$\big{|}E_{\mu_{k,l,j}}\left[V_{k,k^{\prime},l,A}\right]\big{|}\leq\|G\|_{\infty}\Big{|}E_{\nu_{0,l,j}}[g(\eta(0))]-E_{\nu_{j/(2l+1)}}[g(\eta(0))]\Big{|}.$

By equivalence of ensembles (cf. p.355, [18]), the right hand side of the
above disaplay vanishes as $l\to\infty$, uniformly for
$\rho=j/(2l+1)\in[0,A]$. The lemma now is proved by letting $\gamma\to\infty$.
∎

###  8.2. Local $2$-block estimate

We now detail the local $2$-block estimate following the outline of the local
$1$-block estimate. Recall the notation $\Lambda_{k,l}$ from the $1$-block
estimate and let
$\Lambda_{k,k^{\prime},l}=\Lambda_{k,l}\cup\Lambda_{k^{\prime},l}$ for
$|k-k^{\prime}|>l$. Define the generator $L_{k,k^{\prime},l}$ on
$\Omega_{k,k^{\prime},l}=\mathbb{N}_{0}^{\Lambda_{k,k^{\prime},l}}$:

$\begin{split}L_{k,k^{\prime},l}f(\eta)&=L_{k,l}f(\eta)+L_{k^{\prime},l}f(\eta)\\\
&+g(\eta(k+l))\big{[}f\big{(}\eta^{k+l,k^{\prime}-l}\big{)}-f(\eta)\big{]}+g(\eta(k^{\prime}-l))\big{[}f\big{(}\eta^{k^{\prime}-l,k+l}\big{)}-f(\eta)\big{]}.\end{split}$

When $|k-k^{\prime}|$ is large, the process governed by $L_{k,k^{\prime},l}$
in effect treats the blocks $\Lambda_{k,l}$ and $\Lambda_{k^{\prime},l}$ as
adjacent, with a connecting bond.

Let
$\Omega_{k,k^{\prime},l,j}:=\\{\eta\in\Omega_{k,k^{\prime},l}:\sum_{x\in\Lambda_{k,k^{\prime},l}}\eta(x)=j\\}$.
As before, the localized measure $\nu^{\rho}_{k,k^{\prime},l}$ defined by
$\nu_{\rho}$ limited to sites in $\Lambda_{k,k^{\prime},l}$, as well as
$\nu_{k,k^{\prime},l,j}$, the canonical measure of $\nu^{\rho}_{k,k;,l}$ on
$\Omega_{k,k^{\prime},l,j}$, are both invariant and reversible with respect to
$L_{k,k^{\prime},l}$.

The corresponding Dirichlet form, with measure $\kappa$ given by
$\mu_{k,k^{\prime},l}$ or $\mu_{k,k^{\prime},l,j}$, is given by

$\begin{split}E_{\kappa}\left[f(-L_{k,k^{\prime},l}f)\right]=&\sum_{x,x+1\in\Lambda_{k,k^{\prime},l}}E_{\kappa}\left[g(\eta(x))\left[f\left(\eta^{x,x+1}\right)-f(\eta)\right]^{2}\right]\\\
&+E_{\kappa}\Big{[}g(\eta(k+l))\left[f\left(\eta^{k+l,k^{\prime}-l}\right)-f(\eta)\right]^{2}\Big{]}.\end{split}$
(8.6)

For $l,j\geq 1$, let $b_{l,l,j}$ be the spectral gap of $-L_{k,k^{\prime},l}$
on $\Omega_{k,k^{\prime},l,j}$ ( cf. (8.2)). As $b_{l,l,j}$ is strictly
postive, we have the following Poincaré inequality (cf. (8.3)): for all $f\in
L^{2}(\nu_{k,k^{\prime},l,j})$

${\rm Var}_{\nu_{k,k^{\prime},l,j}}(f)\leq
C_{l,l,j}E_{\nu_{k,k^{\prime},l,j}}\left[f(-L_{k,k^{\prime},l}f)\right]$ (8.7)

where $C_{l,l,j}:=b_{l,l,j}^{-1}$ for $j\geq 1$ and $C_{l,l,0}=0$.

We now state and show a local $2$-blocks estimate. The scheme is similar to
that of the local $1$-block estimate.

######  Lemma 8.2 (Local $2$-block estimate).

We have

$\limsup_{l\to\infty}\limsup_{\theta\to
0}\limsup_{N\to\infty}\sup_{k,k^{\prime\prime}}\mathbb{E}_{N}\Big{[}\int_{0}^{T}\left|\Phi(\eta_{s}^{l}(k))-\Phi(\eta_{s}^{\theta
N}(k^{\prime\prime}))\right|ds\Big{]}=0$ (8.8)

where the $\sup$ is taken over all $k$ and $k^{\prime\prime}$ such that
$\Lambda_{k,l}\subset\Lambda_{k^{\prime\prime},\theta N}$ and
$\Lambda_{k^{\prime\prime},\theta N}\cap{\mathfrak{D}}_{N}=\emptyset$.

###### Proof.

We separate the argument into steps.

Step 1. Since $\Phi(\cdot)$ is Lipschitz, to prove the lemma, it suffices to
show (8.8) with $\Phi(\eta_{s}^{l}(k))-\Phi(\eta_{s}^{\theta
N}(k^{\prime\prime}))$ replaced by $\eta_{s}^{l}(k)-\eta_{s}^{\theta
N}(k^{\prime\prime})$. We may further replace $\eta_{s}^{\theta
N}(k^{\prime\prime})$ by $(2\theta
N+1)^{-1}\sum_{k^{\prime}\in\Lambda_{k^{\prime\prime},\theta
N-l}}\eta^{l}(k^{\prime})$. Indeed, the replacement error is

$\mathbb{E}_{N}\Big{[}\Big{|}\eta_{s}^{\theta
N}(k^{\prime\prime})-\dfrac{1}{2\theta
N+1}\sum_{k^{\prime}\in\Lambda_{k^{\prime\prime},\theta
N-l}}\eta_{s}^{l}(k^{\prime})\Big{|}\Big{]}\leq\dfrac{2l+1}{2\theta
N+1}\mathbb{E}_{N}\big{[}\eta_{s}^{l}(k^{\prime\prime}-\theta
N+l)+\eta_{s}^{l}(k^{\prime\prime}+\theta N-l)\big{]}.$

By attractiveness (as explained in Step 1 of the proof of the local 1-block
Lemma 8.1), for either case $g(n)\sim n^{\alpha}$ or $g$ bounded, the
expectation term $\mathbb{E}_{N}$ is bounded uniformly in $t$, $l$, and $N$,
the right-hand side of the above display vanishes as $N\uparrow\infty$ first.

Therefore, the lemma will follow if we show

$\limsup_{l\to\infty}\limsup_{\theta\to
0}\limsup_{N\to\infty}\sup_{k,k^{\prime}}\mathbb{E}_{N}\Big{[}\int_{0}^{T}\Big{|}\eta_{s}^{l}(k)-\eta_{s}^{l}(k^{\prime})\Big{|}ds\Big{]}=0$
(8.9)

where the $\sup$ is taken over all $k,k^{\prime}$ such that
$\Lambda_{k,k^{\prime},l}\cap{\mathfrak{D}}_{N}=\emptyset$ and $2l+1\leq
k^{\prime}-k\leq\theta N$.

