Interacting particle mean field and vortex blob error — Examples

For $r\mathrm{\ne }0$, their scalar denominators are $\mathrm{\mid }r{\mathrm{\mid }}^3$ and $(\mathrm{\mid }r{\mathrm{\mid }}^2+{\epsilon }^2{)}^{3\mathrm{/}2}$, which differ because $\epsilon >0$. Thus $F_{\epsilon }(r)\mathrm{\ne }F(r)$ for every nonzero $r$.

At $r=0$, the unsoftened formula is undefined, whereas the softened formula gives $F_{\epsilon }(0)=0$. The denominator $\mathrm{\mid }r{\mathrm{\mid }}^2+{\epsilon }^2$ is smooth and strictly positive, and $u\mapsto u^{-3\mathrm{/}2}$ is smooth for $u>0$; hence the softened field is $C^{\mathrm{\infty }}$ on all of ${\mathbb{R}}^3$. Its smooth extension therefore supplies a convention for its own regularized model, not a collision law for the singular Newtonian system.

Let $S=\{Y_1,\dots ,Y_N\}$ be the empirical support. It contains at most $N$ vertices, so ${\mu }_M(V_M\setminus S)\ge 1-N\mathrm{/}{2}^M$. Distinct vertices of $V_M$ have Euclidean distance at least one.

In any coupling of ${\mu }_M$ to ${\nu }_N$, every point of $V_M\setminus S$ must send its mass to a point of $S$, at cost at least one per unit mass. Hence every coupling costs at least $1-N\mathrm{/}{2}^M$, and taking the infimum in [F1] proves the bound. If $2^M\ge 2N$, it is at least $1\mathrm{/}2$; choosing $N$ arbitrarily large and then such an $M$ rules out a uniform dimension-free $C{N}^{-1\mathrm{/}2}$ bound.

Among any countable partition, at most two cells have nonzero signed mass, each of magnitude one; the partition into the two singleton atoms and their complement attains the sum two. Thus $\mathrm{\mid }\omega \mathrm{\mid }({\mathbb{R}}^2)=2$ by [F3], while its signed mass is $1-1=0$. Therefore $\omega$ is neither a nonnegative probability measure nor normalizable by its net mass, so the domain condition in [F1] fails.

The function $\varphi (x)=\min (1,\mathrm{\mid }x\mathrm{\mid })$ is bounded by one and 1-Lipschitz by [F4], with $\varphi (0)=0$ and $\varphi (1,0)=1$. Hence $\mathrm{\mid }\int \varphi d\omega \mathrm{\mid }=1$, so $d_{BL}(\omega ,0)\ge 1$. This finite signed dual quantity remains meaningful by [F2], whereas probability $W_1(\omega ,0)$ is undefined.

For all $x,y,x^{\mathrm{\prime }},y^{\mathrm{\prime }}$, $\mathrm{\mid }K(x,y)-K(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\mathrm{\mid }=a\mathrm{\mid }(y-y^{\mathrm{\prime }})-(x-x^{\mathrm{\prime }})\mathrm{\mid }\le a(\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }+\mathrm{\mid }y-y^{\mathrm{\prime }}\mathrm{\mid })$, so the global Lipschitz constant may be taken as $L=a$ in [F1].

Substitution into [F2] gives ${\dot{x}}_i=a(\bar{x}-x_i)$, where $\bar{x}=N^{-1}{\sum }_j{x}_j$. Summing over $i$ yields $\dot{\bar{x}}=0$, and hence $x_i(t)=\bar{x}(0)+e^{-at}(x_i(0)-\bar{x}(0))\mathrm{.}$ For $a>0$, the formula tends to the constant mean as $t\to \mathrm{\infty }$; for $a=0$, it equals $x_i(0)$. Thus the empirical measure is an exact weak solution by [F3]; for probability solutions the stability factor is $e^{2at}$ by [F4].

Write the exact Euler characteristic velocity as the mollified interaction plus residual $(K-K_{\epsilon })\ast {\omega }_t$, whose norm is at most $C_{\rho }\epsilon W$ by [F2]. The quadrature particle field is the mollified interaction plus a common residual of norm at most ${\eta }_q$ by [F3]. Fix any initial coupling with cost $D_0\le E_0+\delta$, where $\delta >0$. Transport it by the Euler flow and the exact quadrature-force particle flow. By [F4], the resulting coupling has cost at time $T$ at most $e^{2L_{\epsilon }T}(E_0+\delta +T(C_{\rho }\epsilon W+{\eta }_q))\mathrm{.}$

Use the same initial particle label in the exact quadrature-force path and the RK4 path. The RK4 state error in [F5] is Euclidean on the full particle state, so each particle-block displacement is at most $C_{RK}{h}^4$. Since the initial nodes are distinct, partition the coupling by its second coordinate and label each part by its unique particle index. Push this labelled coupling through the exact Euler flow in the first coordinate and the RK4 particle map in the second. The resulting coupling from the Euler law to the RK4 empirical law has cost at most the step 1.1 bound plus $C_{RK}{h}^4$. Thus it bounds their $W_1$. Since $\delta >0$ was arbitrary, let $\delta \downarrow 0$ to obtain $W_1({\omega }_T,{\omega }_{N,h}^{\epsilon ,q}(T))\le e^{2L_{\epsilon }T}(E_0+T(C_{\rho }\epsilon W+{\eta }_q))+C_{RK}{h}^4\mathrm{.}$ This same-sign estimate uses probability transport and has no AC hypothesis; it makes no uniform claim as $\epsilon \to 0$.