} + \frac{44 \sqrt{5 \sqrt{5} + 25}}{5} + 20 \sqrt{\sqrt{5} + 5}}{- 15 \sqrt{\sqrt{5} + 5} + 15 \sqrt{5 - \sqrt{5}} + 29 \sqrt{5} \sqrt{5 - \sqrt{5}} + 29 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 7 \sqrt{2} x - 3 \sqrt{10} x + 3 \sqrt{10} + 7 \sqrt{2}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 10 \sqrt{5} \sqrt{5 - \sqrt{5}} + 10 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{4 \sqrt{2} y}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\\\displaystyle \frac{4 \sqrt{2} \left(- x + 1 + \sqrt{5}\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8)), (-sqrt(5)/4 - 1/4, 0)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y - 3 \sqrt{5} \sqrt{5 - \sqrt{5}} - 3 \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} - \sqrt{2}}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{\frac{3 y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{7 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{7 y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{2 \sqrt{5}}{5} + 1}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{7 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{7 x \sqrt{10 - 2 \sqrt{5}}}{10} - \frac{3 x \sqrt{50 - 10 \sqrt{5}}}{10} - \frac{31 \sqrt{10 \sqrt{5} + 50}}{40} - \frac{69 \sqrt{2 \sqrt{5} + 10}}{40} - \frac{3 \sqrt{50 - 10 \sqrt{5}}}{10} - \frac{13 \sqrt{10 - 2 \sqrt{5}}}{20}}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8)), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y + \frac{8 \sqrt{25 - 5 \sqrt{5}}}{5} + 4 \sqrt{5 - \sqrt{5}}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x - 14 \sqrt{10} - 30 \sqrt{2}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle - \frac{y \sqrt{10 \sqrt{5} + 50}}{50} + \frac{y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{1}{5}\\\displaystyle \frac{- 2 \sqrt{10} x + 2 \sqrt{2} x - \sqrt{10} + 3 \sqrt{2}}{5 \left(-1 + \sqrt{5}\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8)), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{2 \sqrt{2} \cdot \left(1 - x\right)}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{3} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{2 \sqrt{2} \cdot \left(1 - x\right)}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle - \frac{3 \sqrt{5} y \sqrt{2 \sqrt{5} + 10}}{25} - \frac{\sqrt{5} y \sqrt{10 \sqrt{5} + 50}}{25} + \frac{y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{1}{5}\\\displaystyle \frac{- 6 \sqrt{2} x - 2 \sqrt{10} x + \sqrt{2} + \sqrt{10}}{5 \left(\sqrt{5} + 3\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{- \frac{y \sqrt{10 \sqrt{5} + 50}}{10} - \frac{y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{1}{2} + \frac{\sqrt{5}}{10}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{3 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{x \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{\sqrt{10 \sqrt{5} + 50}}{40} - \frac{\sqrt{50 - 10 \sqrt{5}}}{40} - \frac{\sqrt{2 \sqrt{5} + 10}}{40} + \frac{\sqrt{10 - 2 \sqrt{5}}}{40}}{-10 - 2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8)), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 8 \sqrt{5 - \sqrt{5}} - \frac{16 \sqrt{25 - 5 \sqrt{5}}}{5} + \frac{8 \sqrt{5 \sqrt{5} + 25}}{5} + 4 \sqrt{\sqrt{5} + 5}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x - 34 \sqrt{10} - 70 \sqrt{2}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{2 \cdot \left(2 \sqrt{2} y \left(1 + \sqrt{5}\right) + 3 \sqrt{5 - \sqrt{5}} + 3 \sqrt{5} \sqrt{5 - \sqrt{5}}\right)}{5 \left(1 + \sqrt{5}\right)^{2} \sqrt{5 - \sqrt{5}}}\\\displaystyle \frac{\sqrt{2} \left(- 4 x - \sqrt{5} - 1\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8)), (-sqrt(5)/4 - 1/4, 0)))\\\left(\begin{array}{c}\displaystyle \frac{8 y}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{2} + \sqrt{10}\right)}\\\displaystyle \frac{4 \left(- \sqrt{2} x + \sqrt{2} + \sqrt{10}\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{\frac{44 \sqrt{2} y}{5} + 4 \sqrt{10} y - 4 \sqrt{5 \sqrt{5} + 25} - \frac{44 \sqrt{\sqrt{5} + 5}}{5} - 4 \sqrt{25 - 5 \sqrt{5}} - \frac{44 \sqrt{5 - \sqrt{5}}}{5}}{7 \sqrt{5} \sqrt{5 - \sqrt{5}} + 17 \sqrt{5 - \sqrt{5}} + 5 \sqrt{5} \sqrt{\sqrt{5} + 5} + 13 \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 260 \sqrt{2} x - 116 \sqrt{10} x + 116 \sqrt{10} + 260 \sqrt{2}}{500 \sqrt{5 - \sqrt{5}} + 230 \sqrt{5} \sqrt{5 - \sqrt{5}} + 325 \sqrt{\sqrt{5} + 5} + 155 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8)), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 4 \sqrt{\sqrt{5} + 5} - \frac{8 \sqrt{5 \sqrt{5} + 25}}{5} + \frac{16 \sqrt{25 - 5 \sqrt{5}}}{5} + 8 \sqrt{5 - \sqrt{5}}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 40 \sqrt{2} x - 16 \sqrt{10} x - 34 \sqrt{10} - 70 \sqrt{2}}{- 25 \sqrt{\sqrt{5} + 5} + 25 \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{5 - \sqrt{5}} + 55 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle - \frac{y \sqrt{10 \sqrt{5} + 50}}{50} + \frac{y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{2}{5}\\\displaystyle \frac{2 \left(- \sqrt{10} x + \sqrt{2} x - 2 \sqrt{10} + 4 \sqrt{2}\right)}{5 \left(-1 + \sqrt{5}\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8)), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} + \frac{1}{5}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{2} + \sqrt{10}}{10 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{4} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} + \frac{1}{5}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} + 9 \sqrt{2}}{10 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{2 \left(- 3 \sqrt{2} x - \sqrt{10} x + \sqrt{10} + 3 \sqrt{2}\right)}{5 \left(\sqrt{5} + 3\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, sqrt(sqrt(5)/8 + 5/8)/2), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle \frac{8 \sqrt{2} y - 4 \sqrt{\sqrt{5} + 5}}{- 45 \sqrt{\sqrt{5} + 5} - 30 \sqrt{5 - \sqrt{5}} + 20 \sqrt{5} \sqrt{5 - \sqrt{5}} + 25 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{\frac{4 \sqrt{10} x}{5} - \sqrt{2} + \frac{\sqrt{10}}{5}}{- 5 \sqrt{\sqrt{5} + 5} - 5 \sqrt{5 - \sqrt{5}} + \sqrt{5} \sqrt{5 - \sqrt{5}} + \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, sqrt(sqrt(5)/8 + 5/8)), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{16 \sqrt{10} y}{5} + 8 \sqrt{2} y - 4 \sqrt{5 - \sqrt{5}} - \frac{8 \sqrt{25 - 5 \sqrt{5}}}{5}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{5 - \sqrt{5}} + 11 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 12 \sqrt{2} x - 4 \sqrt{10} x - 4 \sqrt{10} - 8 \sqrt{2}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 15 \sqrt{5} \sqrt{5 - \sqrt{5}} + 15 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, sqrt(5/8 - sqrt(5)/8)/2 + sqrt(sqrt(5)/8 + 5/8)/2), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y + \sqrt{5 - \sqrt{5}} + \sqrt{5} \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{\sqrt{2} \left(- 4 x - \sqrt{5} - 1\right)}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, sqrt(5/8 - sqrt(5)/8)), (-sqrt(5)/4 - 1/4, 0)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y + 2 \sqrt{10} y + 3 \sqrt{5 - \sqrt{5}} + 3 \sqrt{5} \sqrt{5 - \sqrt{5}}}{5 \sqrt{5 - \sqrt{5}} \left(\sqrt{5} + 3\right)}\\\displaystyle \frac{- 4 \sqrt{2} x - \sqrt{10} - \sqrt{2}}{5 \cdot \left(1 + \sqrt{5}\right) \sqrt{5 - \sqrt{5}}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8))))\\\left(\begin{array}{c}\displaystyle \frac{\frac{3 y \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{7 y \sqrt{10 - 2 \sqrt{5}}}{10} + \frac{3 y \sqrt{10 \sqrt{5} + 50}}{10} + \frac{7 y \sqrt{2 \sqrt{5} + 10}}{10} + \frac{11}{2} + \frac{5 \sqrt{5}}{2}}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\\\displaystyle \frac{- \frac{7 x \sqrt{2 \sqrt{5} + 10}}{10} - \frac{3 x \sqrt{10 \sqrt{5} + 50}}{10} - \frac{7 x \sqrt{10 - 2 \sqrt{5}}}{10} - \frac{3 x \sqrt{50 - 10 \sqrt{5}}}{10} + \frac{19 \sqrt{10 - 2 \sqrt{5}}}{40} + \frac{9 \sqrt{50 - 10 \sqrt{5}}}{40} + \frac{51 \sqrt{2 \sqrt{5} + 10}}{40} + \frac{23 \sqrt{10 \sqrt{5} + 50}}{40}}{2 \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + \sqrt{5} \sqrt{5 - \sqrt{5}} \sqrt{\sqrt{5} + 5} + 10 + 5 \sqrt{5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-sqrt(5)/4 - 1/4, -sqrt(5/8 - sqrt(5)/8)), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\frac{44 \sqrt{10} y}{5} + 20 \sqrt{2} y - 20 \sqrt{\sqrt{5} + 5} - \frac{44 \sqrt{5 \sqrt{5} + 25}}{5} - 20 \sqrt{5 - \sqrt{5}} - \frac{44 \sqrt{25 - 5 \sqrt{5}}}{5}}{- 15 \sqrt{\sqrt{5} + 5} + 15 \sqrt{5 - \sqrt{5}} + 29 \sqrt{5} \sqrt{5 - \sqrt{5}} + 29 \sqrt{5} \sqrt{\sqrt{5} + 5}}\\\displaystyle \frac{- 7 \sqrt{2} x - 3 \sqrt{10} x + 3 \sqrt{10} + 7 \sqrt{2}}{- 5 \sqrt{\sqrt{5} + 5} + 5 \sqrt{5 - \sqrt{5}} + 10 \sqrt{5} \sqrt{5 - \sqrt{5}} + 10 \sqrt{5} \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4, -sqrt(sqrt(5)/8 + 5/8)/2 - sqrt(5/8 - sqrt(5)/8)/2), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8))))\\\left(\begin{array}{c}\displaystyle - \frac{y \sqrt{10 \sqrt{5} + 50}}{50} + \frac{y \sqrt{2 \sqrt{5} + 10}}{10} - \frac{2}{5}\\\displaystyle \frac{2 \left(- \sqrt{10} x + \sqrt{2} x - \sqrt{10} - \sqrt{2}\right)}{5 \left(-1 + \sqrt{5}\right) \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/4 + sqrt(5)/4, -sqrt(sqrt(5)/8 + 5/8)), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2)))\\\left(\begin{array}{c}\displaystyle \frac{2 \sqrt{2} y}{5 \sqrt{\sqrt{5} + 5}} + \frac{2}{5}\\\displaystyle \frac{- 2 \sqrt{2} x - 3 \sqrt{2} + \sqrt{10}}{5 \sqrt{\sqrt{5} + 5}}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (sqrt(5)/8 + 3/8, -sqrt(sqrt(5)/8 + 5/8)/2), (1, 0)))\end{cases}$$