Step 2. By a similar coupling argument as in the Step 1 of the proof of local
$1$-block Lemma 8.1, we may apply a cutoff of large densities. Therefore, to
prove the lemma, it suffices to show

$\limsup_{l\to\infty}\limsup_{\theta\to
0}\limsup_{N\to\infty}\sup_{k,k^{\prime}}\mathbb{E}_{N}\Big{[}\int_{0}^{T}\Big{|}\eta_{s}^{l}(k)-\eta_{s}^{l}(k^{\prime})\Big{|}\mathbbm{1}_{\\{\eta_{s}^{l}(k,k^{\prime})\leq
A\\}}ds\Big{]}=0$

where $\eta_{s}^{l}(k,k^{\prime})=\eta_{s}^{l}(k)+\eta_{s}^{l}(k^{\prime})$
and the $\sup$ is over $k,k^{\prime}$ as in (8.9).

Let
$U_{k,k^{\prime},l,A}(\eta):=\left|\eta^{l}(k)-\eta^{l}(k^{\prime})\right|\mathbbm{1}_{\\{\eta^{l}(k,k^{\prime})\leq
A\\}}$. Following the proof of Lemma 8.1, for fixed $l,\theta,N,k,k^{\prime}$,
in order to estimate
$\mathbb{E}_{N}\big{[}\int_{0}^{T}U_{k,k^{\prime},l,A}(\eta_{s})ds\big{]}$, it
suffices to bound

$(\gamma
N)^{-1}\lambda_{N,l}=\sup_{f}\left\\{E_{\mathscr{R}_{c_{0}}^{N}}\left[U_{k,k^{\prime},l,A}f\right]-\gamma^{-1}NE_{\mathscr{R}_{c_{0}}^{N}}\left[\sqrt{f}(-L_{N}\sqrt{f})\right]\right\\}.$
(8.10)

where the supremum is over all $f$ which are densities with respect to
$\mathscr{R}_{c_{0}}^{N}$.

Step 3. Recall the generator $L_{k,k^{\prime},l}$ and its Dirichlet form
defined in the beginning of this subsection. We now argue the following
Dirichlet form inequality

$E_{\mathscr{R}_{c_{0}}^{N}}\big{[}\sqrt{f}(-L_{k,k^{\prime},l}\sqrt{f})\big{]}\leq\theta
NE_{\mathscr{R}_{c_{0}}^{N}}\big{[}\sqrt{f}(-L_{N}\sqrt{f})\big{]}.$ (8.11)

The Dirichlet form with respect to the full generator $L_{N}$ under
$\mathscr{R}_{c_{0}}^{N}$ is given by

$E_{\mathscr{R}_{c_{0}}^{N}}\left[f(-L_{N}f)\right]=\sum_{k\in\mathbb{T}_{N}}E_{\mathscr{R}_{c_{0}}^{N}}\Big{[}g_{k,N}(\eta(k))\big{(}f(\eta^{k,k+1})-f(\eta)\big{)}^{2}\Big{]}.$
(8.12)

First, writing out the Dirichlet form in (8.6), in terms of the product
measure $\mathscr{R}_{c_{0}}^{N}$, we have

$\begin{split}E_{\mathscr{R}_{c_{0}}^{N}}\left[f(-L_{k,k^{\prime},l}f)\right]=&\sum_{x,x+1\in\Lambda_{k,k^{\prime},l}}E_{\mathscr{R}_{c_{0}}^{N}}\Big{[}g(\eta(x))\big{(}f(\eta^{x,x+1})-f(\eta)\big{)}^{2}\Big{]}\\\
&+E_{\mathscr{R}_{c_{0}}^{N}}\Big{[}g(\eta(k+l))\big{(}f(\eta^{k+l,k^{\prime}-l})-f(\eta)\big{)}^{2}\Big{]}.\end{split}$

Next, by adding and subtracting at most $\theta N$ terms, we have

$\begin{split}\left[f\left(\eta^{k+l,k^{\prime}-l}\right)-f(\eta)\right]^{2}\leq(k^{\prime}-k-2l)\sum_{q=0}^{k^{\prime}-k-2l-1}\left[f\left(\eta^{k+l,k+l+q+1}\right)-f\left(\eta^{k+l,k+l+q}\right)\right]^{2}.\end{split}$

By the change of variables $\xi=\eta^{k+l,k+l+q}$, which takes away a particle
at $k+l$ and adds one at $k+l+q$, we have
$\mathscr{R}_{c_{0}}^{N}(\eta)=\dfrac{g(\eta(k+l+q)+1)}{g(\eta(k+l))}\mathscr{R}_{c_{0}}^{N}(\xi)$.
Then

$\displaystyle
E_{\mathscr{R}_{c_{0}}^{N}}\left[g(\eta(k+l))\left[f\left(\eta^{k+l,k+l+q+1}\right)-f\left(\eta^{k+l,k+l+q}\right)\right]^{2}\right]$
$\displaystyle=$
$\displaystyle\sum_{\xi}\mathscr{R}_{c_{0}}^{N}(\eta)g(\eta(k+l))\left[f\left(\xi^{k+l+q,k+l+q+1}\right)-f\left(\xi\right)\right]^{2}$
$\displaystyle=$ $\displaystyle
E_{\mathscr{R}_{c_{0}}^{N}}\Big{[}g(\eta(k+l+q))\Big{[}f\left(\eta^{k+l+q,k+l+q+1}\right)-f\left(\eta\right)\Big{]}^{2}\Big{]}.$

From these observations, (8.11) follows.

Step 4. Let $\mu_{k,k^{\prime},l}$ be the restriction of
$\mu=\mathscr{R}_{c_{0}}^{N}$ to $\Lambda_{k,k^{\prime},l}$. Clearly,
$\mu_{k,k^{\prime},l}=\nu_{k,k^{\prime},l}$. Inputting (8.11) into (8.10), and
considering the conditional expectation of $f$ with respect to
$\Omega_{k,k^{\prime},l}$ as in the $1$-block estimate proof, we have

$(\gamma
N)^{-1}\lambda_{N,l}\leq\sup_{f_{k.k^{\prime},l}}\Big{\\{}E_{\mu_{k,k^{\prime},l}}\left[U_{k,k^{\prime},l,A}f_{k,k^{\prime},l}\right]-\dfrac{1}{\theta\gamma}E_{\mu_{k,k^{\prime},l}}\Big{[}\sqrt{f_{k,k^{\prime},l}}(-L_{k,k^{\prime},l}\sqrt{f_{k,k^{\prime},l}})\Big{]}\Big{\\}},$

where the supremum is over densities $f_{k,k^{\prime},l}$ with respect to
$\mu_{k,k^{\prime},l}$.

Again, as in the proof of the $1$-block estimate, decomposing
$f_{k,k^{\prime},l}d\mu_{k,k^{\prime}.l}$ along configurations with common
total number $j$, we need only to bound

$\sup_{0\leq j\leq
A(2l+1)}\sup_{f}\left\\{E_{\nu_{k,k^{\prime},l,j}}\left[U_{k,k^{\prime},l,A}f\right]-\dfrac{1}{\theta\gamma}E_{\nu_{k,k^{\prime},l,j}}\left[\sqrt{f}(-L_{k,k^{\prime},l}\sqrt{f})\right]\right\\},$

where the supremum is over densities $f$ with respect to
$\nu_{k,k^{\prime},l,j}$.

