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Deterministic reciprocal particle-mesh Ewald potential, energy, and force bounds

Statement

Assume Countable Choice and neutral finite source data on the cell $A Z^{3}$ from Neutral point sources and Fourier data on a three-dimensional torus. Fix $α > 0$ , integers $M_{r} \geq 1$ , an order $P \geq 2$ , and a finite symmetric set $K \subset Z^{3} ∖ {0}$ whose signed representatives are injective modulo $(M_{1}, M_{2}, M_{3})$ and satisfy $∣ k_{r} ∣ < M_{r} / 2$ . Require $D (k) \neq 0$ for every $k \in K$ , and let $c_{k} = 4 π e^{- ∣ κ_{k} ∣^{2} / (4 α^{2})} / (V ∣ κ_{k} ∣^{2})$ be the positive Ewald coefficient. Use the assignment factors $a_{k}, b_{k}, ρ_{k}, z_{k}$ of Periodic cardinal-spline assignment and unitary mesh transform, and put $Q = \sum_{j} ∣ q_{j} ∣$ .

For a query point $x$ write $t = A^{- 1} x$ modulo $Z^{3}$ and define the retained exact and mesh reciprocal potentials by $U_{K} (x) = \sum_{k \in K} c_{k} Re (ρ_{k} \overline{a_{k} (t)}), U_{M, K} (x) = \sum_{k \in K} c_{k} Re (z_{k} \overline{b_{k} (t)}) .$ Define $E_{K} = \frac{1}{2} \sum_{k \in K} c_{k} ∣ ρ_{k} ∣^{2}$ and $E_{M, K} = \frac{1}{2} \sum_{k \in K} c_{k} ∣ z_{k} ∣^{2}$ . For each mode let $A_{k}$ be the source-alias controller in Exact three-dimensional spline assignment aliases and FFT normalization, and let $B_{k}$ be the derivative-alias controller in Derivative-window alias bounds including the P=2 knot branch.

Let a complex-vector derivative/gather value $d_{k} (t)$ satisfy $∥ d_{k} (t) - d_{k} (t) ∥_{2} \leq H_{k}$ for a stated $H_{k} \geq 0$ , where $d_{k}$ is the declared derivative branch of Periodic cardinal-spline assignment and unitary mesh transform and $\nabla_{x} = A^{- T} \nabla_{t}$ . The uncorrected mesh and exact reciprocal forces are $F_{M, i}^{d} = - q_{i} \sum_{k \in K} c_{k} Re (\overline{z_{k}} d_{k} (t_{i})), F_{K, i} = - q_{i} \sum_{k \in K} c_{k} Re (\overline{ρ_{k}} \nabla_{x} a_{k} (t_{i})) .$ Then for every query point $x$ , $∣ U_{M, K} (x) - U_{K} (x) ∣ \leq Q \sum_{k \in K} c_{k} A_{k} (2 + A_{k}), ∣ E_{M, K} - E_{K} ∣ \leq \frac{Q^{2}}{2} \sum_{k \in K} c_{k} A_{k} (2 + A_{k}),$ and for every particle $i$ , $∥ F_{M, i}^{d} - F_{K, i} ∥_{2} \leq ∣ q_{i} ∣ Q \sum_{k \in K} c_{k} [∣ κ_{k} ∣ A_{k} + (1 + A_{k}) B_{k} + (1 + A_{k}) H_{k}] .$

For the declared energy-gradient gather $d_{k} = d_{k}$ (so $H_{k} = 0$ ), the mesh force is $- \nabla_{x_{i}} E_{M, K}$ wherever the classical derivative exists. This holds everywhere for $P \geq 3$ and off the assignment knots for $P = 2$ ; at a $P = 2$ knot the selected one-sided derivative or its average is a procedure value, with the same bound but without a differentiability claim.

