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The L2 projection and Ritz projection give different initial data

Example

Let $Ω = (0, 1)^{2}$ and partition its closure into four triangles by joining the center $c = (1 / 2, 1 / 2)$ to the four corners. Let $V_{h}$ be the continuous piecewise-linear functions on this mesh with zero boundary values. Its standard nodal basis consists of the center hat $ϕ$ , equal on each triangle to the barycentric coordinate of $c$ . For $u_{0} = ϕ^{2}$ , the $L^{2}$ projection into $V_{h}$ and the Dirichlet Ritz projection differ. This gives a nonzero initial error term when the semidiscrete heat equation is initialized by the $L^{2}$ projection instead of the Ritz projection.

Use $V = H_{0}^{1} (Ω)$ with energy norm $∥ v ∥_{V} = ∥ \nabla v ∥_{L^{2} (Ω)}$ and $H = L^{2} (Ω)$ . Poincare's inequality on the unit square makes this an evolution triple and makes the Dirichlet form bounded and coercive in the stated norm.

Facts & Assumptions

Given: Countable Choice; the four-triangle fan mesh and its continuous zero-boundary P1 space; the energy-norm evolution triple; $u_{0} = ϕ^{2}$ ; and the Dirichlet form.

[F1]

A matching affine simplicial mesh consists of finitely many nondegenerate simplices meeting along whole faces, with the open domain the interior of their union (Matching affine simplicial meshes and shape regularity).

[F2]

The continuous matching piecewise-polynomial space and its zero-boundary subspace are defined by facet matching and zero boundary-facet polynomial traces (Broken and conforming polynomial spaces on a matching mesh).

[F3]

A real zero-boundary conforming piecewise-polynomial function on a matching affine mesh lies in $W_{0}^{1, 2} (Ω)$ under Countable Choice (Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare).

[F4]

Poincare's inequality on the unit square gives $∥ v ∥_{L^{2}} \leq ∥ \nabla v ∥_{L^{2}}$ for $v \in H_{0}^{1} (Ω)$ ; hence the energy norm is Hilbert, $V ↪ H$ is continuous, and the Dirichlet form has $M = α = 1$ (Integer-order Sobolev spaces and their norms, The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction, Bounded, coercive and symmetric sesquilinear forms).

[F5]

The Ritz equation has a unique solution characterized by $a (R_{h} u, v_{h}) = a (u, v_{h})$ for every $v_{h} \in V_{h}$ (The elliptic Ritz projection for a conforming finite-element space, Existence, uniqueness, linearity, and boundedness of the Ritz projection).

[A1]

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

Verification

1.1A1F1F2F3F4given

The four triangles each have area $1 / 4$ , have disjoint interiors, and meet in whole edges or vertices; their union is the closed unit square. The piecewise barycentric coordinate $ϕ$ is zero on the boundary and agrees on shared edges. An affine polynomial on each triangle is determined by its three vertex values; matching traces make shared vertex values agree. Thus a global continuous P1 function is determined by the five mesh vertex values. The four boundary values vanish, leaving only the center value, so $V_{h} = span {ϕ}$ . By [F3], $ϕ \in H_{0}^{1} (Ω)$ ; $u_{0} = ϕ^{2}$ is also a continuous zero-boundary piecewise polynomial and lies in $H_{0}^{1} (Ω)$ . The norm and form hypotheses follow from [F4].

1.2F1givenalgebra

On a triangle $K$ of area $A = 1 / 4$ , the affine map from the reference triangle has determinant magnitude $2 A$ . If $λ$ is the barycentric coordinate of the center vertex, direct integration gives $\int_{K} λ^{m} d x = 2 A \int_{0}^{1} \int_{0}^{1 - x} (1 - x - y)^{m} d y d x = 2 A / ((m + 1) (m + 2))$ for $m = 1, 2, 3$ . Summing over the four triangles yields $\int_{Ω} ϕ = 1 / 3$ , $\int_{Ω} ϕ^{2} = 1 / 6$ , and $\int_{Ω} ϕ^{3} = 1 / 10$ . The altitude from the center to each opposite boundary edge is $1 / 2$ , so the barycentric coordinate has gradient magnitude $2$ and $∣ \nabla ϕ ∣^{2} = 4$ on each triangle. Hence $a (ϕ, ϕ) = 4$ and $a (ϕ^{2}, ϕ) = 2 \int_{Ω} ϕ ∣ \nabla ϕ ∣^{2} = 8 / 3$ .

2.1F5step 1.2givenalgebra

Write the orthogonal projection as $P_{h} u_{0} = p ϕ$ . Its defining equation $(u_{0} - p ϕ, ϕ)_{L^{2}} = 0$ gives $p = (1 / 10) / (1 / 6) = 3 / 5$ . By [F5], the Ritz projection is $R_{h} u_{0} = r ϕ$ with $a (u_{0} - r ϕ, ϕ) = 0$ , hence $r = (8 / 3) / 4 = 2 / 3$ .

3.1step 2.1givenalgebra∎

Therefore $P_{h} u_{0} - R_{h} u_{0} = (3 / 5 - 2 / 3) ϕ = - ϕ / 15 \neq 0$ , since $ϕ (c) = 1$ . If $u_{h} (0) = P_{h} u_{0}$ , the error-splitting variable $θ (0) = u_{h} (0) - R_{h} u_{0}$ is exactly this nonzero function.

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