Step 5. Consider the centered object

$\widehat{U}_{k,k^{\prime},l,A}=U_{k,k^{\prime},l,A}-E_{\nu_{k,k^{\prime},l,j}}\left[U_{k,k^{\prime},l,A}\right].$

Recall $C_{l,l,j}$, the inverse spectral gap of $L_{k,k^{\prime},l}$ from
(8.7) and note that $\|\widehat{U}_{k,k^{\prime},l,A}\|_{\infty}\leq A$. Using
the Rayleigh expansion (cf. p.375, [18]), we have

$\begin{split}&E_{\nu_{k,k^{\prime},l,j}}\big{[}\widehat{U}_{k,k^{\prime},l,A}f\big{]}-(\theta\gamma)^{-1}E_{\nu_{k,k^{\prime},l,j}}\big{[}\sqrt{f}(-L_{k,k^{\prime},l}\sqrt{f})\big{]}\\\
\leq&\dfrac{\theta\gamma}{1-2AC_{l,l,j}\,\theta\gamma}E_{\nu_{k,k;,l,j}}\big{[}\widehat{U}_{k,k^{\prime},l,A}(-L_{k,k^{\prime},l})^{-1}\widehat{U}_{k,k^{\prime},l,A}\big{]}\\\
\leq&\dfrac{\theta\gamma
C_{l,l,j}}{1-2AC_{l,l,j}\,\theta\gamma}E_{\nu_{k,k^{\prime},l,j}}\big{[}\widehat{U}_{k,k^{\prime},l,A}^{2}\big{]}\to
0\text{ as }\theta\to 0.\end{split}$

Step 6. Recall the definition of $U_{k,k^{\prime},l,A}$ in Step 2. To finish,
we still need to estimate the centering term
$E_{\nu_{k,k^{\prime},l,j}}\left[U_{k,k^{\prime},l,A}\right]$. By adding and
subtracting $j/(4l+2)$, we need only bound
$E_{\nu_{k,k^{\prime},l,j}}\left[\big{|}\eta^{l}(k)-j/(4l+2)\big{|}\right]$.
By exchangeability and an equivalence of ensemble estimate (cf. p. 355 [18]),
the canonical variance

$\displaystyle
E_{\nu_{k,k^{\prime},l,j}}\left[\big{|}\eta^{l}(k)-j/(4l+2)\big{|}^{2}\right]=O(l^{-1})E_{\nu_{k,k^{\prime},l,j}}\left[(\eta(k)-j/(4l+2))^{2}\right]$
$\displaystyle\ \ \ \ \ \ \ \ \ \
+O(1)E_{\nu_{k,k^{\prime},l,j}}\left[(\eta(k)-j/(4l+2))(\eta(k+1)-j/(4l+2))\right]$

and is further bounded by $C(A){\rm
Var}_{\nu^{j/(4l+2)}_{k,k^{\prime},l}}\left(\eta^{l}(k)\right)$ for some
constant $C(A)$ depending only on $A$. This variance is of order $O(l^{-1})$,
since the single site variance ${\rm
Var}_{\nu^{j/(4l+2)}_{k,k^{\prime},l}}\left(\eta(k)\right)$ is uniformly
bounded for $j/(4l+2)\leq A$. Hence, $\sup_{0\leq j\leq
A(4l+2)}E_{\nu_{k,k^{\prime},l,j}}\left[V_{k,k^{\prime},l,A}\right]$ is of
order $O(l^{-1/2})$, vanishing as $l\uparrow\infty$. This finishes the proof.
∎

######  Remark 8.3.

We comment that Lemmas 8.1 and 8.2 with the choice
$k=k^{\prime}=k^{\prime\prime}$ gives the replacement of $g(\eta_{s}(k))$ by
$\Phi\left(\eta_{s}^{\theta N}(k)\right)$ which will be used in the ‘bulk’
replacement below. Moreover, choosing $k^{\prime}=x+1$, $k=x+l+1$, and
$k^{\prime\prime}=x+\theta N+1$, we may replace $g(x+1)$ by
$\Phi\left(\eta_{s}^{\theta N,+}(x)\right)$. Here
$\eta^{l,+}(x)=\frac{1}{2l+1}\sum_{y=x+1}^{x+2l+1}(x)$ is the average number
of particles over the $2l+1$ neighboring sites to the right of $x$. Such an
average will be used to treat ‘replacement’ near the boundary of a defect site
in Section 9.

###  8.3. Bulk Replacement Lemma

Let $G(t,x)$ be a bounded function on $[0,T]\times\mathbb{T}$ with compact
support on $[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}})$. As in Remark 8.3,
Lemma 8.1 implies that

$\limsup_{l\to\infty}\limsup_{N\to\infty}\mathbb{E}_{N}\Big{[}\dfrac{1}{N}\sum_{k\in\mathbb{T}_{N}}\Big{|}\int_{0}^{T}G(s,k/N)\left(g(\eta_{s}(k))-\Phi(\eta_{s}^{l}(k))\right)ds\Big{|}\Big{]}=0$

and by Lemma 8.2,

$\limsup_{l\to\infty}\limsup_{\theta\to
0}\limsup_{N\to\infty}\mathbb{E}_{N}\Big{[}\dfrac{1}{N}\sum_{k\in\mathbb{T}_{N}}\int_{0}^{T}G(s,k/N)\Big{|}\Phi\left(\eta_{s}^{l}(k)\right)-\Phi\left(\eta_{s}^{\theta
N}(k)\right)\Big{|}ds\Big{]}=0.$

By Markov’s inequality and triangle inequality, we obtain

######  Lemma 8.4 (Bulk Replacement Lemma).

For each bounded function $G(t,x)$ on $[0,T]\times\mathbb{T}$ with compact
support on $[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}})$, and $\delta>0$,
we have

$\begin{split}\limsup_{\theta\to
0}\limsup_{N\to\infty}\mathbb{P}_{N}\Big{[}\Big{|}\dfrac{1}{N}\sum_{k\in\mathbb{T}_{N}}\int_{0}^{T}G\big{(}s,\frac{k}{N}\big{)}\Big{(}g\left(\eta_{s}(k)\right)-\Phi\left(\eta_{s}^{\theta
N}(k)\right)\Big{)}ds\Big{|}>\delta\Big{]}=0.\end{split}$

######  Remark 8.5.

We comment, in the $g(n)\sim n^{\alpha}$ setting (where ’FEM’ as stated in
[18] holds), the “attractiveness” assumption used in the local $1$ and
$2$-block estimates to introduce cutoffs of large densities in a local region,
may be dropped in the statement of the ‘global average’ Lemma 8.4.

##  9\. Replacement at the boundary

In this section, we show a local replacement near the defect sites, used in
Subsection 7.2, in two steps. In Lemma 9.1, we show that the jump rate
$g_{k,N}(\eta_{s}(k))$ at any site $k\in\mathbb{T}_{N}$ is close to
$g_{k+1,N}(\eta_{s}(k+1))$, that of its neighbor site $k+1$.

In later use, when $k$ is a defect site in ${\mathfrak{D}}_{N}$, this neighbor
site will be a non-defect or regular site for $N$ large. Then, we may apply
local $1$ and $2$-blocks estimates from last section to obtain our local
replacement Lemma 9.2 near the defect sites.

In both lemmas, the replacements will hold in the $g(n)\sim n^{\alpha}$ and
$g$ bounded settings.

