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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: \(\begin{cases}
- 2 x - 2 y + 1&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\), \(\begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\2 x - 1&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\), \(\begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\2 y - 1&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\), \(\begin{cases}
2 x&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\- 2 x - 2 y + 2&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\1 - 2 y&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\), \(\begin{cases}
2 y&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\- 2 x - 2 y + 2&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\1 - 2 x&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\), \(\begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\2 y&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\2 x&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\2 x + 2 y - 1&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
- \(\mathcal{L}=\{l_0,...,l_{5}\}\)
- Functionals and basis functions:
\(\displaystyle l_{0}:v\mapsto v(0,0)\)
\(\displaystyle \phi_{0} = \begin{cases}
- 2 x - 2 y + 1&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with vertex 0 of the reference element.
\(\displaystyle l_{1}:v\mapsto v(1,0)\)
\(\displaystyle \phi_{1} = \begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\2 x - 1&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with vertex 1 of the reference element.
\(\displaystyle l_{2}:v\mapsto v(0,1)\)
\(\displaystyle \phi_{2} = \begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\2 y - 1&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\0&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with vertex 2 of the reference element.
\(\displaystyle l_{3}:v\mapsto v(\tfrac{1}{2},\tfrac{1}{2})\)
\(\displaystyle \phi_{3} = \begin{cases}
0&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\2 y&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\2 x&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\2 x + 2 y - 1&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with edge 0 of the reference element.
\(\displaystyle l_{4}:v\mapsto v(0,\tfrac{1}{2})\)
\(\displaystyle \phi_{4} = \begin{cases}
2 y&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\0&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\- 2 x - 2 y + 2&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\1 - 2 x&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with edge 1 of the reference element.
\(\displaystyle l_{5}:v\mapsto v(\tfrac{1}{2},0)\)
\(\displaystyle \phi_{5} = \begin{cases}
2 x&\text{in }\operatorname{Triangle}(((0, 0), (1/2, 0), (0, 1/2)))\\- 2 x - 2 y + 2&\text{in }\operatorname{Triangle}(((1, 0), (1/2, 1/2), (1/2, 0)))\\0&\text{in }\operatorname{Triangle}(((0, 1), (0, 1/2), (1/2, 1/2)))\\1 - 2 y&\text{in }\operatorname{Triangle}(((0, 1/2), (1/2, 1/2), (1/2, 0)))\end{cases}\)
This DOF is associated with edge 2 of the reference element.