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Forward Euler has first-order global error and preserves a margin tube for the regularized SPH ODE

Statement

Use the state norm ||z||#=max_i{|q_i|,|v_i|,|e_i|} on the finite-dimensional state space, and dist#(z,Kᶜ)=inf_{y∉K}||z−y||#. Let T>0 and let z(t), 0≤t≤T, be the unique solution from Local well-posedness of the regularized finite-particle SPH system. Let K⊂U be a nonempty compact convex set, with dist#(z(t),Kᶜ)≥δ>0 for every t. Assume B,L≥0, ||F(w)||#≤B for every w∈K, and Lip#(F|K)≤L. For an integer n≥1 set h=T/n, Z₀=z(0), and Z_{k+1}=Z_k+hF(Z_k). Put E_h=(Bh/2)(exp(LT)−1), which is zero when L=0. If E_h<δ, then every Euler state Z_k lies in K and

max⁡0≤k≤n∥Zk−z(kh)∥#≤Eh.

The tube margin and exact-path hypothesis are explicit. This result makes no energy- or momentum-conservation claim for the Euler time discretization.

Facts & Assumptions

Given: The regularized SPH vector field F, an exact solution on [0,T] with the stated margin in K, finite B,L, a positive step h=T/n, and exact initial data.

[F1]

Local well-posedness of the regularized finite-particle SPH system: F has a unique local solution on the positive-density open state domain.

[F2]

A first-order initial value problem is equivalent to its Volterra integral equation: a C¹ solution of ż=F(z) satisfies z(t+s)=z(t)+∫₀ˢF(z(t+r))dr.

[F3]

Explicit local Lipschitz bound for the regularized SPH vector field: F is L-Lipschitz on K and has the stated bound B when assumed in the Given clause.

[F6]

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation: each Euclidean position/velocity block norm satisfies the triangle and homogeneity axioms; the real internal-energy blocks have the same properties, and taking a finite maximum preserves them. Thus ||·||# is a norm.

[F7]

For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable: applying the integral norm inequality to each Euclidean state block, then taking the finite maximum, gives ||∫g||#≤∫||g||# on a time interval.

Proof

1.1F2F3F7givenalgebra

The local defect from an exact start is z(t+h)−z(t)−hF(z(t))=∫₀ʰ(F(z(t+s))−F(z(t)))ds by [F2]. Since z(t)∈K and ||F(w)||#≤B for every w∈K, ||ż(t)||#≤B; applying [F7] to the Volterra formula gives ||z(t+s)−z(t)||#≤Bs. Then [F3] bounds the defect integrand by LBs. A second application of [F7] and integration of LBs give the local defect norm bound LBh²/2.

2.1F1F3F6step 1.1algebra

Suppose Z_k∈K. By convexity of K the segment from Z_k to z(kh) lies in K, so [F3] gives ||Z_k+hF(Z_k)−(z(kh)+hF(z(kh)))||#≤(1+hL)||Z_k−z(kh)||#. The triangle and homogeneity properties in [F6], together with the defect from step 1.1, yield e_{k+1}≤(1+hL)e_k+LBh²/2, where e_k=||Z_k−z(kh)||#.

3.1F4F5step 2.1algebra

Starting with e₀=0 and iterating step 2.1 gives e_k≤(Bh/2)((1+hL)^k−1) when L>0, by the finite geometric-sum identity; when L=0 the defect is zero and e_k=0. Since 1+hL≤exp(hL) by [F4] and exp(khL)=exp(Lkh) by repeated [F5], e_k≤E_h for k≤n.

4.1givenstep 2.1step 3.1induction∎

Initially Z₀=z(0)∈K. If Z₀,…,Z_k∈K, steps 2.1–3.1 give e_{k+1}≤E_h<δ. The exact state z((k+1)h) is at distance at least δ from Kᶜ, so Z_{k+1}∈K. Induction proves all Euler states remain in K and the bound holds for every k.

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