Nonuniqueness of Solutions to the Euler Equations with Vorticity in a Lorentz Space

Abstract

For the two dimensional Euler equations, a classical result by Yudovich states that solutions are unique in the class of bounded vorticity; it is a celebrated open problem whether this uniqueness result can be extended in other integrability spaces. We prove in this note that such uniqueness theorem fails in the class of vector fields u with uniformly bounded kinetic energy and vorticity in the Lorentz space $L^{1, \infty }$.

1 Introduction

Let us consider the 2-dimensional Euler equations

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u + \mathrm{{div}}\,(u \otimes u ) + \nabla p =0 \\ \mathrm{{div}}\,u = 0 \end{array}\right. } \end{aligned}$$

(1)

where $u: [0,1] \times {{\mathbb {T}}}^2 \rightarrow {\mathbb {R}}^2$ is the velocity of a fluid and $p: [0,1] \times {{\mathbb {T}}}^2 \rightarrow {\mathbb {R}}$ the pressure. This system can be equivalently rewritten as the two dimensional Euler system in vorticity formulation, which is a transport equation for the vorticity $\omega = \textrm{curl} (u)$, i.e.

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \omega + u \cdot \nabla \omega =0 \\ u = \nabla ^\perp \Delta ^{-1} \omega . \end{array}\right. \qquad \text{ in } {{\mathbb {T}}}^2 \times [0,1] . \end{aligned}$$

(2)

In the latter formulation, it is clear that $L^p$ norms of the vorticity are formally conserved for any $p \in [1,\infty ]$. For $p>1$, this was used in [11] to prove the existence of distributional solutions starting from an initial datum with vorticity in $L^p$. A similar existence result is much more involved for $p=1$, and it was obtained by Delort [10] (see also [11, 12, 22]), improving the existence theory up to measure initial vorticities in $H^{-1}$ (this latter condition guarantees finiteness of the energy) whose positive (or negative) part is absolutely continuous. As regards uniqueness, the classical result of Yudovich [15, 16] (see also the proof in [17]) states that, given an initial datum $\omega _0 \in L^\infty$, there exists a unique bounded solution to (2) starting from $\omega _0$. However, the classical problem raised by Yudovich about the sharpness of his result is still open. Let $u_0$ be an initial datum in $L^2$ with ${{\,\textrm{curl}\,}}u_0$ in some function space X. Is the solution of the Euler equations in vorticity formulation unique in the class $L^\infty (X)$?

The main result of this paper provides a negative answer when X is the Lorentz space $L^{1,\infty }$.

Theorem 1.1

There exists a nontrivial solution $u\in C^0([0,1]; L^2({{\mathbb {T}}}^2))$ to (1) satisfying

• (i) $\omega = {{\,\textrm{curl}\,}}u\in C^0([0,1]; L^{1,\infty }({{\mathbb {T}}}^2))$;

• (ii) $u(0,\cdot ) = 0$.

Moreover, $u\in C^0([0,1]; W^{s,p}({{\mathbb {T}}}^2))$ for any $s\in (0,1)$ and $p\in (1, \frac{2}{1+s})$.

Remark 1.2

The conclusion (i) in Theorem 1.1 has to be interpreted as follows. There exist a function ${{\,\textrm{curl}\,}}u\in C^0([0,1]; L^{1,\infty }({{\mathbb {T}}}^2))$ and a sequence $u_n\in C^\infty ([0,1]\times {{\mathbb {T}}}^2)$ solving the Euler equations with an error term $R_n$ in the right hand side (see (3)) such that

$$\begin{aligned} \Vert R_n \Vert _{C^0(L^1)} + \Vert u_n - u \Vert _{C^0(L^2)} + \Vert {{\,\textrm{curl}\,}}u_n - {{\,\textrm{curl}\,}}u \Vert _{C^0(L^{1,\infty })} \rightarrow 0 , \end{aligned}$$

as $n \rightarrow 0$.

Recently, there have been formidable attempts to disprove this conjecture for $X= L^p$, none of which has by now fully solved it. Vishik [23, 24], see also [1], proposed a complex line of approach to this problem, which however has the price of showing nonuniqueness only with an additional degree of freedom, namely a forcing term in the right-hand side of the equation (2) in the integrability space $L^1(L^p)$. The nonuniqueness suggested by this work is of symmetry breaking type and, in contrast with the ideas of this paper, his nonuniqueness stems from the linear part of the equation, by carefully choosing an initial datum that sees the instability directions of a linearized operator.

A second attempt has been pursued by Bressan and Shen [2], based on numerical experiments which share the symmetry breaking type of nonuniqueness of Vishik. Their work is a first step in the direction of a computer assisted proof.

Our approach is instead of different nature and stems from the convex integration technique. The latter was introduced by De Lellis and Székelyhidi [9] in the context of nonlinear PDEs, inspired by the work of Nash on isometric embeddings [20], which found striking applications in recent years to different PDEs (see for instance [5,6,7, 14, 18, 19] and the references quoted therein). As such, our proof would probably be less constructive with respect to the strategies of [2, 23, 24], where an initial datum for which nonuniqueness is expected is described fairly explicitly as well as the mechanism for the creation of two different singularities. Conversely, the latter approaches see the drawbacks described above and are by no means “generic” in the initial data, whereas it is known (see for instance [8, 21]) that convex integration methods yield not only the lack of uniqueness/smoothness for certain specific initial data, but also that solutions are typical (in the Baire category sense).

1.1 Strategy of proof

The guiding thread of this construction is an iterative procedure, where one starts from a solution $(u_q, p_q, R_q)$ of the Euler equations with an error term in the right-hand side, namely

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _t u_q + \mathrm{{div}}\,(u_q \otimes u_q ) + \nabla p_q = \mathrm{{div}}\,R_q \\ \mathrm{{div}}\,u_q = 0, \end{array}\right. } \end{aligned}$$

(3)

and iteratively corrects this error by adding a fastly oscillating perturbation to the approximate solution. The nonlinear interaction of this perturbation with itself generates a resonance which allows for the cancellation of the previous error; the other terms are mainly seen as new error terms, with smaller size with respect to the previous error. More precisely, we define the new solution $(u_{q+1}, p_{q+1}, R_{q+1})$ by setting

$$\begin{aligned} u_{q+1}(x) =u_q(x) + a_q(x) w_{\lambda _{q+1}}(x) , \quad w_{\lambda _{q+1}}(x) := w_{q+1}(\lambda _{q+1} x) , \, \, \, \, \, \, \lambda _{q+1}\in \mathbb {Z} , \end{aligned}$$

where $\lambda _{q+1} \gg \lambda _q$ is a higher frequency with respect to the typical frequencies in $u_q$, $w_{q+1}$ is called building block of the construction and enjoys suitable integrability properties, $a_q$ is a slowly varying coefficient.

