We study the vanishing viscosity/diffusivity limit for the transport equation of a passive scalar $f(x,t)\in\mathbb{R}$ along a divergence-free vector field $u(x,t)\in\mathbb{R}^2$, given by $\frac{\partial f}{\partial t} + \nabla\cdot (u f) = 0$; and the associated advection-diffusion equation of $f$ along $u$ for positive viscosity/diffusivity parameter $\nu>0$, expressed by $\frac{\partial f}{\partial t} + \nabla\cdot (u f) -\nu\Delta f = 0$. We demonstrate failure of the vanishing viscosity limit of the advection-diffusion equation to select unique solutions, or to select entropy-admissible solutions, to transport along $u$.

First, we construct a bounded divergence-free vector field $u$ which admits, for each (non-constant) initial datum, two weak solutions to the initial value problem for the transport equation. Moreover, we show that both these solutions are renormalised weak solutions, and are strong limits along different subsequences of vanishing viscosity of solutions to the corresponding advection-diffusion equation.

Second, we construct a second bounded divergence-free vector field $u$ admitting, for any initial datum, a weak solution to the transport equation which is perfectly mixed to its spatial average, and after some delay in time, it unmixes to its initial state. Moreover, we show that this entropy-inadmissible unmixing is the unique weak vanishing viscosity limit of the corresponding advection-diffusion equation.

§ INTRODUCTION

§.§ Background

We are concerned with the fundamental problem of selection among non-unique weak solutions in fluid mechanics. Consider a divergence-free vector field $u:\mathbb{T}^d\times[0,T]\to\mathbb{R}^d$, on a $d$-dimensional spatial torus $\mathbb{T}^d$ and a time interval $[0,T]$. For a passive scalar $f:\mathbb{T}^d\times[0,T]\to\mathbb{R}$, we may write the passive transport of $f$ along $u$ in conservation-form as

\begin{equation*}
\frac{\partial f}{\partial t} + \nabla\cdot (u f) = 0. \tag{TE}
\end{equation*}

This equation is understood in the distributional sense, and may exhibit non-uniqueness of weak solutions, see e.g. diperna1989ordinary,depauw2003non, and also modena2018non,cheskidov2021nonuniqueness.

Meanwhile, the advection-diffusion of $f$ along $u$ with viscosity/diffusivity parameter $\nu>0$ is given by

\begin{equation*}
\frac{\partial f}{\partial t} + \nabla \cdot (uf) - \nu\Delta f = 0. \tag{$\nu$-ADE}
\end{equation*}

In contrast to the transport equation, parabolic equations of this type are well-posed, see e.g. evans2010partial,flandoli2010well. Therefore, one may ask what the behaviour of the limiting solution is, and in particular which among the general class of non-unique solutions are selected as viscosity vanishes.

In the general context of fluid mechanics selection by vanishing viscosity has a long history of study for various fluid models, see for example dafermos2005hyperbolic,Bianchini2005,bardos2012vanishing,bardos2013stability,Nussenzveig_Lopes_2021. Notably, these works all support the hypothesis that vanishing viscosity selects uniquely the physical solution to the system despite a-priori non-uniqueness to the inviscid model.

The inviscid limit $\nu\to0$ is also of great interest to many other problems in fluid mechanics, including anomalous dissipation in turbulence (both hydrodynamic kolmogorov1941local,brue2023anomalous,brue2022onsager, and passive scalar corrsin1951spectrum,obukhov1949,drivas2019anomalous,armstrong2023anomalous), enhanced dissipation constantin2008diffusion,bedrossian2021almost,elgindi2023optimal, and the general study of passive scalar mixing, see e.g. depauw2003non,bressan2003lemma,lunasin2012optimal,alberti2019exponential. We focus primarily on its use as a physical selection mechanism for the inviscid problem, though our results also inform the study of anomalous dissipation, and passive scalar mixing.

To date there have been very few counter examples to uniqueness or physical admissibility of the vanishing viscosity limit. The first non-uniqueness results for the passive scalar transport model were announced recently, in <cit.> and also <cit.>, through analysis of anomalous dissipation of energy. In contrast, the general framework we develop for controlling the vanishing viscosity limit relies on intermittency in frequency space of small scale/high frequency structures in the fluid. With this we construct in Theorem <ref> below a example of a bounded divergence-free vector field exhibiting non-uniqueness of the vanishing viscosity limit for all (non-constant) initial data, with strong convergence of the vanishing viscosity limit to multiple renormalised weak solutions, and without any anomalous dissipation in the vanishing viscosity limit.

Our second construction aims to tackle the fundamental question of whether vanishing viscosity solutions are indeed physical. This has generally been overlooked since one argues by the presence of molecular diffusion that vanishing viscosity should be taken as the definition of physical admissibility. However, one also concludes from the second law of thermodynamics that any physical solution should satisfy certain entropy-inequalities, namely those introduced in <cit.>. Since energy (or entropies) are weakly lower semi-continuous, one proves that the vanishing viscosity limit may only dissipate energy/entropy. It is therefore generally believed that these two selection criteria coincide, and indeed all previously mentioned works support this hypothesis.

We show instead, in Theorem <ref> below, that entropy-admissibility of the vanishing viscosity limit to the passive scalar transport model fails for a particular choice of vector field. This hold for all (non-constant) initial datum, with uniqueness of the vanishing viscosity limit. For each initial datum, this unique vanishing viscosity solution is mixed to its spatial average, and after a short delay perfectly unmixes to its original state, thus exhibiting a dissipation of energy/entropy and subsequently a reverse of this dissipation. This new method of proof for uniqueness of the vanishing viscosity limit despite non-uniqueness to the inviscid model relies on a particular kind of spatial intermittency of small scale structures.

We highlight the importance of such a phenomenon more generally. One may ask if a similar phenomenon is possible for active scalar models, such as the incompressible Euler equations for which entropy-inadmissible solutions are often termed wild solutions. Such solutions have a long history of study, starting with the early works of scheffer1993inviscid,shnirelman1997nonuniqueness,constantin1994onsager, and also a large body of recent progress owing to the convex-integration method, see e.g. de2013dissipative,buckmaster2017onsager,buckmaster2019nonuniqueness. It is widely believed that these wild solutions should not be the limit, let alone unique limit, of strong or weak Leray-Hopf <cit.> solutions of the incompressible Navier-Stokes equations, and so should not be relevant to physical hydrodynamic turbulence.

Our construction for the passive scalar transport model highlights that this conclusion may be unfounded. While weak lower semi-continuity of energy ensures that energy may only dissipate in the vanishing viscosity limit, it fails to ensure that the amount of dissipation increases with time. It is therefore possible that a vanishing viscosity solution exhibits, as in our construction, an energy cascade to infinite frequencies only for short time, before cascading back to finite frequencies. This highlights that there is no known mechanism preventing wild solutions to the Euler equations, in particular those constructed by the convex-integration machinery (see e.g. de2013dissipative,buckmaster2017onsager,buckmaster2019nonuniqueness), arising as the late-time behaviour of vanishing viscosity limits of strong solutions to the incompressible Navier-Stokes equations, given sufficient initial energy.

We highlight also that the failure of vanishing viscosity solutions to be entropy-admissible stresses that the general existence of any entropy-admissible solution to the passive scalar transport model (along an arbitrary vector field and for arbitrary initial datum) is an important open problem.

§.§ Main results

We introduce in Section <ref> standard notation of function spaces, and rigorous definitions of weak solutions of the transport equation (Definition <ref>, (<ref>)), weak renormalised solutions of the transport equation (Definition <ref>), and weak solutions of the advection-diffusion equation with viscosity $\nu>0$ (Definition <ref>, (<ref>)). We also highlight Theorem <ref> in Section <ref> for a proof of well-posedness of the advection-diffusion equation, and in particular the uniqueness stated in the below Theorems.

There exists a divergence-free vector field $u \in L^\infty([0,1];L^\infty(\mathbb{T}^2;\mathbb{R}^2))$, and a sequence $\{\nu_n\}_{n\in\mathbb{N}}$ with $\nu_n > 0$ and $\nu_n\xrightarrow{n\to\infty}0$, such that for any initial data $f_0\in  L^\infty(\mathbb{T}^2)$, and for $f^\nu$ the unique solution to (<ref>) along $u$ with initial data $f_0$, one has

\begin{gather*}
f^{\nu_{2n}}\xrightarrow{n\to\infty}f^\mathrm{even}, \\
\end{gather*}

with the above convergence in weak-$*$ $L^\infty([0,1];L^\infty(\mathbb{T}^2))$, and strong in $L^p([0,1];L^p(\mathbb{T}^2))$ for all $p \in [1,\infty)$. Furthermore, the limit functions $f^\mathrm{even}, f^\mathrm{odd}$ are renormalised weak solutions to (<ref>) along $u$ with initial data $f_0$.

[One may show as a corollary that uncountably many vanishing viscosity solutions exist. See the footnote for the result as stated in Section <ref>.]Moreover, if $f_0$ is not constant, then $f^\mathrm{even}\ne f^\mathrm{odd}$.

There exists a divergence-free vector field $u \in L^\infty([0,100];L^\infty(\mathbb{T}^2;\mathbb{R}^2))$, such that for any initial data $f_0\in L^\infty(\mathbb{T}^2)$, and for $f^\nu$ the unique solution to (<ref>) along $u$ with initial data $f_0$, one has

\begin{equation*}
\end{equation*}

with the above convergence in weak-$*$ $L^\infty([0,100];L^\infty(\mathbb{T}^2))$, strong in $L^p([0,42];L^p(\mathbb{T}^2))$, $C^0([0,42-\epsilon];L^p(\mathbb{T}^2))$, $L^p([58,100];L^p(\mathbb{T}^2))$, and $C^0([58+\epsilon,100];L^p(\mathbb{T}^2))$ for all $p \in [1,\infty)$, and all $\epsilon >0$. The limit function $f \in C_{\mathrm{weak-}*}^0([0,100];L^\infty(\mathbb{T}^2))$ is a weak solution to (<ref>) along $u$ with initial data $f_0$.

Moreover, for all $t \in [42,58]$

\begin{equation*}
f(\cdot, t) \equiv \int_{\mathbb{T}^2}f_0(y)\;dy,
\end{equation*}

is perfectly mixed to its spatial average.

Furthermore, for all $t \in [0,100]$, $f(\cdot, t)=f(\cdot,100-t)$ and in particular,

\begin{equation*}
f(\cdot, 100) = f_0,
\end{equation*}

is perfectly unmixed. In particular, if $f_0$ is not constant, any $L^p(\mathbb{T}^2)$ norms of $f(\cdot, t)$ (for $p\in(1,\infty]$) increase after $t=58$, contrary to the entropy-admissibility criterion of Dafermos in <cit.>.

§.§ Outline of paper

Section <ref> contains the notation and definitions used throughout this paper.

Section <ref> contains the necessary uniqueness, existence, and regularity theory for the transport equation and advection-diffusion equation.

In Section <ref> we develop the machinery (and intuition) for controlling the vanishing viscosity limits in the proofs of Theorems <ref>, <ref>.

Section <ref> is then devoted to the construction and proof of Theorem <ref>.

Section <ref> is instead devoted to the construction and proof of Theorem <ref>, and is independent of Section <ref>.

§ NOTATION AND DEFINITIONS

§.§ Notation

Denote by $\mathbb{T}^d = \mathbb{R}^d / \mathbb{Z}^d$ the $d$-dimensional unit torus. Throughout this paper we work on the spatial domain $\mathbb{T}^d$, or on the spatio-temporal domain $\mathbb{T}^{d}\times [0,T]$ for some $T>0$. The gradient operator $\nabla$ and Laplacian $\Delta$ will act on spatial coordinates only.

If the dimension $d$ and time interval $[0,T]$ are implicit we use the shorthand notation $L^pL^q$ for $L^p\left([0,T];L^q(\mathbb{T}^d)\right)$ and similar spaces. Meanwhile, $L^q(\mathbb{T}^d)$ is shorthand for scalar functions $L^q(\mathbb{T}^d;\mathbb{R})$.

For integrability exponents $p,q,... \in [1,\infty]$ (which are allowed to be infinite unless otherwise stated), denote by $p^*$ the Hölder conjugate.

For a positive integer $n\in\mathbb{N}$ we denote the homogeneous Sobolev space

\begin{equation*}
\dot{H}^n(\mathbb{T}^d)=\left\{f\in L^2(\mathbb{T}^d):\int_{\mathbb{T}^d}f(x)\;dx=0, \hbox{ and } (-\Delta)^{\frac{n}{2}}f\in L^2(\mathbb{T}^d)\right\},
\end{equation*}

where the fractional Laplacian $(-\Delta)^{\frac{n}{2}}$ is defined in terms of the Fourier transform/series $\mathcal{F}$ by $\mathcal{F}((-\Delta)^{\frac{n}{2}}f) = |\xi|^n \mathcal{F}(f)$. We denote the norm in $\dot{H}^n(\mathbb{T}^d)$ by

\begin{equation*}
\left\|f\right\|_{\dot{H}^n(\mathbb{T}^d)} = \left\|(-\Delta)^{\frac{n}{2}}f\right\|_{L^2(\mathbb{T}^d)},
\end{equation*}

for all $f \in \dot{H}^n(\mathbb{T}^d)$. Moreover, we define the Sobolev space

\begin{equation*}
H^n(\mathbb{T}^d) = \left\{f\in L^2(\mathbb{T}^d):(-\Delta)^{\frac{n}{2}}f\in L^2(\mathbb{T}^d)\right\},
\end{equation*}

with the norm

\begin{equation*}
\left\|f\right\|_{H^n(\mathbb{T}^d)} = \left\|f\right\|_{L^2(\mathbb{T}^d)} + \left\|(-\Delta)^{\frac{n}{2}}f\right\|_{L^2(\mathbb{T}^d)}.
\end{equation*}

Furthermore, we denote by $H^{-n}(\mathbb{T}^d)$ the dual space of $H^n(\mathbb{T}^d)$.

For each $p \in [1,\infty]$ we define also the space

\begin{equation*}
C^0L^p = \left\{f:[0,T]\to L^p(\mathbb{T}^d) \text{ such that $f$ is continuous}\right\} \subset L^\infty L^p,
\end{equation*}

with the norm

\begin{equation*}
\left\|f\right\|_{C^0L^p} = \sup_{0\le t \le T} \left\|f(\cdot,t)\right\|_{L^p(\mathbb{T}^d)}.
\end{equation*}

Moreover, we define the linear space

\begin{equation*}
C_{\mathrm{weak}}^0L^p = \left\{f:[0,T]\to L^p(\mathbb{T}^d) \text{ such that $f$ is continuous in $L^p_{\mathrm{weak}}$}\right\} \subset L^\infty L^p,
\end{equation*}

and for $p \in (1,\infty]$ the linear space

\begin{equation*}
C_{\mathrm{weak-}*}^0L^p = \left\{f:[0,T]\to L^p(\mathbb{T}^d) \text{ such that $f$ is continuous in $L^p_{\mathrm{weak-}*}$}\right\} \subset L^\infty L^p.
\end{equation*}

For $p \in (1,\infty)$, $L^p(\mathbb{T}^d)$ is reflexive and so these two definitions coincide. However, we commonly take $p=\infty$, hence we choose the notation $C_{\mathrm{weak-}*}^0L^p$ and take care of the case when $p=1$.

All results will be stated and proved for the compact domains $\mathbb{T}^d$ and $[0,T]$. The exact statements do not necessarily hold when we replace $\mathbb{T}^d$ with $\mathbb{R}^d$, or $[0,T]$ with $[0,\infty)$, though analogues can surely be found.

Consider a vector field $u\in L^1\left([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d)\right)$ with $\nabla\cdot u=0$ in the distributional sense.

We say $f\in L^1\left([0,T];L^1(\mathbb{T}^d;\mathbb{R})\right)$ with $uf \in L^1L^1$ is a weak solution to the transport equation along $u$

\begin{equation}\label{eqTE}
\frac{\partial f}{\partial t} + \nabla \cdot (u f) = 0, \tag{TE}
\end{equation}

with initial data $f_0 \in L^1(\mathbb{T}^d)$ if, [Equivalently we may take $\phi$ to be Lipschitz in time and space with compact support in $\mathbb{T}^d \times [0,T)$. This is done by finding a sequence of smooth functions $\phi_n \in C_c^\infty(\mathbb{T}^d\times[0,T))$ such that $\phi_n$ are bounded in $C^1$ by the Lipschitz norm of $\phi$, and $\frac{\partial \phi_n}{\partial t} \xrightarrow{n\to\infty} \frac{\partial \phi}{\partial t}$, $\nabla \phi_n \xrightarrow{n\to\infty} \nabla \phi$ pointwise almost everywhere.]for any $\phi \in C_c^\infty(\mathbb{T}^d\times[0,T))$,

\begin{equation*}
\int_{\mathbb{T}^d\times[0,T)} f \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi\right) \; dxdt = - \int_{\mathbb{T}^d} f_0 \phi_0 \; dx,
\end{equation*}

where $\phi_0(x) = \phi(x, 0)$.

Meanwhile, we say the transport equation is satisfied on an open interval $I\subset(0,T)$ if, for any $\phi \in C_c^\infty(\mathbb{T}^d\times I)$,

\begin{equation*}
\int_{\mathbb{T}^d\times I} f \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi\right) \; dxdt = 0.
\end{equation*}

Following the definition introduced in <cit.>, suppose $f$ is a weak solution to (<ref>) along $u$ with initial data $f_0$.

If, for any $\beta \in C_b^0(\mathbb{R})$, $\beta(f)$ is a weak solution to (<ref>) along $u$ with initial data $\beta(f_0)$, then we say $f$ is a renormalised weak solution of (<ref>).

This definition is well motivated by the expression $\left( \frac{\partial}{\partial t} + u \cdot \nabla \right) \beta(f) = \beta'(f)\left( \frac{\partial}{\partial t} + u \cdot \nabla \right) f$ when $\beta$ is a differentiable function. Indeed some authors require $\beta \in C^1(\mathbb{R})$ as in <cit.>, or even with decay at infinity as in <cit.>. In our case ($\nabla \cdot u = 0$) it is straightforward to show that these give equivalent definitions.

For a divergence-free vector field $u \in L^1([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d))$, we say a family ($t \in [0,T])$ of Lebesgue-measure preserving bijections $y_t : \mathbb{T}^d \to \mathbb{T}^d$ is a Lagrangian flow along $u$ if for a.e. $x\in \mathbb{T}^d$ the map $t \mapsto y_t(x)$ is absolutely continuous and the derivative satisfies $\frac{d y_t(x)}{dt} = u(y_t(x),t)$ as a function class in $L^1([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d))$.

[Notice, without loss of generality we may take $y_0=\mathrm{Id}$.]We say that a function $f \in L^1([0,T]; L^1(\mathbb{T}^d;\mathbb{R}))$ is a Lagrangian solution to (<ref>) along $u$ (with initial data $f_0 \circ y_0^{-1}$) if $f(\cdot, t) = f_0 \circ y_t^{-1}$ for $f_0\in L^1(\mathbb{T}^d)$, and $\{y_t\}_{t\in[0,T]}$ a Lagrangian flow along $u$.