The optional diagonal self correction is $C_{M} = \frac{1}{2} \sum_{i} q_{i}^{2} \sum_{k \in K} c_{k} (∣ b_{k} (t_{i}) ∣^{2} - 1), E_{M, K}^{c o r r} = E_{M, K} - C_{M} .$ For its finite cyclic implementation, set $h_{k} = c_{k} / ∣ D (k) ∣^{2}$ on $K$ , $γ_{d} = \sum_{k \in K} h_{k} e^{- 2 π i \sum_{r} k_{r} d_{r} / M_{r}}, S_{i} = \sum_{ℓ, m \in G} w_{ℓ} (t_{i}) w_{m} (t_{i}) γ_{ℓ - m} .$ Then $γ$ is real and even, $S_{i} = \sum_{k \in K} c_{k} ∣ b_{k} (t_{i}) ∣^{2}$ , and, on every differentiable branch (or as the same declared one-sided branch at a $P = 2$ knot), $\nabla_{x_{i}} (S_{i} / 2) = \sum_{ℓ, m \in G} w_{ℓ} (t_{i}) \nabla_{x} w_{m} (t_{i}) γ_{ℓ - m} .$ Consequently this local cyclic-kernel expression computes $C_{M}$ and its derivative exactly in real arithmetic. The finite tensor-transform gathers also have the exact real-arithmetic forms $U_{M, K} (x) = Re \sum_{ℓ \in G} w_{ℓ} (t_{x}) u_{ℓ}, u_{ℓ} = \sum_{k \in K} \frac{c_{k} z_{k}}{D (k)} e^{2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}},$ $F_{M, i} = - q_{i} Re \sum_{ℓ \in G} λ_{ℓ} \nabla_{x} w_{ℓ} (t_{i}), λ_{ℓ} = \sum_{k \in K} \frac{c_{k} \overline{z_{k}}}{\overline{D (k)}} e^{- 2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}} .$ These identities hold with the declared derivative branch, including the selected one-sided $P = 2$ knot value. It satisfies $∣ E_{M, K}^{c o r r} - E_{K} ∣ \leq \frac{Q^{2} - \sum_{i} q_{i}^{2}}{2} \sum_{k \in K} c_{k} A_{k} (2 + A_{k}) .$ In the energy-gradient branch its exactly differentiated force is $F_{M, i}^{c o r r} = - q_{i} \sum_{k \in K} c_{k} Re (\overline{z_{k} - q_{i} b_{k} (t_{i})} d_{k} (t_{i})),$ and $∥ F_{M, i}^{c o r r} - F_{K, i} ∥_{2} \leq ∣ q_{i} ∣ (Q - ∣ q_{i} ∣) \sum_{k \in K} c_{k} [∣ κ_{k} ∣ A_{k} + (1 + A_{k}) B_{k}] .$ The corrected force is the negative classical energy gradient in the same differentiable cases; at a $P = 2$ knot it is only the declared branch value. No pointwise error estimate here replaces an RMS estimate, and no claim is made that a different spectral differentiation rule is the gradient of the mesh energy.

Facts & Assumptions

Given: Countable Choice; a neutral finite source configuration, $A, V$ , $κ_{k}$ , and $a_{k} (t) = e^{- i κ_{k} \cdot x}$ ; an order- $P$ periodic mesh assignment; a finite symmetric retained mode set with $D (k) \neq 0$ ; and a derivative/gather residual $H_{k}$ .

[A1]

Countable Choice is assumed by the Gaussian Poisson and mean-zero periodic Ewald suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The cell coordinates obey $x = A t$ , $κ_{k} = 2 π A^{- T} k$ , the source coefficient is $ρ_{k} = \sum_{j} q_{j} e^{- i κ_{k} \cdot x_{j}}$ , and the charges are real (Neutral point sources and Fourier data on a three-dimensional torus).

[F2]

The positive coefficient $c_{k} = 4 π e^{- ∣ κ_{k} ∣^{2} / (4 α^{2})} / (V ∣ κ_{k} ∣^{2})$ and exact retained reciprocal energy are those in the alpha-independent Ewald formula (The alpha-independent Ewald potential and renormalized energy).

[F3]

For each retained mode, $∣ b_{k} - a_{k} ∣ \leq A_{k}$ , $∣ b_{k} ∣ \leq 1 + A_{k}$ , and $∣ z_{k} - ρ_{k} ∣ \leq Q A_{k}$ ; $A_{k}$ has finite p-series remainder controllers (Exact three-dimensional spline assignment aliases and FFT normalization).

[F4]

The declared derivative branch obeys $∥ d_{k} - \nabla_{x} a_{k} ∥_{2} \leq B_{k}$ , including the finite selected branch at $P = 2$ knots (Periodic cardinal-spline assignment and unitary mesh transform, Derivative-window alias bounds including the P=2 knot branch).

[F5]

$\nabla_{x} a_{k} = - i κ_{k} a_{k}$ ; the derivative follows from the sine/cosine derivatives and the chain rule (Neutral point sources and Fourier data on a three-dimensional torus, The derivatives of sine and cosine are cosine and minus sine, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F6]

The exact retained reciprocal potential is $\sum_{k \in K} c_{k} ρ_{k} e^{i κ_{k} \cdot x}$ , and the reciprocal energy is $\frac{1}{2} \sum_{k \in K} c_{k} ∣ ρ_{k} ∣^{2}$ (The alpha-independent Ewald potential and renormalized energy).

