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A conservative positive-pressure SPH system can close a particle gap

Statement refuted

In a conservative SPH system with positive pressure, every adjacent particle gap has nonnegative initial relative acceleration. This is false: a third-particle pressure force can close a neighboring gap even while the semidiscrete invariants remain exact.

Facts & Assumptions

Given: Let d=1, N=3, m_i=1, ε>0, q_1=0, q_2=9ε/10, q_3=13ε/10, and v_i=0 initially. Use W(s)=(35/32)(1−s²)³ for |s|≤1 and zero otherwise, the smooth ideal-gas pressure map P(ρ,e)=(γ−1)ρe with γ>1, and e_i=p_0ρ_i(q)/(γ−1) initially, where p_0>0.

[F1]

Fixed-width regularized conservative SPH particle system: A_i=P(ρ_i,e_i)/ρ_i² and the particle acceleration is the symmetric pressure-gradient sum.

[F2]

SPH pressure forces are equal-opposite and central: the paired pressure forces are equal-opposite and central.

[F3]

Linear and angular momentum are conserved by semidiscrete SPH: total linear and angular momenta are constant along solutions.

[F4]

The matched internal-energy equation conserves total semidiscrete energy: H=Σ_i(|v_i|²/2+e_i) is constant along solutions for unit masses.

[F5]

Local well-posedness of the regularized finite-particle SPH system: a unique local solution exists from every positive-density state.

Counterexample

1.1F1F5givenalgebra

The polynomial kernel is nonnegative and even. At s=±1 its value and first two inside derivatives vanish, so extension by zero is C²; its integral is (35/32)·2(1−1+3/5−1/7)=1. Each density is positive because its self term W_ε(0)>0. The chosen internal energies give p_i=(γ−1)ρ_ie_i=p_0ρ_i²>0 and A_i=p_0 at t=0. By [F1] and [F5], a unique local solution exists; pressure positivity persists for a short interval by continuity.

1.2givenalgebra

For 0<s<1, |W′(s)|=(105/16)s(1−s²)². Direct substitution gives |W′(2/5)|/|W′(9/10)|=28224/3249>2. Also |q_3−q_1|=13ε/10>ε, so the 1–3 kernel-gradient term vanishes.

2.1F1step 1.2algebra

Let r(t)=q_2(t)−q_1(t). From [F1] and step 1.2, the accelerations at t=0 are q̈_1=−2p_0ε^(−2)|W′(9/10)| and q̈_2=2p_0ε^(−2)(|W′(9/10)|−|W′(2/5)|). Therefore r̈(0)=2p_0ε^(−2)(2|W′(9/10)|−|W′(2/5)|)<0.

3.1F5step 2.1algebra

Since ṙ(0)=0 and r̈ is continuous, ṙ(t)<0 and r(t)<r(0) for all sufficiently small t>0. Thus the adjacent gap closes initially under strictly positive pressure.

4.1F2F3F4step 1.1step 3.1∎

The exact pair antisymmetry and matched internal-energy identity apply by [F2] and [F4], so total momentum and H are conserved throughout this same local motion. The example therefore refutes the claimed gap-stability implication while retaining those invariants.

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