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Explicit Euler gives a second-order local backward-foot defect

Statement

Let b∈C1([0,T]×Rd;Rd) be periodic in space, and set M:=sup⁡∥b∥2,Lx:=sup⁡∥Dxb∥,Lt:=sup⁡∥∂tb∥2, where the suprema are over one full periodic time-space cylinder. For 0≤tn<tn+1≤T, Δt=tn+1−tn, and an arrival lift x, let X(tn;tn+1,x) be the exact backward characteristic foot. The explicit Euler foot x−Δt b(tn+1,x) satisfies ∥X(tn;tn+1,x)−(x−Δt b(tn+1,x))∥2≤12(LxM+Lt)(Δt)2. The estimate is local for every positive Δt; it requires no relation between displacement and grid width.

Facts & Assumptions

Given: The periodic C1 field, its finite derivative bounds, a time interval [tn,tn+1], and an arrival point x.

[F1]

Explicit Euler characteristic backtracking with multilinear interpolation: the algorithm uses the lifted explicit backward foot x−Δt b(tn+1,x).

[F2]

A periodic C1 velocity has a global characteristic flow on a finite horizon: the exact characteristic exists on the whole interval, and ∥b∥2≤M on its periodic tube.

[F3]

A first-order initial value problem is equivalent to its Volterra integral equation: the exact characteristic increment is the oriented integral of b(s,X(s)).

Proof

1.1givenF2F3F4

By [F2] and [F3], for s∈[tn,tn+1] the backward characteristic satisfies X(s)=x−∫stn+1b(r,X(r)) dr. Applying [F4] and the velocity bound gives ∥X(s)−x∥2≤M(tn+1−s).

2.1givenF1F3F4step 1.1algebra∎

Subtracting the explicit Euler foot from the exact foot and using [F3] writes the difference as the integral from t_n to t_{n+1} of b(t_{n+1},x)-b(s,X(s)). The derivative bounds make its integrand norm at most L_x times the state difference plus L_t times (t_{n+1}-s); by step 1.1 the state difference is at most M(t_{n+1}-s). Integrating the resulting bound (L_x M+L_t)(t_{n+1}-s) gives one half (L_x M+L_t)(Delta t)^2.

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