The cancellation of error happens because the low frequency term in $a^2_q w_{\lambda _{q+1}} \otimes w_{\lambda _{q+1}}$ satisfies

$$\begin{aligned} a^2_q \int w_{\lambda _{q+1}} \otimes w_{\lambda _{q+1}} \sim R_q \, . \end{aligned}$$

This forces us to require that

$$\begin{aligned} \begin{aligned}&\int _{{{\mathbb {T}}}^2} |w_{q+1}|^2 \sim 1\\&a^2_q \sim |R_q| . \end{aligned} \end{aligned}$$

(4)

On the contrary, we wish to control the quantity $\Vert D u_q\Vert _{L^{1, \infty }}$ and to this end we need

$$\begin{aligned} \Vert D (u_{q+1} - u_q) \Vert _{L^{1,\infty }} \sim \Vert a_q \Vert _{L^\infty } \Vert \lambda _{q+1} w_{\lambda _{q+1}} \Vert _{L^{1, \infty }} , \end{aligned}$$

arbitrarily small. This will be achieved by designing a new family of intermittent building blocks with gradients small in Lorentz spaces, see Sect. 1.3 below.

1.2 Limitations of current convex integration schemes

We now justify why, with the current method, getting a vorticity in $L^p$, $p \ge 1$, cannot be expected. The new error $R_{q+1}$, generated after correcting the old error $R_q$ with the low frequency term in $a^2_q w_{\lambda _{q+1}} \otimes w_{\lambda _{q+1}}$, contains the high frequencies of $a^2_q w_{\lambda _{q+1}} \otimes w_{\lambda _{q+1}}$, hence its size is at least

$$\begin{aligned} \Vert R_{q+1} \Vert _{L^1}&\ge \Big \Vert \nabla (a_q^2) \textrm{div}^{-1}\Big (w_{\lambda _{q+1}}\otimes w_{\lambda _{q+1}} - \int w_{\lambda _{q+1}}\otimes w_{\lambda _{q+1}}\big )\Big \Vert _{L^1} \\&\ge \lambda _{q+1}^{-1}\Vert \nabla R_q \Vert _{L^1} \ge \lambda _{q+1}^{-1}\lambda _q \Vert R_q \Vert _{L^1}. \end{aligned}$$

Here we used (4), i.e. $a^2\sim |R_q|$ and that the intermittent term $w_{\lambda _{q+1}}$ has unitary norm in $L^2$. In particular, we have that $\Vert R_q \Vert _{L^1} \lambda _q$ is a nondecreasing sequence in q and hence

$$\begin{aligned} \Vert R_q \Vert _{L^1} > rsim \lambda _q^{-1} . \end{aligned}$$

(5)

In order to control $\Vert \nabla u_q\Vert _{L^1}$ we need $\Vert \nabla (u_{q+1} - u_q) \Vert _{L^1}$ arbitrarily small, but the Sobolev inequality gives that

$$\begin{aligned} \Vert \nabla w_{\lambda _{q+1}} \Vert _{L^{1}} = \lambda _{q+1} \Vert \nabla w_{q+1} \Vert _{L^{1}} > rsim \lambda _{q+1} \Vert w_{q+1} \Vert _{L^2} \sim \lambda _{q+1} . \end{aligned}$$

Hence,

$$\begin{aligned} \Vert \nabla (u_{q+1} - u_q) \Vert _{L^1} \sim \Vert a_q \Vert _{L^\infty } \Vert \lambda _{q+1} w_{\lambda _{q+1}} \Vert _{L^1} > rsim \lambda _{q+1} \sqrt{\Vert R_q \Vert }_{L^1} > rsim \lambda _{q+1}^{1/2}, \end{aligned}$$

(6)

where we used (5) in the last step.

In particular, with the current way to cancel the error in the iteration, we cannot expect $\nabla u \in L^1$.

Note that the inequality (6), where we end up with a term of size $\sim \sqrt{\lambda _{q+1}}$, is in agreement with a known limitation of the convex integration scheme, see [5, Section 2.4.1]. In fact, it is shown there that the quadratic error term allows to control at most half of the derivative of the solution u. The only way to overcome this is to exploit the intermittency, but in our context we cannot because of the Sobolev inequality.

1.3 Building blocks with Lorentz integrability

To push the convex integration scheme to its boundaries and obtain $X=L^{1,\infty }$, we need to introduce a new family of building blocks $\{W_i\}_{ i=1, \ldots , 4}$. The latter is the most important novelty of this paper, and its construction requires a new idea. In a nutshell, we design $W_i$ so that its atomic decomposition, as a Lorentz function, is made up of “almost solutions” to the Euler equations. To this aim, we bundle together a family of intermittent jets [4] with different sizes and characteristic velocities. This structure allows sharpening the intermittency mechanism reaching the critical $L^{1,\infty }$ integrability of $\nabla W_i$.

To put forward this idea, there are several technical challenges to overcome, let us mention a few.

The high velocity of each jet, needed to force the bundle to almost solve the Euler equations, makes the term $\partial _t W_i$ big. The latter should be treated as an error, hence we have to make its anti-divergence small. To do so, we exploit a special structure: we build the profiles of our jets in such a way that $\partial _t W_i = \mathrm{{div}}\,(A_i)$ where $A_i$ is a small symmetric potential.

The bundle structure and the 2-dimensional constraint make it very difficult to keep the supports of $W_i$ disjoint in space-time. It requires a new, delicate, combinatorial argument.

We refer the reader to Sect. 4 for the precise construction and more explanations on our choice of building blocks.