[F7]

The mesh factors are finite sums of the assignment weights, $\nabla_{x} = A^{- T} \nabla_{t}$ , and the declared branch is classical wherever the derivative exists; the chain rule differentiates these factors componentwise there (Periodic cardinal-spline assignment and unitary mesh transform, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[F8]

The periodic assignment weights are real, $b_{k} (t) = (\sum_{ℓ \in G} w_{ℓ} (t) e^{- 2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}}) / \overline{D (k)}$ , and $D (- k) = \overline{D (k)}$ (Periodic cardinal-spline assignment and unitary mesh transform).

Proof

technique · direct

1.1givenA1F1F2F3F6algebra

Under the Countable Choice assumption [A1], the Ewald supplier gives the reciprocal potential and energy formulas [F6]. Because the charges and assignment weights are real, $c_{- k} = c_{k}$ , $ρ_{- k} = \overline{ρ_{k}}$ , $a_{- k} = \overline{a_{k}}$ , $D (- k) = \overline{D (k)}$ , $b_{- k} = \overline{b_{k}}$ , and $z_{- k} = \overline{z_{k}}$ . Thus the sums over symmetric $K$ are real and their displayed real-part forms agree with the reciprocal potentials. By [F3], $∣ ρ_{k} ∣ \leq Q$ , $∣ z_{k} - ρ_{k} ∣ \leq Q A_{k}$ , and $∣ z_{k} ∣ \leq Q (1 + A_{k})$ . Since $e^{i κ_{k} \cdot x} = \overline{a_{k} (t)}$ , [F6] gives the exact reciprocal potential. For each mode, $∣ z_{k} \overline{b_{k}} - ρ_{k} \overline{a_{k}} ∣ \leq ∣ z_{k} - ρ_{k} ∣ ∣ a_{k} ∣ + ∣ z_{k} ∣ ∣ b_{k} - a_{k} ∣ \leq Q A_{k} + Q (1 + A_{k}) A_{k} = Q A_{k} (2 + A_{k})$ using $∣ a_{k} ∣ = 1$ . Sum with $c_{k} > 0$ to obtain the potential bound.

1.2F2F3algebra

The identity $∣ ∣ z_{k} ∣^{2} - ∣ ρ_{k} ∣^{2} ∣ \leq ∣ z_{k} - ρ_{k} ∣ (∣ z_{k} ∣ + ∣ ρ_{k} ∣)$ and [F3] give $∣ ∣ z_{k} ∣^{2} - ∣ ρ_{k} ∣^{2} ∣ \leq Q A_{k} (Q (1 + A_{k}) + Q) = Q^{2} A_{k} (2 + A_{k})$ . Multiplying by $c_{k} / 2$ and summing proves the energy bound.

1.3F3F4F5F6F7algebra

The mode sums are finite. For $P \geq 3$ , and for $P = 2$ off the assignment knots, [F4] supplies the classical derivatives. Since $z_{k} = \sum_{j} q_{j} b_{k} (t_{j})$ by [F3], [F7] gives $\nabla_{x_{i}} z_{k} = q_{i} \nabla_{x} b_{k} (t_{i})$ . Differentiating $\frac{1}{2} c_{k} ∣ z_{k} ∣^{2}$ yields $q_{i} c_{k} Re (\overline{z_{k}} \nabla_{x} b_{k} (t_{i}))$ , so negating and summing proves $F_{M, i}^{\nabla} = - \nabla_{x_{i}} E_{M, K}$ in those differentiable cases. Also $\nabla_{x_{i}} ρ_{k} = q_{i} \nabla_{x} a_{k} (t_{i})$ by [F5]; differentiating $\frac{1}{2} c_{k} ∣ ρ_{k} ∣^{2}$ gives $q_{i} c_{k} Re (\overline{ρ_{k}} \nabla_{x} a_{k} (t_{i}))$ , so the negative derivative is exactly $F_{K, i}$ . At a $P = 2$ knot the chosen derivative is a procedure value, not a classical derivative.

1.4givenF3F4F5algebra

Decompose the force integrand as $\overline{z_{k}} d_{k} - \overline{ρ_{k}} \nabla_{x} a_{k} = \overline{(z_{k} - ρ_{k})} \nabla_{x} a_{k} + \overline{z_{k}} (d_{k} - \nabla_{x} a_{k}) + \overline{z_{k}} (d_{k} - d_{k})$ . By [F3], [F4], the given residual bound $H_{k}$ , and $∥ \nabla_{x} a_{k} ∥_{2} = ∣ κ_{k} ∣$ from [F5], its norm is at most $Q [∣ κ_{k} ∣ A_{k} + (1 + A_{k}) B_{k} + (1 + A_{k}) H_{k}]$ . Multiply by $∣ q_{i} ∣ c_{k}$ and sum to prove the uncorrected force bound.

