Meshfree kernel approximation and conservative sph

By the integration-compatible partition in [F1] and linearity in [F3], subtracting Q_hg cell by cell gives ∫Ωg−Q_hg=Σ{j∈J_g}∫{C_j}(g(y)−g(x_j))dy, using ∫{C_j}g(x_j)=V_jg(x_j). On a bounding rectangle for each cell, [F3] gives |∫{C_j}(g−g(x_j))|≤∫{C_j}|g−g(x_j)|≤V_j osc_{C_j}(g). Summing proves the first inequality.

For y,z∈C_j, [F1] gives |y−x_j|≤h and |z−x_j|≤h; the triangle inequality in [F2] gives |y−z|≤2h. Hence |g(y)−g(z)|≤2hL_g and osc_{C_j}(g)≤2hL_g. Substituting this bound in step 1.1 proves the second inequality.

For each coordinate k, g_k(z)=z_kW(z) is continuous and compactly supported by [F1], hence Riemann integrable on a bounding box by [F3]. It satisfies g_k(−z)=−g_k(z). Apply [F2] to T(z)=−z, whose determinant has absolute value one: ∫g_k(z)dz=∫g_k(−z)dz=−∫g_k(z)dz. Hence ∫z_kW(z)dz=0.

For each k, z_kW_ε(z) is continuous and compactly supported, so [F3] makes it integrable on a bounding box. Apply [F2] to T(u)=εu: ∫z_kW_ε(z)dz=ε∫u_kW(u)du=0 by step 1.1. This holds for every k, so both vector first moments vanish.

Choose a box with the support of W in its interior. Every integrand below is continuous on this box by [F1] and integrable by [F5]. Fix all coordinates except z_k. Applying [F4] with F=W and G=1 gives ∫∂kW dz_k=0 because W vanishes near the two endpoints. Applying it with G=z_j gives ∫z_j∂kW dz_k=−δ{jk}∫W dz_k. Integrating the section identities over the remaining coordinates by [F3] gives the first identity and ∫z_j∂kW dz=−δ{jk}∫W dz=−δ{jk}.

With G=z_jz_l, [F4] gives ∫z_jz_l∂kW dz=−δ{jk}∫z_lW dz−δ_{lk}∫z_jW dz after [F3] integrates the other coordinates. Both moments vanish by [F2]. Finally |z|³|∇W| is continuous and compactly supported, so [F5] makes its integral finite; this is m₃.

If W_ε(z)≠0 then z/ε∈supp W, so |z|≤Rε. The chain rule in [F5] applied to ε^(−d)W(z/ε) gives the displayed pointwise first and second derivative formulas; taking suprema and substituting u=z/ε gives the claimed derivative sup norms.

The functions W_ε, zW_ε, and |z|²W_ε are continuous and compactly supported, hence integrable on a bounding box by [F4]. In the integrals, the affine substitution z=εu has Jacobian determinant ε^d. Thus ∫W_ε=∫W=1, ∫zW_ε(z)dz=ε∫uW(u)du=0 by [F2], and ∫|z|²W_ε(z)dz=ε²∫|u|²W(u)du=ε²m₂.

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.

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.

For each x,z apply [F2] with increment −z. It gives f(x−z)=f(x)−∇f(x)·z+R(x,z), where |R(x,z)|≤(1/2)||D²f||∞|z|².

Integrate this identity against the nonnegative W_ε. Linearity in [F3] and [F1] make the constant term integrate to f(x) and the linear term zero. The absolute-value estimate and monotonicity in [F3], together with W_ε≥0, bound the remainder by (1/2)||D²f||∞∫|z|²W_ε=(m₂/2)ε²||D²f||∞.

Let z,z′∈K, δ=||z−z′||#, and z(s)=(1−s)z+sz′ for 0≤s≤1; convexity keeps this segment in K. Differentiating ρ_i(q(s)) using [F5] and the finite-sum rule [F6] gives Σ_jm_j∇W_ε(q_i(s)−q_j(s))·(q_i′−q_i−q_j′+q_j), whose absolute value is at most 2MK₁δ by [F1], [F2], and Cauchy-Schwarz [F7]. Along the same segment, A_i(s)=a(ρ_i(q(s)),e_i(s)) has derivative bounded by 2MK₁A_ρδ+A_eδ=A_Lipδ, by [F5]. The one-dimensional mean value inequality [F3] on s∈[0,1] yields |A_i−A_i′|≤A_Lipδ.

For the acceleration component, subtract its two finite sums. The coefficient difference contributes at most 2MA_LipK₁δ, and the kernel-gradient difference at most 4MA₀K₂δ because the triangle inequality [F7] gives |(q_i−q_j)−(q_i′−q_j′)|≤2δ and [F3] bounds the change in ∇W_ε by K₂ times its argument change. Thus |v̇_i−v̇_i′|≤M(2A_LipK₁+4A₀K₂)δ.

For ė_i, the coefficient difference contributes at most 2MVA_LipK₁δ; the relative-velocity difference at most 2MA₀K₁δ by the triangle inequality [F7]; and the kernel-gradient difference at most 4MA₀VK₂δ by [F3]. Also |q̇_i−q̇_i′|≤δ. Taking the maximum of the three component bounds gives the stated L.

Differentiate P using the finite-sum rule [F3]. The self terms vanish because ∇W_ε(0)=0 by [F2], and grouping the remaining terms by unordered pairs gives Ṗ=Σ_{i<j}(F_ij+F_ji)=0.

