an encyclopedia of finite element definitions

# Degree 1 rotated Buffa–Christiansen on a dual polygon

◀ Back to rotated Buffa–Christiansen definition page
In this example:
• $$R$$ is the reference dual polygon. The following numbering of the subentities of the reference is used:
• Basis functions:
$$\displaystyle \boldsymbol{\phi}_{0} = \begin{cases} \left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{1} = \begin{cases} \left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{2} = \begin{cases} \left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$
$$\displaystyle \boldsymbol{\phi}_{3} = \begin{cases} \left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1, 0), (1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 1/2), (0, 1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, 1), (-1/2, 1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{1}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, 1/2), (-1, 0)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle \tfrac{1}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1, 0), (-1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle \tfrac{3}{8} - \tfrac{x}{4}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (-1/2, -1/2), (0, -1)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} - \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{3}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (0, -1), (1/2, -1/2)))\\\left(\begin{array}{c}\displaystyle \tfrac{y}{4} + \tfrac{3}{8}\\\displaystyle - \tfrac{x}{4} - \tfrac{1}{8}\end{array}\right)&\text{in }\operatorname{Triangle}(((0, 0), (1/2, -1/2), (1, 0)))\end{cases}$$