[We reduce the problem to bounded $f$ and $f_0$ by the point-wise approximation with $f 1_{|f|\le k}$, so that $u f 1_{|f|\le k} \xrightarrow{k\to\infty} uf$ in $L^1L^1$ by dominated convergence. Observe then that $f 1_{|f|\le k}$ are already Lagrangian solutions for the initial data $f_0 1_{|f_0|\le k}$, since $(f_0 \circ y_t^{-1}) 1_{|f_0 \circ y_t^{-1}| \le k} = (f_0 1_{|f_0|\le k}) \circ y_t^{-1}$.]It is straightforward to show that, if additionally $uf \in L^1 L^1$, then a Lagrangian solution to (<ref>) along $u$ with initial data $f_0$ is also a weak solution in the sense of Definition <ref>. The result for bounded $f$ and $f_0$ follows from changing variables in the integral $\int_{\mathbb{T}^d\times[0,T)} f_0(y_t^{-1}(x)) \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi\right) \; dxdt$, and using that $\left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi\right)(y_t(x)) = \frac{\partial}{\partial t}(\phi(y_t(x), t))$ by the chain rule for absolutely continuous functions.

Moreover, since for any $\beta \in C_b^0(\mathbb{R})$ we can rewrite $\beta(f_0 \circ y_t^{-1}) = \beta(f_0) \circ y_t^{-1}$, these solutions are then also renormalised weak solutions in the sense of Definition <ref>.

Consider a vector field $u\in L^1\left([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d)\right)$ with $\nabla\cdot u=0$ in the distributional sense, and some positive constant viscosity $\nu > 0$ (also called diffusivity).

We say $f\in L^1\left([0,T];L^1(\mathbb{T}^d;\mathbb{R})\right)$ with $uf \in L^1L^1$ is a weak solution to the advection-diffusion equation along $u$

\begin{equation}\label{eqADE}
\frac{\partial f}{\partial t} + \nabla \cdot (u f) - \nu \Delta f = 0 \tag{$\nu$-ADE},
\end{equation}

with initial data $f_0 \in L^1(\mathbb{T}^d)$, if for any $\phi \in C_c^\infty(\mathbb{T}^d\times[0,T))$,

\begin{equation*}
\int_{\mathbb{T}^d\times[0,T]} f \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi + \nu \Delta \phi\right) \; dxdt = - \int_{\mathbb{T}^d} f_0 \phi_0 \; dx,
\end{equation*}

where $\phi_0(x) = \phi(x,0)$.

Meanwhile, we say the advection-diffusion equation (<ref>) is satisfied on an open interval $I\subset(0,T)$ if, for any $\phi \in C_c^\infty(\mathbb{T}^d\times I)$,

\begin{equation*}
\int_{\mathbb{T}^d\times I} f \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi + \nu\Delta\phi \right) \; dxdt = 0.
\end{equation*}

Consider a vector field $u\in L^1\left([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d)\right)$ with $\nabla\cdot u=0$ in the distributional sense.

We say a weak solution $f$ to (<ref>) along $u$ with initial data $f_0 \in L^1(\mathbb{T}^d)$ is a vanishing viscosity solution, if there exists a positive sequence $\nu_n\to0$ and corresponding weak solutions $f^{(n)}$ to (<ref>) with viscosity $\nu_n$ along $u$ with initial data $f_0$, such that

\begin{equation*}
f^{(n)} \xrightharpoonup{n \to \infty} f,
\end{equation*}

converges as distributions in $\mathcal{D}'(\mathbb{T}^d\times[0,T))$.

In the literature $f^{(n)}$ is often taken to have, in addition, non-constant initial data $f_0^{(n)}$, where $f_0^{(n)}$ converges to $f_0$ in a suitable topology. We do not consider such a more general definition here.

§ NECESSARY BACKGROUND

The following is by no means a full exposition of standard existence, uniqueness, and regularity theory for (<ref>) and (<ref>), but contains some of the more salient points, and in particular those relevant to this paper.

The exact statements are adapted for this paper, with similar results, and similar methods of proof, found in the literature.

We fix a finite time interval $[0,T]$ throughout.

Suppose $f$ is a weak solution to (<ref>) or (<ref>) (i.e. $\nu=0$) along $u \in L^1 L^1$ with initial data $f_0$.

Then for any $\phi\in C^\infty(\mathbb{T}^d\times[0,T])$, for a.e. $t \in [0,T]$,

\begin{multline}
\text{(Trace Formula)\quad} \int_{\mathbb{T}^d} f(\cdot, t) \phi(\cdot, t) \;dx \\ = \int_{\mathbb{T}^d} f_0 \phi_0 \; dx + \int_{\mathbb{T}^d\times[0,t]} f\left(\frac{\partial\phi}{\partial t}+u\cdot\nabla\phi +\nu\Delta \phi\right) dxdt, \label{eqtrace}
\end{multline}

[An analogous result holds when $p=1$ if we additionally assume $f(\cdot,t): [0,T] \to L^1(\mathbb{T}^d)$ is uniformly integrable in the indexing variable $t$ (for a.e. $t\in[0,T]$).]Suppose further that $f \in L^\infty L^p$ for $p \in (1, \infty]$, then there is a (unique) representation of $f \in C_{\mathrm{weak-}*}^0L^p$, such that (<ref>) holds for all $t \in [0,T]$.

In particular $f(\cdot, 0) = f_0$ in $L^p(\mathbb{T}^d)$.

For any weak solution of (<ref>) or (<ref>) we have $f, uf \in L^1L^1$.

Fix some $\phi \in C^\infty\left(\mathbb{T}^d \times [0,T]\right)$ and consider the following $L^1([0,T])$ function of $t \in [0,T]$,

\begin{equation}\label{eq23}
\int_{\mathbb{T}^d} f(\cdot, t) \phi(\cdot, t) \; dx.
\end{equation}

Then for all $\psi \in C_c^\infty([0,T))$, by Definitions <ref>, <ref>,

\begin{multline*}
\int_{\mathbb{T}^d \times [0,T)} f \phi \frac{d\psi}{ds} \; dxds \\
\begin{aligned}[t]
& = \int_{\mathbb{T}^d \times [0,T)} f \phi \frac{d\psi}{ds} \; dxds - \left(\int_{\mathbb{T}^d} f_0 \phi_0 \psi_0 \; dx + \int_{\mathbb{T}^d\times[0,T)} f\left(\frac{\partial}{\partial s}+u\cdot\nabla + \nu \Delta \right)(\phi \psi(s)) \; dxds \right) \\
& = - \int_{\mathbb{T}^d} f_0 \phi_0 \psi_0 \; dx - \int_{\mathbb{T}^d\times[0,T)} f\left(\frac{\partial\phi}{\partial s}+u\cdot\nabla\phi + \nu \Delta \phi\right) \psi(s) \; dxds,
\end{aligned}
\end{multline*}

where $\psi_0=\psi(0)$ and $\phi_0(x)=\phi(x,0)$.

And so the function defined in (<ref>) is an absolutely continuous function of $t \in [0,T]$, with derivative $\int_{\mathbb{T}^d} f(x, t)\left(\frac{\partial\phi}{\partial s}+u\cdot\nabla\phi + \nu \Delta \phi\right)(x, t) \; dx$, and the initial value at $t=0$ is $\int_{\mathbb{T}^d} f_0 \phi_0 \; dx$, see Chapter 3, Lemma 1.1 in <cit.>. That is (<ref>) holds for a.e. $t \in [0,T]$.

Suppose now, for some $p \in (1,\infty]$, that $f \in L^\infty L^p$. Define for each $t \in [0,T]$ the distribution $F_t\in\mathcal{D}'(\mathbb{T}^d)$ acting on test functions $\chi \in C^\infty(\mathbb{T}^d)$, given by

\begin{equation*}
\left\langle F_t,\chi\right\rangle = \int_{\mathbb{T}^d} f_0 \chi \; dx + \int_{\mathbb{T}^d\times[0,t]} f\left(u\cdot\nabla\chi +\nu\Delta \chi\right) dxdt.
\end{equation*}

Then thanks to (<ref>), for a.e. $t \in [0,T]$, $F_t = f(\cdot, t)$ as distributions in $\mathcal{D}'(\mathbb{T}^d)$, which is assumed uniformly bounded in $L^p(\mathbb{T}^d)$. We therefore have for a.e. $t \in [0,T]$, for all $\chi \in C^\infty(\mathbb{T}^d)$, the bound $|F_t(\chi)| \le \left\|f\right\|_{L^\infty L^p}\left\|\chi\right\|_{L^{p^*}}$. By the absolute continuity of $t\mapsto \left\langle F_t,\chi\right\rangle$ the bound in fact holds for all $t \in [0,T]$. Since $p\in(1,\infty]$ this implies further that $F_t$ can be extended to a function $\bar{f}(\cdot, t) \in L^p(\mathbb{T}^d)$ for all $t \in [0,T]$, and the absolute continuity of $\left\langle F_t, \chi \right\rangle$ implies $\bar{f} \in C_{\mathrm{weak-}*}^0L^p$ with (<ref>) holding for the representative $\bar{f}$ now for all $t \in [0,T]$, as required.

An important application of this is that for $t \in [0, T]$, and any $\phi \in C_c^\infty\left(\mathbb{T}^d \times [t,T)\right)$,

\begin{equation*}
\int_{\mathbb{T}^d\times[t,T]} f\left(\frac{\partial\phi}{\partial t}+u\cdot\nabla\phi +\nu\Delta \phi\right) dxdt = - \int_{\mathbb{T}^d} f(\cdot, t) \phi(\cdot, t) \;dx.
\end{equation*}

So if $f\in C_{\mathrm{weak-}*}^0L^p$ is a weak solution to (<ref>)/(<ref>) along $u$ with initial data $f_0$, then it is also a weak solution to (<ref>)/(<ref>) along $u$ on the time interval $[t, T]$ with initial data $f(\cdot, t)$.

This in particular allows us to say that $f\in C_{\mathrm{weak-}*}^0L^p$ is a weak solution to (<ref>)/(<ref>) with initial data $f_0$ if and only if it is a weak solution to (<ref>)/(<ref>) on $[0,t]$ with initial data $f_0$, and on $[t, T]$ with initial data $f(\cdot, t)$. That is we have transitivity of the transport equation.

Suppose $f_0 \in L^\infty(\mathbb{T}^d)$, and $u\in L^1([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d))$ is divergence-free in the distributional sense, then there exists a weak solution $f \in C_{\mathrm{weak-}*}^0 L^\infty$ to (<ref>)/(<ref>) along u with initial data $f_0$. Moreover, for all $p \in [1,\infty]$, $t \in [0,T]$,

\begin{gather}
\text{(Initial $L^p$-Inequality)\quad} \left\|f(\cdot, t)\right\|_{L^p(\mathbb{T}^d)} \le \left\|f_0\right\|_{L^p} \label{eqinitineq}, \\
\text{(Conservation of Mass)\quad} \int_{\mathbb{T}^d} f(\cdot, t) \; dx = \int_{\mathbb{T}^d} f_0 \; dx \label{eqconsmass}.
\end{gather}

More generally, for $f_0 \in L^q(\mathbb{T}^d)$, $u \in L^1([0,T];L^{q^*}(\mathbb{T}^d;\mathbb{R}^d))$, existence of a weak solution $f$ with $\left\|f\right\|_{L^\infty L^p} \le \left\|f_0\right\|_{L^p}$ (for all $p\in[1,\infty]$, permitting infinite values of the norms) follows from a standard approximation scheme (regularisation of $u$ and $f_0$), see <cit.> for details; though the proof considers only $\nu = 0$ (i.e. (<ref>)), and the spatial domain $\mathbb{R}^d$ instead of $\mathbb{T}^d$, an identical argument goes through here.

By assumption $u \in L^1L^1$, and so we may apply the above result with $q=\infty$.

By Theorem <ref> the solution is in $C_{\mathrm{weak-}*}^0 L^\infty$, and so the bound $\left\|f\right\|_{L^\infty L^p} \le \left\|f_0\right\|_{L^p}$ implies the Initial $L^p$-Inequality (<ref>).

Conservation of Mass (<ref>) follows from the Trace Formula (<ref>) in Theorem <ref> with $\phi \equiv 1$ on $\mathbb{T}^d \times [0,T]$.

In fact Conservation of Mass (<ref>) will hold for a.e. $t \in [0,T]$ for every weak solution on the torus, and not only those constructed in Theorem <ref>. This does not hold in $\mathbb{R}^d$.

Though a large topic of interest, we do not mention further regularity results for the transport equation, as they require further assumptions on $u$. Instead we point to <cit.> for an exposition of the Cauchy-Lipschitz theory, and other more standard results, including the DiPerna-Lions theory of renormalised weak solutions (introduced in <cit.>), and Ambrosio's extension of this well-posedness class to $u \in L^1([0,T];BV)$ (originally in <cit.>).

Next, we give the following well-posedness and regularity result for (<ref>). To illustrate the stark contrast between (<ref>) and its regularisation (<ref>) we show well-posedness for any $f_0 \in L^1(\mathbb{T}^d)$. When in addition $f_0$ is more integrable we may obtain further regularity. Therefore, when stating our main results, we shall later assume $f_0\in L^\infty$.

Suppose $u\in L^\infty L^\infty$, then for any initial data $f_0 \in L^1(\mathbb{T}^d)$ any weak solution (in the class $L^1 L^1$) to (<ref>) along $u$ with initial data $f_0$ is unique. Moreover this solution exists, $f\in C^0L^1$ with $\left\|f\right\|_{C^0L^1} \le \left\|f_0\right\|_{L^1(\mathbb{T}^d)}$ and becomes immediately bounded, $f \in C^0([\epsilon,T];C^0(\mathbb{T}^d))$ for all $\epsilon\in(0,T]$.

Furthermore, when $f_0 \in L^\infty$ we have the following additional regularity (for all $p\in[1,\infty)$)

\begin{gather}
f \in (C^0 L^p) \cap (L^2 H^1), \label{eqADEcont} \\
f \in C_{\mathrm{weak-}*}^0L^\infty, \label{eqADEcont2} \\
\text{($L^p$-Inequality)\quad} 0 \le s \le t \implies \left\|f(\cdot, t)\right\|_{L^p(\mathbb{T}^d)} \le \left\|f(\cdot, s)\right\|_{L^p(\mathbb{T}^d)}, \label{eqlpineq} \\
\text{(Energy Identity)\quad} \int_{\mathbb{T}^d} \left|f(x, t)\right|^2 \;dx + 2\nu \int_{\mathbb{T}^d \times [0,t]} \left|\nabla f(x,s)\right|^2\;dxds = \int_{\mathbb{T}^d} \left|f_0(x)\right|^2\;dx, \label{eqenergy} \\
\text{(Equicontinuity)\quad} \left\|\frac{\partial f}{\partial t}\right\|_{L^2H^{-1}} \le \left(\left\|u\right\|_{L^2 L^\infty} + \sqrt{\frac{\nu}{2}}\right)\left\|f_0\right\|_{L^2}. \label{eqequicont}
\end{gather}

The standard well-posedness result is for $f_0\in L^2$, and is done via energy estimates, see for example Chapter 7 in <cit.>. However, if we wish to obtain uniqueness in the class $L^1 L^1$ we must be more careful. [Given by its spatial Fourier series $\mathcal{F}(K_\nu(\cdot,t))(\xi) = e^{-\nu\xi^2 t}$.]Denote by $K_\nu \in C^\infty(\mathbb{T}^d\times(0,\infty))\cap L^\infty((0,\infty); L^1)$ the heat kernel for the heat equation with diffusivity $\nu$, that is for any $\phi \in C_c^\infty(\mathbb{T}^d \times (-\infty, \infty))$ we have

\begin{equation*}
\left(\frac{\partial}{\partial t} - \nu\Delta\right)(\phi *_{x,t} (K_\nu 1_{t>0})) = \phi,
\end{equation*}

where $*_{x,t}$ denotes convolution is space and time. Denoting by $\bar{K}_\nu \in C^\infty(\mathbb{T}^d\times(-\infty,0))$ the backwards heat kernel $\bar{K}_\nu(x,t) = K_\nu(-x,-t)$, then we have for the backwards heat equation, for any $\phi \in C_c^\infty(\mathbb{T}^d \times (-\infty, \infty))$

\begin{equation*}
\left(\frac{\partial}{\partial t} + \nu\Delta\right)(\phi *_{x,t} (\bar{K}_\nu 1_{t<0})) = -\phi.
\end{equation*}

For $\phi \in C_c^\infty(\mathbb{T}^d\times[0,T))$ we may take $\phi *_{x,t} (\bar{K}_\nu 1_{t<0})$ as a test function in (<ref>), which (by expanding out all convolutions) can be rewritten as

\begin{align*}
\int_{\mathbb{T}^d\times[0,T]} f\phi \; dx dt & = \int_{\mathbb{T}^d} f_0(x) \left(\phi *_{x,t} (\bar{K}_\nu 1_{t<0})\right)(x,0) \; dx + \int_{\mathbb{T}^d\times[0,T]} fu\cdot \left(\phi *_{x,t} (\nabla\bar{K}_\nu 1_{t<0})\right) \; dx dt \\
& = \int_{\mathbb{T}^d} \phi(x,t) \left(f_0 *_x (K_\nu(\cdot, t)) - (fu 1_{t \in [0,T]}) *_{x,t} (\nabla K_\nu 1_{t>0})\right)(x,t) \; dx dt,
\end{align*}

where $*_x$ denotes convolution over the spatial variable $x \in \mathbb{T}^d$ only.

That is we have shown, indeed for any weak solution $f$ to (<ref>) along $u$ with initial data $f_0 \in L^1(\mathbb{T}^d)$,

\begin{equation}
f = f_0 *_x K_\nu(\cdot, t) - (fu1_{t\in[0,T]}) *_{x,t} (\nabla K_\nu 1_{t>0}). \label{eqduhamel}
\end{equation}

When $f_0 = 0$ we have by Young's convolution inequality,

\begin{equation*}
\left\|f1_{t\in[0,\epsilon]}\right\|_{L^1 L^1} \le \left\|f1_{t\in[0,\epsilon]}\right\|_{L^1 L^1} \left\|u\right\|_{L^\infty L^\infty} \left\| \nabla K_\nu 1_{t\in(0,\epsilon]}\right\|_{L^1 L^1}.
\end{equation*}

It is straightforward to check that $\nabla K_\nu 1_{t \in (0,T]} \in L^1 L^1$, and so for $\epsilon$ small enough (depending only on $\left\|u\right\|_{L^\infty L^\infty}$ and $\nu$) the above implies that $f1_{t\in[0,\epsilon]} = 0$. Repeating the argument then shows that for all $n \in \mathbb{N}$, $f1_{[0,n\epsilon]}=0$, and so indeed $f=0$, proving uniqueness.

Existence of a weak solution $f \in L^\infty L^1$ with $\left\|f\right\|_{L^\infty L^1} \le \left\|f_0\right\|_{L^1}$ follows by solving the equation for mollified $u$ and $f_0$ as in Theorem <ref>. It can be checked that this produces a sequence of smooth functions $f_n$ converging to the solution $f$ weakly, and moreover, by the argument in <cit.>, with an a-priori bound on $\left\|f_n\right\|_{L^\infty([\epsilon,T];L^\infty(\mathbb{T}^d))}$ depending only on $\left\|f_0\right\|_{L^1(\mathbb{T}^d)}$, $\left\|u\right\|_{L^1L^\infty}$, $\epsilon > 0$, and $\nu$. We therefore have $f \in L^\infty L^1$ and $f\in L^\infty([\epsilon,T];L^\infty(\mathbb{T}^d))$ for any $\epsilon>0$. [We do not elaborate further on this point as it is not essential for the remainder of the paper.]The further regularity $f \in C^0 L^1$, and $f \in C^0([\epsilon,T];C^0(\mathbb{T}^d))$ for all $\epsilon\in(0,T]$, then follow from the formula (<ref>) and the regularity of $K_\nu$.

