Particle in cell and particle mesh kinetic methods — Examples

Since $6\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}5)$, [F2] gives $e^{i6x_j}=e^{i{x}_j}$ for each $j$. Taking real parts yields $\rho (x_j)=\cos (x_j)$, whose unique odd-grid interpolant is $\cos x$ by [F1]. On the five-mode space $\mathrm{\mid }k\mathrm{\mid }\le 2$, the collocation operator $-{\mathrm{\partial }}_x^2$ multiplies mode $k$ by $k^2$, so its only zero mode is the constant. Therefore ${\varphi }_N=\cos x$ is the unique zero-mean collocation solution to these sampled data.

Direct differentiation gives $-\frac{d^2}{d{x}^2}(\cos (6x)\mathrm{/}36)=\cos (6x)$ and $-\frac{d}{dx}(\cos (6x)\mathrm{/}36)=\sin (6x)\mathrm{/}6$. Since ${\varphi }_N=\cos x$, exact collocation differentiation [F3], applied twice to its interpolant, gives computed field $E_N=\sin x$. At each node, $\sin (6x_j)=\sin (x_j+2\pi j)=\sin x_j$, so $E_N(x_j)=6E(x_j)$.

The deposited source has total grid charge $h{\sum }_j{s}_j={\sum }_j{w}_j(X)=1$ by [F3]. Since $L=1$, its grid mean is $\bar{s}=L^{-1}h{\sum }_j{s}_j=1$.

By [F2], every solvable source for $A_h\varphi =s$ must have zero mean. Step 1.1 gives $\bar{s}=1$, so this unneutralized Poisson equation is incompatible. Subtracting the uniform background $1\mathrm{/}L=1$ would instead make the source mean zero.

Since $h^{-1}=5$, the two nonzero assigned densities are $5\cdot 2\cdot (2\mathrm{/}3)=20\mathrm{/}3$ and $5\cdot 2\cdot (1\mathrm{/}3)=10\mathrm{/}3$. The background is $Q\mathrm{/}L=2$, so the neutralized values are $14\mathrm{/}3,4\mathrm{/}3,-2,-2,-2$; their sum is $6-6=0$, hence their cell-weighted sum is zero.

For any point in a periodic cell $[x_j,x_{j+1}]$, with its right endpoint interpreted modulo $L$, the CIC formula gives $w_j=(x_{j+1}-x)\mathrm{/}h$ and $w_{j+1}=(x-x_j)\mathrm{/}h$; all other weights vanish. Both are nonnegative and sum to one. At a grid node this reduces to the single weight one. Hence the stated CIC family is a nonnegative normalized assignment shape.

The pre-neutralized cell-weighted sum is $h(20\mathrm{/}3+10\mathrm{/}3)=2$, verifying charge conservation directly; [F3] also applies because step 1.2 proves the shape partition. By [F4], the potential pairing equals $2(\frac{2}{3}{\varphi }_0+\frac{1}{3}{\varphi }_1)-2\bar{\varphi }$. Since $\bar{\varphi }=\frac{1}{5}{\sum }_{j=0}^4{\varphi }_j$, the formula includes exactly the contribution of the uniform background.

Near node $x_0=0$, the periodic distance formula gives $w_0(x)=1+x\mathrm{/}h$ for $-h<x<0$ and $w_0(x)=1-x\mathrm{/}h$ for $0<x<h$. The one-sided derivatives at the node are $1\mathrm{/}h$ and $-1\mathrm{/}h$, so the CIC weight is continuous but not differentiable at this grid crossing. Therefore CIC regularity alone does not supply the $C^4$ vector-field hypothesis of the RK4 error theorem; the page's time-error claim uses the separate smooth fixed-$\epsilon$ branch.

There are $2\cdot 63=126$ labels. Their positive charges sum to $126\mathrm{/}126=1=Q$ and their positive masses sum to $126\mathrm{/}126=1=M$. Within each fixed $\sigma$, every velocity equals $\sigma \mathrm{/}2$, so the two populations are cold counter-streaming beams.

