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Nitsche coercivity above the inverse-trace threshold

Statement

Assume Countable Choice and the hypotheses of Symmetric and nonsymmetric Nitsche boundary forms and Global boundary-flux estimate on a matching mesh. Put $C_{\partial}$ as in the latter item and let $θ \in {0, 1}$ . If $γ_{0} > \frac{(1 + θ)^{2} C_{\partial}^{2}}{4},$ then for every $v_{h} \in V_{h}$ , $a_{h}^{θ} (v_{h}, v_{h}) \geq λ_{θ} ∥ v_{h} ∥_{E, h}^{2}, λ_{θ} : = \frac{1 + γ_{0} - \sqrt{(1 - γ_{0})^{2} + (1 + θ)^{2} C_{\partial}^{2}}}{2} > 0.$ Thus the sufficient threshold is $C_{\partial}^{2} / 4$ for the nonsymmetric form and $C_{\partial}^{2}$ for the symmetric form. These are thresholds certified by the displayed inverse-trace bound, not asserted to be necessary for each individual mesh.

Facts & Assumptions

Given: Countable Choice; a mesh, the forms and energy seminorm of Symmetric and nonsymmetric Nitsche boundary forms; $C_{\partial}$ and its boundary cross-term estimate from Global boundary-flux estimate on a matching mesh; and $γ_{0} > (1 + θ)^{2} C_{\partial}^{2} / 4$ .

[F1]

The diagonal form is $a_{h}^{θ} (v_{h}, v_{h}) = ∥ \nabla v_{h} ∥_{Ω}^{2} - (1 + θ) \sum_{F} \int_{F} (\partial_{n} v_{h}) v_{h} + γ_{0} \sum_{F} h_{F}^{- 1} ∥ v_{h} ∥_{F}^{2}$ (Symmetric and nonsymmetric Nitsche boundary forms).

[F2]

The absolute boundary cross term is bounded by $C_{\partial} ∥ \nabla v_{h} ∥_{Ω} (\sum_{F} h_{F}^{- 1} ∥ v_{h} ∥_{F}^{2})^{1 / 2}$ (Global boundary-flux estimate on a matching mesh).

[A1]

$γ_{0} > (1 + θ)^{2} C_{\partial}^{2} / 4$ .

Proof

Given: Countable Choice; $v_{h} \in V_{h}$ ; and [F1], [F2], [A1].

1.1F1F2givenalgebra

Set $x = ∥ \nabla v_{h} ∥_{Ω}$ and $y = (\sum_{F} h_{F}^{- 1} ∥ v_{h} ∥_{F}^{2})^{1 / 2}$ . By [F1] and [F2], $a_{h}^{θ} (v_{h}, v_{h}) \geq x^{2} - (1 + θ) C_{\partial} x y + γ_{0} y^{2} .$ This is the quadratic form of the real symmetric matrix $M_{θ} = (\begin{array}{cc} 1 & - (1 + θ) C_{\partial} / 2 \\ - (1 + θ) C_{\partial} / 2 & γ_{0} \end{array})$ .

2.1A1step 1.1algebra∎

The eigenvalues of $M_{θ}$ are $\frac{1 + γ_{0} \pm \sqrt{(1 - γ_{0})^{2} + (1 + θ)^{2} C_{\partial}^{2}}}{2}$ , by the quadratic formula for its characteristic polynomial. Its trace is $1 + γ_{0} > 0$ and its determinant is $γ_{0} - (1 + θ)^{2} C_{\partial}^{2} / 4 > 0$ by [A1]; hence both eigenvalues are positive and the smaller is $λ_{θ}$ in the Statement. Therefore $x^{2} - (1 + θ) C_{\partial} x y + γ_{0} y^{2} \geq λ_{θ} (x^{2} + y^{2}) = λ_{θ} ∥ v_{h} ∥_{E, h}^{2}$ . Combine with step 1.1.

Depends on

Used by

Exact affine data under weak imposition Example
Penalties below and above the Nitsche coercivity threshold Example
Nitsche consistency and energy quasi-optimality Theorem

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Chapter 37, Nitsche boundary penalty method (standard reference, not scraped)
Mika Juntunen and Rolf Stenberg, Nitsche's method for general boundary conditions, Mathematics of Computation 78 (2009), 1353–1374 (standard reference, not scraped)