When in addition $f_0 \in L^2(\mathbb{T}^d)$ the regularised sequence $f_n$ can further be shown to converge in $(C^0L^2) \cap (L^2H^1)$. Statements (<ref>) and (<ref>) then follow from their counterparts for smooth $f_0$ and $u$. If $f_0 \in L^\infty(\mathbb{T}^d)$ we have, say by (<ref>), $f \in L^\infty L^\infty$, and hence by Theorem <ref> we prove (<ref>). The continuity $f \in C^0L^p$ then follows from $f\in C^0L^2$, for $p\in [1,2)$ by compactness of $\mathbb{T}^d$, and for $p\in(2,\infty)$ by interpolation with $f\in L^\infty L^\infty$. It remains to show (<ref>). From (<ref>) we have the bounds $\left\|f\right\|_{L^\infty L^2}, \sqrt{2\nu}\left\|\nabla f\right\|_{L^2L^2} \le \left\|f_0\right\|_{L^2}$, and hence for any $\phi \in C_c^\infty(\mathbb{T}^d\times(0,T))$, by the equation (<ref>),

\begin{align*}
\left|\int_{\mathbb{T}^d\times(0,T)}f\frac{\partial \phi}{\partial t} \; dxdt\right| & = \left|\int_{\mathbb{T}^d\times(0,T)}f \left(u\cdot\nabla\phi + \nu \Delta \phi\right) \; dxdt\right| \\
& \le \left\|f\right\|_{L^\infty L^2}\left\|u\right\|_{L^2L^\infty}\left\|\nabla\phi\right\|_{L^2L^2} + \nu\left\|\nabla f\right\|_{L^2L^2}\left\|\nabla\phi\right\|_{L^2L^2} \\
& \le \left\|f_0\right\|_{L^2}\left\|u\right\|_{L^2L^\infty}\left\|\nabla\phi\right\|_{L^2L^2} + \sqrt{\frac{\nu}{2}}\left\|f_0\right\|_{L^2}\left\|\nabla\phi\right\|_{L^2L^2},
\end{align*}

which proves (<ref>).

It is likely possible to extend the first part of Theorem <ref> to the Prodi-Serrin class <cit.>, $u \in L^pL^q$, $\frac{2}{p} + \frac{d}{q} \le 1$ ($d<q$), though we do not attempt to prove such a result here as it is immaterial to this paper.

In fact, such a result is already available in the stochastic setting due to Flandoli, see <cit.>.

When $f_0 \in L^1$ explicit decay rates of the $L^\infty(\mathbb{T}^d)$ norm (and other $L^p$ norms) are provable. See <cit.> for details when working on the whole space $\mathbb{R}^d$.	For general existence and regularity results concerning similar parabolic PDEs, we point to <cit.>, and also Chapter 7 in <cit.>.

§ CONTROLLING THE VANISHING VISCOSITY LIMIT

The main purpose of this section is to prove Theorem <ref> below, which allows us to construct bounded divergence-free vector fields in a way that permits control of the corresponding vanishing viscosity limit of (<ref>). This relies on two Propositions.

The first, Proposition <ref> below, gives a general criterion for which the vanishing viscosity limit of (<ref>) converges strongly to (<ref>). That is, for a suitable divergence-free vector field $u:\mathbb{T}^d\times[0,T]\to\mathbb{R}^d$, and small viscosity $\nu>0$, that solutions of (<ref>) along $u$ are well-approximated by a weak solution of (<ref>) along $u$. This result is a generalisation of the Selection Theorem in <cit.>.

The second, Proposition <ref> below, uses a similar argument to show, for fixed viscosity $\nu>0$, how solutions of (<ref>) depend little on the small spatial scales of the vector field $u$. We quantify these scales through the weak-$*$ topology of vector fields in $L^\infty([0,T];L^\infty(\mathbb{T}^d;\mathbb{R}^d))$. The intuition is that the viscosity `blurs' these small spatial scales.

The key idea of Theorem <ref> is then that solving (<ref>) with reduced viscosity $\nu>0$, is akin to adding small spatial scales while solving (<ref>).

The advantage of solving (<ref>) is then that Lagrangian solutions (Definition <ref>) can be designed rather explicitly.

Consider a vector field $u \in L^1\left( [0,T]; L^1(\mathbb{T}^d;\mathbb{R}^d)\right)$ with $\nabla\cdot u=0$ in the distributional sense. Fix some initial data $f_0 \in L^\infty(\mathbb{T}^d)$.

[Since TE is linear this implies uniqueness for all initial data in $L^\infty$.]Suppose that there is a unique weak solution $f$ (in the class $L^\infty L^\infty$) to (<ref>) along $u$ with initial data $f_0$, [This is not uncommon, see Remark 4.3 in <cit.>, and is true for example when $u \in L^1([0,T];BV)$ in <cit.>.]and that additionally $f$ is a renormalised weak solution (Definition <ref>).

For each $\nu > 0$ denote by $f^\nu$ any weak solution to (<ref>) along $u$ with initial data $f_0$. Suppose in addition that $f^\nu \in C_{\mathrm{weak-}*}^0L^\infty$ and satisfies the Initial $L^p$-Inequality (<ref>) for all $p \in [1,\infty]$. (This would be the case if say $u \in L^\infty L^\infty$, see Theorem <ref>.)

Then, for each $p \in [1,\infty)$, $f^\nu\xrightarrow{\nu \to 0}f$ converges strongly in $L^p L^p$ and also in weak-$*$ $L^\infty L^\infty$.

[By compactness of $\mathbb{T}^d$, and interpolation with the bound in $L^\infty L^\infty$, $f \in C^0L^1$ is equivalent to $f \in C^0L^p$ for any $p \in [1,\infty)$.]If additionally $f \in C^0 L^1$, then for each $p \in [1,\infty)$, $f^\nu\xrightarrow{\nu \to 0}f$ also converges strongly in $L^\infty L^p$.

Suppose that $f^\nu$ does not converge in weak-$*$ $L^\infty L^\infty$ to $f$ as $\nu\to0$. Then there exists some $g \in L^1 L^1$, and a sequence $\nu_i \xrightarrow{i\to\infty}0$ and $c>0$ such that for all $i\in\mathbb{N}$

\begin{equation}
\left|\int_{\mathbb{T}^d\times[0,T]}(f^{\nu_i}-f)g\;dxdt\right|\ge c. \label{eqcontr1}
\end{equation}

By the Initial $L^p$-Inequality (<ref>) $f^{\nu_i}$ is uniformly bounded for all $i\in\mathbb{N}$ in $L^\infty L^\infty$, and so by taking a subsequence if necessary, we may assume $f^{\nu_i}\xrightharpoonup{i\to\infty}\bar{f}$ converges in weak-$*$ $L^\infty L^\infty$ to some $\bar{f} \in L^\infty L^\infty$. Then, for any $\phi \in C_c^\infty(\mathbb{T}^d \times [0,T))$,

\begin{align*}
& \int_{\mathbb{T}^d\times[0,T)} \bar{f} \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi\right) \; dxdt \\
& = \lim_{i\to\infty} \int_{\mathbb{T}^d\times[0,T)} f^{\nu_i} \left(\frac{\partial \phi}{\partial t} + u\cdot\nabla \phi + \nu_i \Delta \phi\right) \; dxdt \\
& = - \int_{\mathbb{T}^d} f_0 \phi_0 \; dx,
\end{align*}

and so the limit $\bar{f}$ is a weak solution to (<ref>) along $u$ with initial data $f_0$. Moreover, it is in $L^\infty L^\infty$ so by assumption must be the unique weak solution $f$, contradicting (<ref>). Therefore $f^\nu \xrightarrow{\nu\to0}f$ converges in weak-$*$ $L^\infty L^\infty$.

Recall by Theorem <ref> that $f, f^\nu \in C_{\mathrm{weak-}*}^0L^\infty$.

Since by assumption $f^\nu$ satisfies the Initial $L^p$-Inequality (<ref>) for all $p \in [1,\infty]$, we may bound

\begin{equation}\label{eq10.5}
\left\|f^\nu\right\|_{L^p L^p} \le T^{\frac{1}{p}}\left\|f_0\right\|_{L^p}.
\end{equation}

If $p\in(1,\infty]$, weak-$*$ convergence in $L^\infty L^\infty$ implies weak-$*$ convergence in $L^p L^p$. [This is a standard result following from uniform convexity of $L^p(\mathbb{T}^d)$ for $p \in (1,\infty)$.]Whenever $p \in (1,\infty)$, weak-$*$ convergence in $L^pL^p$ is also strong in $L^pL^p$ if and only if $\underset{\nu\to0}{\limsup}\left\|f^\nu\right\|_{L^pL^p} \le \left\|f\right\|_{L^pL^p}$.

To show that this is satisfied we use that $f \in C_{\mathrm{weak-}*}^0L^\infty$ is a renormalised weak solution to (<ref>) (Definition <ref>). [Here $a\wedge b = \min\left\{a,b\right\}$.]Let $M \in \mathbb{N}$ and $\beta(x) = M\wedge|x|^p$, then $\beta(f)$ is a weak solution to (<ref>) along $u$ with initial data $\beta(f_0) \in L^\infty$. Taking $\phi \equiv 1$ in the Trace Formula (<ref>) shows that there exists a subset $E_M\subset[0,T]$ with zero Lebesgue-measure, such that for all $t \in [0,T]\setminus E_M$ we have,

\begin{equation*}
\int_{\mathbb{T}^d} M \wedge |f(\cdot, t)|^p \; dx = \int_{\mathbb{T}^d} M \wedge |f_0|^p \; dx.
\end{equation*}

In particular, the above holds for all $t \in [0,T]\setminus \bigcup_{M\in\mathbb{N}}E_M$. By the Lebesgue monotone convergence Theorem, taking $M\to\infty$ shows that $\left\|f(\cdot, t)\right\|_{L^p(\mathbb{T}^d)} = \left\|f_0\right\|_{L^p}$ for all $t \in [0,T]\setminus \bigcup_{M\in\mathbb{N}}E_M$, which implies $\left\|f\right\|_{L^p L^p} = T^{\frac{1}{p}}\left\|f_0\right\|_{L^p}$. Combined with (<ref>) this implies that $\underset{\nu\to0}{\limsup}\left\|f^\nu\right\|_{L^pL^p} \le \left\|f\right\|_{L^pL^p}$, and hence $f^\nu \xrightarrow{\nu\to0} f$ in $L^pL^p$ as required.

Convergence of $f^\nu \xrightarrow{\nu\to0} f$ in $L^1 L^1$ follows from convergence in $L^p L^p$ for any $p \in (1,\infty)$ and the compactness of the domain $\mathbb{T}^d \times [0,T]$.

We now assume that $f \in C^0 L^1$ and wish to upgrade to convergence in $L^\infty L^p$ for each $p \in [1,\infty)$. The idea is to use the Trace Formula (<ref>) to show for each $t \in [0,T]$ that $f^\nu(\cdot, t) \xrightharpoonup{\nu\to0} f(\cdot, t)$ converges in weak-$*$ $L^\infty$, and then upgrade to (uniform in time) strong convergence in $L^p(\mathbb{T}^d)$ by convergence of the norm $\left\|f^\nu(\cdot,t)\right\|_{L^p(\mathbb{T}^d)}$ for $p \in (1,\infty)$.

[Convergence in $L^\infty L^p$ for $p \in (1,\infty)$ can also be shown directly, though is more technical. This is useful if say $f_0 \notin L^2(\mathbb{T}^d)$.]Since the $L^2(\mathbb{T}^d)$-inner product makes life easier, we will only prove convergence in $L^\infty L^2$, and notice that convergence in $L^\infty L^p$ follows for $p \in [1,2)$ by compactness of $\mathbb{T}^d$, and for $p\in(2,\infty)$ by interpolation with the existing uniform bound in $L^\infty L^\infty$.

We have already shown that $f^\nu \xrightarrow{\nu\to0} f$ in $L^1L^1$, with uniform bound in $L^\infty L^\infty$. Suppose that $uf^\nu$ does not converge strongly in $L^1 L^1$ to $uf$ as $\nu\to0$. Then there exists a sequence $\nu_i \xrightarrow{i\to\infty}0$ and $c>0$ such that for all $i\in\mathbb{N}$

\begin{equation}
\left\|uf^{\nu_i} - uf\right\|_{L^1 L^1} \ge c \label{eqcontr2}.
\end{equation}

Now $f^{\nu_i} \xrightarrow{i\to\infty} f$ strongly in $L^1 L^1$, and so by taking a further subsequence if necessary, we may assume that $f^{\nu_i} \xrightarrow{i\to\infty} f$ point-wise a.e. in $\mathbb{T}^d\times[0,T]$. But then $uf^{\nu_i} \xrightarrow{i\to\infty} uf$ converge point-wise a.e. in $\mathbb{T}^d\times[0,T]$, and are also uniformly bounded for all $i\in\mathbb{N}$ in $L^1 L^1$ by $\left\|u\right\|_{L^1 L^1}\left\|f_0\right\|_{L^\infty(\mathbb{T}^d)}$, and so the dominated convergence Theorem yields a contradiciton to (<ref>). Therefore the product $uf^\nu \xrightarrow{\nu\to0} uf$ converges strongly in $L^1 L^1$.

We now use that $f \in C^0 L^1$, and therefore $f\in C^0 L^2$ (by interpolation with the existing bound $f \in L^\infty L^\infty$), to take a smooth approximation $\phi_\epsilon \in C^\infty(\mathbb{T}^d\times[0,T])$, such that $\left\|\phi_\epsilon-f\right\|_{L^\infty L^2} \le \epsilon$. Then by the Trace Formula (<ref>), and that $\left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)} \le \left\|f_0\right\|_{L^2} = \left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}$ for a.e. $t\in[0,T]$, we may write for a.e. $t \in [0,T]$

\begin{align*}
& \left\|f^\nu(\cdot, t) - f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 \\
& \qquad = \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 - 2 \int_{\mathbb{T}^d} f^\nu(\cdot, t) f(\cdot, t)\;dx \nonumber \\
& \qquad \le 2\left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2(\mathbb{T}^d)} - 2 \int_{\mathbb{T}^d} f^\nu(\cdot, t) \phi_\epsilon(\cdot, t)\;dx \nonumber \\
& \qquad \le
\begin{aligned}[t]
& 2\left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2(\mathbb{T}^d)} - 2\int_{\mathbb{T}^d}f_0\phi_\epsilon(\cdot, 0)\;dx \\
& - 2\int_{\mathbb{T}^d \times [0,t]}f^\nu \left(\frac{\partial \phi_\epsilon}{\partial t} + u \cdot \nabla \phi_\epsilon + \nu \Delta \phi_\epsilon \right) \;dxdt
\end{aligned} \nonumber \\
& \qquad \le
\begin{aligned}[t]
& 2\left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2(\mathbb{T}^d)} - 2\int_{\mathbb{T}^d}f_0\phi_\epsilon(\cdot, 0)\;dx \\
& - 2\int_{\mathbb{T}^d \times [0,t]}f \left(\frac{\partial \phi_\epsilon}{\partial t} + u \cdot \nabla \phi_\epsilon\right) \;dxdt \\
& + 2\left\|f^\nu - f\right\|_{L^1L^1}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^\infty L^\infty} + 2\nu\left\|f^\nu\right\|_{L^1L^1}\left\|\Delta \phi_\epsilon\right\|_{L^\infty L^\infty} \\
& + 2\left\|uf^\nu - uf\right\|_{L^1L^1}\left\|\nabla \phi_\epsilon\right\|_{L^\infty L^\infty}
\end{aligned} \nonumber \\
& \qquad =
\begin{aligned}[t]
& 2\left\|f(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2(\mathbb{T}^d)} - 2\int_{\mathbb{T}^d}f(\cdot, t)\phi_\epsilon(\cdot, t)\;dx \\
& + 2\left\|f^\nu - f\right\|_{L^1L^1}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^\infty L^\infty} + 2\nu\left\|f^\nu\right\|_{L^1L^1}\left\|\Delta \phi_\epsilon\right\|_{L^\infty L^\infty} \\
& + 2\left\|uf^\nu - uf\right\|_{L^1L^1}\left\|\nabla \phi_\epsilon\right\|_{L^\infty L^\infty}
\end{aligned} \nonumber \\
& \qquad \le 4\epsilon\left\|f_0\right\|_{L^2(\mathbb{T}^d)} + C_\epsilon\left(\left\|f^\nu - f\right\|_{L^1L^1} + \nu\left\|f^\nu\right\|_{L^1L^1} + \left\|uf^\nu - uf\right\|_{L^1L^1}\right),
\end{align*}

with $\epsilon>0$ arbitrary, and $C_\epsilon$ a constant that depends only on $\epsilon$ and $f$, and not on $t \in [0,T]$. Then since $f^\nu \xrightarrow{\nu \to 0} f$, and $uf^\nu \xrightarrow{\nu \to 0} uf$ in $L^1 L^1$ we see that $f^\nu \xrightarrow{\nu \to 0} f$ in $L^\infty L^2$ as required.

Consider a sequence of uniformly bounded, divergence-free vector fields $u_n \in L^\infty \left( [0,T];L^\infty(\mathbb{T}^d;\mathbb{R}^d)\right)$, such that $u_n \xrightharpoonup{n\to\infty} u$ converges in weak-$*$ $L^\infty L^\infty$ to some $u \in L^\infty L^\infty$. Fix some initial data $f_0 \in L^\infty(\mathbb{T}^d)$.

For each $n \in\mathbb{N}$, and $\nu > 0$ denote by $f^{n,\nu}$ the unique (by Theorem <ref>) weak solution to (<ref>) along $u_n$ with initial data $f_0$. Similarly denote by $f^\nu$ the unique weak solution to (<ref>) along $u$ with initial data $f_0$. Then for each $0<a\le b$, $p \in [1,\infty)$

\begin{equation*}
\sup_{\nu\in[a,b]} \left\|f^{n,\nu}-f^\nu\right\|_{(L^2H^1) \cap (L^\infty L^p)} \xrightarrow{n \to \infty} 0.
\end{equation*}

The case $p \in (2,\infty)$ follows from the same result for $p=2$ by interpolation with the existing uniform bound on $f^{n,\nu}$, $f^\nu$ in $L^\infty L^\infty$ from the $L^p$-Inequality (<ref>), while for the case $p \in (1,2)$ it follows from the case for $p=2$ by compactness of $\mathbb{T}^d$. Therefore, we may restrict to $p=2$.

Assume to the contrary that for some $0<a\le b$ that there exists $c > 0$ and sequences $\{n_i\}_{i \in \mathbb{N}}$, $\{\nu_i\}_{i \in \mathbb{N}}$ with $n_i \in \mathbb{N}$ increasing, $a\le\nu_i \le b$, such that for all $i \in\mathbb{N}$

\begin{equation}\label{eqcontr3}
\left\|f^{n_i,\nu_i} - f^{\nu_i}\right\|_{L^2H^1} + \left\|f^{n_i,\nu_i} - f^{\nu_i}\right\|_{L^\infty L^2} \ge c.
\end{equation}

By taking a subsequence if necessary we may assume $\nu_i \xrightarrow{i\to\infty} \nu$ for some $\nu\in[a,b]$.

Since $\nu_i$ are bounded above and below, by Energy Identity (<ref>) $f^{n_i, \nu_i}$, $f^{\nu_i}$ are uniformly bounded for all $i\in\mathbb{N}$ in $L^2 H^1$, and by Equicontinuity (<ref>) $\frac{\partial}{\partial t}f^{n_i, \nu_i}$, $\frac{\partial}{\partial t}f^{\nu_i}$ are uniformly bounded for all $i\in\mathbb{N}$ in $L^2 H^{-1}$.

