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SPH pressure forces are equal-opposite and central

Statement

For the system in Fixed-width regularized conservative SPH particle system, set g_ij=∇W_ε(q_i−q_j) and define the force contribution F_ij=−m_im_j(A_i+A_j)g_ij. Radial evenness gives g_ji=−g_ij and g_ii=0, so for i≠j, F_ji=−F_ij; also F_ij is parallel to q_i−q_j (the zero vector is allowed).

Facts & Assumptions

Given: A state in the positive-density domain, positive masses, and a radial C² kernel.

[F1]

Fixed-width regularized conservative SPH particle system: A_i=p_i/ρ_i² and the acceleration sum is −Σ_jm_j(A_i+A_j)∇W_ε(q_i−q_j).

[F2]

Scaled symmetric compactly supported kernel family: W_ε is differentiable and radial, hence even.

[F3]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): the derivative of W_ε composed with r↦−r is obtained by the chain rule.

Proof

1.1F1F2F3algebra

Since W_ε(−r)=W_ε(r), differentiating the identity in r using [F3] gives ∇W_ε(−r)=−∇W_ε(r). The scalar coefficient m_im_j(A_i+A_j) is symmetric in i,j, so F_ji=−F_ij.

2.1F2F3algebra∎

If r≠0 and u·r=0, then |r+tu|²=|r|²+t²|u|², so the derivative at t=0 of W_ε(r+tu), which is w_ε(|r+tu|), is zero by the chain rule [F3]. Thus ∇W_ε(r) is orthogonal to every u orthogonal to r and lies in span(r). At r=0, differentiable evenness gives ∇W_ε(0)=0 by [F3]. Hence each F_ij is central.

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