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The linear Ritz projection need not preserve the nonlinear p-Laplacian residual

Statement

Assume Countable Choice. On $Ω = (0, 1)$ , let $V = H_{0}^{1} (0, 1; R)$ with the full $H^{1}$ norm, let $H = L^{2} (0, 1; R)$ , and use their canonical evolution triple. Let $w (x) = 2 x$ on $[0, 1 / 2]$ and $w (x) = 2 (1 - x)$ on $[1 / 2, 1]$ , and set $V_{h} = span {w}$ . Use the coercive symmetric form $a (f, g) = \int_{0}^{1} f^{'} g^{'} d x$ for the Ritz projection, and let $v (x) = x (1 - x)$ . For the nonlinear $3$ -Laplacian flux $J_{3} (s) = ∣ s ∣ s$ , $\int_{0}^{1} ∣ (R_{h} v)^{'} ∣ (R_{h} v)^{'} w^{'} d x = \frac{1}{2} \neq \frac{2}{3} = \int_{0}^{1} ∣ v^{'} ∣ v^{'} w^{'} d x .$ Thus the linear Ritz projection does not in general preserve the nonlinear $p$ -Laplacian Galerkin residual.

Facts & Assumptions

Given: Countable Choice, the interval, the spaces and functions above, the Ritz form, and the scalar flux $J_{3}$ .

[F1]

For a real evolution triple, the elliptic Ritz projection is defined by the bounded coercive form equation $a (R_{h} z, y_{h}) = a (z, y_{h})$ for every $y_{h} \in V_{h}$ (The elliptic Ritz projection for a conforming finite-element space).

[F2]

The midpoint hat $w$ and quadratic $v = x (1 - x)$ belong to $H_{0}^{1} (0, 1)$ (The two-cell hat and quadratic functions belong to H¹₀).

[F3]

Poincare's inequality on $(0, 1)$ gives $∥ z ∥_{L^{2}} \leq ∥ z^{'} ∥_{L^{2}}$ for $z \in H_{0}^{1} (0, 1)$ , so the form $a (f, g) = \int f^{'} g^{'}$ is bounded and coercive on the full $H^{1}$ norm (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F4]

The real space $H_{0}^{1} (0, 1)$ is a separable Hilbert space continuously and densely included in $L^{2} (0, 1)$ , hence defines the required evolution triple (The zero-boundary Sobolev energy space is a separable Hilbert space).

[A1]

Countable Choice is assumed by the zero-boundary evolution-space supplier (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

1.1A1F1F2F3F4givenalgebra

By [F2], $V_{h} \subset V$ and $v \in V$ . By [F3], $∣ a (f, g) ∣ \leq ∥ f ∥_{H^{1}} ∥ g ∥_{H^{1}}$ and $a (f, f) = ∥ f^{'} ∥_{2}^{2} \geq \frac{1}{2} ∥ f ∥_{H^{1}}^{2}$ , so $a$ is bounded, symmetric and coercive as required by [F1]. The displayed hat has slopes $2$ and $- 2$ . Also $v^{'} (x) = 1 - 2 x$ , and direct integration gives $a (v, w) = 2 \int_{0}^{1 / 2} (1 - 2 x) d x - 2 \int_{1 / 2}^{1} (1 - 2 x) d x = 1, a (w, w) = 4.$ Since $R_{h} v = c w$ , [F1] with test $w$ gives $4 c = 1$ , so $R_{h} v = w / 4$ and its slopes are $1 / 2$ and $- 1 / 2$ .

2.1F2step 1.1givenalgebra

On each half interval, $∣ (R_{h} v)^{'} ∣ (R_{h} v)^{'} w^{'} = 1 / 2$ , so its integral is $1 / 2$ . On the left half, $v^{'} \geq 0$ and $∣ v^{'} ∣ v^{'} w^{'} = 2 (1 - 2 x)^{2}$ , whose integral is $1 / 3$ . On the right half, $v^{'} \leq 0$ and the same integrand $2 (1 - 2 x)^{2}$ also integrates to $1 / 3$ . Therefore $\int_{0}^{1} ∣ v^{'} ∣ v^{'} w^{'} = 2 / 3$ .

3.1step 1.1step 2.1givendischarge-contradiction∎

The two residual pairings are $1 / 2$ and $2 / 3$ , so they are unequal. This explicit conforming Ritz space refutes any general claim that a linear Ritz projection preserves every nonlinear $p$ -Laplacian Galerkin residual.

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