Since $H^1 \Subset L^2 \subset H^{-1}$ we may apply the Aubin-Lions compactness Lemma (see Chapter 3, Theorem 2.1 in <cit.>) to deduce that the set $\{f \in L^2H^1 : \left\|\frac{\partial f}{\partial t}\right\|_{L^2 H^{-1}} \le C\}$ is compactly embedded into $L^2 L^2$.

Hence, again taking a subsequence if necessary, we may assume that both $f^{n_i, \nu_i}$, $f^{\nu_i}$ converge, to some limits, strongly in $L^2L^2$ (and weakly in $L^2H^1$) as $i\to\infty$.

We next show that both $f^{n_i, \nu_i}$, and $f^{\nu_i}$ converge to $f^\nu$ strongly in $L^2H^1$ and $L^\infty L^2$ as $i\to\infty$, contradicting the assumption (<ref>).

We will only show the required convergence for $f^{n_i, \nu_i} \xrightarrow {i \to \infty} f^{\nu}$, since the same result for $f^{\nu_i}$ follows the same proof by additionally assuming that $u_{n_i}$ are a constant sequence $u_{n_i} = u$ for all $i \in \mathbb{N}$.

Since $u_{n_i} \xrightharpoonup{i \to \infty} u$ are uniformly bounded and converge in weak-$*$ $L^\infty L^\infty$, and as we have already shown that $f^{n_i,\nu_i}$ converges to some $\bar{f}$ strongly in $L^2L^2$ as $i\to\infty$, then the product $f^{n_i, \nu_i}u_{n_i}$ converges to $\bar{f}u$ weakly in $L^2 L^2$ as $i\to\infty$.

Then by the weak formulation of (<ref>), for any $\phi \in C_c^\infty(\mathbb{T}^d \times [0,T))$, one has

\begin{multline*}
\int_{\mathbb{T}^d\times[0,T]} \bar{f} \left( \frac{\partial \phi}{\partial t} + u\cdot\nabla \phi + \nu \Delta \phi\right) \; dxdt \\
\begin{aligned}
& = \lim_{i\to\infty} \int_{\mathbb{T}^d\times[0,T]} f^{n_i,\nu_i}\left(\frac{\partial \phi}{\partial t} + u_{n_i}\cdot\nabla \phi + \nu_i \Delta \phi\right) \; dxdt \\
& = - \int_{\mathbb{T}^d} f_0 \phi_0 \; dx.
\end{aligned}
\end{multline*}

Consequently, we see that $\bar{f}$ is a weak solution to (<ref>) along $u$ with initial data $f_0$. By Theorem <ref> this solution is unique and so must be $f^\nu$.

We are left to upgrade the convergence $f^{n_i,\nu_i} \xrightarrow{i\to\infty} \bar{f}$ from strong in $L^2 L^2$ and weak in $L^2 H^1$, to strong in $L^2 H^1$ and strong in $L^\infty L^2$.

In light of the Trace Formula (<ref>), and the uniform bound on $f^{n_i,\nu_i}$ for all $i\in\mathbb{N}$ in $C^0L^2$ (by Theorem <ref>), strong convergence $f^{n_i,\nu_i} \xrightarrow{i\to\infty} f^\nu$ in $L^2 L^2$ implies $f^{n_i,\nu_i}(\cdot, T) \xrightharpoonup{i\to\infty} f^\nu(\cdot, T)$ converges weakly in $L^2(\mathbb{T}^d)$ (note that $f^{n_i,\nu_i}$, $f^{\nu}$ are uniformly bounded in $C^0L^2$ by Theorem <ref>). In particular

\begin{equation*}
\liminf_{i\to\infty} \int_{\mathbb{T}^d} \left|f^{n_i,\nu_i}(\cdot, T)\right|^2 \; dx \ge \int_{\mathbb{T}^d} \left|f^\nu(\cdot, T)\right|^2 \; dx.
\end{equation*}

In light of the Energy Identity (<ref>) this implies

\begin{equation*}
\limsup_{i\to\infty} \int_{\mathbb{T}^d \times [0,T]} \left|\nabla f^{n_i,\nu_i}\right|^2 \; dx dt \le \int_{\mathbb{T}^d \times [0,T]} \left|\nabla f^\nu\right|^2 \; dx dt.
\end{equation*}

Since also $f^{n_i,\nu_i} \xrightharpoonup{i\to\infty} f^\nu$ weakly in $L^2H^1$ we have

\begin{equation*}
\liminf_{i\to\infty} \int_{\mathbb{T}^d \times [0,T]} \left( \left|f^{n_i,\nu_i}\right|^2 + \left|\nabla f^{n_i,\nu_i}\right|^2 \right) \; dx dt \ge \int_{\mathbb{T}^d \times [0,T]} \left( \left|f^\nu\right|^2 + \left|\nabla f^\nu\right|^2 \right) \; dx dt.
\end{equation*}

However, since $f^{n_i,\nu_i} \xrightarrow{i\to\infty} f^\nu$ strongly in $L^2L^2$, from the above we must have convergence of the norms $\left\|f^{n_i,\nu_i}\right\|_{L^2H^1}^2 \xrightarrow{i\to\infty} \left\|f^\nu\right\|_{L^2H^1}^2$. Thus the weak convergence $f^{n_i,\nu_i} \xrightharpoonup{i\to\infty} f^\nu$ in $L^2H^1$ is in fact strong, as required.

To extend to strong convergence in $L^\infty L^2$, we use the fact that $f^\nu \in C^0L^2$ to take a smooth approximation $\phi_\epsilon \in C^\infty(\mathbb{T}^d\times [0,T])$, such that $\left\|\phi_\epsilon - f^\nu\right\|_{L^\infty L^2} \le \epsilon$. Then by the Trace Formula (<ref>), and the Energy Identity (<ref>), for all $t \in [0,T]$ we see that

\begin{align*}
& \left\|f^{n_i,\nu_i}(\cdot, t) - f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 \\
& \qquad =
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 - 2 \int_{\mathbb{T}^d} f^{n_i,\nu_i}(\cdot, t) f^\nu(\cdot, t)\;dx
\end{aligned} \\
& \qquad \le
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2} - 2 \int_{\mathbb{T}^d} f^{n_i,\nu_i}(\cdot, t) \phi_\epsilon(\cdot, t)\;dx
\end{aligned} \\
& \qquad =
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2} - 2 \int_{\mathbb{T}^d} f_0 \phi_\epsilon(\cdot, 0)\;dx \\
& - 2\int_{\mathbb{T}^d \times [0,t]}f^{n_i,\nu_i} \left(\frac{\partial \phi_\epsilon}{\partial t} + u_{n_i} \cdot \nabla \phi_\epsilon + \nu_i \Delta \phi_\epsilon \right) \;dxdt
\end{aligned} \\
& \qquad \le
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2} - 2 \int_{\mathbb{T}^d} f_0 \phi_\epsilon(\cdot, 0)\;dx \\
& - 2\int_{\mathbb{T}^d \times [0,t]}f^\nu \left(\frac{\partial \phi_\epsilon}{\partial t} + u \cdot \nabla \phi_\epsilon + \nu \Delta \phi_\epsilon \right) \;dxdt \\
& + 2\left\|f^{n_i,\nu_i} - f^\nu\right\|_{L^2L^2}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^2 L^2} + 2|\nu_i-\nu|\left\|f^{n_i,\nu_i}\right\|_{L^2L^2}\left\|\Delta \phi_\epsilon\right\|_{L^2 L^2} \\
& + 2\left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right|
\end{aligned} \\
& \qquad =
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 2\epsilon\left\|f_0\right\|_{L^2} - 2 \int_{\mathbb{T}^d} f^\nu(\cdot, t) \phi_\epsilon(\cdot, t)\;dx \\
& + 2\left\|f^{n_i,\nu_i} - f^\nu\right\|_{L^2L^2}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^2 L^2} + 2|\nu_i-\nu|\left\|f^{n_i,\nu_i}\right\|_{L^2L^2}\left\|\Delta \phi_\epsilon\right\|_{L^2 L^2} \\
& + 2\left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right|
\end{aligned} \\
& \qquad \le
\begin{aligned}[t]
& \left\|f^{n_i,\nu_i}(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 - \left\|f^\nu(\cdot, t)\right\|_{L^2(\mathbb{T}^d)}^2 + 4\epsilon\left\|f_0\right\|_{L^2} \\
& + 2\left\|f^{n_i,\nu_i} - f^\nu\right\|_{L^2L^2}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^2 L^2} + 2|\nu_i-\nu|\left\|f^{n_i,\nu_i}\right\|_{L^2L^2}\left\|\Delta \phi_\epsilon\right\|_{L^2 L^2} \\
& + 2\left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right|
\end{aligned} \\
& \qquad =
\begin{aligned}[t]
& 4\epsilon\left\|f_0\right\|_{L^2} - \int_{\mathbb{T}^d \times [0,t]} \left(2\nu_i \left|\nabla f^{n_i,\nu_i}\right|^2 - 2\nu \left|\nabla f^\nu\right|^2\right) \;dxdt  \\
& + 2\left\|f^{n_i,\nu_i} - f^\nu\right\|_{L^2L^2}\left\|\frac{\partial \phi_\epsilon}{\partial t}\right\|_{L^2 L^2} + 2|\nu_i-\nu|\left\|f^{n_i,\nu_i}\right\|_{L^2L^2}\left\|\Delta \phi_\epsilon\right\|_{L^2 L^2} \\
& + 2\left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right|
\end{aligned} \\
& \qquad \le
\begin{aligned}[t]
& 4\epsilon\left\|f_0\right\|_{L^2} + 2\left\|{\nu_i}^\frac{1}{2}f^{n_i,\nu_i} - {\nu}^\frac{1}{2}f^\nu\right\|_{L^2 H^1}\left({\nu_i}^\frac{1}{2}\left\|f^{n_i,\nu_i}\right\|_{L^2 H^1} + {\nu}^\frac{1}{2}\left\|f^\nu\right\|_{L^2 H^1}\right) \\
& + C_\epsilon\left(\left\|f^{n_i,\nu_i} - f^\nu\right\|_{L^2L^2} + |\nu_i-\nu|\left\|f^{n_i,\nu_i}\right\|_{L^2L^2}\right) \\
& + 2\left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right|,
\end{aligned}
\end{align*}

with $\epsilon>0$ arbitrary, and $C_\epsilon$ a constant that depends only on $\phi_\epsilon$, and not on $t \in [0,T]$.

Therefore by the convergence $\nu_i \xrightarrow{i\to\infty} \nu$, and $f^{n_i,\nu_i} \xrightarrow{i\to\infty} f^\nu$ strongly in $L^2H^1$, to show strong convergence of $f^{n_i,\nu_i}\xrightarrow{i\to\infty}f^\nu$ in $L^\infty L^2$ we need only show, for fixed $\phi_\epsilon$

\begin{equation*}
\sup_{t \in[0,T]} \left|\int_{\mathbb{T}^d \times [0,t]} (f^{n_i,\nu_i}u_{n_i} - f^\nu u)\cdot \nabla \phi_\epsilon \; dxdt \right| \xrightarrow{i \to \infty} 0.
\end{equation*}

Suppose not, then there exists a sequence $t_i \in [0,T]$ for $i\in\mathbb{N}$ such that

\begin{equation}
\left|\int_{\mathbb{T}^d\times[0,t_i]}\left(f^{n_i,\nu_i}u_{n_i} - f^\nu u\right)\cdot\nabla\phi_\epsilon \; dxdt \right| \ge c, \label{eqcontr}
\end{equation}

for some $c>0$.

By compactness of $[0,T]$, and taking a subsequence if necessary, we may further assume that $t_i \xrightarrow{i\to\infty} \bar{t}$ for some $\bar{t}\in[0,T]$. However, there is strong convergence $1_{t\in[0,t_i]}f^{n_i,\nu_i}, \; 1_{t\in[0,t_i]}f^\nu \xrightarrow{i\to\infty} 1_{t\in[0,\bar{t}]}f^\nu$ in $L^2 L^2$, and boundedness of the sequence $u_{n_i}$ in $L^\infty L^\infty$, and weak-$*$ convergence $u_{n_i} \xrightharpoonup{i\to\infty} u$ in $L^\infty L^\infty$. Therefore, the products $1_{t\in[0,t_i]}f^{n_i,\nu_i} u_{n_i} - 1_{t\in[0,t_i]}f^\nu u \xrightharpoonup{i\to\infty} 0$ converge weakly in $L^2 L^2$, contradicting (<ref>), and hence proving the claim.

We now give the following corollary of the previous two propositions, which allows us to construct bounded divergence-free vector fields $u_\infty$ in a way that permits control of the corresponding vanishing viscosity limit of (<ref>).

Fix some $M>0$, and a metric $d_*$ inducing the weak-$*$ topology on

\begin{equation*}
X = \left\{u\in L^\infty \left( [0,T];L^\infty(\mathbb{T}^d;\mathbb{R}^d)\right) : \left\|u\right\|_{L^\infty L^\infty}\le M \right\}.
\end{equation*}

[Note that $Y$ is non-empty, e.g. $0\in Y$ and any divergence-free, in the distribution sense, vector field in $X$ and $L^1W^{1,1}$ is in $Y$, see <cit.>. Moreover, by the work of Ambrosio in <cit.>, we may relax $W^{1,1}$ to the space of bounded variation $BV$.]Let $Y\subset X$ be the set of all divergence-free vector fields admitting a unique renormalised weak solution (unique in the class of all $L^\infty L^\infty$ weak solutions) to (<ref>) for any initial data $f_0 \in L^\infty(\mathbb{T}^d)$.

Let $\left\{u_i\right\}_{i\in\mathbb{N}}\subset Y$, $f_0 \in L^\infty(\mathbb{T}^d)$. For each $n\in\mathbb{N}$, and $\nu > 0$, denote by $f^{n,\nu}$, respectively $f^n$, the unique weak solution to (<ref>), respectively (<ref>), along $u_n$ with initial data $f_0$. Then,

*  For all $n\in\mathbb{N}$, there exists $\nu_n>0$, $\epsilon_n>0$ depending only on $\left\{u_i\right\}_{i=1}^n$ (and in particular not on $f_0$), with $\nu_n\xrightarrow{n\to\infty}0$ monotonically, such that the following hold true:

*  For all $p\in[1,\infty)$,

\begin{equation*}
\sup_{0<\nu\le\nu_n}\left\|f^{n,\nu}-f^n\right\|_{L^\infty L^p}\xrightarrow{n\to\infty}0.
\end{equation*}

*  If $d_*(u_{n+1},u_n) \le \epsilon_n$ for all $n\in\mathbb{N}$, then $u_n \xrightharpoonup{n\to\infty} u_\infty$ converges in weak-$*$ $L^\infty L^\infty$ to some divergence-free vector field $u_\infty \in L^\infty L^\infty$,

*  and if we denote by $f^{\infty,\nu}$ the unique weak solution to (<ref>) along $u_\infty$ with initial data $f_0$, then for all $p \in [1,\infty)$,

\begin{equation*}
\sup_{\nu_n \le \nu\le\nu_1}\left\|f^{n,\nu}-f^{\infty,\nu}\right\|_{L^\infty L^p}\xrightarrow{n\to\infty}0.
\end{equation*}

*  In particular, for all $p\in[1,\infty)$,

\begin{equation*}
\left\|f^{\infty,\nu_n}-f^n\right\|_{L^\infty L^p}\xrightarrow{n\to\infty}0.
\end{equation*}

By separability of $L^1(\mathbb{T}^d)$, take a countable dense sequence $\{f_0^m\}_{m \in\mathbb{N}}$ in $L^1(\mathbb{T}^d)$, satisfying $f_0^m \in L^\infty(\mathbb{T}^d)$.

For all $m\in\mathbb{N}$, $n \in \mathbb{N}$, and $\nu > 0$, denote by $f^{m;n,\nu}$, respectively $f^{m;n}$, the unique weak solution to (<ref>), respectively (<ref>), along $u_n$ with initial data $f^m_0$. If in addition $u_n \xrightharpoonup{n\to\infty} u_\infty$ in weak-$*$ $L^\infty L^\infty$, denote by $f^{m;\infty,\nu}$ the unique weak solution to (<ref>) along $u_\infty$ with initial data $f_0^m$.

For the given $f_0 \in L^\infty(\mathbb{T}^d)$, for each $n \in \mathbb{N}$, and $\nu > 0$, denote by $f^{n,\nu}$, respectively $f^n$, the unique weak solution to (<ref>), respectively (<ref>), along $u_n$ with initial data $f_0$. If in addition $u_n \xrightharpoonup{n\to\infty} u_\infty$ in weak-$*$ $L^\infty L^\infty$, denote by $f^{\infty,\nu}$ the unique weak solution to (<ref>) along $u_\infty$ with initial data $f_0$.

Notice, by linearity of (<ref>), (<ref>), and the Initial $L^p$-Inequality (<ref>), that for each $m \in \mathbb{N}$, $n\in\mathbb{N}$, $\nu >0$, we have

\begin{gather}
\left\|f^{n,\nu}-f^{m;n,\nu}\right\|_{L^\infty L^1} \le\left\|f_0 - f^m_0\right\|_{L^1}, \label{eqdensity1} \\
\left\|f^{n} - f^{m;n}\right\|_{L^\infty L^1} \le\left\|f_0 - f^m_0\right\|_{L^1}, \label{eqdensity2} \\
\left\|f^{\infty,\nu}-f^{m;\infty,\nu}\right\|_{L^\infty L^1}\le\left\|f_0 - f^m_0\right\|_{L^1}. \label{eqdensity3}
\end{gather}

We will use this and $L^1(\mathbb{T}^d)$-density to only consider the countable set of initial data $\{f_0^m\}_{m\in\mathbb{N}}$.

Given $n\in\mathbb{N}$ and $\{u_i\}_{i=1}^n\subset Y$ we need to find $\nu_n>0$, $\epsilon_n>0$ such that (<ref>)-(<ref>) hold. For the purpose of the proof we additionally find $\delta_n>0$ such that, for all $m\in\{1,...,n\}$, $\nu\in[\nu_n,\nu_1]$, and $w\in Y$ with $d_*(w,u_n)\le\delta_n$,

\begin{equation}\label{eqdelta}
\left\|f^{m;n,\nu} - g^{m;\nu}\right\|_{L^\infty L^1} \le \frac{1}{n},
\end{equation}

where $g^{m,\nu}$ is the unique weak solution to (<ref>) along $w$ with initial data $f_0^m$.

We do so by induction. Given $\{u_i\}_{i=1}^n$ as in the statement of the theorem assume $\{(\nu_i, \epsilon_i, \delta_i)\}_{i=1}^{n-1}$ are already chosen (an empty set if $n=1$).

By applying Proposition <ref> for each initial data $\{f_0^m\}_{m=1}^n$ there exists some $\nu_n > 0$ such that for all $m\in\{1,...,n\}$, $\nu \in (0,\nu_n]$,

\begin{equation}\label{eqnu}
\left\|f^{m;n,\nu} - f^{m;n} \right\|_{L^\infty L^1} \le \frac{1}{n}.
\end{equation}

One may choose $\nu_n$ to be smaller, so that $\nu_n \le \frac{1}{n}$, and if $n\ne1$, $\nu_n < \nu_{n-1}$.