######  Lemma 9.1.

Let $G(\cdot):[0,T]\mapsto\mathbb{R}$ be bounded. Then, we have

$\limsup_{N\to\infty}\sup_{k\in\mathbb{T}_{N}}\mathbb{E}_{N}\Big{[}\Big{|}\int_{0}^{T}G(s)\big{(}g_{k,N}(\eta_{s}(k))-g_{k+1,N}(\eta_{s}(k+1))\big{)}ds\Big{|}\Big{]}=0.$

###### Proof.

Let

$U_{s,k}(\eta)=\,2N^{-1}(G(s))^{2}\big{(}g_{k,N}(\eta(k))+g_{k+1,N}(\eta(k+1))\big{)}.$

By Lemma 5.2,
$\lim_{N\to\infty}\sup_{k\in\mathbb{T}_{N}}\mathbb{E}_{N}\big{[}\int_{0}^{T}U_{s,k}(\eta_{s})ds\big{]}=0$.
Let

$V_{s,k}(\eta)=\,G(s)\big{(}g_{k,N}(\eta(k))-g_{k+1,N}(\eta(k+1))\big{)}.$

Then, to prove the lemma, it suffices to show that

$\limsup_{\kappa\to\infty}\limsup_{N\to\infty}\sup_{k\in\mathbb{T}_{N}}\mathbb{E}_{N}\Big{[}\Big{|}\int_{0}^{T}V_{s,k}(\eta_{s})ds\Big{|}-\kappa\int_{0}^{T}U_{s,k}(\eta_{s})ds\Big{]}=0.$
(9.1)

Recall the initial entropy bound $H(\mu^{N}|\mathscr{R}_{c_{0}}^{N})\leq CN$.
Then, for $c=c_{0}$ and $\kappa>0$, by the entropy inequality, the expectaton
in the previous display is bounded from above by

$\dfrac{C}{\kappa}+\dfrac{1}{\kappa
N}\ln\mathbb{E}_{\mathscr{R}_{c}^{N}}\Big{[}\exp\Big{\\{}\kappa
N\Big{|}\int_{0}^{T}V_{s,k}(\eta_{s})ds\Big{|}-\kappa^{2}N\int_{0}^{T}U_{s,k}(\eta_{s})ds\Big{\\}}\Big{]}.$

The absolute value in the right hand side of last inequality can be dropped by
using $e^{|x|}\leq e^{x}+e^{-x}$. By Feynman-Kac formula (cf. p.336, [18]), we
have

$\begin{split}\dfrac{1}{\kappa
N}\ln\mathbb{E}_{\mathscr{R}_{c}^{N}}\Big{[}\exp\Big{\\{}\kappa
N\int_{0}^{T}(V_{s,k}-\kappa U_{s,k})ds\Big{\\}}\Big{]}\leq\dfrac{1}{\kappa
N}\int_{0}^{T}\lambda_{N,k}(s)ds\end{split}$

where $\lambda_{N,k}(s)$ is the largest eigenvalue of $N^{2}L_{N}+\kappa
N(V_{s,k}(\eta)-\kappa U_{s,k}(\eta))$. Fix $s\in[0,T]$ and note the
variational formula for $\lambda_{N,k}$:

$(\kappa
N)^{-1}\lambda_{N,k}=\sup_{f}\left\\{E_{\mathscr{R}_{c}^{N}}\left[(V_{s,k}-\kappa
U_{s,k})f\right]-\kappa^{-1}NE_{\mathscr{R}_{c}^{N}}\big{[}\sqrt{f}(-L_{N}\sqrt{f})\big{]}\right\\}$

where the supremum is over all $f$ which are densities with respect to
$\mathscr{R}_{c}^{N}$ (cf. [18], p. 377). Thus, to prove (9.1), it remains to
show, for any density $f$,

$E_{\mathscr{R}_{c}^{N}}[V_{s,k}f]\leq E_{\mathscr{R}_{c}^{N}}[\kappa
U_{s,k}f]+\kappa^{-1}NE_{\mathscr{R}_{c}^{N}}\big{[}\sqrt{f}(-L_{N}\sqrt{f})\big{]}.$
(9.2)

By the product structure of $\mathscr{R}_{c}^{N}$, we have

$g_{k,N}(\eta(k))\mathscr{R}_{c}^{N}(\eta)=g_{k+1,N}(\eta(k+1)+1)\mathscr{R}_{c}^{N}(\eta^{k,k+1}).$

Thus, we compute that

$\begin{split}E_{\mathscr{R}_{c}^{N}}\left[V_{s,k}f\right]=&E_{\mathscr{R}_{c}^{N}}\Big{[}G(s)\Big{(}g_{k,N}(\eta(k))-g_{k+1,N}(\eta(k+1))\Big{)}f(\eta)\Big{]}\\\
=&E_{\mathscr{R}_{c}^{N}}\Big{[}G(s)g_{k,N}(\eta(k))\Big{(}f(\eta)-f(\eta^{k,k+1})\Big{)}\Big{]}\\\
=&E_{\mathscr{R}_{c}^{N}}\Big{[}G(s)g_{k,N}(\eta(k))\Big{(}\sqrt{f(\eta)}-\sqrt{f(\eta^{k,k+1})}\Big{)}\Big{(}\sqrt{f(\eta)}+\sqrt{f(\eta^{k,k+1})}\Big{)}\Big{]}.\end{split}$

By Cauchy-Schwarz, for any $A>0$, the above display is estimated from above by

$\begin{split}&AE_{\mathscr{R}_{c}^{N}}\Big{[}g_{k,N}(\eta(k))\Big{(}\sqrt{f(\eta)}-\sqrt{f(\eta^{k,k+1})}\Big{)}^{2}\Big{]}\\\
&\qquad\qquad\qquad+A^{-1}E_{\mathscr{R}_{c}^{N}}\Big{[}G(s)^{2}g_{k,N}(\eta(k))\Big{(}\sqrt{f(\eta)}+\sqrt{f(\eta^{k,k+1})}\Big{)}^{2}.\end{split}$

Notice that the first expectation in the above display is bounded by
$E_{\mathscr{R}_{c}^{N}}\big{[}\sqrt{f}(-L_{N}\sqrt{f})\big{]}$ (cf. (8.12)).
Take $A=\kappa^{-1}N$. The second summand is estimated from above by

$\begin{split}&2\kappa
N^{-1}E_{\mathscr{R}_{c}^{N}}\Big{[}(G(s))^{2}g_{k,N}(\eta(k))\big{(}f(\eta)+f(\eta^{k,k+1})\big{)}\Big{]}\\\
=\,&2\kappa
N^{-1}E_{\mathscr{R}_{c}^{N}}\Big{[}(G(s))^{2}\big{(}g_{k,N}(\eta(k))+g_{k+1,N}(\eta(k+1))\big{)}f(\eta)\Big{]}\\\
=\,&E_{\mathscr{R}_{c}^{N}}\big{[}\kappa U_{s,k}f\big{]}.\end{split}$

Retracing the terms, we obtain (9.2), finishing the proof. ∎

We now finish this section with a local replacement lemma at defect sites:

######  Lemma 9.2 (Local replacement at defect sites).

Let $G(\cdot):[0,T]\mapsto\mathbb{R}$ be bounded. Then, for each defect site
$k_{j,N}\in{\mathfrak{D}}_{N}$, we have

$\lim_{\theta\to
0}\limsup_{N\to\infty}\mathbb{E}_{N}\left[\left|\int_{0}^{T}G(s)\left(g_{k_{j,N},N}(\eta_{s}(k_{j,N}))-\Phi(\eta_{s}^{\theta
N,+}(k_{j,N}))\right)ds\right|\right]=0$

where $\eta^{l,+}(k):=(2l+1)^{-1}\sum_{k+1\leq x\leq k+2l+1}\eta(x)$.