Remark 1.3

The proof of Theorem 1.1 is flexible enough, due to the exponential convergence of the iterative sequence, to give $\omega \in L^{1,q}$ for some $q \gg 1$. A technical refinement of the current proof, based on Remark 4.4, would give $q>4$.

2 Iteration and Euler–Reynolds System

We consider the system of equations (3) in $[0,1] \times {{\mathbb {T}}}^2$, where R is a traceless symmetric tensor.

As already remarked, our solution to (1) is obtained by passing to the limit solutions of (3) with suitable constraints on u and R. The latter are built by means of an iterative procedure based on the following.

Proposition 2.1

There exists $M>0$ such that the following holds. For any smooth solution $(u_0, p_0, R_0)$ of (3), there exists another smooth solution $(u_1,p_1,R_1)$ of (3) such that

• (i) $\Vert R_1 \Vert _{L^\infty (L^1)} \le \frac{1}{3}\Vert R_0 \Vert _{L^\infty (L^1)}$;

• (ii) $\Vert u_1 - u_0 \Vert _{C^0(L^2)} + \Vert {{\,\textrm{curl}\,}}(u_1 - u_0) \Vert _{C^0(L^{1,\infty })} \le M \Vert R_0 \Vert _{L^\infty (L^1)}$;

• (iii) for any $s\in (0,1)$, $p\in (1, \frac{2}{1+s})$ there exists $c(p,s)>0$ s.t.

  $$\begin{aligned} \Vert D^s(u_1 - u_0) \Vert _{C^0(L^p)} \le (M \Vert R_0 \Vert _{L^\infty (L^1)})^{c(p,s)}\,; \end{aligned}$$

• (iv) if $R_0(\cdot ,t)=0$ in $[0,t_0]$, then $R_1(\cdot ,t )=0$ and $u_1(\cdot ,t) = u_0(\cdot ,t)$ in $[0,t_0/2]$.

Proof of Theorem 1.1, given Proposition 2.1

Fix $\lambda >0$. We start the iteration scheme with

$$\begin{aligned} u_0(x,t) := \chi (t) \sin (x_2 \lambda ) e_1 \end{aligned}$$

where $\chi \in C_c^\infty ([0,1])$, $\chi =0$ in [0, 1/2] and $\chi = 1$ in [3/2, 1]. Notice that $-\mathrm{{div}}\,R_0 = \chi '(t) \sin (x_2 \lambda ) e_1 + \nabla p$, hence we can choose a traceless symmetric tensor $R_0$ such that $\Vert R_0 \Vert _{L^1} \le C\lambda ^{-1}$.

Applying iteratively Proposition 2.1 with $t_0 = 1/2$ we build a sequence $\{(u_n, p_n, R_n)\,: \, n\in {\mathbb {N}}\}$ of smooth solutions to (1) such that, for any $n\ge 0$, it holds

$$\begin{aligned} \Vert R_n \Vert _{L^\infty (L^1)}&\le C 3^{-n} \lambda ^{-1} , \quad \Vert u_{n+1} - u_n \Vert _{C^0(L^2)} + \Vert {{\,\textrm{curl}\,}}(u_{n+1} - u_n) \Vert _{C^0(L^{1,\infty })}\\ {}&\le CM 3^{- n + 1} \lambda ^{-1} , \end{aligned}$$

and $u_n(\cdot , t) = 0$ for any $t\in [0,2^{-n-1}]$. Moreover, for any $s\in (0,1)$ and $p\in \left( 1, \frac{2}{1+s}\right)$ it holds

$$\begin{aligned} \Vert D^s(u_{n+1} - u_n) \Vert _{C^0(L^p)} \le C(M,\lambda ) 3^{-n c(p,s)} . \end{aligned}$$

(7)

It follows that $R_n \rightarrow 0$ in $L^\infty (L^1)$ and $u_n \rightarrow u$ in $C^0(L^2)$, where u satisfies the assumptions of Theorem 1.1. Moreover $u\in C^0(W^{s,p})$ for $s\in (0,1)$ and $p\in (1, \frac{2}{1+s})$ as a consequence of (7).

We now prove that there exists ${{\,\textrm{curl}\,}}u\in C^0(L^{1,\infty })$ with the property that $\Vert {{\,\textrm{curl}\,}}u_n - {{\,\textrm{curl}\,}}u\Vert _{C^0(L^{1,\infty })} \rightarrow 0$ as $n\rightarrow \infty$. A bit of care is needed since only the weak triangle inequality $\Vert f + g \Vert _{L^{1,\infty }} \le 2 \Vert f\Vert _{L^{1,\infty }} + 2 \Vert g \Vert _{L^{1,\infty }}$ holds true. However, the latter is enough for our purposes

$$\begin{aligned} \Vert {{\,\textrm{curl}\,}}u_N \Vert _{C^0(L^{1,\infty })}&= \Big \Vert {{\,\textrm{curl}\,}}u_0 + {{\,\textrm{curl}\,}}\left( \sum _{n=0}^{N-1} u_{n+1} - u_n\right) \Big \Vert _{C^0(L^{1,\infty })}\\&\le 2 \Vert {{\,\textrm{curl}\,}}u_0 \Vert _{C^0(L^{1,\infty })} + \sum _{n=0}^{N-1} 2^{n + 1} \Vert {{\,\textrm{curl}\,}}(u_{n+1} - u_n) \Vert _{C^0(L^{1,\infty })}\\&\le 2 \Vert {{\,\textrm{curl}\,}}u_0 \Vert _{C^0(L^{1,\infty })} + CM\lambda ^{-1}\sum _{n=0}^{N-1} 2^{n + 1}3^{-n + 1} < \infty , \end{aligned}$$

hence setting ${{\,\textrm{curl}\,}}u:= {{\,\textrm{curl}\,}}u_0 + \sum _{n=0}^\infty ({{\,\textrm{curl}\,}}u_{n+1} - {{\,\textrm{curl}\,}}u_n)$ we get the sought conclusion. $\square$

The remaining part of this note is devoted to the proof of Proposition 2.1. In Sect. 4 we introduce the building blocks of our construction, in Sect. 5 we use them to define the perturbation $u_1-u_0$, finally in Sect. 6, we introduce the new error term $R_1$ and show that it can be made arbitrarily small.