The stated positions are points of ${\mathbb{T}}_{2\pi }$ after reduction modulo $2\pi$, and $N=63$ is odd as required by the grid convention. At each of the finitely many RK4 stages, the smooth-shape deposition conserves charge $Q=1$; subtracting the uniform background $Q\mathrm{/}L=1\mathrm{/}(2\pi )$ makes the source compatible by [F2]. By [F3], the exact finite Fourier Poisson solve therefore returns a potential at every stage, so this exact-arithmetic experiment is defined. No physical conclusion follows from this specification.

Since $0<a<1$, the density $(1+a\cos y)\mathrm{/}(2\pi )$ is nonnegative and integrates to one. Convolving its constant and first Fourier modes with [F1] leaves the constant coefficient $1\mathrm{/}(2\pi )$ and multiplies the cosine coefficient by ${\alpha }_{\epsilon }$. Subtracting the neutralizing background gives the displayed source. Since $-\frac{d^2}{d{x}^2}\cos x=\cos x$, the unique zero-mean potential is ${\varphi }_{\epsilon }=(a{\alpha }_{\epsilon }\mathrm{/}(2\pi ))\cos x$, and differentiation yields the displayed $E_{\epsilon }$. A second convolution multiplies its sine mode by ${\alpha }_{\epsilon }$, giving ${\eta }_{\epsilon }\ast E_{\epsilon }=a{\alpha }_{\epsilon }^2(2\pi {)}^{-1}\sin x$.

Apply [F2] to this smooth source on an odd $N$ grid. Since $d=1$, the power $h^{2-d\mathrm{/}2}$ equals $h^{3\mathrm{/}2}$, and $L^{d\mathrm{/}2}=(2\pi {)}^{1\mathrm{/}2}$, giving the displayed estimate. The upper bound ${\gamma }_{N,2\pi }\le \pi \mathrm{/}2$ is uniform in $N$, and all fixed-$\epsilon$ derivative constants are finite by [F1]; both powers $h^{3\mathrm{/}2}$ and $h^2$ tend to zero. No limit as $\epsilon \to 0$ is taken.

The half kick gives $p_{1\mathrm{/}2}=p_0-(\mathrm{\Delta }t\mathrm{/}2)q_0=-1\mathrm{/}4$. The drift gives $q_1=q_0+\mathrm{\Delta }t{p}_{1\mathrm{/}2}=1-(1\mathrm{/}2)(1\mathrm{/}4)=7\mathrm{/}8$. The final half kick gives $p_1=p_{1\mathrm{/}2}-(\mathrm{\Delta }t\mathrm{/}2)q_1=-1\mathrm{/}4-7\mathrm{/}32=-15\mathrm{/}32$.

For the oscillator, $\dot{q}=p$ and $\dot{p}=-q$, so $\frac{d}{dt}H=qp-pq=0$. The kick and drift matrices for the test are $K_{1\mathrm{/}4}=\left(\begin{array}{cc}1& 0\\ -1\mathrm{/}4& 1\end{array}\right),D_{1\mathrm{/}2}=\left(\begin{array}{cc}1& 1\mathrm{/}2\\ 0& 1\end{array}\right)\mathrm{.}$ Each has determinant one. Direct multiplication gives $A^{T}JA=(\det A)J=J$ for every real $2\times 2$ matrix $A$, where $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so their composition is symplectic. The shears have inverses $K_{-1\mathrm{/}4}$ and $D_{-1\mathrm{/}2}$; reversing the palindromic composition with these inverses proves the step reversible. The exact Hamiltonian at the updated point is $H(q_1,p_1)=\frac{1}{2}(\frac{49}{64}+\frac{225}{1024})=\frac{1}{2}\frac{1009}{1024}=\frac{1009}{2048},$ while $H(q_0,p_0)=1\mathrm{/}2=1024\mathrm{/}2048$. Their difference is $-15\mathrm{/}2048$. Therefore this reversible symplectic step does not preserve its Hamiltonian exactly. The separate PIC results [F2] and [F3] concern their own variational force and tube hypotheses and are not transferred to this scalar test.