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A Taylor-Hood local pressure mode on a right triangle

Statement

On the reference right triangle $K = {(x, y) : x \geq 0, y \geq 0, x + y \leq 1}$ , let $q (x, y) = x - 1 / 3$ and $v (x, y) = (x^{2}, 0)$ . Then $q$ has zero cell mean and the local Stokes pairing is $- \int_{K} q div v = - 1 / 18$ . This is a local mode-detection calculation only; it proves no global Taylor–Hood stability theorem.

Facts & Assumptions

Given: The right triangle, the displayed local polynomials, and $b_{K} (v, q) : = - \int_{K} q div v$ .

Example

The pressure has zero cell mean and its local Stokes pairing is computed exactly.

Verification

technique · direct integration

1.1givenalgebraconstruct

The triangle area is $1 / 2$ , and $\int_{K} x = \int_{0}^{1} x (1 - x) d x = 1 / 6$ , so $\int_{K} q = 1 / 6 - (1 / 3) (1 / 2) = 0$ . Also $div v = 2 x$ .

2.1givenalgebradischarge-constructstep 1.1∎

Direct integration gives $\int_{K} x^{2} = \int_{0}^{1} x^{2} (1 - x) d x = 1 / 12$ ; hence $b_{K} (v, q) = - 2 (\int_{K} x^{2} - \frac{1}{3} \int_{K} x) = - 2 (\frac{1}{12} - \frac{1}{18}) = - \frac{1}{18}$ . This nonzero local pairing detects that polynomial pressure mode against this polynomial velocity; no boundary condition or global mesh stability is asserted.

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Chapter 54 (complete course notes) (standard reference, not scraped)
Daniele Boffi, Franco Brezzi, Michel Fortin, Mixed Finite Element Methods and Applications (complete textbook) (standard reference, not scraped)