• $$R$$ is the reference dual polygon. The following numbering of the subentities of the reference is used:
• Basis functions:
$$\displaystyle \boldsymbol{\phi}_{0} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 15\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(7 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(2 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 3\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 3\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(2 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(7 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 15\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{1} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 15\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(2 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 5\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 3\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(5 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(7 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{2} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 2\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(15 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 7\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 5\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 1\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(3 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(5 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{3} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(3 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 1\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 2\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 7\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(15 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(15 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 7\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 2\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 1\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(3 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{4} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(5 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(3 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 1\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 5\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 7\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(15 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \left(- x - 2\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{5} = \begin{cases} \left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \cdot \left(7 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(5 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, sqrt(3)/4), (1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, sqrt(3)/2), (0, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, sqrt(3)/2), (-1/2, sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 3\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, sqrt(3)/2), (-3/4, sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 5\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, sqrt(3)/4), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{4}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \cdot \left(1 - 4 x\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-3/4, -sqrt(3)/4), (-1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{1}{6}\\\displaystyle \frac{\sqrt{3} \cdot \left(2 - x\right)}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -sqrt(3)/2), (0, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{6}\\\displaystyle - \frac{\sqrt{3} x}{9}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -sqrt(3)/2), (1/2, -sqrt(3)/2)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} - \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 15\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -sqrt(3)/2), (3/4, -sqrt(3)/4)))\\\left(\begin{array}{c}\displaystyle \frac{\sqrt{3} y}{9} + \frac{5}{12}\\\displaystyle \frac{\sqrt{3} \left(- 4 x - 1\right)}{36}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (3/4, -sqrt(3)/4), (1, 0)))\end{cases}$$

References

• Buffa, Annalisa and Christiansen, Snorre H. A dual finite element complex on the barycentric refinement, Mathematics of Computation 76, 1743–1769, 2007. [BibTeX]

DefElement stats

 Element added 24 January 2021 Element last updated 10 February 2022