3 Preliminary Lemmas

3.1 Lorentz spaces

For every measurable function $f:{{\mathbb {T}}}^d \rightarrow {\mathbb {R}}$ we recall the definition

$$\begin{aligned} \Vert {f}\Vert _{L^{r,q}} := r^{1/q} \big \Vert \lambda {{\mathscr {L}}}^{d}(\{ |f| \ge \lambda \})^{1/r} \Vert _{L^q\left( (0,{\infty }), \frac{\, d \lambda }{\lambda } \right) } , \end{aligned}$$

(see e.g. [13]) and we define the Lorentz space $L^{r,q}$ with $r\in [1,\infty )$, $q\in [1,\infty ]$, as the space of those functions f such that $\Vert f\Vert _{L^{r,q}}<\infty$. Note that, in spite of the notation, $\Vert \cdot \Vert _{L^{r,q}}$ is in general not a norm but for $(r,q)\ne (1,\infty )$ the topological vector space $L^{r,q}$ is locally convex and there exists a norm $|||\cdot |||_{r,q}$ which is equivalent to $\Vert \cdot \Vert _{L^{r,q}}$ in the sense that the inequality $C^{-1} |||f|||_{r,q}\le \Vert f\Vert _{L^{r,q}} \le C |||f|||_{r,q}$ holds.

3.2 Improved Hölder inequality

We recall the following improved Hölder inequality, stated as in [18, Lemma 2.6] (see also [3, Lemma 3.7]). If $\lambda \in {\mathbb {N}}$ and $f,g:{{\mathbb {T}}}^2 \rightarrow {\mathbb {R}}$ are smooth functions, then we have

$$\begin{aligned} \Vert f(x) g(\lambda x) \Vert _{L^p} \le \Vert f \Vert _{L^p} \Vert g \Vert _{L^p} + C(p) \lambda ^{-1/p} \Vert f \Vert _{C^1} \Vert g\Vert _{L^p} \, \quad \text {for any}\, p\in [1, \infty ] . \nonumber \\ \end{aligned}$$

(8)

When $\int _{{{\mathbb {T}}}^2} g = 0$, then

$$\begin{aligned} \Big | \int _{{{\mathbb {T}}}^2} f(x) g(\lambda x) \, dx \Big | \le C \lambda ^{-1} \Vert f \Vert _{C^1} \Vert g\Vert _{L^1} . \end{aligned}$$

(9)

3.3 Anti-divergence operators

Let now us introduce the anti-divergence operator

$$\begin{aligned}&\mathcal {R}_0 : C^\infty ({{\mathbb {T}}}^2;{\mathbb {R}}^2) \rightarrow C^\infty ({{\mathbb {T}}}^2;\text {Sym}_2), \\&\mathcal {R}_0 (v) := (D\Delta ^{-1} + (D \Delta ^{-1})^{T} - I \cdot \mathrm{{div}}\,\Delta ^{-1} )\Big (v - \int _{{{\mathbb {T}}}^2}v\Big ) . \end{aligned}$$

Here $\text {Sym}_2$ denotes the space of symmetric matrices in $\mathbb {R}^2$. It is simple to check that $\mathrm{{div}}\,(\mathcal {R}_0(v)) = v-\int _{{{\mathbb {T}}}^2} v$, and that $D \mathcal {R}_0$ is a Calderon-Zygmund operator, in particular it holds

$$\begin{aligned}{} & {} \Vert \mathcal {R}_0(v) \Vert _{L^p} \le C \Vert \Delta ^{-1/2} v \Vert _{L^p} \quad \text {for any}\, p\in (1, \infty )\,, \end{aligned}$$

(10)

$$\begin{aligned}{} & {} \Vert \mathcal {R}_0(v) \Vert _{L^p} \le C(p) \Vert v \Vert _{L^p} \quad \text {for any}\, p\in [1, \infty ] \,. \end{aligned}$$

(11)

Notice that (10) and (11) imply

$$\begin{aligned} \Vert \mathcal {R}_0(v_\lambda ) \Vert _{L^p} \le C(p) \lambda ^{-1} \Vert v \Vert _{L^p} \quad \text {for any}\, p\in [1, \infty ], \end{aligned}$$

(12)

where $v_\lambda (x):= v(\lambda x)$ for some $\lambda \in {\mathbb {N}}$. The latter is immediate for $p\in (1,\infty )$, since

$$\begin{aligned} \Vert \mathcal {R}_0(v_\lambda ) \Vert _{L^p} \le C \Vert \Delta ^{-1/2} v_\lambda \Vert _{L^p} \le C \lambda ^{-1} \Vert v \Vert _{L^p} , \end{aligned}$$

in the case $p=1$ and $p=\infty$ we need to take advantage of the Sobolev embedding theorem:

$$\begin{aligned} \Vert \mathcal {R}_0(v_\lambda ) \Vert _{L^\infty }&= \lambda ^{-1} \Vert \mathcal {R}_0(v)\Vert _{L^\infty } \le C \lambda ^{-1}\Vert \nabla \mathcal {R}_0(v) \Vert _{L^{3/2}} \\ {}&\le C\lambda ^{-1} \Vert v \Vert _{L^{3/2}} \le C\lambda ^{-1} \Vert v \Vert _{L^\infty } = C\lambda ^{-1} \Vert v \Vert _{L^\infty } . \end{aligned}$$

The same argument applies to $\mathcal {R}_0^*$, the adjoint of $\mathcal {R}_0$, then case $p=1$ follows by duality.

