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A nonsymmetric coercive form attains the M over alpha factor

Example

On $V = R^{2}$ with the Euclidean norm, let $A = (\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array}), a (x, y) : = (A x) \cdot y .$ Take the trial and test subspace $V_{h} = span {e_{1}}$ , the exact solution $u = e_{2}$ , and the functional $F (v) = a (e_{2}, v)$ . The form is coercive with $α = 1$ and bounded with $M = \sqrt{2}$ . The Galerkin solution is $u_{h} = - e_{1}$ , so $\frac{∥ u - u_{h} ∥_{2}}{\inf_{v_{h} \in V_{h}} ∥ u - v_{h} ∥_{2}} = \sqrt{2} = \frac{M}{α} .$ Thus the Céa factor $M / α$ is attained even though the form is not symmetric.

Facts & Assumptions

Given: Countable Choice; the matrix $A$ , the Euclidean inner product, $V_{h} = span {e_{1}}$ , $u = e_{2}$ , and $F (v) = a (u, v)$ .

[F1]

Under Countable Choice, the Céa theorem asserts $∥ u - u_{h} ∥_{V} \leq (M / α) \inf_{v_{h} \in V_{h}} ∥ u - v_{h} ∥_{V}$ (Céa quasi-optimality for coercive conforming Galerkin problems).

Verification

1.1givenalgebra

For $x = (x_{1}, x_{2})$ , $A x = (x_{1} - x_{2}, x_{1} + x_{2})$ , so $∥ A x ∥_{2}^{2} = 2 ∥ x ∥_{2}^{2}$ and $∣ a (x, y) ∣ \leq \sqrt{2} ∥ x ∥_{2} ∥ y ∥_{2}$ . Also $a (x, x) = x_{1}^{2} + x_{2}^{2}$ , so the coercivity constant is $α = 1$ . Finally, $a (e_{1}, e_{2}) = 1$ and $a (e_{2}, e_{1}) = - 1$ , so the form is not symmetric.

1.2givenalgebra

Since $u = e_{2}$ and $F (v) = a (e_{2}, v)$ , the continuous variational equation holds for every $v \in R^{2}$ . Write $u_{h} = c e_{1}$ . Testing the discrete equation with $e_{1}$ gives $c a (e_{1}, e_{1}) = a (e_{2}, e_{1})$ , hence $c = - 1$ because $a (e_{1}, e_{1}) = 1$ and $a (e_{2}, e_{1}) = - 1$ . Thus $u_{h} = - e_{1}$ .

2.1F1step 1.1step 1.2algebra∎

For $v_{h} = t e_{1}$ , $∥ u - v_{h} ∥_{2}^{2} = ∥ e_{2} - t e_{1} ∥_{2}^{2} = 1 + t^{2}$ , whose infimum is $1$ at $t = 0$ . Meanwhile $u - u_{h} = e_{1} + e_{2}$ , so $∥ u - u_{h} ∥_{2} = \sqrt{2}$ . Therefore the ratio is $\sqrt{2} = M / α$ , and the constant in [F1] cannot be decreased in general.

Depends on

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