We next find $\delta_n > 0$ satisfying (<ref>). Assume to the contrary that no such $\delta_n$ exists, then there is a sequence of divergence-free vector fields $\{w_k\}_{k \in \mathbb{N}} \subset Y$ with $w_k \xrightharpoonup{k\to\infty} u_n$ converging in weak-$*$ $L^\infty L^\infty$, which violate Proposition <ref> for at least one of the initial data $\{f_0^m\}_{m=1}^n$, where we have substituted $a=\nu_n$, $b = \nu_1$ into Proposition <ref>.

One may choose $\delta_n$ smaller so that $\delta_n \le \frac{1}{n}$. Finally take

\begin{equation}\label{eqepsilon}
\epsilon_n = \min_{k\in\{0,...,n-1\}}\{\delta_{n-k}2^{-k-1}\}.
\end{equation}

(<ref>) follows immediately from our choice of $\nu_n$. Meanwhile (<ref>) with $p=1$ follows from (<ref>), (<ref>), (<ref>). Interpolation with the existing uniform bound on $f^{n,\nu}$, $f^n$ in $L^\infty L^\infty$ then gives (<ref>) for $p\in(1,\infty)$.

We are now left to show (<ref>)-(<ref>) are satisfied when in addition $d_*(u_{n+1},u_n) \le \epsilon_n$ for all $n\in\mathbb{N}$.

By (<ref>), $\sum_{i=n}^\infty \epsilon_i \le \delta_n$, and also $\delta_n \to 0$, so $\{u_i\}_{i\in\mathbb{N}}$ is a $d_*$-Cauchy sequence, uniformly bounded in $L^\infty L^\infty$ by $M$, and so converges in weak-$*$ $L^\infty L^\infty$ to some limit $u_\infty$, proving (<ref>).

Moreover, for all $n\in\mathbb{N}$, $d_*(u_\infty, u_n) \le \delta_n$, so taking in (<ref>) $w=u_\infty$, $g^{m;\nu}=f^{m;\infty,\nu}$ then (<ref>) follows from (<ref>), (<ref>), and interpolation with the existing uniform bound on $f^{n,\nu}$, $f^{\infty,\nu}$ in $L^\infty L^\infty$.

Finally (<ref>) is a simple corollary of (<ref>) and (<ref>).

§ FRACTAL SHEAR FLOW - NON-UNIQUENESS

§.§ Notation

For $x \in \mathbb{R}^2$ we set $x=(x_1, x_2)$, and define the corresponding unit vectors $e_1 = (1,0)$, $e_2 = (0,1)$. For $i \in \{1,2\}$ an index, we define by $\hat{i}$ the other index, that is $\hat{i} \in \{1,2\}\setminus\{i\}$.

When working on the 2-torus, $[x] \in \mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$ shall instead always be written in terms of a representative $x \in \mathbb{R}^2$, usually $x \in [0,1)^2$. For notational convenience functions on the torus $\mathbb{T}^2$ will often be considered periodic functions on $\mathbb{R}^2$, and vice versa. Integrals over $\mathbb{T}^2$ are then strictly speaking over $[0,1)^2$, such as when taking norms.

First, we define the shear flows that form the building block of our construction. We shall later apply these shears in a carefully designed order, reminiscent of a cantor-subset of $[0,T]$. The arising self-cancellation of these shear flows will produce non-uniqueness of renormalised weak solutions (Definition <ref>) for any initial data to (<ref>). Moreover, the small spatial-scale of the shears, see (<ref>) below, will allow us to approximate the vanishing viscosity limit of (<ref>).

For $L\in\mathbb{N}$, we divide $\mathbb{R}$ into disjoint intervals $\bigcup_{m \in \mathbb{Z}} \left[\frac{m}{2L},\frac{m+1}{2L}\right)$. We define for each $i\in\{1,2\}$, $L\in \mathbb{N}$, the periodic vector field $u^{(i;L)}: \mathbb{T}^2 \to \mathbb{R}^2$,

\begin{equation*}
u^{(i;L)}(x) = \begin{cases} e_i & \text{if } x_{\hat{i}} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for even $m$}, \\ -e_i & \text{if } x_{\hat{i}} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for odd $m$}, \end{cases}
\end{equation*}

where, since $2L$ is even, the definition is a periodic function of $x\in\mathbb{R}^2$, and so is well-defined on $\mathbb{T}^2$. $u^{(i;L)}$ is bounded and divergence-free in the distributional sense since it is of the form $u^{(1;L)}(x)=(g(x_2),0)$, or $u^{(2;L)}(x)=(0,g(x_1))$ for some $g \in L^\infty(\mathbb{T}^2)$. We refer to this vector field as the $(i;L)$-Lagrangian shear.

[Strictly speaking a Lagrangian flow along a divergence-free vector field $u\in L^1([0,T];L^1(\mathbb{T}^d;\mathbb{R}^d))$ need not be unique, though it is in this case, e.g. by Proposition <ref>, below.]We shall denote by $\big\{y_t^{(i;L)}\big\}_{t \in (-\infty, \infty)}$ the corresponding Lagrangian flow (Definition <ref>) for the transport problem along $u^{(i;L)}$,

\begin{equation*}
y_t^{(i;L)} : \begin{array}[t]{rcl} \mathbb{T}^2 & \to & \mathbb{T}^2 \\ x & \mapsto & \begin{cases} x+t e_i \mod \mathbb{Z}^2 & \text{if } x_{\hat{i}} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for even $m$}, \\ x-t e_i \mod \mathbb{Z}^2 & \text{if } x_{\hat{i}} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for odd $m$}. \end{cases} \end{array}
\end{equation*}

As per Definition <ref> of Lagrangian flows, this preserves the Lebesgue-measure on $\mathbb{T}^2$, e.g. for $i=1$ we can decompose a Lebesgue-measureable subset $A \subset \mathbb{T}^2$ into $A_m = A \cap \left(\mathbb{T} \times \left[\frac{m}{2L},\frac{m+1}{2L}\right)\right)$, for which each $\big(y_t^{(1;L)}\big)^{-1}(A_m)$ is a translation. Moreover, it is invertible with inverse given by $\big(y_t^{(i;L)}\big)^{-1}(x) = y_{-t}^{(i;L)}(x)$, and is absolutely continuous since with respect to $t$ since it is differentiable with derivative $\frac{\partial}{\partial t} y_t^{(i;L)} (x) = u^{(i;L)}(x) = u^{(i;L)}(y_t^{(i;L)}(x))$. Therefore, $\big\{y_t^{(i;L)}\big\}_{t \in (-\infty, \infty)}$ is a Lagrangian flow along $u^{(i;L)}$.

Finally, for each $i \in \{1,2\}$, we show

\begin{equation}\label{eqshearweakconv}
u^{(i;L)} \xrightharpoonup{L\to\infty} 0,
\end{equation}

with convergence in weak-$*$ $L^\infty(\mathbb{T}^2)$.

The proof of (<ref>) is similar to that of the Riemann-Lebesgue lemma. Notice that $u^{(i;L)}\left(x+\frac{1}{2L}e_{\hat{i}}\right) = - u^{(i;L)}(x)$ and so for any test function $f \in L^1(\mathbb{T}^2)$, by changing variables we have that

\begin{equation*}
\int_{\mathbb{T}^2} f(x)u^{(i;L)}(x) \; dx = \int_{\mathbb{T}^2} \frac{f(x) - f\left(x+\frac{1}{2L}e_{\hat{i}}\right)}{2} u^{(i;L)}(x) \; dx.
\end{equation*}

Now $f\left(x+\frac{1}{2L}e_{\hat{i}}\right) \xrightarrow{L\to\infty} f$ converges strongly in $L^1(\mathbb{T}^2)$. Together with the uniform bound $\left\| u^{(i;L)} \right\|_{L^\infty} \le 1$ this implies the above integral converges to zero as $L\to\infty$, as required.

Next, we prove well-posedness of (<ref>) along these vector fields.

We follow the notation introduced in Definition <ref>.

Suppose $f\in L^1L^1$ is a weak solution to (<ref>) along $u^{(i;L)}$ on an open interval $I\subset(0,T)$. Then there exists some $g \in L^1(\mathbb{T}^2)$ such that $f$ is (a.e. in $\mathbb{T}^2\times I$) equal to the following Lagrangian solution (Definition <ref>) associated with $g$,

\begin{equation*}
f(\cdot,t)=g \circ \big(y^{(i;L)}_t\big)^{-1}.
\end{equation*}

In particular $f$ is a renormalised weak solution to (<ref>) (Definition <ref>). Moreover, $g \circ \big(y^{(i;L)}_t\big)^{-1}\in C^0((-\infty,\infty);L^1(\mathbb{T}^2))$.

Without loss of generality suppose $i=1$, and use the shorthand $y_t=y_t^{(1;L)}$ for each $t \in (-\infty, \infty)$, that is

\begin{equation*}
y_t(x) = \begin{cases} x+t e_1 & \text{if } x_{2} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for even $m$}, \\ x-t e_1 & \text{if } x_{2} \in \left[\frac{m}{2L},\frac{m+1}{2L}\right) \text{ for odd $m$}. \end{cases}
\end{equation*}

Since the $e_2$-component of $y_t(x)$ is unchanged, that is $[y_t(x)]_2 = x_2$, we see

\begin{equation*}
u^{(1;L)}(y_t(x),t) = u^{(1;L)}(x,t).
\end{equation*}

Fix $m \in \{0,1,...2L-1\}$. Take a test function $\phi \in C_c^\infty\left(\left(\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right)\right) \times I\right)$, and let $\psi(x,t)=\phi(y_t^{-1}(x), t)$. This is well-defined since $y_t^{-1}$ preserves the set $\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right)$.

[Notice also that $\psi$ is Lipschitz (moreover it is smooth). This will not be true for a general test function $\phi\in C_c^\infty(\mathbb{T}^2\times I)$. For an arbitrary Lagrangian flow this issue is why existence of a Lagrangian flow does not prove uniqueness of weak solutions to (<ref>) .]Then $\psi \in C_c^\infty\left(\left(\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right)\right) \times I\right)$ is bounded, and supported and smooth on $\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right) \times I$ since $y_t^{-1}$ is smooth there. By chain rule, the following point-wise equality holds

\begin{equation*}
\left(\frac{\partial \psi}{\partial t} + u^{(1;L)}\cdot\nabla \psi\right)(x,t) = \frac{\partial \phi}{\partial t}(y_t^{-1}(x),t).
\end{equation*}

Since $f \in L^1L^1$ is assumed to be a weak solution to (<ref>) along $u^{(1;L)}$ on the open interval $I\subset (0,T)$,

\begin{equation*}
\int_{\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right) \times I} f(x,t) \left(\frac{\partial \psi}{\partial t} + u^{(1;L)}\cdot\nabla \psi\right)(x,t) \; dxdt = 0,
\end{equation*}

which implies

\begin{equation*}
\int_{\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right) \times I} f(x,t) \frac{\partial \phi}{\partial t}(y_t^{-1}(x),t) \; dxdt = 0,
\end{equation*}

and after changing variables,

\begin{equation*}
\int_{\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right) \times I} f(y_t(x), t) \; \frac{\partial \phi}{\partial t}(x,t) \; dxdt = 0.
\end{equation*}

[See for example Theorem 3.1.4' in <cit.>]This holds for all $\phi \in C_c^\infty\left(\left(\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right)\right) \times I\right)$, and so as a distribution $f(y_t(x), t) \in L^1L^1$ is independent of $t\in I$, so there must exists some $g_m \in L^1(\mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right))$ such that (for a.e. $t \in I$)

\begin{equation*}
f(y_t(x), t) = g_m(x) \text{ a.e. on } \mathbb{T} \times \left(\frac{m}{2L},\frac{m+1}{2L}\right).
\end{equation*}

Repeating for each $m \in \{0,1,...2L-1\}$ gives some $g \in L^1(\mathbb{T}^2)$ such that,

\begin{equation*}
f(y_t(x), t) = g(x) \text{ a.e. on } \mathbb{T}^2.
\end{equation*}

And so $f(\cdot, t) = g \circ y_t^{-1}$ is a Lagrangian solution to (<ref>) (Definition <ref>), and hence also a renormalised weak solution (Definition <ref>, Remark <ref>).

To prove that $g\circ y_t^{-1} \in C^0((-\infty,\infty);L^1(\mathbb{T}^2))$ we shall use the fact that $y_t$ is bijective, measure preserving, and $1$-Lipschitz in time, i.e. for all $x \in \mathbb{T}^2$, $t,s \in \mathbb{R}$

\begin{equation*}
\left|y_t(x) - y_s(x)\right| \le |t-s|,
\end{equation*}

and so, by replacing $x$ with $y_s^{-1}(x)$,

\begin{equation}\label{eqshearlip}
\left|y_t \circ y_s^{-1}(x) - x\right| \le |t-s|.
\end{equation}

The following follows a standard argument. We fix $t \in I$ and write

\begin{equation*}
g\circ y_s^{-1} = \left( g\circ y_t^{-1}\right) \circ y_t \circ y_s^{-1}.
\end{equation*}

Now mollify $g\circ y_t^{-1} $ in $L^1(\mathbb{T}^2)$, that is for each $\epsilon > 0$ take some $\phi_\epsilon \in C^\infty(\mathbb{T}^2)$ such that

\begin{equation*}
\left\|\phi_\epsilon - g\circ y_t^{-1}\right\|_{L^1(\mathbb{T}^2)} \le \epsilon.
\end{equation*}

But also, since $y_t$, $y_s^{-1}$ are measure preserving,

\begin{equation*}
\left\|\phi_\epsilon \circ y_t \circ y_s^{-1} - g\circ y_s^{-1}\right\|_{L^1(\mathbb{T}^2)} \le \epsilon.
\end{equation*}

Then since $\phi_\epsilon$ is uniformly continuous on $\mathbb{T}^2$, by (<ref>) there exists $\delta \in (0,1)$ such that if $|t-s| < \delta$ then $\|\phi_\epsilon \circ y_t \circ y_s^{-1} - \phi_\epsilon\|_{L^1} \le \epsilon$, and so

\begin{equation*}
\|g\circ y_t^{-1} - g\circ y_s^{-1}\|_{L^1(\mathbb{T}^2)} \le 3 \epsilon,
\end{equation*}

which completes the proof.

Next, we demonstrate how these shear flows self-cancel. It is this property that gives rise to non-uniqueness in our later construction.

Let $L_1, L_2 \in \mathbb{N}$, and $i_1, i_2 \in \{1,2\}$ such that $i_1\ne i_2$. Suppose that $2\tau_2 = \frac{1}{2L_1}$, and that $2L_2\tau_1$ is an odd integer.

[Here we denote by $f^2 := f \circ f$.]Then the composition

\begin{gather*}
\left(y_{\tau_2}^{(i_2;L_2)}\right)^2 \circ y_{\tau_1}^{(i_1;L_1)} \circ \left(y_{\tau_2}^{(i_2;L_2)}\right)^2 \circ y_{\tau_1}^{(i_1;L_1)} = \mathrm{Id}.
\end{gather*}

Without loss of generality suppose $i_1=1,i_2=2$.

Since $\left(y_{\tau_2}^{(2;L_2)}\right)^2 = y_{2\tau_2}^{(2;L_2)}$ we instead prove the more general result

\begin{gather*}
y_{\tau_2}^{(2;L_2)}\circ y_{\tau_1}^{(1;L_1)}\circ y_{\tau_2}^{(2;L_2)}\circ y_{\tau_1}^{(1;L_1)} = \mathrm{Id},
\end{gather*}

whenever both $2L_1\tau_2$ and $2L_2\tau_1$ are odd integers.

To this end we divide $\mathbb{T}^2$ into tiles as follows. Given $x \in [0,1)^2$ we find the unique integers $m_1, m_2 \in \mathbb{Z}$ such that

\begin{equation}\label{eqequivclass}
x \in \left[\frac{m_1}{2L_2},\frac{m_1+1}{2L_2}\right) \times \left[\frac{m_2}{2L_1},\frac{m_2+1}{2L_1}\right).
\end{equation}

We are concerned with the parity of $m_1$ and $m_2$ (even or odd), that is we have four `colours' on our tiling, two for each coordinate.

This defines two equivalence relations $\left[\cdot\right]_{m_1}, \left[\cdot\right]_{m_2}$ on $\mathbb{T}^2$, each with two equivalence classes corresponding to the parity of either $m_1$ or $m_2$ in (<ref>).

Since $2L_1\tau_2$, $2L_2\tau_1$ are odd integers, the action of $y_{\tau_1}^{(1;L_1)}$ changes the parity of $m_1$ and not $m_2$, while the action of $y_{\tau_2}^{(2;L_2)}$ changes the parity of $m_2$ and not $m_1$.

Introduce the shorthand $x=:x^{(0)}$, $y_{\tau_1}^{(1;L_1)}(x^{(0)})=:x^{(1)}$, $y_{\tau_2}^{(2;L_2)}(x^{(1)})=:x^{(2)}$, $y_{\tau_1}^{(1;L_1)}(x^{(2)})=:x^{(3)}$, $y_{\tau_2}^{(2;L_2)}(x^{(3)})=:x^{(4)}$.

Then by the above discussion,

\begin{gather*}
\left[x^{(2)}\right]_{m_2} \ne \left[x^{(0)}\right]_{m_2}, \\
\left[x^{(3)}\right]_{m_1} \ne \left[x^{(1)}\right]_{m_1}.
\end{gather*}

From the definitions of $y_{\tau_1}^{(1;L_1)}, y_{\tau_2}^{(2;L_2)}$ this implies

\begin{gather*}
x^{(3)} - x^{(2)} = -\left(x^{(1)} - x^{(0)}\right), \\
x^{(4)} - x^{(3)} = -\left(x^{(2)} - x^{(1)}\right),
\end{gather*}

and hence $x^{(4)} = x^{(0)}$ as required.

Next, we define the times at which to apply these shear flows $y_\tau^{(i;L)}$. We use a double index $(k,m)\in\mathcal{D}\subset\mathbb{Z}^2$, where $k$ shall index a Lagrangian shear $y_{\tau_k}^{(i_k;L_k)}$, while $m$ denotes the $m$th occurrence of that shear.