Lemma 3.1

Let $\lambda \in {\mathbb {N}}$ and $f \in C^\infty ( {{\mathbb {T}}}^2; {\mathbb {R}})$, $v\in C^\infty ( {{\mathbb {T}}}^2; {\mathbb {R}}^2)$ with $\int _{{{\mathbb {T}}}^2} v = 0$, and $v_\lambda = v(\lambda x)$. If we set

$$\begin{aligned} {{\mathcal {R}}}(f v_\lambda ) = f \mathcal {R}_0 v_\lambda -\mathcal {R}_0 \left( \nabla f \cdot \mathcal {R}_0 v_\lambda + \int _{{{\mathbb {T}}}^2} f v_\lambda \right) \in C^\infty ({{\mathbb {T}}}^2; \text {Sym}_2) \end{aligned}$$

then,

$$\begin{aligned}{} & {} \textrm{div} {{\mathcal {R}}}(f v_\lambda ) = f v_\lambda -\int _{{{\mathbb {T}}}^2} f v_\lambda \,, \nonumber \\{} & {} \Vert {{\mathcal {R}}}(f v_\lambda )\Vert _{L^p} \le C(p) \lambda ^{-1} \Vert f\Vert _{C^{1}} \Vert v\Vert _{L^p} \qquad \text{ for } \text{ every } \, p\in [1,\infty ] \,. \end{aligned}$$

(13)

Proof

The verification of $\mathrm{{div}}\,{{\mathcal {R}}}(f v_\lambda ) = f v_\lambda -\int _{{{\mathbb {T}}}^2} fv_\lambda$ is immediate. To prove (13) we use (12) and (9):

$$\begin{aligned}{} & {} \Vert f \mathcal {R}_0 v_\lambda \Vert _{L^p} \le \Vert f \Vert _{C^0} \Vert \mathcal {R}_0 v_\lambda \Vert _{L^p} \le C \lambda ^{-1} \Vert f \Vert _{C^0} \Vert v\Vert _{L^p} \,,\\{} & {} \Big \Vert \mathcal {R}_0 \left( \nabla f \cdot \mathcal {R}_0 v_\lambda +\int _{{{\mathbb {T}}}^2} f v_\lambda \right) \Big \Vert _{L^p} \\{} & {} \quad \le C \Big \Vert \nabla f \cdot \mathcal {R}_0 v_\lambda +\int _{{{\mathbb {T}}}^2} f v_\lambda \Big \Vert _{L^p} \le C\lambda ^{-1} \Vert f \Vert _{C^1} \Vert v \Vert _{L^p} + C\lambda ^{-1} \Vert f\Vert _{C^1} \Vert v \Vert _{L^1} \,. \end{aligned}$$

$\square$

Remark 3.2

The operator $\mathcal {R}$ can be also defined on scalar functions $f: {{\mathbb {T}}}^2 \rightarrow {\mathbb {R}}$, $v: {{\mathbb {T}}}^2\rightarrow {\mathbb {R}}$ as

$$\begin{aligned} {{\mathcal {R}}}(f v_\lambda ) = f \nabla \Delta ^{-1} v_\lambda -\nabla \Delta ^{-1}\left( \nabla f \cdot \mathcal {R}_0 v_\lambda + \int _{{{\mathbb {T}}}^2} f v_\lambda \right) \in C^\infty ({{\mathbb {T}}}^2; {\mathbb {R}}^2) , \end{aligned}$$

and arguing as in Lemma 3.1 we can easily show that $\textrm{div} {{\mathcal {R}}}(f v_\lambda ) = f v_\lambda -\int _{{{\mathbb {T}}}^2} f v_\lambda$ and

$$\begin{aligned} \Vert {{\mathcal {R}}}(f v_\lambda )\Vert _{L^p} \le C(p) \lambda ^{-1} \Vert f\Vert _{C^{1}} \Vert v\Vert _{L^p} \qquad \text{ for } \text{ every } \, p\in [1,\infty ] . \end{aligned}$$

Lemma 3.3

For any $a\in C^\infty ({{\mathbb {T}}}^2)$ and $A\in C^\infty ({{\mathbb {T}}}; {\mathbb {R}}^{2\times 2})$ with $\int _{{{\mathbb {T}}}^2} A =0$, it holds

$$\begin{aligned} \Vert \mathcal {R}_0\mathcal {R}(\nabla a \cdot \textrm{div}\, A)\Vert _{L^1} \le C(\Vert a \Vert _{C^3}) \Vert A \Vert _{L^1} . \end{aligned}$$

(14)

Proof

Set $T(A):= \mathcal {R}(\nabla a \cdot \mathrm{{div}}\,A)$. By duality, it suffices to show that

$$\begin{aligned} \Vert T^* \mathcal {R}_0^*(B)\Vert _{L^\infty } \le C(\Vert a \Vert _{C^3}) \Vert B \Vert _{L^\infty } , \end{aligned}$$

where $T^*$ and $\mathcal {R}_0^*$ denote the adjoint of T and $\mathcal {R}_0$, respectively. To this aim we employ the Sobolev embedding and the fact that $D T^* \mathcal {R}_0^*(B)$ maps $L^p$ into $L^p$ for any $p\in (1,\infty )$:

$$\begin{aligned} \Vert T^* \mathcal {R}_0^*(B) \Vert _{L^\infty } \le C \Vert D T^* \mathcal {R}_0^*(B)\Vert _{L^3} \le C(\Vert a \Vert _{C^3}) \Vert B \Vert _{L^3} \le C(\Vert a \Vert _{C^3}) \Vert B \Vert _{L^\infty } . \end{aligned}$$

$\square$

4 Building Blocks

In this section we introduce the building blocks of our construction. They will be employed in Sect. 5 to define the principal term of $u_1 - u_0$ in Proposition 2.1.

Proposition 4.1

(Building blocks) Set $\xi _1:= e_1$, $\xi _2:= e_2$, $\xi _3:= e_1 + e_2$ and $\xi _4: = e_1 - e_2$. Then, for any ${\varepsilon }>0$ there exist $W^p_{i}, W^c_{i}, Q_i \in C^\infty ((-1,1)\times {{\mathbb {T}}}^2; {\mathbb {R}}^2)$, $A_i \in C^\infty ((-1,1)\times {{\mathbb {T}}}^2; \textrm{Sym}_2)$ for $i=1,\ldots , 4$, such that

• (i) $\mathrm{{div}}\,(W_i^p + W_i^c) =0$, $\partial _t Q_i = \mathrm{{div}}\,(W_i^p\otimes W_i^p)$, and $\partial _t (W^p_i + W^c_i) = \mathrm{{div}}\,(A_i)$;

