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Backward Euler strongly damps high periodic modes

Example

Fix $Δ t > 0$ and consider even periodic one-dimensional grids with $h = L / N \to 0$ . Backward Euler multiplies the Nyquist mode by $1 / (1 + 4 κ Δ t / h^{2})$ , which tends to zero, while its weighted grid $L^{2}$ norm never increases.

Verification

Given: $κ > 0$ , a fixed $Δ t > 0$ , and an even periodic grid with spacing $h = L / N$ .

[F1] Fully discrete θ heat finite-difference scheme: backward Euler is the θ update with $θ = 1$ and zero source.

[F2] Periodic heat-stencil eigenvalues and θ amplification factor: the Nyquist eigenvalue is $λ = 4 κ / h^{2}$ , and its multiplier is $g_{θ} = (1 - (1 - θ) Δ t λ) / (1 + θ Δ t λ)$ .

1.1F1F2givenalgebra

Setting $θ = 1$ and $λ = 4 κ / h^{2}$ in [F2] gives $g = 1 / (1 + 4 κ Δ t / h^{2})$ .

2.1step 1.1algebra

As $h \to 0$ with $κ, Δ t > 0$ fixed, the denominator tends to infinity, so $g \to 0$ . Thus this high-frequency mode is strongly damped.

3.1F2step 1.1algebra∎

For every mode $λ \geq 0$ , the backward-Euler factor is $1 / (1 + Δ t λ) \in (0, 1]$ , so the periodic Fourier basis gives nonincrease of the full weighted grid $L^{2}$ norm.

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