###### Proof.

Lemma 9.1 shows we may replace $g_{k_{j,N},N}(\eta_{s}(k_{j,N}))$ by
$g_{k_{j,N}+1,N}(\eta_{s}(k_{j,N}+1))$. To finish, notice
$k_{j,N}+1\in\mathbb{T}_{N}\setminus{\mathfrak{D}}_{N}$ for $N$ large. Then,
we may further replace $g_{k_{j,N}+1,N}(\eta_{s}(k_{j,N}+1))$ by
$\Phi(\eta_{s}^{\theta N,+}(k_{j,N}))$ using Lemma 8.1 and Lemma 8.2; see
Remark 8.3. The proof is now complete. ∎

##  10\. Energy estimate

By Lemma 7.1, we know that $Q$ is supported on paths $\pi_{t}(dx)$ which can
be decomposed into an absolute continuous part $\rho(t,x)dx$ and atoms
$\sum_{j\in J_{c}}{\mathfrak{m}}_{j}(t)\delta_{x_{j}}(dx)$. In this section,
we prove an energy estimate for $\rho(t,x)$.

######  Proposition 10.1.

$Q$ is supported on paths $\pi_{t}(dx)$ such that $\Phi(\rho(t,x))$ is weakly
differentiable with respect to $x$ on $[0,T]\times\mathbb{T}$ and
$\partial_{x}\Phi(\rho(t,x))$, the weakly derivative, satisfies

$\int_{0}^{T}\int_{\mathbb{T}}\dfrac{(\partial_{x}\Phi(\rho(t,x)))^{2}}{\Phi(\rho(t,x)}dx\,dt<\infty.$
(10.1)

######  Remark 10.2.

As $\rho(t,x)\in L^{\infty}([0,T]\times\mathbb{T})$ (see Lemma 7.1) and
$\Phi(\cdot)$ is Lipschitz, it follows from (10.1) that
$\partial_{x}\Phi(\rho(t,x))\in L^{2}([0,T]\times\mathbb{T})$.

###### Proof of Proposition 10.1.

The proof presented here is based on the one of Theorem 7.1, p. 102, [18].
However, because of the presence of the defect sites and the difference in the
underlying topology, many details are different in subtle ways.

Let $\\{H_{j}\\}_{j\in\mathbb{N}}$ be a dense sequence in
$C_{c}^{0,1}([0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}))$ under the norm
$\|H\|_{\infty}+\|\partial_{x}H\|_{\infty}$. We split the proof into steps. In
the first of two steps, we show there is a constant $K_{0}$ such that, for all
$\epsilon$ small,

$\limsup_{N\to\infty}\mathbb{E}_{N}\Big{[}\max_{1\leq j\leq
m}\Big{\\{}\int_{0}^{T}W_{N}(\epsilon,H_{j}(s,\cdot),\eta_{s})ds\Big{\\}}\Big{]}\leq
K_{0}$ (10.2)

where

$\begin{split}W_{N}(\epsilon,H(\cdot),\eta):=&\sum_{x\in\mathbb{T}_{N}}\dfrac{H(x/N)}{\epsilon
N}\left(g_{x,N}(\eta(x))-g_{x+\epsilon N,N}(\eta(x+\epsilon N))\right)\\\
&-\dfrac{2}{N}\sum_{x\in\mathbb{T}_{N}}\dfrac{H^{2}(x/N)}{\epsilon
N}\sum_{0\leq k\leq\epsilon N}g_{x+k,N}(\eta(x+k)).\end{split}$

Step 1. Let $c=c_{0}$. By the entropy inequality, the expectation in (10.2) is
bounded from above by

$\begin{split}\dfrac{H(\mu^{N}|\mathscr{R}_{c_{0}}^{N})}{2N}+\dfrac{1}{2N}\ln\mathbb{E}_{\mathscr{R}_{c_{0}}^{N}}\Big{[}\exp\Big{\\{}\max_{1\leq
j\leq
m}\Big{\\{}2N\int_{0}^{T}W_{N}(\epsilon,H_{j}(s,\cdot),\eta_{s})ds\Big{\\}}\Big{\\}}\Big{]}.\end{split}$

For convenience, we set $c=c_{0}$. Using $H(\mu^{N}|\mathscr{R}_{c}^{N})\leq
CN$ and $e^{\max a_{j}}\leq\sum e^{a_{j}}$, the limsup in $N$ of the previous
display is estimated from above by

$\begin{split}\dfrac{C}{2}+\max_{1\leq j\leq
m}\limsup_{N\to\infty}\dfrac{1}{2N}\ln\mathbb{E}_{\mathscr{R}_{c}^{N}}\Big{[}\exp\Big{\\{}2N\int_{0}^{T}W_{N}(\epsilon,H_{j}(s,\cdot),\eta_{s})ds\Big{\\}}\Big{]}.\end{split}$

By Feynman-Kac formula, for any fixed index $j$, the limsup term in the
previous display is less than or equal to

$\limsup_{N\to\infty}\int_{0}^{T}\sup_{f}\left\\{E_{\mathscr{R}_{c}^{N}}\left[W_{N}(\epsilon,H_{j}(s,\cdot),\eta)f(\eta)\right]-\dfrac{N}{2}E_{\mathscr{R}_{c}^{N}}\Big{[}\sqrt{f}(-L_{N}\sqrt{f})\Big{]}\right\\}ds$

where the supremum is over all $f$ which are densities with respect to
$\mathscr{R}_{c}^{N}$.

Step 2. To show (10.2), it now remains to show, for all $H$ in
$C_{c}^{1}(\mathbb{T}\setminus{\mathfrak{D}})$, that

$\begin{split}E_{\mathscr{R}_{c}^{N}}\left[W_{N}(\epsilon,H(\cdot),\eta)f(\eta)\right]-NE_{\mathscr{R}_{c}^{N}}\Big{[}\sqrt{f}(-L_{N}\sqrt{f})\Big{]}\leq
0.\end{split}$ (10.3)

We first compute that
$E_{\mathscr{R}_{c}^{N}}\left[W_{N}(\epsilon,H(\cdot),\eta)f(\eta)\right]$
equals

$\begin{split}&E_{\mathscr{R}_{c}^{N}}\Big{[}\sum_{x\in\mathbb{T}_{N}}\dfrac{H(x/N)}{\epsilon
N}(g_{x,N}(\eta(x))-g_{x+\epsilon N,N}(\eta(x+\epsilon N)))f(\eta)\Big{]}\\\
&-\dfrac{2}{N}E_{\mathscr{R}_{c}^{N}}\Big{[}\sum_{x\in\mathbb{T}_{N}}\dfrac{H^{2}(x/N)}{\epsilon
N}\sum_{0\leq k\leq\epsilon N}g_{x+k,N}(\eta(x+k))f(\eta)\Big{]}.\end{split}$
(10.4)

Let $\delta_{x}$ be the configuration with the only particle at $x$ and
$\eta+\delta_{x}$ be the configuration obtaining from adding one particle at
$x$ to $\eta$ By the definition of $\mathscr{R}_{c}^{N}$, we have, for each
$x$,

$E_{\mathscr{R}_{c}^{N}}[g_{x,N}(\eta(x))f(\eta)]=\varphi
E_{\mathscr{R}_{c}^{N}}[f(\eta+\delta_{x})]$ (10.5)

where $\varphi=\Phi(c)$. Then, the first expectation in (10.4) is written as

$\begin{split}\sum_{x\in\mathbb{T}_{N}}\dfrac{\varphi H(x/N)}{\epsilon
N}E_{\mathscr{R}_{c}^{N}}\Big{[}f(\eta+\delta_{x})-f(\eta+\delta_{x+\epsilon
N})\Big{]}.\end{split}$ (10.6)

which is rewritten as

$\begin{split}&E_{\mathscr{R}_{c}^{N}}\Big{[}\sum_{x\in\mathbb{T}_{N}}\sum_{0\leq
k\leq\epsilon N-1}\dfrac{\varphi H(x/N)}{\epsilon
N}\big{(}\sqrt{f(\eta+\delta_{x+k})}+\sqrt{f(\eta+\delta_{x+k+1})}\big{)}\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\big{(}\sqrt{f(\eta+\delta_{x+k})}-\sqrt{f(\eta+\delta_{x+k+1})}\big{)}\Big{]}.\end{split}$
(10.7)