• (ii) $\int _{{{\mathbb {T}}}^2} A_i =0$, $\int _{{{\mathbb {T}}}^2} W_i^p = \int _{{{\mathbb {T}}}^2} W^c_i = 0$, and $W_i^p$, $W_i^c$, $A_i$ are $\lambda ^{-1}$-periodic functions for some $\lambda \in \mathbb {Z}$ with $\lambda \ge {\varepsilon }^{-1}$;

• (iii) $\int _{{{\mathbb {T}}}^2} W_i^p \otimes W_i^p = \frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|}$;

• (iv) the following estimates hold

  $$\begin{aligned}{} & {} {\varepsilon }\Vert W_i^p \Vert _{L^2} + \Vert W_i^p \Vert _{L^1} + \Vert W_i^c \Vert _{L^2} \le {\varepsilon }\,, \nonumber \\{} & {} \quad \Vert D (W_i^p + W_i^c) \Vert _{L^{1,\infty }} + \Vert Q_i \Vert _{L^2} + \Vert D Q_i \Vert _{L^{1,\infty }} + \Vert A_i \Vert _{L^1} \le {\varepsilon }\,, \nonumber \\{} & {} \quad \Vert D^s (W_i^p + W_i^c) \Vert _{L^p} + \Vert D^s Q_i \Vert _{L^p} \le {\varepsilon }^{c(p,s)} \quad \text {for any}\, s\in (0,1)\,\text { and}\, p\in \left( 1,\frac{2}{1+s}\right) \,; \nonumber \\ \end{aligned}$$

  (15)

• (v) for $i\ne i'$ the union of the supports of $W^p_i$, $W_i^c$, $Q_i$, is disjoint in space-time from the union of the supports of $W^p_{i'}$, $W_{i'}^{c}$, $Q_{i'}$.

The velocity field $W^p_i$ is the principal term, it has zero mean, high frequency $\lambda \ge {\varepsilon }^{-1}$, is controlled in the relevant norms (cf. (iv)), and satisfies the fundamental property (iii): the quadratic interaction $W_i^p\otimes W_i^p$ produces the lower order term $\frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|}$. The latter, combined with slow coefficients $a_i\in C^\infty ({{\mathbb {T}}}^2)$, is used to cancel the error $R_0$ out. To achieve the crucial bound $\Vert D W_i^p \Vert _{L^{1,\infty }}$ we design the principal term as

$$\begin{aligned} W_i^p(x,t) =W_{\xi _i,K,n_0}^p(x,t) := \frac{1}{K^{1/2}} \sum _{k=n_0 + 1}^{K + n_0} W^k_{(\xi _i)}(x,t) , \end{aligned}$$

(16)

where $K, n_0 \gg 1$ are big parameters and $\xi _i$ is one of the four directions appearing in the statement of Proposition 4.1. In a first stage, we build $W_i^p(x,t)$ for a fixed parameter i, ignoring the issue that, for different parameters, such functions will not have disjoint support; only in Sect. 4.6 we make sure to suitably time-translate them, making substantial use of their special structure, to guarantee that Proposition 4.1 (v) holds. The vector fields $W_{(\xi _i)}^k(x,t)$, $k=n_0+1, \ldots , n_0 + K$, are the 2-dimensional counterpart of the intermittent jets introduced in [4]. They have $L^2$ norm equal to 1, and are supported on disjoint balls of radius $2^{-k} r$, for some $r\ll 1$, which move in direction $\xi _i$ with speed $\mu 2^k$, where $\mu \gg 1$. The fast time translation is used to make $W_{(\xi )}^k$ “almost divergence free” and “almost solutions to the Euler equations”. In more rigorous terms, it means that there exist vector fields $(W_{(\xi )}^{k})^c$, $(Q_{(\xi )}^{k})^c$, that are smaller than $W_{(\xi )}^k$ satisfying $\mathrm{{div}}\,(W_{(\xi )}^k + (W_{(\xi )}^{k})^c)=0$ and $\partial _t (Q_{(\xi )}^{k})^c = \mathrm{{div}}\,(W_{(\xi )}^k \otimes W_{(\xi )}^k)$. The vector fields $W_i^c$ and $Q_i$ are defined bundling together $(W_{(\xi )}^{k})^c$ and $(Q_{(\xi )}^{k})^c$ as we did in (16).

Another important property we need is that $W_i^p \otimes W_j^p =0$ when $i\ne j$. It is ensured by (v) in Proposition 4.1, which builds upon a delicate combinatorial lemma presented in Sect. 4.6.

We finally explain the role of the matrix $A_i$ in our construction. Let us begin by noticing that the principal term $W_i^p$ has big time derivative, being fast translating in time. Hence, the term $\partial _t W_i^p$ cannot be treated as an error. To overcome this difficulty we impose an extra structure on $W^p_i$ and $W^c_i$. We construct them in order to have the identity $\partial _t(W_i^p + W_i^c) = \mathrm{{div}}\,(A_i)$, for some symmetric matrix $A_i$ which has small $L^1$-norm. The latter can be added to the new error term $R_1$.

4.1 General notation

Given a velocity field $u:=(u_1, u_2): {\mathbb {R}}^2 \rightarrow {\mathbb {R}}^2$ we write

$$\begin{aligned} u^{\perp } := (-u_2, u_1), \quad \text {curl}(u) := \partial _{1} u_2 - \partial _{2} u_1 \, \quad \text {div}(u) := \partial _1 u_1 + \partial _2 u_2. \end{aligned}$$

Let us fix $r_\perp \ll r_\parallel \ll 1$ and $k\in {\mathbb {N}}$, $k\ge 1$. We adopt the following convention: given any $\rho : \mathbb {R} \rightarrow \mathbb {R}$ supported in $(-1,1)$ we write

$$\begin{aligned} \rho _{r_\perp }^k(x):= & {} \left( \frac{1}{2^{-k} r_\perp }\right) ^{1/2} \rho \left( \frac{x - 2^{2-k}r_\perp }{2^{-k}r_\perp } \right) \,,\\ \rho _{r_\parallel }^k(x):= & {} \left( \frac{1}{2^{-k}r_{\parallel }}\right) ^{1/2} \rho \left( \frac{ x }{2^{-k}r_{\parallel }}\right) \,. \end{aligned}$$

Notice that $\textrm{supp}(\rho _{r_\perp }^k) \subset ( 3 \cdot 2^{-k} r_\perp , 5\cdot 2^{-k}r_\perp )$, in particular

$$\begin{aligned} \textrm{supp}(\rho ^k_{r_\perp }) \cap \textrm{supp}(\rho ^{k'}_{r_\perp }) = \emptyset \quad \text {for}\, k\ne k', \end{aligned}$$

and

$$\begin{aligned} \bigcup _{k \ge 1} \textrm{supp}(\rho ^k_{r_\perp }) \subset (0, 5 r_\perp 2^{-n_0}) . \end{aligned}$$

With a slight abuse of notation we keep denoting by $\rho _{r_\perp }^k, \rho _{r_\parallel }^k: {{\mathbb {T}}}\rightarrow {\mathbb {R}}$ their periodized version.

