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Refining time alone leaves interpolation diffusion on a fixed grid

Example

Fix an integer $N \geq 4$ , the unit torus, and $h = 1 / N$ . Let $b = 1$ and use final time $T = h$ , but split it into $K$ steps of length $Δ t = h / K$ . The smooth periodic initial function $u_{0} (x) : = \prod_{j = 1}^{N - 1} \frac{\sin^{2} (π (x - j / N))}{\sin^{2} (- π j / N)}$ has grid samples $U_{0}^{0} = 1$ and $U_{j}^{0} = 0$ for $1 \leq j < N$ . Exact transport over $T = h$ shifts those samples by exactly one node. The semi-Lagrangian method, however, performs $K$ fractional linear interpolations on the same fixed spatial grid.

Facts & Assumptions

Given: The unit grid, constant velocity, smooth initial function, and update rule in the Example.

[F1]

Explicit Euler characteristic backtracking with multilinear interpolation: for a positive constant velocity and a one-cell-or-smaller backward displacement $α h$ , the update is $U_{j}^{n + 1} = (1 - α) U_{j}^{n} + α U_{j - 1}^{n}$ .

[F2]

For every real $x$ , $(1 + x / n)^{n} \to \exp x$ : for real $x$ , $(1 + x / K)^{K} \to e^{x}$ as integers $K \to \infty$ .

Proof

1.1givenF1algebra

Each denominator in $u_{0}$ is nonzero because $j / N \notin Z$ , each factor is smooth and $1$ -periodic, and at node $x_{0} = 0$ every factor equals $1$ . At node $x_{m} = m / N$ with $1 \leq m < N$ , the factor with $j = m$ vanishes. Thus the grid samples are exactly the stated unit vector. The exact characteristic is $X (t; s, x) = x + (t - s)$ , so at $T = h$ the exact grid vector has value $1$ at node $1$ and $0$ at node $2$ .

1.2givenF1inductionalgebra

For $K \geq 2$ , put $α = Δ t / h = 1 / K$ . Repeated application of [F1] is convolution with the two coefficients $(1 - 1 / K, 1 / K)$ ; induction using Pascal's identity shows that after $K$ steps the contribution from exactly two one-node shifts at node $2$ is $(\binom{K}{2}) K^{- 2} (1 - 1 / K)^{K - 2}$ . Periodic wrap can add other nonnegative contributions, so $U_{2}^{K} \geq (\binom{K}{2}) K^{- 2} (1 - 1 / K)^{K - 2}$ .

2.1givenF2step 1.1step 1.2algebra∎

Since $(\binom{K}{2}) K^{- 2} \to 1 / 2$ and $(1 - 1 / K)^{K - 2} = (1 - 1 / K)^{K} / (1 - 1 / K)^{2} \to e^{- 1}$ by [F2], we have ${lim inf}_{K \to \infty} U_{2}^{K} \geq e^{- 1} / 2 > 0$ , whereas the exact value at node $2$ is zero. Thus reducing $Δ t$ at fixed $h$ does not remove this interpolation error. The sufficient condition $h^{2} / Δ t \to 0$ in the global error theorem cannot be dropped from that bound: here $h^{2} / Δ t = h K \to \infty$ .

Remarks

This is a fixed-grid time-refinement example, not a failure of convergence under simultaneous refinement. It exhibits the per-step interpolation effect that the $h^{2} / Δ t$ term controls in Periodic semi-Lagrangian error is controlled by local foot and interpolation defects.

Depends on

Used by

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Sources

Lukas Einkemmer and Alexander Ostermann, On the error propagation of semi-Lagrange and Fourier methods for advection problems, Computers & Mathematics with Applications 69 (2015), 170–179 (standard reference, not scraped)