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One-dimensional P1 SUPG is consistent and coercive

Statement

Assume the one-dimensional model of One-dimensional stationary convection–diffusion model and uniform P1 space with homogeneous data $g_{0} = g_{1} = 0$ , and the form of One-dimensional residual SUPG form with the exact cell parameter. Let $u \in C^{2} ([0, 1])$ solve $- ϵ u^{''} + b u^{'} = f$ with $u (0) = u (1) = 0$ . Then $a_{h} (u, v_{h}) = F_{h} (v_{h})$ for every $v_{h} \in V_{h}$ . For every $v_{h} \in V_{h}$ , $a_{h} (v_{h}, v_{h}) = ϵ ∥ v_{h}^{'} ∥_{L^{2} (0, 1)}^{2} + τ b^{2} \sum_{K \in T_{h}} ∥ v_{h}^{'} ∥_{L^{2} (K)}^{2} .$ The right-hand side is a squared norm on $V_{h}$ . Consequently the finite-dimensional SUPG system has a unique solution.

Facts & Assumptions

Given: A uniform interval partition, $ϵ > 0$ , $b \neq 0$ , $f \in C ([0, 1])$ , the homogeneous problem above, and the SUPG form and parameter.

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: $V_{h}$ consists of continuous piecewise-affine functions vanishing at both endpoints; the exact solution satisfies $- ϵ u^{''} + b u^{'} = f$ .

[F2]

One-dimensional residual SUPG form with the exact cell parameter: $a_{h} (w, v_{h})$ pairs $- ϵ w^{''} + b w^{'}$ against $τ b v_{h}^{'}$ cellwise and $F_{h}$ pairs $f$ against the same test perturbation; for P1 $w_{h}$ , $w_{h}^{''} = 0$ .

[F3]

The one-dimensional SUPG parameter is positive and bounded: $τ > 0$ for $ϵ > 0$ , $b \neq 0$ , and $h > 0$ .

Proof

1.1F1givenalgebra

For $v_{h} \in V_{h}$ , continuity and piecewise absolute continuity give $\int_{0}^{1} b v_{h}^{'} v_{h} d x = \frac{b}{2} \sum_{K} [v_{h}^{2}]_{\partial K} = \frac{b}{2} (v_{h} (1)^{2} - v_{h} (0)^{2}) = 0$ . The endpoint values vanish by [F1].

1.2F1F2givenalgebra

Integration by parts, with $v_{h} (0) = v_{h} (1) = 0$ , gives $ϵ (u^{'}, v_{h}^{'}) + (b u^{'}, v_{h}) = (- ϵ u^{''} + b u^{'}, v_{h}) = (f, v_{h})$ . The stabilization contribution to $a_{h} (u, v_{h})$ is $\sum_{K} τ (- ϵ u^{''} + b u^{'}, b v_{h}^{'})_{K}$ , which equals $\sum_{K} τ (f, b v_{h}^{'})_{K}$ by the strong equation. This is exactly the stabilization contribution to $F_{h}$ , so $a_{h} (u, v_{h}) = F_{h} (v_{h})$ .

2.1F2F3step 1.1algebra

Substituting step 1.1 into [F2] with $w_{h} = v_{h}$ yields the displayed identity. Its right side is nonnegative by [F3] and $ϵ > 0$ . If it is zero, then $v_{h}^{'} = 0$ on every cell; continuity makes $v_{h}$ constant on $[0, 1]$ , and its zero endpoints force $v_{h} = 0$ . Hence this quadratic form is a norm squared.

3.1step 2.1givenalgebra

In a basis of the finite-dimensional space $V_{h}$ , the discrete form defines a square linear map $V_{h} \to V_{h}^{*}$ . If a vector is in its kernel, testing with that vector and using step 2.1 gives that it is zero. Thus the map is injective, hence bijective in finite equal dimensions, so each discrete load has exactly one solution.

4.1step 1.2step 3.1algebra∎

Subtracting the discrete equation from the consistency identity in step 1.2 gives $a_{h} (u - u_{h}, v_{h}) = 0$ for every $v_{h} \in V_{h}$ ; the solution exists by step 3.1.

Depends on

Used by

One-dimensional SUPG energy error bound with explicit tau dependence Theorem

Dependency tree · one level

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Sources

Volker John, Numerical Methods for Convection-Dominated Problems, lecture notes, Chapter 5 (standard reference, not scraped)
Hans Petter Langtangen, INF5620: Numerical Methods for Partial Differential Equations, lecture slides (standard reference, not scraped)