Using $2ab\leq a^{2}+b^{2}$, for any $A>0$, (10.7) is bounded from above by

$\begin{split}&E_{\mathscr{R}_{c}^{N}}\Big{[}\sum_{x\in\mathbb{T}_{N}}\sum_{0\leq
k\leq\epsilon N-1}\dfrac{\varphi H^{2}(x/N)}{2\epsilon
NA}\big{(}\sqrt{f(\eta+\delta_{x+k})}+\sqrt{f(\eta+\delta_{x+k+1})}\big{)}^{2}\Big{]}\\\
&\qquad+E_{\mathscr{R}_{c}^{N}}\Big{[}\sum_{x\in\mathbb{T}_{N}}\sum_{0\leq
k\leq\epsilon N-1}\dfrac{\varphi A}{2\epsilon
N}\big{(}\sqrt{f(\eta+\delta_{x+k})}-\sqrt{f(\eta+\delta_{x+k+1})}\big{)}^{2}\Big{]}:=I_{1}+I_{2}.\end{split}$
(10.8)

The $I_{2}$ term in (10.8) is rewritten as
$\sum_{x\in\mathbb{T}_{N}}\dfrac{\varphi
A}{2}E_{\mathscr{R}_{c}^{N}}\big{[}\big{(}\sqrt{f(\eta+\delta_{x})}-\sqrt{f(\eta+\delta_{x+1})}\big{)}^{2}\big{]}$
which, by the change of variable formula (10.5), is further recognized as the
Dirichelet form
$\dfrac{A}{2}E_{\mathscr{R}_{c}^{N}}\big{[}\sqrt{f}(-L_{N}\sqrt{f})\big{]}$,
cf. (8.12). For the first expectation $I_{1}$ in (10.8), using first
$(\sqrt{a}+\sqrt{b})^{2}\leq 2(a+b)$ and then (10.5), it is bounded from above
by

$\sum_{x\in\mathbb{T}_{N}}\dfrac{2H^{2}(x/N)}{\epsilon NA}\sum_{0\leq
k\leq\epsilon N}E_{\mathscr{R}_{c}^{N}}\Big{[}g_{x+k,N}(x+k)f(\eta)\Big{]}.$
(10.9)

Notice that the summation of $k$ is ranging from $0\leq k\leq\epsilon N$
instead of $0\leq k\leq\epsilon N-1$. Now, we set $A=N$. Putting together
(10.4) and (10.9), we obtain (10.3) with $K_{0}=C/2$.

Step 3. Recall that $H_{j}$’s have compact support in
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}})$. Let
$\iota_{\delta}=(2\delta)^{-1}\mathbbm{1}_{[-\delta,\delta]}$. Applying the
Bulk Replacement Lemma (Lemma 8.4) to (10.2) and taking $N\to\infty$, we
obtain

$\begin{split}&\limsup_{\delta\to 0}E_{Q}\Big{[}\max_{1\leq j\leq
m}\Big{\\{}\int_{0}^{T}\int_{\mathbb{T}}H_{j}(s,x)\epsilon^{-1}\big{(}\Phi(\iota_{\delta}*\rho_{s}(x))-\Phi(\iota_{\delta}*\rho_{s}(x+\epsilon))\big{)}dxds\\\
&\qquad\qquad\qquad\qquad-2\int_{0}^{T}\int_{\mathbb{T}}H_{j}^{2}(s,x)\epsilon^{-1}\int_{x}^{x+\epsilon}\Phi(\iota_{\delta}*\rho_{s}(u))dudxds\Big{\\}}\Big{]}\leq
K_{0},\end{split}$

Here
$\iota_{\delta}*\rho_{s}(\cdot)=\int_{\mathbb{T}}\iota_{\delta}(x-\cdot)\rho(s,x)dx$
and $\rho(s,x)\in L^{1}([0,T]\times\mathbb{T})$ is the absolute continuous
part of each limit path $\pi_{s}$ (cf. Lemma 7.1).

Sending $\delta\to 0$, applying a discrete integration by parts, and then
taking $\epsilon\to 0$, we have

$\begin{split}&E_{Q}\Big{[}\max_{1\leq j\leq
m}\Big{\\{}\int_{0}^{T}\int_{\mathbb{T}}\partial_{x}H_{j}(s,x)\Phi(\rho(s,x))dxds\\\
&\qquad\qquad\qquad\qquad\qquad\qquad-2\int_{0}^{T}\int_{\mathbb{T}}H_{j}^{2}(s,x)\Phi(\rho(s,x))dxds\Big{\\}}\Big{]}\leq
K_{0},\end{split}$

By monotone convergence, the $\max_{1\leq j\leq m}$ above can be replaced by
$\max_{1\leq j<\infty}$. Furthermore, as $H_{j}$ is dense in
$C_{c}^{0,1}([0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}))$ with respect to
the norm $\|H\|_{\infty}+\|\partial_{x}H\|_{\infty}$, we conclude

$E_{Q}\Big{[}\sup_{H}\Big{\\{}\int_{0}^{T}\int_{\mathbb{T}}\partial_{x}H(s,x)\Phi(\rho(s,x))dxds-2\int_{0}^{T}\int_{\mathbb{T}}H^{2}(s,x)\Phi(\rho(s,x))dxds\Big{\\}}\Big{]}\leq
K_{0},$ (10.10)

where the $\sup$ is over $H\in
C_{c}^{0,1}([0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}))$.

Step 4. As a result of (10.10), for $Q$-a.e. path $\pi_{t}(dx)$, there exists
$B=B(\pi_{t})$ such that, for all $H\in
C_{c}^{0,1}([0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}))$,

$\begin{split}\int_{0}^{T}\int_{\mathbb{T}}\partial_{x}H(s,x)\Phi(\rho(s,x))dxds-2\int_{0}^{T}\int_{\mathbb{T}}H^{2}(s,x)\Phi(\rho(s,x))dxds\leq
B.\end{split}$

Define on $C_{c}^{0,1}([0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}))$ a
linear functional
$l(H):=\int_{0}^{T}\int_{\mathbb{T}}\partial_{x}H(s,x)\Phi(\rho(s,x))dxds$.
Also define
$\|H\|_{2,\rho}:=\left(\int_{0}^{T}\int_{\mathbb{T}}H^{2}(s,x)\Phi(\rho(s,x))dxds\right)^{1/2}$.
Then we have, for all $a\in\mathbb{R}$

$a\,l(H)-2a^{2}\left(\|H\|_{2,\rho}\right)^{2}\leq B.$

Maximizing the left hand side over $a\in\mathbb{R}$, we obtain $l(\cdot)$ is a
bounded linear functional under the norm $\|\cdot\|_{2,\rho}$. By Riesz
representation theorem, there exists $F$ such that $\|F\|_{2,\rho}<\infty$ and
it holds that $l(H)=\langle
H,F\rangle_{\rho}:=\int_{0}^{T}\int_{\mathbb{T}}H(s,x)F(s,x)\Phi(\rho(s,x))dxds$.
Define $\partial_{x}\Phi(\rho(t,x)):=-F(t,x)\Phi(\rho(t,x))$. Then, we have
shown that $\Phi(\rho(t,x))$ is weakly differentialble with respect to $x$ on
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}})$ and the weak derivative
$\partial_{x}\Phi(\rho(t,x))$ satisfies (10.1).