To exploit Cancellation <ref>, we shall ensure that there are exactly two occurances of $y_{\tau_{k+1}}^{(i_{k+1};L_{k+1})}$ between each occurance of $y_{\tau_k}^{(i_k;L_k)}$. If we denote by $t_{k,m}\in[0,1]$ the times at which the shears are first applied we may illustrate this construction in the following diagram:

\begin{equation*}
\begin{tikzpicture}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=12]
\draw[] (0,0) -- (1,0);
\node[anchor=east] (t0) at (0,0) {$t=0$};
\node[anchor=west] (t1) at (1,0) {$t=1$};
\node[yshift=20mm] (t10) at (0,0) {${t_{1,0}}$};
\draw[<-] ([yshift=6/12mm]0,0) -- (t10);
\node[anchor=base,yshift=20mm] (t11) at (1/2,0) {${t_{1,1}}$};
\draw[<-] ([yshift=6/12mm]1/2,0) -- (t11);
\node[anchor=base,yshift=15mm] (t20) at (2/8,0) {${t_{2,0}}$};
\draw[<-] ([yshift=6/12mm]2/8,0) -- (t20);
\node[anchor=base,yshift=15mm] (t21) at (3/8,0) {${t_{2,1}}$};
\draw[<-] ([yshift=6/12mm]3/8,0) -- (t21);
\node[anchor=base,yshift=15mm] (t22) at (6/8,0) {${t_{2,2}}$};
\draw[<-] ([yshift=6/12mm]6/8,0) -- (t22);
\node[anchor=base,yshift=15mm] (t23) at (7/8,0) {${t_{2,3}}$};
\draw[<-] ([yshift=6/12mm]7/8,0) -- (t23);
\node[anchor=base,yshift=10mm] (t305) at (10.5/32,0) {${t_{3,0}}\;t_{3,1}$};
\node[anchor=base,yshift=5mm] (n1) at (10.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]10/32,0) -- ([shift={(-0.5/32,0)}]t305.south);
\node[anchor=base,yshift=5mm] (n1) at (11.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]11/32,0) -- ([shift={(0.5/32,0)}]t305.south);
\node[anchor=base,yshift=10mm] (t325) at (14.5/32,0) {${t_{3,2}}\;t_{3,3}$};
\node[anchor=base,yshift=5mm] (n1) at (14.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]14/32,0) -- ([shift={(-0.5/32,0)}]t325.south);
\node[anchor=base,yshift=5mm] (n1) at (15.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]15/32,0) -- ([shift={(0.5/32,0)}]t325.south);
\node[anchor=base,yshift=10mm] (t345) at (26.5/32,0) {${t_{3,4}}\;t_{3,5}$};
\node[anchor=base,yshift=5mm] (n1) at (26.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]26/32,0) -- ([shift={(-0.5/32,0)}]t345.south);
\node[anchor=base,yshift=5mm] (n1) at (27.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]27/32,0) -- ([shift={(0.5/32,0)}]t345.south);
\node[anchor=base,yshift=10mm] (t365) at (30.5/32,0) {${t_{3,6}}\;t_{3,7}$};
\node[anchor=base,yshift=5mm] (n1) at (30.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]30/32,0) -- ([shift={(-0.5/32,0)}]t365.south);
\node[anchor=base,yshift=5mm] (n1) at (31.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]31/32,0) -- ([shift={(0.5/32,0)}]t365.south);
\foreach \a in {0,1}{
\fill[] (\a/2,0) circle (1/12pt);
\foreach \b in {0,1}{
\fill[] (\a/2+2/8+\b/8,0) circle (1/12pt);
\foreach \c in {0,1}{
\fill[] (\a/2+2/8+\b/8+2/32+\c/32,0) circle (1/12pt);
\foreach \d in {0,1}{
\fill[] (\a/2+2/8+\b/8+2/32+\c/32+2/128+\d/128,0) circle (1/12pt);
\foreach \e in {0,1}{
%\fill[] (\a/2+2/8+\b/8+2/32+\c/32+2/128+\d/128+2/512+\e/512,0) circle (1/18pt);
\node[anchor=base,yshift=-20pt] (n1) at (0,0) {$f_0$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (1/4,0) -- (0,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_1}^{(i_1;L_1)}$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (3/4,0) -- (2/4,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_1}^{(i_1;L_1)}$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (5/16,0) -- (4/16,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_2}^{(i_2;L_2)}$};
\node[anchor=base,yshift=-20pt] (n1) at (5.5/16,0) {$...$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (7/16,0) -- (6/16,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_2}^{(i_2;L_2)}$};
\node[anchor=base,yshift=-20pt] (n1) at (7.5/16,0) {$...$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (13/16,0) -- (12/16,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_2}^{(i_2;L_2)}$};
\node[anchor=base,yshift=-20pt] (n1) at (13.5/16,0) {$...$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (15/16,0) -- (14/16,0) node[anchor=base,pos=0.5,yshift=-20pt,black,font=\tiny]{$\circ y_{-\tau_2}^{(i_2;L_2)}$};
\node[anchor=base,yshift=-20pt] (n1) at (15.5/16,0) {$...$};
\node[anchor=base west,yshift=-20pt] (n1) at (1,0) {$=f(\cdot, 1)$};
\end{scope}
\end{tikzpicture}
\end{equation*}

If we include only the shears $y_{\tau_k}^{(i_k;L_k)}$ for $k\le K$, it can now be seen how Cancellation <ref> may create two different behaviours of the trace $f(\cdot, 1)$ as $2K \to \infty$, and $2K+1 \to \infty$.

The frequencies of the shears, $L_k\in\mathbb{N}$, will later be chosen to grow sufficiently quickly, so that advection-diffusion (<ref>) along a viscosity subsequence $\nu_k>0$ will only include the effect of the first $k$ shears, as per Theorem <ref>.

To this end, we first define a total order $<_\mathrm{time}$ on the indexing set $\mathcal{D}$, and then define $t_{k,m}$ so that they respect this ordering.

We define the following set of `dyadic' pairs

\begin{equation*}
\mathcal{D}=\{(k,m):k\in \mathbb{N}, m \in \mathbb{Z}, 0\le m<2^k\},
\end{equation*}

and define a total order $<_\mathrm{time}$ on $\mathcal{D}$ via

\begin{equation}\label{eqdyord}
(k_1,m_1) <_\mathrm{time} (k_2,m_2) \text{ if and only if }\begin{cases}
m_1 2^{-k_1} < m_2 2^{-k_2}, \text{ or} \\
m_1 2^{-k_1} = m_2 2^{-k_2} \text{ and } k_1<k_2.
\end{cases}
\end{equation}

Define also for each $K \in \mathbb{N}$ the finite subset

\begin{equation*}
\mathcal{D}_K = \{(k,m) \in \mathcal{D} : k \le K\},
\end{equation*}

which inherits the total order $(\mathcal{D},<_\mathrm{time})$, which is now a finite order.

[The bound on $t_{k,m}$ can be found by direct calculation of $\sum_{k\in\mathbb{N}}2^k2^{-2k}=1$.]We define for each $k \in \mathbb{N}$, and $m \in \mathbb{Z}$ with $0\le m < 2^k$,

\begin{equation}\label{eqtkm}
t_{k,m}  = \sum_{(k',m')<_\mathrm{time}(k,m)} 2^{-2k'} < 1,
\end{equation}

where an empty sum is zero. We illustrate this in the following diagram:

\begin{equation}\label{eqdydiag}
\begin{tikzpicture}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=12]
\draw[] (0,0) -- (1,0);
\node[anchor=east] (t0) at (0,0) {$t=0$};
\node[anchor=west] (t1) at (1,0) {$t=1$};
\node[yshift=20mm] (t10) at (0,0) {${t_{1,0}}$};
\draw[<-] ([yshift=6/12mm]0,0) -- (t10);
\node[anchor=base,yshift=20mm] (t11) at (1/2,0) {${t_{1,1}}$};
\draw[<-] ([yshift=6/12mm]1/2,0) -- (t11);
\node[anchor=base,yshift=15mm] (t20) at (2/8,0) {${t_{2,0}}$};
\draw[<-] ([yshift=6/12mm]2/8,0) -- (t20);
\node[anchor=base,yshift=15mm] (t21) at (3/8,0) {${t_{2,1}}$};
\draw[<-] ([yshift=6/12mm]3/8,0) -- (t21);
\node[anchor=base,yshift=15mm] (t22) at (6/8,0) {${t_{2,2}}$};
\draw[<-] ([yshift=6/12mm]6/8,0) -- (t22);
\node[anchor=base,yshift=15mm] (t23) at (7/8,0) {${t_{2,3}}$};
\draw[<-] ([yshift=6/12mm]7/8,0) -- (t23);
\node[anchor=base,yshift=10mm] (t305) at (10.5/32,0) {${t_{3,0}}\;t_{3,1}$};
\node[anchor=base,yshift=5mm] (n1) at (10.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]10/32,0) -- ([shift={(-0.5/32,0)}]t305.south);
\node[anchor=base,yshift=5mm] (n1) at (11.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]11/32,0) -- ([shift={(0.5/32,0)}]t305.south);
\node[anchor=base,yshift=10mm] (t325) at (14.5/32,0) {${t_{3,2}}\;t_{3,3}$};
\node[anchor=base,yshift=5mm] (n1) at (14.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]14/32,0) -- ([shift={(-0.5/32,0)}]t325.south);
\node[anchor=base,yshift=5mm] (n1) at (15.5/32,0) {$...$};
\draw[<-] ([yshift=6/12mm]15/32,0) -- ([shift={(0.5/32,0)}]t325.south);
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Next, we construct the `fractal' vector fields exploiting the above.

Consider a sequence of tuples $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ with $i_k \in \{1,2\}$, $L_k \in \mathbb{N}$, and $\tau_k > 0$.

* We say $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ satisfies the Finiteness Condition if for all $k\in\mathbb{N}$ we have

\begin{equation}\label{eqfincond}
\text{(Finiteness Condition)\quad} \tau_k < 2^{-2k}.
\end{equation}

* We say $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ satisfies the Cancellation Conditions if for all $k\in\mathbb{N}$ we have

\begin{equation}\label{eqcancond}
\text{(Cancellation Conditions)\quad}\begin{gathered}
i_{k+1} \ne i_k, \\
2\tau_{k+1} = \frac{1}{2L_k}, \\
2L_{k+1}\tau_k \text{ an odd integer}.
\end{gathered}
\end{equation}

Following the notation of Definition <ref>, in particular (<ref>), when $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ satisfy the Finiteness Condition (<ref>), we may define for each $K \in \mathbb{N}$ the following fractal shear flows on $\mathbb{T}^2 \times [0, 1] \to \mathbb{R}^2$,

\begin{gather*}
u_K^{\{(i_k, L_k, \tau_k)\}_{k=1}^K}(x,t) = \begin{cases} u^{(i_k;L_k)}(x) & \text{if $t \in [t_{k,m}, t_{k,m} + \tau_k]$ for some $(k,m) \in \mathcal{D}_K$}, \\ 0 & \text{otherwise}, \end{cases} \\
u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}(x,t) = \begin{cases} u^{(i_k;L_k)}(x) & \text{if $t \in [t_{k,m}, t_{k,m} + \tau_k]$ for some $(k,m) \in \mathcal{D}$}, \\ 0 & \text{otherwise}. \end{cases}
\end{gather*}

By (<ref>), this is well-defined since the Finiteness Condition (<ref>) implies that the intervals $[t_{k,m}, t_{k,m} + \tau_k]$ are disjoint subsets of $[0,1]$.

Next, we show how this construction exploits Cancellation <ref>, to create non-uniqueness of renormalised solutions to (<ref>).

We follow the notation introduced in Definitions <ref>, <ref>.

Let $f_0 \in L^\infty(\mathbb{T}^2)$, and suppose we have an infinite sequence of tuples $\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}$ with $i_k \in \{1,2\}$, $L_k \in \mathbb{N}$, and $\tau_k > 0$, satisfying the Finiteness Condition (<ref>). Then for each $K \in \mathbb{N}$ there exists a unique weak solution $f^K$ to (<ref>) along $u_K^{\{(i_k,L_k,\tau_k)\}_{k=1}^K}$ with initial data $f_0$. Moreover $f^K\in (C^0L^1) \cap (L^\infty L^\infty)$, is a Lagrangian solution, and hence also a renormalised weak solution (Definitions <ref>, <ref>).

If in addition $\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}$ satisfy the Cancellation Conditions (<ref>) then

\begin{gather*}
f^{2K}\xrightarrow{K\to\infty}f^{\mathrm{even}}, \\
\end{gather*}

[Moreover, the sequences $f^{2K}\in C^0L^1$ and $f^{2K+1}\in C^0L^1$ are eventually constant when restricted to the spatio-temporal domain $\mathbb{T}^2\times[t_{k,m},t_{k,m}+2^{-2k}]$ for any $(k,m)\in\mathcal{D}$, or to the spatio-temporal domain $\mathbb{T}^2\times\{1\}$, as $K\to\infty$.]with the above convergence in weak-$*$ $L^\infty L^\infty$, and strong in $L^p L^\infty$ for any $p\in[1,\infty)$. Furthermore, the limit functions $f^\mathrm{even}, f^\mathrm{odd}$ are renormalised weak solutions to (<ref>) along $u_\infty^{\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}}$ with initial data $f_0$.

Moreover, if $f_0$ is not constant, then $f^\mathrm{even}\ne f^\mathrm{odd}$. In particular,

\begin{equation}\label{eqnonuniq}
\begin{gathered}
f^{\mathrm{even}}(\cdot, 1) = f_0, \\
f^{\mathrm{odd}}(\cdot, 1) = f_0 \circ y_{-2\tau_1}^{(i_1;L_1)}.
\end{gathered}
\end{equation}

Recall the language and notation introduced in Definitions <ref>, <ref>, as well as Definitions <ref>, <ref> of renormalised and Lagrangian solutions to (<ref>).

By Theorems <ref>, <ref>, there exists some weak solution $f^K \in C_{\mathrm{weak-}*}^0L^\infty$ to (<ref>) along $u_K^{\{(i_k,L_k,\tau_k)\}_{k=1}^K}$ with initial data $f_0$.

We aim to show $f \in C^0 L^1$ with

\begin{equation}\label{eqlagpiece}
f^K(\cdot, t) = f_0 \circ \big(\tilde{y}_t^K\big)^{-1},
\end{equation}

for $\big\{\tilde{y}_t^K\big\}_{t\in[0,1]}$ a Lagrangian flow along $u_K^{\{(i_k,L_k,\tau_k)\}_{k=1}^K}$ which does not depend on $f_0$, thus proving uniqueness.

From the definition of $u_K^{\{(i_k,L_k,\tau_k)\}_{k=1}^K}$, finiteness of the set $\mathcal{D}_K$, and disjointness of the intervals $[t_{k,m}, t_{k,m}+\tau_k]$ for all $(k,m) \in \mathcal{D}_K$, we may piecewise apply Proposition <ref> to $f^K$. [We also use that any weak solution to (<ref>) along $u\equiv 0$ on an open time interval $I$ is a constant function of $t\in I$, see Theorem 3.1.4' in <cit.>.]We use that $f^K \in C_{\mathrm{weak-}*}^0L^\infty$ to glue together the pieces. This constructs for each $K \in \mathbb{N}$ a Lagrangian flow $\tilde{y}_t^K$ satisfying (<ref>), and moreover gives an expression for $\tilde{y}_t^K$ in terms of the Lagrangian shear flows $y_t^{(i_k;L_k)}$, see (<ref>) below.

For any $(k,m) \in \mathcal{D}_K$, define

\begin{equation*}
t_{\mathrm{suc}_K(k,m)} = \begin{cases}
1 & \text{if $(k,m)$ is maximal in $(\mathcal{D}_K,<_\mathrm{time})$, else} \\
t_{\mathcal{S}} & \text{for $\mathcal{S}\in\mathcal{D}_K$ the successor of $(k,m)$ in $(\mathcal{D}_K,<_\mathrm{time})$}.
\end{cases}
\end{equation*}

Note from (<ref>), and the Finiteness Condition (<ref>), that

\begin{equation}\label{eqsucbound2}
\tau_k < 2^{-2k} \le t_{\mathrm{suc}_K(k,m)} - t_{k,m}.
\end{equation}

Therefore, for each $t \in [t_{k,m},t_{\mathrm{suc}_K(k,m)}]$,

\begin{equation}\label{eqlagint}
\tilde{y}_t^K \circ\big(\tilde{y}_{t_{k,m}}^K\big)^{-1} = \begin{cases}
y_{t-t_{k,m}}^{(i_k;L_k)} & \text{if } t \in [t_{k,m},t_{k,m}+\tau_k], \\
y_{\tau_k}^{(i_k;L_k)} & \text{if } t \in [t_{k,m}+\tau_k,t_{\mathrm{suc}_K(k,m)}].
\end{cases}
\end{equation}

This expression defines $\tilde{y}_t^K$ for all $t \in [0,1]$, since $t_{1,0} = 0$ and $\tilde{y}_0^K = \mathrm{Id}$. This completes the proof of uniqueness.

Next, by considering (<ref>), (<ref>), we may show that (see the illustration (<ref>))

\begin{equation}\label{eqsuc}
t_{\mathrm{suc}_K(k,m)} = \begin{cases}
t_{k,m}+2^{-2k} & \text{if } k < K, \\
t_{k,m} + 2^{1-2k} & \text{if } k=K. \\
\end{cases}
\end{equation}

Moreover, for all $(k,m) \in \mathcal{D}$,

\begin{equation}\label{eqfracstruc}
\begin{aligned}
t_{k,m}+2^{-2k} & = t_{k+1,2m}, \\
t_{k,m}+2^{1-2k} & = t_{k,m+1}\quad \text{when $m$ even}.
\end{aligned}
\end{equation}

Assume that additionally $\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}$ satisfy the Cancellation Conditions (<ref>).

Fix some $K \in \mathbb{N}$. We claim that, for each $(k,m) \in \mathcal{D}_K$, and for all $\left.t \in [t_{k,m},t_{k,m}+2^{-2k}]\right.$, one has as maps $\mathbb{T}^2\to\mathbb{T}^2$,

\begin{equation}\label{eqparity}
\tilde{y}_t^{K+2} = \tilde{y}_t^K.
\end{equation}

By (<ref>), (<ref>), it is sufficient to show, for all $(k,m)\in\mathcal{D}_K$, that as maps $\mathbb{T}^2\to\mathbb{T}^2$,

\begin{equation*}
\tilde{y}_{t_{\mathrm{suc}_K(k,m)}}^{K+2} \circ \big(\tilde{y}_{t_{\mathrm{suc}_{K+2}(k,m)}}^{K+2}\big)^{-1} = \tilde{y}_{t_{\mathrm{suc}_K(k,m)}}^K \circ \big(\tilde{y}_{t_{\mathrm{suc}_{K+2}(k,m)}}^K\big)^{-1}.
\end{equation*}

By (<ref>), for all $k<K$ this is immediate. Meanwhile, by (<ref>), (<ref>), for $k=K$ we may rewrite this, and now need to prove that for all $m \in \mathbb{Z}$ with $0\le m<2^K$,

\begin{equation*}
\tilde{y}_{t_{K+1,2m}+2^{-2K}}^{K+2} \circ \big(\tilde{y}_{t_{K+1,2m}}^{K+2}\big)^{-1} = \tilde{y}_{t_{K+1,2m}+2^{-2K}}^{K} \circ \big(\tilde{y}_{t_{K+1,2m}}^{K}\big)^{-1}.
\end{equation*}

In particular, it is certainly sufficient to show, for any $(k,m) \in \mathcal{D}_{K+1}$ with $m$ even, that as maps $\mathbb{T}^2\to\mathbb{T}^2$,

\begin{equation}\label{eqshearcanc}
\tilde{y}_{t_{k,m}+2^{2-2k}}^{K} \circ \big(\tilde{y}_{t_{k,m}}^{K}\big)^{-1} \begin{aligned}[t]
& = \begin{cases}
y_{2\tau_k}^{(i_k;L_k)} & \text{if } k = K\; (\mathrm{mod}\;2), \\
\mathrm{Id} & \text{otherwise}.
\end{cases} \\
& = : Y_{k,m}^K,
\end{aligned}
\end{equation}

where we have defined the shorthand $Y_{k,m}^K:\mathbb{T}^2\to\mathbb{T}^2$.

Fixing an even $m\in\mathbb{Z}$ with $0\le m < 2^{K+1}$, we prove this by induction. For $k=K+1$, by (<ref>), (<ref>) we see that as maps $\mathbb{T}^2\to\mathbb{T}^2$,

\begin{equation*}
\tilde{y}_{t_{\mathrm{suc}_K(k,m)}}^K \circ \big(\tilde{y}_{t_{\mathrm{suc}_{K+2}(k,m)}}^K\big)^{-1} = \mathrm{Id}.
\end{equation*}

After rewriting this using (<ref>), (<ref>) this proves (<ref>) for $k=K+1$.