1.5F2F3algebra

Expanding $∣ z_{k} ∣^{2}$ with $z_{k} = \sum_{j} q_{j} b_{k} (t_{j})$ shows that subtracting $C_{M}$ replaces the diagonal terms $q_{i}^{2} ∣ b_{k} (t_{i}) ∣^{2}$ by $q_{i}^{2}$ . In $∣ ρ_{k} ∣^{2}$ , the diagonal terms are $q_{i}^{2} ∣ a_{k} (t_{i}) ∣^{2} = q_{i}^{2}$ . The remaining difference is a sum over $i \neq j$ of $q_{i} q_{j} (b_{k} (t_{i}) \overline{b_{k} (t_{j})} - a_{k} (t_{i}) \overline{a_{k} (t_{j})}) / 2$ . Each parenthesized difference is bounded by $A_{k} (1 + A_{k}) + A_{k} = A_{k} (2 + A_{k})$ using [F3]; and $\sum_{i \neq j} ∣ q_{i} q_{j} ∣ = Q^{2} - \sum_{i} q_{i}^{2}$ . Multiplication by $c_{k}$ and summation proves the corrected energy estimate.

1.6F2F8algebra

Because $K$ is symmetric and $h_{- k} = h_{k}$ , the finite sum defining $γ$ is real and satisfies $γ_{- d} = γ_{d}$ . Put $B_{k} (t) = \sum_{ℓ \in G} w_{ℓ} (t) e^{- 2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}}$ . Expanding the finite sums gives $S_{i} = \sum_{k \in K} h_{k} ∣ B_{k} (t_{i}) ∣^{2} = \sum_{k \in K} c_{k} ∣ b_{k} (t_{i}) ∣^{2}$ by [F8]. Subtracting $\frac{1}{2} \sum_{i} q_{i}^{2} \sum_{k} c_{k}$ therefore gives exactly $C_{M}$ .

1.7F7F8algebra

By [F8], $\overline{b_{k} (t)} = \sum_{ℓ} w_{ℓ} (t) e^{2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}} / D (k)$ . Also differentiating the finite sum for $b_{k}$ gives $d_{k} (t) = \sum_{ℓ} \nabla_{x} w_{ℓ} (t) e^{- 2 π i \sum_{r} k_{r} ℓ_{r} / M_{r}} / \overline{D (k)}$ on a classical branch and on the declared one-sided branch. Substituting these identities into the finite mode formulas for $U_{M, K}$ and $F_{M, i}$ and interchanging the finite mode and grid sums yields exactly the two tensor-transform gather forms in the Statement.

2.1F7step 1.6algebra

Differentiate the finite sum $S_{i}$ on any declared derivative branch. Symmetry $γ_{ℓ - m} = γ_{m - ℓ}$ makes the two product-rule sums equal, so $\nabla_{x_{i}} (S_{i} / 2) = \sum_{ℓ, m} w_{ℓ} (t_{i}) \nabla_{x} w_{m} (t_{i}) γ_{ℓ - m}$ . The identity is algebraic in the one-sided polynomial derivatives as well, so it holds for the declared $P = 2$ knot branch values.

3.1F3F4F5F7step 1.3step 1.4step 2.1algebra∎

Where the energy-gradient derivative exists, [F7] gives $\nabla_{x_{i}} C_{M} = q_{i}^{2} \sum_{k} c_{k} Re (\overline{b_{k} (t_{i})} \nabla_{x} b_{k} (t_{i}))$ . By step 2.1 this is also the finite cyclic-kernel expression displayed in the Statement. Subtracting this gradient from that of $E_{M}$ yields the displayed corrected force. At a $P = 2$ knot the same algebra defines the selected corrected branch, without claiming it is a gradient. Put $Q_{i} = Q - ∣ q_{i} ∣ = \sum_{j \neq i} ∣ q_{j} ∣$ , $z_{k}^{'} = z_{k} - q_{i} b_{k} (t_{i})$ , and $ρ_{k}^{'} = ρ_{k} - q_{i} a_{k} (t_{i})$ . Since $Re (\overline{a_{k}} \nabla_{x} a_{k}) = Re (- i κ_{k}) = 0$ , the exact force is unchanged when $ρ_{k}$ is replaced by $ρ_{k}^{'}$ . The finite sums and [F3] give $∣ z_{k}^{'} - ρ_{k}^{'} ∣ \leq Q_{i} A_{k}$ and $∣ z_{k}^{'} ∣ \leq Q_{i} (1 + A_{k})$ . Repeating the force decomposition in step 1.4 with these primed coefficients and $H_{k} = 0$ proves the corrected force bound.

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Sources

Markus Deserno and Christian Holm, How to mesh up Ewald sums (I): A theoretical and numerical comparison of various particle mesh routines (1998) (standard reference, not scraped)
Hark Lee and Wei Cai, Ewald Summation for Coulomb Interactions in a Periodic Supercell (2009 lecture notes) (standard reference, not scraped)