To finish, we argue that the weak differentiability of $x$ may be extended
from $\mathbb{T}\setminus{\mathfrak{D}}$ to $\mathbb{T}$ with the same weak
derivative $\partial_{x}\Phi(\rho(t,x))$. The weak differentialbility on
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}})$ implies that, for almost all
$t\in[0,T]$, $\Phi(\rho(t,\cdot))$ is absolutely continuous on
$\mathbb{T}\setminus{\mathfrak{D}}$. It suffices to show that, for almost all
$t\in[0,T]$, $\Phi(\rho(t,\cdot))$ is continuous on the defects
${\mathfrak{D}}$. Notice that, as $(\partial_{x}\Phi(\rho))^{2}/\Phi(\rho)$
and $\Phi(\rho)$ are both in $L^{1}([0,T]\times\mathbb{T})$, by Cauchy-
Schwarz, we have $\partial_{x}\Phi(\rho)$ is in $L^{1}([0,T]\times\mathbb{T})$
as well. Then, for almost all $t\in[0,T]$, we have that the limits $\lim_{x\to
x_{j}\pm}\Phi(\rho(t,x))$ exist and are finite for all
$x_{j}\in{\mathfrak{D}}$ (cf. Remark 4.6). Moreover, by Lemma 9.2, the left
and right limits match. Therefore, we conlude the continuity of
$\Phi(\rho(t,\cdot))$ on ${\mathfrak{D}}$, finishing the proof. ∎

##  11\. Uniqueness

In this section, we present the uniqueness of the weak solutions to equations
(4.3) and (4.5). The proof is based on an energy argument (cf. [17]). Recall
Definitions 4.4 and 4.5.

######  Theorem 11.1.

There exists at most one weak solution to (4.3).

###### Proof.

Let $\pi^{(1)}_{t}$ and $\pi^{(2)}_{t}$ be two weak solutions of (4.3) such
that $\pi^{(i)}_{t}=\rho_{i}(t,x)dx+\sum_{j\in
J_{c}}{\mathfrak{m}}^{(i)}_{j}(t)\delta_{x_{j}}(dx)$ for $i=1,2$. As
${\mathfrak{m}}^{(i)}_{j}(t)=\big{[}\lambda_{j}\Phi(\rho_{i}(t,x_{j}))\big{]}^{1/\alpha}$,
to prove the theorem, it suffices to show $\rho_{1}=\rho_{2}$ for almost all
$t$.

Step 1. Define
$\overline{\Phi}(t,x):=\Phi(\rho_{1}(t,x))-\Phi(\rho_{2}(t,x))$. By Definition
4.4, we have that $\overline{\Phi}$ is weakly differentiable with respect to
$x\in\mathbb{T}$ and $\overline{\partial}_{x}\Phi$ (and therefore
$\overline{\Phi}$) is in $L^{2}([0,T]\times\mathbb{T})$. As
$\Phi(\rho_{i}(t,x_{j}))=\Phi(c_{0})$ for each $i=1,2$ and $j\in J_{s}$, we
have $\overline{\Phi}(t,x_{j})=0$ for for all $j\in J_{s}$. We first show that
$\overline{\Phi}$ can be approximated ‘well’ by a sequence
$\\{\overline{\Phi}_{\varepsilon}\\}_{\varepsilon>0}$ in the sense that (1)
$\overline{\Phi}_{\varepsilon}$ is smooth and compactly supported on
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}_{s})$; (2)
$\overline{\Phi}_{\varepsilon}\to\overline{\Phi}$ and
$\partial_{x}\overline{\Phi}_{\varepsilon}\to\partial_{x}\overline{\Phi}$ in
$L^{2}([0,T]\times\mathbb{T})$; and (3)
$\overline{\Phi}_{\varepsilon}(t,x_{j})\to\overline{\Phi}(t,x_{j})$ in
$L^{2}[0,T]$ for all $j\in J_{c}$ as $\varepsilon\to 0$.

Indeed, to verify this approximation, for $\delta>0$, let
${\mathfrak{D}}_{s}^{\delta}=\cup_{j\in J_{s}}(x_{j}-\delta,x_{j}+\delta)$. We
define $F_{\delta}$ be such that $F_{\delta}(t,x)=0$ on
$[0,T]\times{\mathfrak{D}}_{s}^{\delta}$ and
$F_{\delta}(t,x)=\overline{\Phi}(t,x)$ on
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}_{s}^{2\delta})$. For
$(t,x)\in{\mathfrak{D}}_{s}^{2\delta}\setminus{\mathfrak{D}}_{s}^{\delta}$,
define

$F_{\delta}(t,x)=\begin{cases}\overline{\Phi}(t,2x-x_{j}-2\delta)&{\rm on\
}[0,T]\times[x_{j}+\delta,x_{j}+2\delta),\\\
\overline{\Phi}(t,2x-x_{j}+2\delta)&{\rm on\
}[0,T]\times(x_{j}-2\delta,x_{j}-\delta].\\\ \end{cases}$

We note that $F_{\delta}$ has compact support in
$[0,T]\times(\mathbb{T}\setminus{\mathfrak{D}}_{s})$. It is easy to check that
$F_{\delta}$ and $\partial_{x}F_{\delta}$ approximate $\overline{\Phi}$ and
$\partial_{x}\overline{\Phi}$ in $L^{2}([0,T]\times\mathbb{T})$ respectively.
Since $F_{\delta}$ and $\overline{\Phi}$ match at $x=x_{j}$ for all $j\in
J_{c}$, to find a desired sequence $\\{\overline{\Phi}_{\varepsilon}\\}$, it
suffices to show that, for each small $\delta$, there exists
$\\{F_{\delta,\varepsilon}\\}$ that approximates $F_{\delta}$ ‘well’.

To this end, let $\tau_{\varepsilon}(x)$ be a standard mollifier supported on
$[-\varepsilon,\varepsilon]$. With $F_{\delta}(t,x)$ extended to be $0$ for
$t\notin[0,T]$, we define

$F_{\delta,\varepsilon}(t,x):=\int_{\mathbb{R}}\int_{\mathbb{T}}F_{\delta}(t-s,x-u)\tau_{\varepsilon}(s)\tau_{\varepsilon}(u)\,duds.$

Notice that $F_{\delta,\varepsilon}\in
C_{c}^{\infty}(\mathbb{R}\times(\mathbb{T}\setminus{\mathfrak{D}}_{s}))$ for
$\varepsilon<\delta$. When restricted on $[0,T]\times\mathbb{T}$, it is
standard that $\overline{F}_{\delta,\varepsilon}$ and
$\partial_{x}F_{\delta,\varepsilon}$ approximates $F_{\delta}$ and
$\partial_{x}F_{\delta}$ respectively in $L^{2}([0,T]\times\mathbb{T})$ as
$\varepsilon\to 0$. It remains to show that $F_{\delta,\varepsilon}(t,x_{j})$
approximates $F_{\delta}(t,x_{j})$ in $L^{2}[0,T]$. Write that

$\displaystyle\int_{0}^{T}\left(F_{\delta,\varepsilon}(t,x_{j})-F_{\delta}(t,x_{j})\right)^{2}dt$
$\displaystyle\ \ \ \ \ \ \
=\int_{0}^{T}\Big{[}\int_{\mathbb{R}}\int_{\mathbb{T}}\left(F_{\delta}(t-s,x_{j}-u)-F_{\delta}(t,x_{j})\right)\tau_{\varepsilon}(s)\tau_{\varepsilon}(u)duds\Big{]}^{2}dt.$