4.2 Construction of the principal block

We consider $\Phi , \psi : {\mathbb {R}}\rightarrow {\mathbb {R}}$ supported in $(-1,1)$, we set $\phi := -\Phi '''$ and assume $\int \psi ^2 = \int \phi ^2 =1$. Given $r_\perp \ll r_\parallel \ll 1$ and $k\in {\mathbb {N}}$ we have

$$\begin{aligned} \textrm{supp}(\phi ^k_{r_\perp }) \cap \textrm{supp}(\phi ^{k'}_{r_\perp })&= \textrm{supp}((\Phi ')^k_{r_\perp }) \cap \textrm{supp}((\Phi ')^{k'}_{r_\perp }) \\&=\textrm{supp}((\Phi '')^k_{r_\perp }) \cap \textrm{supp}((\Phi '')^{k'}_{r_\perp }) = \emptyset \quad \text {for}\, k\ne k', \end{aligned}$$

and

$$\begin{aligned} \bigcup _k \textrm{supp}(\phi ^k_{r_\perp }),\, \bigcup _k \textrm{supp}(\Phi ^k_{r_\perp }) \subset (0, 5 r_\perp 2^{-n_0}) \, . \end{aligned}$$

(17)

We periodize $(\Phi ')_{r_\perp }^k$, $(\Phi '')_{r_\perp }^k$, $\phi _{r_\perp }^k$, $\psi _{r_\parallel }^k$ keeping the same notation.

Given a vector $\xi \in \mathbb {Q}^2$, and parameters $\lambda , \mu \gg 1$ we set

$$\begin{aligned} (\Phi ')_{(\xi )}^k(x):= & {} (\Phi ')_{r_\perp }^k ( \lambda x\cdot \xi ^{\perp } ) \,, \quad (\Phi '')_{(\xi )}^k(x):= (\Phi '')_{r_\perp }^k ( \lambda x\cdot \xi ^{\perp } ) \,, \\ \phi _{(\xi )}^k(x):= & {} \phi _{r_\perp }^k ( \lambda x\cdot \xi ^{\perp } ) \,, \\ \psi _{(\xi )}^k( x, t):= & {} \psi _{r_\parallel }^k ( \lambda ( x\cdot \xi + \mu 2^k t ) ) \,, \\ W_{(\xi )}^k(x,t):= & {} \frac{\xi }{|\xi |} \, \psi _{(\xi )}^k(x, t) \phi _{(\xi )}^k(x) \,. \end{aligned}$$

We finally fix $K, n_0 \in {\mathbb {N}}$, and define the principal block

$$\begin{aligned} W_{\xi ,K,n_0}^p(x,t) := \frac{1}{K^{1/2}} \sum _{k=n_0 + 1}^{K + n_0} W^k_{(\xi )}(x, t + t_k) , \end{aligned}$$

(18)

where $t_k$ are time translations that will be chosen later. The following fundamental identity holds

$$\begin{aligned} \int _{{{\mathbb {T}}}^2} W_{\xi ,K,n_0}^p \otimes W_{\xi ,K,n_0}^p= & {} \frac{1}{K}\sum _{k=n_0+1}^{K+n_0} \int _{{{\mathbb {T}}}^2} W_{(\xi )}^k \otimes W_{(\xi )}^k \nonumber \\= & {} \frac{\xi }{|\xi |} \otimes \frac{\xi }{|\xi |} \int _{{{\mathbb {T}}}^2} (\psi _{(\xi )}^k \phi _{(\xi )}^k)^2 = \frac{\xi }{|\xi |} \otimes \frac{\xi }{|\xi |} . \end{aligned}$$

(19)

4.3 Correction of the divergence

Observe that

$$\begin{aligned} \mathrm{{div}}\,W_{(\xi )}^k (x,t) = \frac{\lambda }{2^{-k}r_\parallel } ({\dot{\psi }})^k_{(\xi )} (x, t) \phi ^k_{(\xi )}(x). \end{aligned}$$

Setting

$$\begin{aligned} (W^k_{(\xi )})^c(x,t) := \frac{r_\perp }{r_\parallel } \, \frac{\xi ^{\perp }}{|\xi |} ({\dot{\psi }})^k_{(\xi )} (x, t) (\Phi '')^k_{(\xi )}(x), \end{aligned}$$

and using the identity $2^{-k} r_\perp \partial _{x_1} (\Phi '')_{r_{\perp }}^k = -\phi _{r_\perp }^k$ we get $\mathrm{{div}}\,(W_{(\xi )} + W^c_{(\xi )})=0$.