Differentiate J using the coordinatewise product rule [F3]; the terms v_i⊗v_i cancel, leaving Σ_i(q_i⊗m_iq̈_i−m_iq̈_i⊗q_i). Grouping each pair gives (q_i−q_j)⊗F_ij−F_ij⊗(q_i−q_j)=0 because [F2] makes F_ij parallel to q_i−q_j. Thus J̇=0. Applying [F4] to P and to each component of J on every closed subinterval shows both are constant.

Differentiate K=Σ_im_i|v_i|²/2 using [F3] coordinatewise and substitute [F1]. Pairing the (i,j) and (j,i) terms using [F2] yields K̇=−Σ_{i<j}m_im_j(A_i+A_j)(v_i−v_j)·g_ij.

Differentiate U=Σ_im_ie_i using [F3] and substitute [F1]. The (i,j) term is m_im_jA_i(v_i−v_j)·g_ij, while the (j,i) term equals m_im_jA_j(v_i−v_j)·g_ij by [F2]. Hence U̇=Σ_{i<j}m_im_j(A_i+A_j)(v_i−v_j)·g_ij.

Steps 1.1 and 1.2 cancel term by term, so Ḣ=K̇+U̇=0. Applying [F4] with derivative bound M=0 on any closed subinterval shows H is constant on each solution interval.

The scalar components of f(y)∇W_ε(x−y) are continuous and compactly supported, so [F5] makes them Riemann integrable on a bounding box. Apply [F4] with y=x−εu, whose absolute Jacobian determinant is ε^d, and use [F1] to obtain D_εf(x)=ε^(−1)∫f(x−εu)∇W(u)du.

For each x,u, [F3] gives f(x−εu)=f(x)−ε∇f(x)·u+(ε²/2)D²f(x)[u,u]+R(x,u), with |R(x,u)|≤(ε³/6)||D³f||∞|u|³.

Integrate the expansion of step 1.2 against ε^(−1)∇W(u). Linearity [F6] and [F2] make the constant and quadratic terms vanish; the linear term in coordinate k equals −Σ_j∂_jf(x)∫u_j∂_kW(u)du=∂_kf(x). Thus the integrated polynomial part is ∇f(x).

Put G(u)=R(x,u)∇W(u), a continuous compactly supported vector function, and v=∫G componentwise. If v≠0 set a=v/||v||₂; then ||a||₂=1 and ||v||₂=∫⟨a,G(u)⟩≤∫|⟨a,G(u)⟩|≤∫||G(u)||₂ by linearity, monotonicity, and [F7]. If v=0 the same upper bound is immediate. The pointwise remainder estimate in step 1.2 and [F2] therefore give ||v||₂≤(ε³/6)||D³f||∞∫|u|³|∇W(u)|du=(ε³m₃/6)||D³f||∞. Multiplication by ε^(−1) from step 1.1 proves the stated bound.

Put g₀(y)=f(y)W_ε(x−y). On any segment y(t)=y₀+t(y₁−y₀), [F6] gives the chain and product rules, and Cauchy-Schwarz [F7] gives |(g₀∘y)'(t)|≤(ε^(−d)M₀||∇f||∞+ε^(−d−1)M₁||f||∞)|y₁−y₀| by [F3]. Applying [F5] on that segment gives the same Lipschitz constant for g₀. Its support is contained in the closed ball \overline B(x,Rε), hence in B(x,(R+1)ε). By [F2], the total volume of cells meeting that support is at most C_L((R+1)ε+h)^d≤C_L(R+2)^dε^d. Apply [F4] and cancel ε^d; the factor 2 is absorbed into C. This gives the first inequality.

Put g₁(y)=f(y)∇W_ε(x−y). On any segment y(t)=y₀+t(y₁−y₀), [F6] gives componentwise chain and product rules, and Cauchy-Schwarz [F7] gives ||(g₁∘y)'(t)||₂≤(ε^(−d−1)M₁||∇f||∞+ε^(−d−2)M₂||f||∞)|y₁−y₀| by [F3]. Applying the vector mean value inequality [F5] bounds its Lipschitz constant by this expression. Its support is contained in the same closed ball, so [F2] gives the active-volume bound from step 1.1. Apply [F4] to each vector component; taking the Euclidean norm costs at most a factor √d, and [F4] contributes a factor 2; both are absorbed into C. Cancelling ε^d proves the second inequality.

Add and subtract S_εf(x): |Q⁰_{ε,h}f−f|≤|Q⁰_{ε,h}f−S_εf|+|S_εf−f|. Apply [F1] and the first bound in [F3] to obtain the first inequality after increasing constants.

Add and subtract D_εf(x): |Q¹_{ε,h}f−∇f|≤|Q¹_{ε,h}f−D_εf|+|D_εf−∇f|. Apply [F2] and the second bound in [F3] to obtain the second inequality.

If ε→0 and h/ε²→0, then ε²→0, h/ε²→0, and h/ε≤h/ε² because ε≤1. Every term on the right sides of steps 1.1 and 1.2 therefore tends to zero uniformly at the stated interior points.

Each ρ_i is continuous by [F1], so U is open as the finite intersection of the inverse images of (0,∞). On U, denominators ρ_i² are nonzero, P is C¹, W_ε is C², and the finite sums defining F are continuous.

Fix z₀∈U. Since U is open, choose a nonempty nondegenerate closed box K about z₀ small enough to lie in U. It is convex and compact by [F4]. Each ρ_i is continuous and positive on K, so [F4] gives a positive minimum for each; taking the minimum over finitely many i gives a positive lower density bound on K. [F2] makes F Lipschitz there. Regard F as time-independent on R×U and apply [F3] to obtain a unique local C¹ solution. No choice principle is used.

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.

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)||#.

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.

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.