By adding and subtracting $F_{\delta}(t-s,x_{j})$, the above is bounded above
by $I_{1}+I_{2}$ where

$\begin{split}I_{1}&:=2\int_{0}^{T}\Big{[}\int_{\mathbb{R}}\int_{\mathbb{T}}\left(F_{\delta}(t-s,x_{j}-u)-F_{\delta}(t-s,x_{j})\right)\tau_{\varepsilon}(s)\tau_{\varepsilon}(u)duds\Big{]}^{2}dt,\\\
I_{2}&:=2\int_{0}^{T}\Big{[}\int_{\mathbb{R}}\left(F_{\delta}(t-s,x_{j})-F_{\delta}(t,x_{j})\right)\tau_{\varepsilon}(s)ds\Big{]}^{2}dt\end{split}$

As $\int_{\mathbb{R}}F_{\delta}(t-s,x_{j})\tau_{\varepsilon}(s)ds$
approximates $F_{\delta}(t,x_{j})$ in $L^{2}[0,T]$, the term $I_{2}$ vanishes
as $\varepsilon\to 0$. For the term $I_{1}$, using
$F_{\delta}(t-s,x_{j}-u)-F_{\delta}(t-s,x_{j})=\int_{x_{j}}^{x_{j}-u}\partial_{x}F_{\delta}(t-s,x)dx$,
we have

$I_{1}\leq
2\int_{0}^{T}\int_{\mathbb{R}}\int_{\mathbb{T}}\Big{(}\int_{x_{j}}^{x_{j}-u}\partial_{x}F_{\delta}(t-s,x)dx\Big{)}^{2}\tau_{\varepsilon}(s)\tau_{\varepsilon}(u)dudsdt$

The square term above is further bounded by
$u^{2}\big{|}\int_{x_{j}}^{x_{j}-u}(\partial_{x}F_{\delta}(t,x))^{2}dx\big{|}$.
As $\tau_{\varepsilon}(u)=0$ for $u\notin[-\varepsilon,\varepsilon]$, we have
that $I_{1}\leq
2\varepsilon^{2}\int_{0}^{T}\int_{\mathbb{T}}(\partial_{x}F_{\delta}(t,x))^{2}dxdt$
which vanishes as $\varepsilon\to 0$.

Step 2. We now proceed to the uniqueness of weak solutions. Let
$\overline{\rho}:=\rho_{1}-\rho_{2}$ and
$\overline{\mathfrak{m}}_{j}(t):={\mathfrak{m}}_{j}^{(1)}(t)-{\mathfrak{m}}_{j}^{(2)}(t)$
for each $j\in J_{c}$. As $\pi^{(1)}_{t}$ and $\pi^{(2)}_{t}$ both satisfy
(4.6), we have, for all $G(t,x)\in
C_{c}^{\infty}\left([0,T)\times(\mathbb{T}\setminus{\mathfrak{D}}_{s})\right)$,

$\begin{split}&\int_{0}^{T}\int_{\mathbb{T}}\partial_{t}G(t,x)\overline{\rho}(t,x)dxdt+\sum_{j\in
J_{c}}\int_{0}^{T}\partial_{t}G(t,x_{j})\overline{\mathfrak{m}}_{j}(t)dt=\int_{0}^{T}\int_{\mathbb{T}}\partial_{x}G(t,x)\partial_{x}\overline{\Phi}(t,x)dxdt.\end{split}$
(11.1)

Let $\overline{\Phi}$ be approximated ‘well’ by some
$\\{\overline{\Phi}_{\varepsilon}\\}$ as in Step $1$. Taking
$G(t,x)=-\int_{t}^{T}\overline{\Phi}_{\varepsilon}(s,x)ds$ and then leting
$\varepsilon\to 0$, we obtain

$\begin{split}&\int_{0}^{T}\int_{\mathbb{T}}\overline{\Phi}(t,x)\overline{\rho}(t,x)dxdt+\sum_{j\in
J_{c}}\int_{0}^{T}\overline{\Phi}(t,x_{j})\overline{\mathfrak{m}}_{j}(t)dt\\\
&\qquad\qquad\qquad\qquad=-\int_{\mathbb{T}}\int_{0}^{T}\Big{[}\int_{t}^{T}\partial_{x}\overline{\Phi}(s,x)ds\Big{]}\partial_{x}\overline{\Phi}(t,x)dtdx.\end{split}$
(11.2)

The right hand side of the above is computed as
$-\dfrac{1}{2}\int_{\mathbb{T}}\big{[}\int_{0}^{T}\partial_{x}\overline{\Phi}(t,x)dt\big{]}^{2}dx\leq
0$. However, for the left hand side, we have
$\overline{\Phi}(t,x)\overline{\rho}(t,x)\geq 0$ and
$\overline{\Phi}(t,x_{j})\overline{\mathfrak{m}}_{j}(t)\geq 0$ for all $t$,
$x$, and $j$. Then, we deduce that
$\int_{0}^{T}\int_{\mathbb{T}}\overline{\Phi}(t,x)\overline{\rho}(t,x)dxdt=0$
which implies $\overline{\Phi}(t,x)\overline{\rho}(t,x)=0$ a.e. and,
therefore, $\overline{\rho}(t,x)=0$ a.e. Since $\overline{\rho}$ is continuous
in $x$ for almost all $t$ (cf. Remark 4.6), the theorem is proved. ∎

######  Theorem 11.2.

There exists at most one weak solution to (4.5).

###### Proof.

For $i=1,2$, let $\pi^{(i)}_{t}=\rho_{i}(t,x)dx+\sum_{j\in
J_{c}}{\mathfrak{m}}^{(i)}_{j}(t)\delta_{x_{j}}(dx)$ be two weak solutions to
(4.5). Notice that
${\mathfrak{m}}^{(i)}_{j}(t)={\mathfrak{m}}^{(i)}_{j}(t)\mathbbm{1}_{\rho(t,x_{j})=c_{j}}$
and $\rho(t,x_{j})\leq c_{j}$ implies
$\overline{\Phi}(t,x_{j})\overline{\mathfrak{m}}_{j}(t)\geq 0$ where
$\overline{\Phi}:=\Phi(\rho_{1})-\Phi(\rho_{2})$ and
$\overline{\mathfrak{m}}_{j}:={\mathfrak{m}}_{j}^{(1)}-{\mathfrak{m}}_{j}^{(2)}$.
Following the proof of Theorem 11.1, we have
$\overline{\rho}:=\rho_{1}-\rho_{2}=0$ for almost all $t$. To conclude, it
remains to show $\overline{\mathfrak{m}}_{j}=0$ for each $j\in J_{c}$ for
almost all $t$. To this end, we fix any $j\in J_{c}$ and take $F\in
C^{\infty}(\mathbb{T})$ such that $F(x_{j})=1$ and $\text{supp}\,F\cap
J_{c}=x_{j}$. Letting $G(t,x)=\int_{t}^{T}h(s)F(x)ds$ in (11.1) for any
$h(t)\in C_{c}^{\infty}(0,T)$ and using $\overline{\rho}=\overline{\Phi}=0$,
we obtain $\int_{0}^{T}h(t)\overline{\mathfrak{m}}_{j}(t)dt=0$, and therefore,
$\overline{\mathfrak{m}}_{j}(t)=0$ for almost all $t$, finishing the proof. ∎

Acknowledgements. SS was partly supported by ARO-W911NF-18-1-0311.

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