Ewald periodic potentials and particle mesh error — Examples

Test a hypothetical periodic distribution solution against the constant smooth periodic function $\phi \equiv 1$. Since $-\mathrm{\Delta }1=0$, the left side is $⟨u,-\mathrm{\Delta }1⟩=0$ by [F1].

The right side would be $4\pi ⟨\rho ,1⟩=4\pi q_1=4\pi$ by [F1], contradicting step 1.1. Therefore no periodic distribution solution exists for this nonneutral source.

The two positions are distinct in ${\mathbb{R}}^3\mathrm{/}{\mathbb{Z}}^3$ and $q_1+q_2=0$. For $k\in {\mathbb{Z}}^3$, [F1] gives ${\rho }_k=1-e^{-2\pi i{k}_1\mathrm{/}2}=1-(-1{)}^{k_1}$; hence ${\rho }_0=0$, ${\rho }_k=0$ when $k_1$ is even, and ${\rho }_k=2$ when $k_1$ is odd.

For any finite symmetric reciprocal mode set $\mathcal{K}\subset {\mathbb{Z}}^3\setminus \{0\}$, the dipole contributes $\frac{1}{2}{\sum }_{k\in \mathcal{K}}{c}_k\mathrm{\mid }{\rho }_k{\mathrm{\mid }}^2=2{\sum }_{k\in \mathcal{K},k_1\text{ odd}}{c}_k$ to the retained reciprocal energy. The remaining real-space and self contributions are exactly the terms in [F2], under its neutrality, positive-parameter and Countable Choice hypotheses [A1]. Thus the zero mode is absent and the reciprocal selection is precisely the odd-$k_1$ subset, not an assumed continuum average.

The configuration is neutral, and its two torus positions are distinct. For $\lambda \in {\mathbb{Z}}^3\setminus \{0\}$, $\mathrm{\mid }\lambda \mathrm{\mid }\ge 1$, while every image of $x_2$ relative to $x_1$ has first coordinate in $\mathbb{Z}+1\mathrm{/}2$ and hence distance at least $1\mathrm{/}2$. Thus for $0<r<1\mathrm{/}4$, every real-space image except $(i,\lambda )=(1,0)$ stays at distance at least $1\mathrm{/}4$ from $x$. By [F3], the finitely many nearby nonself terms and their first derivatives are uniformly bounded there, and the Gaussian shell bound makes the remaining potential and derivative tails uniformly convergent. The reciprocal series and its gradient are bounded by [F4]. Hence the nonself periodic remainder and its gradient are bounded near $x_1$.

Writing $r=\mathrm{\mid }x-x_1\mathrm{\mid }$, [F2] gives the self-potential expansion ${\varphi }_s(r)=1\mathrm{/}r-2\alpha \mathrm{/}\sqrt{\pi }+O(r^2)$. Substituting the stated error-function expansion into the derivative formula in [F2] cancels the $r^{-1}$ terms and gives ${\varphi }_s^{\mathrm{\prime }}(r)=-1\mathrm{/}{r}^2+O(r)$. Since $q_1=1$, its gradient is ${\varphi }_s^{\mathrm{\prime }}(r)(x-x_1)\mathrm{/}r=-(x-x_1)\mathrm{/}{r}^3+O(r)$. Adding the bounded nonself potential and gradient from step 1.1 proves both expansions in the Example.

The magnitude of the leading gradient term is $r^{-2}$, while the remainder is bounded; the potential likewise diverges as $r^{-1}$. Neither has a finite value at the source, so uniform pointwise potential or field error claims for these point sources must exclude collisions.

For $\alpha >0$ set $\mathcal{R}(\alpha )=\frac{Q{C}_R}{\sqrt{\pi }\alpha }{\sum }_{m=M_R}^{\mathrm{\infty }}(m+1{)}^3{e}^{-{\alpha }^2{m}^2}$ and $\mathcal{K}(\alpha )=\frac{4\pi Q{C}_K}{V}{\sum }_{m=M_K}^{\mathrm{\infty }}(m+1{)}^3{e}^{-m^2\mathrm{/}(4{\alpha }^2)}$, where $M_R=\lfloor R\rfloor$ and $M_K=\lfloor K_c\rfloor$. These are exactly the two bounds in [F1], and they are finite by [F2].

If $0<{\alpha }_1<{\alpha }_2$, then for every $m\ge M_R\ge 1$, $e^{-{\alpha }_2^2{m}^2}\mathrm{/}{\alpha }_2<e^{-{\alpha }_1^2{m}^2}\mathrm{/}{\alpha }_1$ because both positive factors strictly decrease. The positive convergent sum therefore gives $\mathcal{R}({\alpha }_2)<\mathcal{R}({\alpha }_1)$.

For every $m\ge M_K\ge 1$, $e^{-m^2\mathrm{/}(4{\alpha }_2^2)}>e^{-m^2\mathrm{/}(4{\alpha }_1^2)}$, since ${\alpha }_2>{\alpha }_1$ makes the negative exponent less negative. Summing positive terms gives $\mathcal{K}({\alpha }_2)>\mathcal{K}({\alpha }_1)$. Hence a fixed-tolerance certificate must balance the real and reciprocal tails; the theorem does not infer an optimal parameter from FFT work or RMS data.

The two positions are distinct and neutral. Each coordinate multiplied by $M_r=4$ is either $1\mathrm{/}2$ or $5\mathrm{/}2$, hence no coordinate is an integer and both particles are off assignment knots. The mode set is symmetric, injective modulo $(4,4,4)$, nonzero, and satisfies $\mathrm{\mid }{k}_r\mathrm{\mid }<2$. Since $\mathrm{sinc}(\pi \mathrm{/}4)\mathrm{\ne }0$, the required window denominators are nonzero.

By [F1], $z_k=q_1{b}_k(t_1)+q_2{b}_k(t_2)$. With $P=3$, the factors $b_k$ are differentiable. The chain rule gives ${\mathrm{\nabla }}_{x_i}{z}_k=q_i{\mathrm{\nabla }}_x{b}_k(t_i)$. Therefore ${\mathrm{\nabla }}_{x_i}(\frac{1}{2}{c}_k\mathrm{\mid }{z}_k{\mathrm{\mid }}^2)=q_i{c}_k\mathrm{Re}(\overline{z_k}{\mathrm{\nabla }}_x{b}_k(t_i))$. Summing over $k\in \mathcal{K}$ and negating yields exactly the stated energy-gradient force.

Differentiating $C_M=\frac{1}{2}{\sum }_j{q}_j^2{\sum }_k{c}_k(\mathrm{\mid }{b}_k(t_j){\mathrm{\mid }}^2-1)$ with respect to $x_i$ gives ${\mathrm{\nabla }}_{x_i}{C}_M=q_i^2{\sum }_k{c}_k\mathrm{Re}(\overline{b_k(t_i)}{\mathrm{\nabla }}_x{b}_k(t_i))$. Hence $-{\mathrm{\nabla }}_{x_i}(E_M-C_M)=-q_i{\sum }_k{c}_k\mathrm{Re}(\overline{z_k-q_i{b}_k(t_i)}{\mathrm{\nabla }}_x{b}_k(t_i))$, the corrected force formula.

Since $P=3$, this instance has classical derivatives at all positions. Applying [F3] to the same admissible data gives the deterministic retained-mode error certificate, including the corrected bound when the correction is selected. A spectral differentiation rule is a different force choice and does not inherit this energy-gradient identity; at $P=2$ knots the assignment definition supplies a branch value but no classical-gradient claim.