Now let $k\le K$. By (<ref>), and since by assumption $m$ is even, we may rewrite (it may be helpful to consider the illustration (<ref>))

\begin{equation*}
Y_{k,m}^K = \begin{aligned}[t]
& \left(\tilde{y}_{t_{k+1,2m+2}+2^{-2k}}^{K} \circ \big(\tilde{y}_{t_{k+1,2m+2}}^{K}\big)^{-1}\right) \\
& \circ \left(\tilde{y}_{t_{k,m+1}+2^{-2k}}^{K} \circ \big(\tilde{y}_{t_{k,m+1}}^{K}\big)^{-1}\right) \\
& \circ \left(\tilde{y}_{t_{k+1,2m}+2^{-2k}}^{K} \circ \big(\tilde{y}_{t_{k+1,2m}}^{K}\big)^{-1}\right) \\
& \circ \left(\tilde{y}_{t_{k,m}+2^{-2k}}^{K} \circ \big(\tilde{y}_{t_{k,m}}^{K}\big)^{-1}\right).
\end{aligned}
\end{equation*}

Then by the definition of $Y_{k+1,m'}^K$, (<ref>), and (<ref>), the right hand side may be rewritten again as

\begin{equation*}
Y_{k,m}^K = Y_{k+1,2m+2}^K \circ y_{\tau_k}^{(i_k;L_k)} \circ Y_{k+1,2m}^K \circ y_{\tau_k}^{(i_k;L_k)}.
\end{equation*}

By Cancellation <ref>, the result (<ref>) for $k$ now follows from the same result for $k+1$. This completes the induction, and thus the proof of (<ref>), and so also the proof of (<ref>).

Next, note by (<ref>), that

\begin{align*}
E & = \left(\bigcup_{(k,m)\in\mathcal{D}} [t_{k,m},t_{k,m}+2^{-2k}]\right) \cup \left\{t_{k,m}+2^{2-2k}:(k,m)\in\mathcal{D} \text{ with $m$ even}\right\} \\
& \subset [0,1],
\end{align*}

has Lebesgue-measure 1. Moreover, by (<ref>), (<ref>), the sequences $f^{2K}$, $f^{2K+1}$ are eventually constant on $\mathbb{T}^2 \times [t_{k,m},t_{k,m}+2^{-2k}]$, and each $\mathbb{T}^2 \times \{t_{k,m}+2^{2-2k}\}$ with $m$ even.

Therefore, for all $t\in E$, and so for a.e. $t \in [0,1]$, we see that

\begin{equation}\label{eqpointconv}
\begin{gathered}
f^{2K}(\cdot, t) \xrightarrow{K\to\infty} f_0 \circ \big(\tilde{y}^\mathrm{even}_t\big)^{-1}, \\
f^{2K+1}(\cdot, t) \xrightarrow{K\to\infty} f_0 \circ \big(\tilde{y}^\mathrm{odd}_t\big)^{-1},
\end{gathered}
\end{equation}

for some Lebesgue-measure preserving $\{\tilde{y}^\mathrm{even}_t\}_{t\in E}$, $\{\tilde{y}^\mathrm{odd}_t\}_{t\in E}$ independent of $f_0$, with the convergence strong in $L^\infty(\mathbb{T}^2)$.

Moreover, by the dominated convergence Theorem, we see that (<ref>) converges strongly in $L^p(E;L^\infty(\mathbb{T}^2))$ for all $p\in[1,\infty)$, and hence also in weak-$*$ $L^\infty(E;L^\infty(\mathbb{T}^2))$.

By Definition <ref>, for all $t\in[0,1]$ (in particular $t\in E$), $u_K^{\{(i_k, L_k, \tau_k)\}_{k=1}^K}(\cdot, t)$ is bounded by 1 in $L^\infty(\mathbb{T}^2;\mathbb{R}^2)$, and as $K\to\infty$ the sequence is eventually equal to $u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}(\cdot,t)$. Therefore it too converges strongly in $L^p(E;L^\infty(\mathbb{T}^2;\mathbb{R}^2))$ for all $p\in[1,\infty)$.

Fix $t \in E$, and apply the Trace Formula (<ref>) to $f^{2K}$, $f^{2K+1}$. Taking now the limit $K\to\infty$ then gives that for all $t\in E$, for any $\phi\in C^\infty(\mathbb{T}^d\times[0,T])$,

\begin{gather*}
\begin{aligned}
& \int_{\mathbb{T}^2} f_0 \circ \big(\tilde{y}^\mathrm{even}_t\big)^{-1} \phi(\cdot, t) \;dx \\
& = \int_{\mathbb{T}^2} f_0 \phi_0 \; dx + \int_{\mathbb{T}^2\times\left(E\cap[0,t]\right)} f_0 \circ \big(\tilde{y}^\mathrm{even}_t\big)^{-1}\left(\frac{\partial\phi}{\partial t}+u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}\cdot\nabla\phi +\nu\Delta \phi\right) dxdt,
\end{aligned} \\
\begin{aligned}
& \int_{\mathbb{T}^2} f_0 \circ \big(\tilde{y}^\mathrm{odd}_t\big)^{-1} \phi(\cdot, t) \;dx \\
& = \int_{\mathbb{T}^2} f_0 \phi_0 \; dx + \int_{\mathbb{T}^2\times\left(E\cap[0,t]\right)} f_0 \circ \big(\tilde{y}^\mathrm{odd}_t\big)^{-1}\left(\frac{\partial\phi}{\partial t}+u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}\cdot\nabla\phi +\nu\Delta \phi\right) dxdt,
\end{aligned}
\end{gather*}

where $\phi_0=\phi(\cdot,0)$.

This implies that $f_0 \circ \big(\tilde{y}^\mathrm{even}_t\big)^{-1}, f_0 \circ \big(\tilde{y}^\mathrm{odd}_t\big)^{-1} \in C^0_{\mathrm{weak-}*}(E;L^\infty(\mathbb{T}^2))$ and so may be extended to $f^\mathrm{even}, f^\mathrm{odd}\in C^0_{\mathrm{weak-}*}([0,1];L^\infty)$, as argued in Theorem <ref>.

For any $\phi \in C_c^\infty(\mathbb{T}^2\times[0,1))$ let $t=1$ in the above (noting that $1=t_{1,0}+2^{2-2}\in E$). This proves that $f^\mathrm{even}$, $f^\mathrm{odd}$ are both weak solutions to (<ref>) along $u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}$ with initial data $f_0$.

Moreover, for any $\beta \in C_b^0(\mathbb{R})$, we may rewrite $\beta\big(f_0\circ\big(\tilde{y}^\mathrm{even}_t\big)^{-1}\big) = \beta(f_0)\circ\big(\tilde{y}^\mathrm{even}_t\big)^{-1}$. By repeating the above arguments we have also that $\beta(f_0)\circ\big(\tilde{y}^\mathrm{even}_t\big)^{-1}$ (can be extended to) a weak solution to (<ref>) along $u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}$ with initial data $\beta(f_0)$. Therefore, we see that $f^\mathrm{even}$ is a renormalised weak solution to (<ref>) along $u_\infty^{\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}}$ with initial data $f_0$. Similarly for $f^\mathrm{odd}$.

Next, observe that by (<ref>), (<ref>), for all $(k,m)\in\mathcal{D}$ with $m$ even, that

\begin{equation}\label{eqparitystruc}
\begin{gathered}
f^\mathrm{even}(\cdot,t_{k,m}+2^{2-2k}) = f^\mathrm{even}(\cdot,t_{k,m}) \circ \begin{cases}
\big(y_{2\tau_k}^{(i_k;L_k)}\big)^{-1} & \text{for even } k, \\
\mathrm{Id} & \text{for odd } k.
\end{cases} \\
f^\mathrm{odd}(\cdot,t_{k,m}+2^{2-2k}) = f^\mathrm{odd}(\cdot,t_{k,m}) \circ \begin{cases}
\mathrm{Id} & \text{for even } k, \\
\big(y_{2\tau_k}^{(i_k;L_k)}\big)^{-1} & \text{for odd } k.
\end{cases}
\end{gathered}
\end{equation}

In particular, since $t_{1,0}=0$, we have proved (<ref>).

It remains to show, when $f_0$ is not constant, that from (<ref>) we deduce $f^\mathrm{even} \ne f^\mathrm{odd}$. Assume to the contrary that they are equal and call this solution $f$. Then by (<ref>) (noting that by (<ref>), (<ref>), $1-2^{2-2k} = t_{k,2^k-2}$, see for example the illustration (<ref>)), we have for all $k\in\mathbb{N}$, that

\begin{gather*}
f(\cdot, 1) = f(\cdot,1-2^{2-2k}), \\
f(\cdot, 1) = f(\cdot,1-2^{2-2k}) \circ \big(y_{2\tau_k}^{(i_k;L_k)}\big)^{-1}.
\end{gather*}

In particular, taking $k=1$, we see $f(\cdot,1)=f_0$. Substituting this back in gives $f_0 = f(\cdot,1-2^{2-2k})$ for all $k\in\mathbb{N}$. Again substituting this back in gives that for all $k\in\mathbb{N}$,

\begin{equation*}
f_0 = f_0 \circ y_{2\tau_k}^{(i_k;L_k)}.
\end{equation*}

In terms of the unit vectors $e_1$, $e_2$, we have that for a.e. $x\in\mathbb{T}^2$, $f_0(x) = f_0 (x+2\tau_k e_{i_k})$. Convolving with a smooth function $\rho \in C^\infty(\mathbb{T}^2)$ then gives that for all $x \in \mathbb{T}^2$,

\begin{equation}\label{eqconst}
(\rho * f_0)(x) = (\rho * f_0)(x+2\tau_k e_{i_k}).
\end{equation}

By the Finiteness and Cancellation Conditions (<ref>), (<ref>), we see that $\tau_k \xrightarrow{k\to\infty} 0$, and $i_{k+1}\ne i_k$ (so equal to both $1$ and $2$ infinitely often). Now $(\rho * f_0) \in C^\infty(\mathbb{T}^2)$, and so by (<ref>) we must have that $\rho*f_0$ is a constant. This holds for all $\rho \in C^\infty(\mathbb{T}^2)$, and therefore implies also that $f_0$ is a constant, reaching the required contradiction.

Finally, for a suitably fast growing sequence $\{L_k\}_{k\in\mathbb{N}}\subset\mathbb{N}$, we apply Theorem <ref> to the sequence of vector fields $u_n^{\{(i_k,L_k,\tau_k)\}_{k=1}^n}(x,t)$. This allows us to control the vanishing viscosity limit along the vector field $u_\infty^{\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}}$.

There exists a divergence-free vector field $u \in L^\infty([0,1];L^\infty(\mathbb{T}^2;\mathbb{R}^2))$, and a sequence $\{\nu_n\}_{n\in\mathbb{N}}$ with $\nu_n > 0$ and $\nu_n\xrightarrow{n\to\infty}0$, such that for any initial data $f_0\in  L^\infty(\mathbb{T}^2)$, and for $f^\nu$ the unique solution to (<ref>) along $u$ with initial data $f_0$, one has

\begin{gather*}
f^{\nu_{2n}}\xrightarrow{n\to\infty}f^\mathrm{even}, \\
\end{gather*}

with the above convergence in weak-$*$ $L^\infty([0,1];L^\infty(\mathbb{T}^2))$, and strong in $L^p([0,1];L^p(\mathbb{T}^2))$ for all $p \in [1,\infty)$. Furthermore, the limit functions $f^\mathrm{even}, f^\mathrm{odd}$ are renormalised weak solutions to (<ref>) along $u$ with initial data $f_0$.

[In this case one may show that uncountably many vanishing viscosity solutions exist. We very briefly sketch this. Work in the weak-$*$ topology on $L^\infty L^\infty$. First, any weak-$*$ limit point of $f^\nu$ as $\nu\to0$ is a vanishing viscosity solution. Next, the set of limit points is bounded, and so a metric space. Moreover, the map $\nu\to f^\nu$ is continuous, which implies the set of limit points is connected. If it is a connected metric space, and not unique, then it is uncountable.]Moreover, if $f_0$ is not constant, then $f^\mathrm{even}\ne f^\mathrm{odd}$.

Recall the language and notation introduced in Definitions <ref>, <ref>, as well as Definitions <ref>, <ref> of renormalised and Lagrangian solutions to (<ref>).

Consider any infinite sequence of tuples $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ with $i_k \in \{1,2\}$, $L_k \in \mathbb{N}$, $\tau_k >0$, and satisfying the Finiteness Condition (<ref>).

By Proposition <ref> there exists a Lagrangian solution $f^n\in C^0L^1$, unique in the class of all weak solutions, to (<ref>) along $u_n^{\{(i_k,L_k,\tau_k)\}_{k=1}^n}$ for any initial data $f_0 \in L^\infty(\mathbb{T}^2)$. Moreover, $u_n^{\{(i_k,L_k,\tau_k)\}_{k=1}^n}$ is bounded by $1$ in $L^\infty([0,1];L^\infty(\mathbb{T}^2;\mathbb{R}^2))$.

Let $d_*$ be a metric inducing the weak-$*$ topology on

\begin{equation*}
X = \left\{u\in L^\infty \left( [0,1];L^\infty(\mathbb{T}^2;\mathbb{R}^2)\right) : \left\|u\right\|_{L^\infty L^\infty}\le 1 \right\}.
\end{equation*}

Let $f_0 \in L^\infty(\mathbb{T}^2)$, and denote for each $n\in\mathbb{N}$, $\nu > 0$, by $f^{n,\nu}$, respectively $f^n$, the unique weak solution to (<ref>), respectively (<ref>), along $u_n$ with initial data $f_0$. Moreover denote by $f^{\infty,\nu}$ the unique weak solution to (<ref>) along $u_\infty^{\{(i_k,L_k,\tau_k)\}_{k\in\mathbb{N}}}$ with initial data $f_0$.

Then by Theorem <ref>,

*  For all $n\in\mathbb{N}$, there exists $\nu_n>0$, $\epsilon_n>0$ depending only on $\{(i_k,L_k,\tau_k)\}_{k=1}^n$ (and in particular not on $f_0$), with $\nu_n\xrightarrow{n\to\infty}0$ monotonically, such that the following holds true:

*  If $d_*(u_{n+1},u_n) \le \epsilon_n$ for all $n\in\mathbb{N}$, then for all $p \in [1,\infty)$

\begin{equation*}
\left\|f^{\infty,\nu_n}-f^n\right\|_{L^\infty L^p}\xrightarrow{n\to\infty}0.
\end{equation*}

We now construct such a sequence $\{(i_k,L_k,\tau_k)\}_{k\in \mathbb{N}}$. Let $(i_1, L_1, \tau_1)=(1,2^2,2^{-2})$. We proceed by induction on $n\in\mathbb{N}$. Assume $\{(i_k,L_k,\tau_k)\}_{k=1}^n$ are given, and satisfy for $k\in\{1,...,n\}$.

\begin{equation*}
\begin{gathered}
i_k \in \{1,2\}, \\
L_k \in \mathbb{N}, \\
L_k \ge 2^{2k}, \\
0 < \tau_k \le 2^{-2k},
\end{gathered}
\end{equation*}

It is straightforward to check that this is satisfied for the base case $(i_1, L_1, \tau_1)=(1,2^2,2^{-2})$.

If $n \ge 2$, we assume in the inductive hypothesis that also, for $k\in\{1,...,n-1\}$,

\begin{equation*}
\begin{gathered}
i_{k+1} \ne i_k, \\
2\tau_{k+1} = \frac{1}{2L_k}, \\
2L_{k+1}\tau_{k}\; \text{is an odd integer}, \\
d_*\left(u_{k+1}^{\{(i_{l},L_{l},\tau_{l})\}_{l=1}^{k+1}},u_{k}^{\{(i_{l},L_{l},\tau_{l})\}_{l=1}^{k}}\right) \le \epsilon_{k}.
\end{gathered}
\end{equation*}

We then choose $(i_{n+1}, L_{n+1}, \tau_{n+1})$ as follows. Let $i_{n+1}\in\{1,2\}\setminus\{i_n\}$. Let $\tau_{n+1}=\frac{1}{4L_n}$ which by the inductive hypothesis satisfies $\tau_{n+1}\le 2^{-2k-2}$. Let

\begin{equation}\label{eqLn}
\end{equation}

for some large $M\in\mathbb{N}$ to be chosen, where when $n=1$ we take $L_0 = 1$. This will ensure $2L_{n+1}\tau_n$ is an odd integer.

By (<ref>) and Definition <ref>, we see for $n\in\mathbb{N}$ fixed, that

\begin{equation*}
\end{equation*}

Therefore, by taking $M\in\mathbb{N}$ large enough in (<ref>), we have that both $L_{n+1}\ge 2^{2n+2}$, and

\begin{equation*}
d_*\left(u_{n+1}^{\{(i_k,L_k,\tau_k)\}_{k=1}^{n+1}},u_n^{\{(i_k,L_k,\tau_k)\}_{k=1}^n}\right) \le \epsilon_n,
\end{equation*}

completing the induction.

We have constructed $\{(i_k, L_k, \tau_k)\}_{k \in \mathbb{N}}$ satisfying the Finiteness and Cancellation Conditions (<ref>), (<ref>), and also (<ref>), (<ref>). The statement of the theorem is now a straightforward corollary of Proposition <ref>.

§ GRADUAL PERFECT MIXING - INADMISSIBILITY

§.§ Notation

We continue with the same notation for the 2-torus introduced in Section <ref>. However we no longer make use of the notation in Definitions <ref>, <ref>, <ref>. In addition, we define the binary expansion of some $x=(x_1,x_2) \in \mathbb{T}^2$ as follows. For the representative $(x_1,x_2)\in[0,1)^2$ denote for $i\in\{1,2\}$, $k\in\mathbb{N}$, by $x_{i,k}$ the $k$th binary digit of the $i$th coordinate of $x$. That is, for each $i \in \{1,2\}$,

\begin{equation}\label{eqbinexp}
x_i = \sum_{k=1}^\infty x_{i,k} 2^{-k},
\end{equation}

where $x_{i,k} \in \{0,1\}$ and $x_{i,k} \centernot{\xrightarrow{k\to\infty}} 1$ (as is standard to ensure uniqueness of the binary expansion).

First, we define vector fields and corresponding Lagrangian flows which swap points in $\mathbb{T}^2$ according to their binary expansion. These `binary swaps' form the building block of our construction. Subsequently we shall give a divergence-free vector field in $L^\infty([0,100];L^\infty(\mathbb{T}^2;\mathbb{R}^2))$ with $L^\infty L^\infty$-norm equal to 1, which perfectly mixes the transported scalar to its spatial average, and subsequently unmixes, any initial data to (<ref>). Our aim is then to show that this behaviour is the unique limit point of vanishing viscosity of the associated solution to (<ref>). In order to ensure uniqueness of the limit points it is necessary to `gradually' perform these binary swaps, that is they must be restricted to gradually smaller regions of space, see (<ref>) below.

Suppose $i \in \{1,2\}$, $k \in \mathbb{N}$, $n\in\{1,...,2^{\left\lfloor k/2\right\rfloor}\}$, $L \in \mathbb{N}$, with $L \ge k+1$.

In the proof below will define a time-dependent, divergence-free vector field, called the $(i,k,n;L)$-binary swap

\begin{equation*}
u^{(i,k,n;L)} : \mathbb{T}^2 \times [0,3\cdot2^{-k}] \to \mathbb{R}^2,
\end{equation*}

and a corresponding Lagrangian flow map (Definition <ref>) $\big\{y^{(i,k,n;L)}_t\big\}_{t \in [0,3\cdot2^{-k}]}$ with the properties (<ref>)-(<ref>) below.

