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# Degree 1 Lagrange on a hexahedron

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In this example:
• $$R$$ is the reference hexahedron. The following numbering of the subentities of the reference is used:
• $$\mathcal{V}$$ is spanned by: $$1$$, $$z$$, $$y$$, $$y z$$, $$x$$, $$x z$$, $$x y$$, $$x y z$$
• $$\mathcal{L}=\{l_0,...,l_{7}\}$$
• Functionals and basis functions:
$$\displaystyle l_{0}:v\mapsto v(0,0,0)$$

$$\displaystyle \phi_{0} = - x y z + x y + x z - x + y z - y - z + 1$$

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

$$\displaystyle \phi_{1} = x \left(y z - y - z + 1\right)$$

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

$$\displaystyle \phi_{2} = y \left(x z - x - z + 1\right)$$

This DOF is associated with vertex 2 of the reference element.
$$\displaystyle l_{3}:v\mapsto v(1,1,0)$$

$$\displaystyle \phi_{3} = x y \left(1 - z\right)$$

This DOF is associated with vertex 3 of the reference element.
$$\displaystyle l_{4}:v\mapsto v(0,0,1)$$

$$\displaystyle \phi_{4} = z \left(x y - x - y + 1\right)$$

This DOF is associated with vertex 4 of the reference element.
$$\displaystyle l_{5}:v\mapsto v(1,0,1)$$

$$\displaystyle \phi_{5} = x z \left(1 - y\right)$$

This DOF is associated with vertex 5 of the reference element.
$$\displaystyle l_{6}:v\mapsto v(0,1,1)$$

$$\displaystyle \phi_{6} = y z \left(1 - x\right)$$

This DOF is associated with vertex 6 of the reference element.
$$\displaystyle l_{7}:v\mapsto v(1,1,1)$$

$$\displaystyle \phi_{7} = x y z$$

This DOF is associated with vertex 7 of the reference element.