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A nonzero CR function on a two-triangle square

Example

Split the unit square along the diagonal from $(0, 0)$ to $(1, 1)$ into $K_{1} = conv {(0, 0), (1, 0), (1, 1)}$ and $K_{2} = conv {(0, 0), (0, 1), (1, 1)}$ . The CR basis function for the diagonal is nonzero and has broken-gradient norm $2 \sqrt{2}$ . Its boundary edge means vanish, although its boundary traces are not pointwise zero.

The figure uses Cartesian coordinates with $x$ increasing rightward and $y$ upward. The red diagonal is the one interior facet; $K_{1}$ is below and to its right, and $K_{2}$ is above and to its left.

\begin{tikzpicture}[scale=2,line width=0.7pt,every node/.style={font=\small}] \fill[blue!8] (0,0)--(1,0)--(1,1)--cycle; \fill[green!8] (0,0)--(0,1)--(1,1)--cycle; \draw[gray] (0,0)--(1,0)--(1,1)--(0,1)--cycle; \draw[red!75!black,very thick] (0,0)--(1,1); \fill (0,0) circle (0.025) (1,0) circle (0.025) (1,1) circle (0.025) (0,1) circle (0.025); \node[below left] at (0,0) { $(0, 0)$ }; \node[below right] at (1,0) { $(1, 0)$ }; \node[above right] at (1,1) { $(1, 1)$ }; \node[above left] at (0,1) { $(0, 1)$ }; \node at (0.72,0.35) { $K_{1}$ }; \node at (0.28,0.65) { $K_{2}$ }; \node[red!75!black,above left] at (0.5,0.5) { $F$ }; \end{tikzpicture}

Facts & Assumptions

Given: The two-triangle matching mesh of $Ω = (0, 1)^{2}$ and the homogeneous CR face basis of Interior face means form the global Crouzeix–Raviart degrees of freedom.

[F1]

The local basis polynomial opposite a facet is $1 - 2 λ_{F}$ , and its mean is one on that facet and zero on the other edges. (Face averages are unisolvent for the local Crouzeix–Raviart element)

[F2]

The global basis function for the only interior facet is supported on its two incident triangles. (Interior face means form the global Crouzeix–Raviart degrees of freedom)

Verification

1.1F1givenalgebra

On $K_{1}$ , the barycentric coordinate of the vertex $(1, 0)$ opposite the diagonal is $λ_{1} (x, y) = x - y$ , so the local polynomial is $v_{1} (x, y) = 1 - 2 x + 2 y$ and $\nabla v_{1} = (- 2, 2)$ . On $K_{2}$ , the opposite vertex is $(0, 1)$ and $λ_{2} (x, y) = y - x$ , giving $v_{2} (x, y) = 1 + 2 x - 2 y$ and $\nabla v_{2} = (2, - 2)$ .

2.1F1F2step 1.1algebra∎

Each triangle has area $1 / 2$ , and both gradient squares have magnitude $8$ . Thus $∥ \nabla_{h} v_{h} ∥_{2}^{2} = \frac{1}{2} (8) + \frac{1}{2} (8) = 8, ∥ \nabla_{h} v_{h} ∥_{2} = 2 \sqrt{2} > 0.$ By [F1] its edge means are one on the shared diagonal and zero on the four boundary edges; hence [F2] gives a nonzero element of $V_{h}^{C R, 0}$ . On the bottom boundary edge of $K_{1}$ , $v_{1} (x, 0) = 1 - 2 x$ , which is not identically zero although its edge mean is zero. The example therefore distinguishes the CR mean boundary condition from pointwise Dirichlet values.

Depends on

Used by

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Part VIII, Chapter 36, Crouzeix–Raviart approximation (standard reference, not scraped)
Joachim Schöberl, Numerical Methods for Partial Differential Equations, Chapter 25, Non-conforming Finite Element Methods (standard reference, not scraped)