Define $J_{k,n}=\left[(n-1)2^{-\left\lfloor k/2\right\rfloor},n2^{-\left\lfloor k/2\right\rfloor}\right]\subset\mathbb{T}$. Then at time $t=3\cdot2^{-k}$, for a.e. $x=(x_1,x_2) \in \mathbb{T}^2$, $y^{(i,k,n;L)}_{3\cdot2^{-k}}$ will swap the $k$th and $(k+1)$th binary digits of $x_i$ if $x_i\in J_{k,n}$.

That is $y_0^{(i,k,n;L)} = \mathrm{Id}$. For a.e. $x=(x_1,x_2)\in\mathbb{T}^2$ denote by $(x'_1,x'_2)=y^{(i,k,n;L)}_{3\cdot2^{-k}}(x)$. Following the notation for binary expansions in (<ref>), for $j\in\{1,2\}$ the coordinate, and for $l\in\mathbb{N}$ the binary digit,

\begin{equation}\label{eqbinaryswap}
\begin{aligned}
x'_j & = x_j \; \text{if} \; j\neq i, \\
x'_{i,l} & = x_{i,l} \; \text{for} \; l\notin\{k,k+1\}, \\
x'_{i,k+1} & = \begin{cases}
x_{i,k} & \text{if } x_i \in J_{k,n}, \\
x_{i,k+1} & \text{otherwise},
\end{cases} \\
x'_{i,k} & = \begin{cases}
x_{i,k+1} & \text{if } x_i \in J_{k,n}, \\
x_{i,k} & \text{otherwise}.
\end{cases}
\end{aligned}
\end{equation}

Additionally, the vector field $u^{(i,k,n;L)}$ will satisfy

\begin{equation}\label{equnifbound}
\left\|u^{(i,k,n;L)}\right\|_{L^\infty L^\infty} \le 1,
\end{equation}

and for all $t\in[0,3\cdot2^{-k}]$, and all $x=(x_1,x_2)\in\mathbb{T}^2$ with $x_i \notin J_{k,n}$,

\begin{equation}\label{eqgradualmixing}
\end{equation}

Moreover, for $i\in\{1,2\}$, $k\in\mathbb{N}$ fixed, as $L \to \infty$,

\begin{equation}\label{eqweakconv}
u^{(i,k,n;L)} \xrightharpoonup{L \to \infty} 0, \; \text{in weak-$*$ $L^\infty L^\infty$}.
\end{equation}

Finally, for any $r, r' \in \{1,...,2^{k-1}\}$, for the spatial intervals $J=[(r-1)2^{1-k},r2^{1-k}]$, and $J'=[(r'-1)2^{1-k},r'2^{1-k}]$, the Lagrangian-flow $y^{(i,k,n;L)}_t$ preserves the squares

\begin{equation}\label{eqrectangle}
y^{(i,k,n;L)}_t:J\times J'\leftrightarrow J\times J'.
\end{equation}

We now construct the above vector field and Lagrangian flow. We shall give the construction for $i=1$, and then define for $i=2$ its coordinate reflection. That is, for each $(x_1,x_2) \in \mathbb{T}^2$, $t \in [0,3\cdot2^{-k}]$

\begin{gather}
u^{(2,k,n;L)}((x_1,x_2),t) = u^{(1,k,n;L)}((x_2,x_1),t), \label{eqi2u} \\
y_t^{(2,k,n;L)}((x_1,x_2)) = y_t^{(1,k,n;L)}((x_2,x_1)). \label{eqi2y}
\end{gather}

so that (<ref>), (<ref>), (<ref>) with $i=2$ follow from the same result for $i=1$.

We shall achieve the required binary-swaps (<ref>) by piecing together particular rotating vector fields which perform half rotations of rectangular regions of $\mathbb{T}^2$. Suppose $W,H>0$, and define the following 1-Lipschitz stream-function

\begin{gather*}
\psi:\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)\subset\mathbb{R}^2 \to \mathbb{R}, \\
\psi((x_1,x_2)) = \min\{W,H\}\cdot\max\left\{\left(\frac{x_1}{W}\right)^2,\left(\frac{x_2}{H}\right)^2\right\}.
\end{gather*}

This defines a 1-bounded, time-independent, divergence-free, vector field $\nabla^\perp\psi = \big(-\frac{\partial\psi}{\partial x_2}, \frac{\partial\psi}{\partial x_1}\big)$ on the open rectangle $\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$. $\nabla^\perp\psi$ has rectangular flow lines, as illustrated in the following diagram:

\begin{equation}\label{eqrectflow}
\begin{tikzpicture}[scale=3/2]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=1/2,decoration={
mark=at position 1/36 with {\arrowreversed[xshift=-2pt]{latex[length=4pt]}},
mark=at position 10/36 with {\arrowreversed[xshift=-2pt]{latex[length=4pt]}},
mark=at position 19/36 with {\arrowreversed[xshift=-2pt]{latex[length=4pt]}},
mark=at position 28/36 with {\arrowreversed[xshift=-2pt]{latex[length=4pt]}},
\draw[dashed,postaction={decorate}] (3,3/8) rectangle (5,5/8);
\draw[dashed,postaction={decorate}] (2,1/4) rectangle (6,3/4);
\draw[dashed,postaction={decorate}] (1,1/8) rectangle (7,7/8);
\draw[thick] (0,0) -- (8,1);
\draw[thick,] (0,1) -- (8,0);
\draw[thick,red] (0,0) rectangle (8,1);
\coordinate (n3) at (0,0);
\coordinate (n4) at (0,1);
\coordinate (n5) at (8,0);
\coordinate (n6) at (8,1);
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,1) -- (8,1) node[pos=0.5,above=10pt,black]{$W$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,0) -- (0,1) node[pos=0.5,left=7pt,black]{$H$};
\end{scope}
\end{tikzpicture}
\end{equation}

Moreover, on each triangular segment, $\nabla^\perp\psi$ is a linear function of space, and each flow line within each segment has time period equal to $\max\{W,H\}$.

[That is for all $t\in(-\infty,\infty)$, $y_t:\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)\leftrightarrow \left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$ is a Lebesgue-measure preserving bijection, $y_0= \mathrm{Id}$, and for all $x\in\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$, and all $t\in(-\infty,\infty)$, $\frac{d}{dt}y_t(x)=\nabla^\perp\psi(y_t(x))$.]Therefore, $\nabla^\perp\psi$ admits a Lagrangian flow $\big\{y_t\big\}_{t\in(-\infty,\infty)}$ on $\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$.

[Vector fields achieving the same half rotation property with more regularity exist in the literature. See for example Section 3 in <cit.>.]Moreover, we have $y_0= \mathrm{Id}$, while $y_{2\max\{W,H\}}$ is exactly a half rotation of $\left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$ around its centre. [This may be easier to see by first using a different metric.]Finally, both $y_t$ and $y_t^{-1}$ are Lipschitz in both $x$ and $t$. That is, for all $x,x' \in \left(-\frac{W}{2},\frac{W}{2}\right)\times\left(-\frac{H}{2},\frac{H}{2}\right)$, and all $t,s \in (-\infty,\infty)$,

\begin{equation}\label{eqlipbound}
\begin{gathered}
\left|y_t(x) - y_s(x')\right| \le C(|x-x'| + |t-s|), \\
\left|y_t^{-1}(x) - y_s^{-1}(x')\right| \le C(|x-x'| + |t-s|),
\end{gathered}
\end{equation}

for some constant $C>0$.

Consider now any open rectangle $Q\subset\mathbb{R}^2$ with width $W$ and height $H$. By translating (<ref>) to $Q$ we have the same vector field on $Q$, which to simplify the following presentation we symbolically notate by

\begin{equation*}
\vcenter{\hbox{\begin{tikzpicture}[scale=1]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=1/4]
\draw[red, thick] (-8,0) rectangle (0,1);
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (-4,1/2) ellipse (2 and 1/4);
\end{scope}
\end{tikzpicture}}} : Q \to \mathbb{R}^2.
\end{equation*}

We iterate that this is a 1-bounded, time-independent, divergence-free vector field, admitting a Lipschitz Lagrangian flow $\big\{y_t\big\}_{t\in(-\infty,\infty)}$, as in (<ref>). Moreover, $y_0=\mathrm{Id}$, while $y_{2\max\{W,H\}}$ is exactly a half rotation of $Q$ around its centre.

We also denote the same vector field multiplied by $-1$ by the same diagram with the arrows reversed,

\begin{equation*}
\vcenter{\hbox{\begin{tikzpicture}[scale=1]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=1/4]
\draw[red, thick] (-8,0) rectangle (0,1);
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrowreversed[scale=1]{latex}},
mark=at position 3/4 with {\arrowreversed[scale=1]{latex}},
}] (-4,1/2) ellipse (2 and 1/4);
\end{scope}
\end{tikzpicture}}} = -
\vcenter{\hbox{\begin{tikzpicture}[scale=1]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=1/4]
\draw[red, thick] (-8,0) rectangle (0,1);
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (-4,1/2) ellipse (2 and 1/4);
\end{scope}
\end{tikzpicture}}}.
\end{equation*}

This admits the Lagrangian flow $\big\{y_t^{-1}\big\}_{t\in(-\infty,\infty)}$, and so enjoys the same properties as above.

Next, we piece these half-rotations together to perform orientation-preserving swaps. This is exactly the building block required to perform the binary swaps (<ref>). Let $Q\subset\mathbb{R}^2$ be again an open rectangle with width $W>0$, and height $H>0$. Let $L \in \mathbb{N}$ and suppose in addition that $H$ is an integer multiple of $2^{1-L}$, and $W \ge 2^{1-L}$.

Subsequently, we define a time-dependent vector field, symbolically notated by $\vcenter{\hbox{\begin{tikzpicture}[scale=1/2]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}
\draw[dashed] (1,0) -- (1,1/2);
\draw[thick,-latex,yshift=3pt] (0.5,1/4) -- (1.5,1/4);
\draw[thick,latex-,yshift=-3pt] (0.5,1/4) -- (1.5,1/4);
\draw[thick,red] (0,0) rectangle (2,1/2);
\end{scope}
\end{tikzpicture}}}: Q \times \left[0,3W\right] \to \mathbb{R}^2$, by

\begin{equation}\label{eqflipdiag}
\vcenter{\hbox{
\begin{tikzpicture}[scale=1]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}
\draw[dashed] (1,0) -- (1,1);
\draw[thick,-latex,yshift=3pt] (0.5,1/2) -- (1.5,1/2);
\draw[thick,latex-,yshift=-3pt] (0.5,1/2) -- (1.5,1/2);
\draw[thick,red] (0,0) rectangle (2,1);
\end{scope}
\end{tikzpicture}}}\;(\cdot,t) = \begin{cases}
\begin{tikzpicture}[scale=2/3]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=2,shift={(0.75,0.5)}]
\foreach \y in {0,...,3}{
%\draw[thick, red] (0,\y/4) rectangle (2,\y/4+1/4);
\pgfmathparse{int(mod(\y,2))}
\ifnum\pgfmathresult=0
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrowreversed[scale=1]{latex}},
mark=at position 3/4 with {\arrowreversed[scale=1]{latex}},
}] (1,\y/4+1/8) ellipse (1/2 and 1/16);
\else
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (1,\y/4+1/8) ellipse (1/2 and 1/16);
\fi
\foreach \y in {1,...,3}{
\draw[dashed] (0,\y/4) -- (2,\y/4);
\foreach \y in {1,...,1}{
\draw[thick,dashed] (0,\y/2) -- (2,\y/2);
\draw[thick,red] (0,0) rectangle (2,1);
\coordinate (n3) at (0,0);
\coordinate (n4) at (0,1);
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,1) --  (2,1) node[pos=0.5,above=5pt,black]{$W$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,0) --  (0,1) node[pos=0.5,left=5pt,black]{$H$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (2,0.25) --  (2,0) node[pos=0.5,right=2pt,black]{$2^{-L}$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (2,1) --  (2,0.5) node[pos=0.5,right=2pt,black]{$2^{1-L}$};
\coordinate[right=10pt] (time0) at (2,1) {};
\end{scope}
\end{tikzpicture} & \text{if $t \in [0,2W)$,} \\
\begin{tikzpicture}[scale=2/3]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=2,shift={(3.25,0.5)}]
\foreach \y in {0,...,3}{
\pgfmathparse{int(mod(\y,2))}
\foreach \x in {0,1}{
%\draw[thick, red] (\x,\y/4) rectangle (\x+1,\y/4+1/4);
\ifnum\pgfmathresult=0
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrowreversed[scale=1]{latex}},
mark=at position 3/4 with {\arrowreversed[scale=1]{latex}},
}] (\x+1/2,\y/4+1/8) ellipse (1/4 and 1/16);
\else
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (\x+1/2,\y/4+1/8) ellipse (1/4 and 1/16);
\fi
\foreach \y in {1,...,3}{
\draw[dashed] (0,\y/4) -- (2,\y/4);
\foreach \y in {1,...,1}{
\draw[thick,dashed] (0,\y/2) -- (2,\y/2);
\draw[thick,dashed] (1,0) -- (1,1);
\draw[thick,red] (0,0) rectangle (2,1);
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,1) -- (1,1) node[pos=0.5,above=5pt,black]{$\frac{1}{2}W$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,0) --  (0,1) node[pos=0.5,left=5pt,black]{$H$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (2,1) --  (2,3/4) node[pos=0.5,right=2pt,black]{$2^{-L}$};
\coordinate[right=10pt] (time1) at (2,1) {};
\end{scope}
\end{tikzpicture} & \text{if $t \in [2W,3W]$.}
\end{cases}
\end{equation}

This has the property that at time $t=3W$ the left and right halves are swapped in an orientation preserving manner. This is illustrated below:

\begin{equation*}
\begin{tikzpicture}[scale=2/3]
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}[scale=2,shift={(-2,0.5)}]
\shade[left color=black!80,right color=white,shading angle=0] (0,0) rectangle (2,1);
\shade[opacity=0.6,left color=black,right color=white,shading angle=90] (0,0) rectangle (2,1);
\draw[thick,red] (0,0) rectangle (2,1);
\end{scope}
\begin{scope}[scale=2,shift={(0.75,0.5)}]
\shade[left color=black!80,right color=white,shading angle=0] (0,0) rectangle (2,1);
\shade[opacity=0.6,left color=black,right color=white,shading angle=90] (0,0) rectangle (2,1);
\foreach \y in {0,...,3}{
%\draw[thick, red] (0,\y/4) rectangle (2,\y/4+1/4);
\pgfmathparse{int(mod(\y,2))}
\ifnum\pgfmathresult=0
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrowreversed[scale=1]{latex}},
mark=at position 3/4 with {\arrowreversed[scale=1]{latex}},
}] (1,\y/4+1/8) ellipse (1/2 and 1/16);
\else
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (1,\y/4+1/8) ellipse (1/2 and 1/16);
\fi
\foreach \y in {1,...,3}{
\draw[dashed] (0,\y/4) -- (2,\y/4);
\draw[thick,red] (0,0) rectangle (2,1);
\coordinate (n3) at (0,0);
\coordinate (n4) at (0,1);
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,1) --  (2,1) node[pos=0.5,above=5pt,black]{$W$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,3/4) --  (0,1) node[pos=0.5,left=2pt,black]{$2^{-L}$};
\coordinate[right=10pt] (time0) at (2,1) {};
\end{scope}
\begin{scope}[scale=2,shift={(3.25,0.5)}]
\foreach \y in {0,...,3}{
\pgfmathparse{\y*80/4}
\let\a\pgfmathresult
\pgfmathparse{\y*80/4+20}
\let\b\pgfmathresult
\shade[left color=black!\a,right color=black!\b,shading angle=0] (0,\y/4) rectangle (2,\y/4+1/4);
\shade[opacity=0.6,left color=white,right color=black,shading angle=90] (0,0) rectangle (2,1);
\foreach \y in {0,...,3}{
\pgfmathparse{int(mod(\y,2))}
\foreach \x in {0,1}{
%\draw[thick, red] (\x,\y/4) rectangle (\x+1,\y/4+1/4);
\ifnum\pgfmathresult=0
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrowreversed[scale=1]{latex}},
mark=at position 3/4 with {\arrowreversed[scale=1]{latex}},
}] (\x+1/2,\y/4+1/8) ellipse (1/4 and 1/16);
\else
\draw[thick,postaction={decorate},decoration={
mark=at position 1/4 with {\arrow[scale=1]{latex}},
mark=at position 3/4 with {\arrow[scale=1]{latex}},
}] (\x+1/2,\y/4+1/8) ellipse (1/4 and 1/16);
\fi
\foreach \y in {1,...,3}{
\draw[dashed] (0,\y/4) -- (2,\y/4);
\draw[dashed] (1,0) -- (1,1);
\draw[thick,red] (0,0) rectangle (2,1);
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (0,1) -- (1,1) node[pos=0.5,above=5pt,black]{$\frac{1}{2}W$};
\draw[thick,decorate,decoration = {
calligraphic brace,
}] (2,1) --  (2,3/4) node[pos=0.5,right=2pt,black]{$2^{-L}$};
\coordinate[right=10pt] (time1) at (2,1) {};
\end{scope}
\begin{scope}[scale=2,shift={(6,0.5)}]
\shade[left color=black!80,right color=white,shading angle=0] (0,0) rectangle (2,1);
\shade[opacity=0.6,left color=black!50,right color=white,shading angle=90] (0,0) rectangle (1,1);
\shade[opacity=0.6,left color=black,right color=black!50,shading angle=90] (1,0) rectangle (2,1);
\draw[thick, red] (0,0) rectangle (2,1);
\coordinate[right=10pt] (time2) at (2,1) {};
\end{scope}
\draw[thick, -latex] (1,0) -- (11,0);
\fill (1,0) circle (2pt);
\fill (6,0) circle (2pt);
\fill (11,0) circle (2pt);
\node[below=5pt] (t0) at (1,0) {$t=0$};
\node[below=5pt] (t1) at (6,0) {$t=2W$};
\node[below=5pt] (t2) at (11,0) {$t=3W$};
\draw[very thick,decorate,decoration = {
calligraphic brace,
}] (1,0) --  (6,0) node[pos=0.5,above=5pt,black]{};
\draw[very thick,decorate,decoration = {
calligraphic brace,
}] (6,0) --  (11,0) node[pos=0.5,above=5pt,black]{};
\end{tikzpicture}
\end{equation*}

Finally, suppose $k \in\mathbb{N}$, $n \in \{1,...,2^{\left\lfloor k/2\right\rfloor}\}$, and $L\in\mathbb{N}$, with $L\ge k+1$.

Let $J_{k,n}=\left[(n-1)2^{-\left\lfloor k/2\right\rfloor},n2^{-\left\lfloor k/2\right\rfloor}\right]\subset\mathbb{T}$.

For $W=2^{-k}$, $H=1$, and $L\ge k+1$ we may use
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{scope}
\draw[dashed] (1,0) -- (1,1/2);
\draw[thick,-latex,yshift=3pt] (0.5,1/4) -- (1.5,1/4);
\draw[thick,latex-,yshift=-3pt] (0.5,1/4) -- (1.5,1/4);
\draw[thick,red] (0,0) rectangle (2,1/2);
\end{scope}
\end{tikzpicture}}}$ to define the time-dependent vector field

\begin{gather*}
u^{(1,k,n;L)} : \mathbb{T}^2 \times [0,3\cdot2^{-k}] \to \mathbb{R}^2, \\
\begin{tikzpicture}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}