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A two-cell SUPG mesh misses the interior shape of an outflow layer

Example

Take $b > 0$ , $ϵ > 0$ , $f = 0$ , and endpoint data $u (0) = 0$ , $u (1) = 1$ . On the two-cell mesh $h = 1 / 2$ , the exact-parameter P1 SUPG solution has the exact nodal value at $x = 1 / 2$ . Nevertheless, at a fixed point inside the outflow cell its error tends to $0.98$ as $ϵ ↓ 0$ . The example separates nodal fitting from resolution of the layer inside an element.

Facts & Assumptions

Given: The model on $(0, 1)$ with $b > 0$ , $ϵ > 0$ , $f = 0$ , endpoint values $0, 1$ , two equal cells, and the exact parameter $τ$ from the one-dimensional SUPG form.

[F1]

One-dimensional stationary convection–diffusion model and uniform P1 space: the equation is $- ϵ u^{''} + b u^{'} = f$ , the two-cell P1 trial function is continuous with the prescribed endpoint values, and $P e = b h / (2 ϵ)$ for $b > 0$ .

[F2]

One-dimensional residual SUPG form with the exact cell parameter: $a_{h} (w_{h}, v_{h}) = ϵ (w_{h}^{'}, v_{h}^{'}) + b (w_{h}^{'}, v_{h}) + τ b^{2} (w_{h}^{'}, v_{h}^{'})$ , with the corresponding residual-weighted load, and $τ = h (2 b)^{- 1} (\coth (P e) - {P e}^{- 1})$ for $b > 0$ .

[F3]

The one-dimensional SUPG parameter is positive and bounded: the reference parameter $τ$ is strictly positive.

Verification

1.1F1givenalgebra

Put $r = b / ϵ$ . The function $u (x) = (e^{r x} - 1) / (e^{r} - 1)$ has endpoint values $0, 1$ , and $- ϵ u^{''} + b u^{'} = (- ϵ r^{2} + b r) e^{r x} / (e^{r} - 1) = 0$ . Hence it is the exact solution.

2.1F1F2F3step 1.1algebra

Let $ϕ$ be the P1 hat function at $x = h = 1 / 2$ , let $χ_{h} \in S_{h} (0, 1)$ have nodal values $(0, 0, 1)$ , and write $u_{h} = U ϕ + χ_{h}$ . Put $D = ϵ + τ b^{2} > 0$ by [F3]. Since $ϕ^{'} = 1 / h$ on the left cell and $- 1 / h$ on the right, $\int_{0}^{1} (ϕ^{'})^{2} d x = 2 / h$ and $\int_{0}^{1} ϕ^{'} ϕ d x = \frac{1}{2} [ϕ^{2}]_{0}^{1} = 0$ , so $a_{h} (ϕ, ϕ) = 2 D / h$ . Also $χ_{h}^{'} = 0$ on the left and $1 / h$ on the right, giving $D \int χ_{h}^{'} ϕ^{'} = - D / h$ and $b \int χ_{h}^{'} ϕ = b / h \int_{h}^{1} ϕ = b / 2$ . Thus $a_{h} (χ_{h}, ϕ) = - D / h + b / 2$ and the one interior equation is $D (2 U - 1) / h + b / 2 = 0$ . By [F2] and the Péclet formula in [F1], $D = ϵ + (b h / 2) (\coth (P e) - 1 / P e) = (b h / 2) \coth (P e)$ , so $U = (1 - \tanh (P e)) / 2 = 1 / (e^{2 P e} + 1)$ . The exact value $u (h) = (e^{b h / ϵ} - 1) / (e^{2 b h / ϵ} - 1)$ is the same number because $2 P e = b h / ϵ$ .

3.1step 1.1step 2.1algebra∎

At $x = 0.99$ , the piecewise-linear value is $u_{h} (0.99) = 0.98 + 0.02 U$ , which tends to $0.98$ as $ϵ ↓ 0$ . The exact value is $(e^{0.99 b / ϵ} - 1) / (e^{b / ϵ} - 1)$ , which tends to zero because $b > 0$ and $0.99 < 1$ . Therefore the pointwise error tends to $0.98$ . The exact midpoint node does not resolve the exponential layer inside the final cell.

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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)