To correct the divergence of $W_{\xi , K,n_0}$ we introduce

$$\begin{aligned} W_{\xi , K ,n_0}^c(x,t) := \frac{1}{K^{1/2}} \sum _{k=n_0 + 1}^{K + n_0} (W^k_{(\xi )})^c(x,t + t_k) , \end{aligned}$$

and set

$$\begin{aligned} W_{\xi , K, n_0}(x,t) := W_{\xi , K ,n_0}^p(x,t) + W_{\xi , K ,n_0}^c(x,t) . \end{aligned}$$

4.4 Time correction

Let us now set

$$\begin{aligned} Q_{(\xi )}^k(x,t):= \frac{1}{2^k\mu } \xi (\psi _{(\xi )}^k(x,t + t_k)\phi _{(\xi )}^k(x))^2 , \end{aligned}$$

and observe that

$$\begin{aligned} \mathrm{{div}}\,( W_{(\xi )}^k \otimes W_{(\xi )}^k)&= 2(W_{(\xi )}^k \cdot \nabla \psi _{(\xi )}^k) \phi _{(\xi )}^k \frac{\xi }{|\xi |} = \frac{1}{2^k\mu }2 \left( W_{(\xi )}^k \cdot \partial _t \psi _{(\xi )}^k\right) \phi _{(\xi )}^k \frac{\xi }{|\xi |} \\&= \frac{1}{2^k\mu } \partial _t \left( \psi _{(\xi )}^k\phi _{(\xi )}^k\right) ^2 \frac{\xi }{|\xi |} = \partial _t Q_{(\xi )}^k . \end{aligned}$$

Hence

$$\begin{aligned} \mathrm{{div}}\,(W_{\xi ,K,n_0}^p \otimes W_{\xi ,K,n_0}^p) = \frac{1}{K}\sum _{k=n_0+1}^{K+n_0} \mathrm{{div}}\,( W_{(\xi )}^k \otimes W_{(\xi )}^k) = \partial _t \left( \frac{1}{K} \sum _{k=n_0 + 1}^{K + n_0} Q_{(\xi )}^k \right) . \end{aligned}$$

(20)

The time corrector is defined as

$$\begin{aligned} Q_{\xi ,K,n_0}(x,t) := \frac{1}{K} \sum _{k=n_0 + 1}^{K + n_0} Q_{(\xi )}^k (x,t) . \end{aligned}$$

4.5 Estimates on building blocks

In this section we collect the relevant estimates on the building blocks. Given $N\ge 0$ we write $D^N = D^{\left\lfloor {N}\right\rfloor } \Delta ^{s/2}$ where $\left\lfloor {N}\right\rfloor$ is the integer part of N, $s:= N - \left\lfloor {N}\right\rfloor$ and $D^{\left\lfloor {N}\right\rfloor }$ is the standard derivative operator.

Lemma 4.2

For any $N\in [0,\infty )$, $M\ge 0$ integer and $p\in [1,\infty ]$ there exists $C=C(N,M,p,|\xi |,\Phi ,\psi )>0$ such that the following hold.

$$\begin{aligned}{} & {} \Vert D^N \partial _t^M \psi _{(\xi )}^k \Vert _{L^p({{\mathbb {T}}})} \le C 2^{k(N+2M+1/2-1/p)} r_\parallel ^{1/p - 1/2} \left( \frac{ \lambda }{r_\parallel } \right) ^N \left( \frac{ \lambda \mu }{r_\parallel } \right) ^M \,, \\{} & {} \Vert D^N (\Phi ')_{(\xi )}^k\Vert _{L^p({{\mathbb {T}}})} + \Vert D^N (\Phi '')_{(\xi )}^k\Vert _{L^p({{\mathbb {T}}})} + \Vert D^N \phi _{(\xi )}^k\Vert _{L^p({{\mathbb {T}}})} \\{} & {} \qquad \le C 2^{k(N+1/2-1/p)} r_\perp ^{1/p-1/2} \left( \frac{ \lambda }{r_\perp } \right) ^N \,, \\{} & {} \Vert D^N \partial _t^M W_{(\xi )}^k \Vert _{L^p({{\mathbb {T}}}^2)} + \frac{r_{\parallel }}{r_\perp } \Vert D^N \partial _t^M (W_{(\xi )}^k)^c \Vert _{L^p({{\mathbb {T}}}^2)} \\{} & {} \qquad \le C 2^{k(N+ 2M +1 -2/p)} (r_{\parallel }r_\perp )^{1/p-1/2} \left( \frac{\lambda }{r_\perp }\right) ^N \left( \frac{ \lambda \mu }{r_\parallel } \right) ^M \,, \\{} & {} 2^k\mu \Vert D^N \partial _t^M Q_{(\xi )}^k \Vert _{L^p({{\mathbb {T}}}^2)} \le C2^{k(N + 2M + 2-2/p)} (r_{\parallel }r_\perp )^{1/p-1} \left( \frac{\lambda }{r_\perp }\right) ^N \left( \frac{ \lambda \mu }{r_\parallel } \right) ^M \,. \end{aligned}$$

The proof of Lemma 4.2 is a simple computation, so we omit it. Let us draw some useful consequence. Summing on k and reminding that then terms in the sum in (18) have disjoint support, we get

(21)

(in particular, this says that the principal part is much smaller than the corrector),

$$\begin{aligned} \Vert Q_{\xi ,K,n_0} \Vert _{L^2({{\mathbb {T}}}^2)} \le \frac{C}{\mu (r_\parallel r_\perp )^{1/2}} , \end{aligned}$$

(22)

and

$$\begin{aligned} \left\| W_{\xi ,K,n_0}^p \right\| _{L^p({{\mathbb {T}}}^2)} + \frac{r_{\parallel }}{r_\perp } \Vert W_{\xi ,K,n_0}^c \Vert _{L^p({{\mathbb {T}}}^2)} \le C \frac{(r_\perp r_\parallel )^{1/p-1/2}}{K^{1/2}} , \quad \text {for any}\, p\in [1,2). \nonumber \\ \end{aligned}$$

(23)

Moreover, for $s\in (0,1)$ and $p<\frac{2}{1 + s}$ it holds

$$\begin{aligned}{} & {} \left\| D^s W_{(\xi )}^k \right\| _{L^p({{\mathbb {T}}}^2)} + \frac{r_{\parallel }}{r_\perp } \Vert D^s (W_{(\xi )}^k)^c \Vert _{L^p({{\mathbb {T}}}^2)} \le C 2^{ - k \gamma } (r_{\parallel }r_\perp )^{1/p-1/2} \left( \frac{\lambda }{r_\perp }\right) ^s \,, \\{} & {} \mu \Vert D^s Q_{(\xi )}^k \Vert _{L^p({{\mathbb {T}}}^2)} \le C2^{ - k\gamma } (r_{\parallel }r_\perp )^{1/p-1} \left( \frac{\lambda }{r_\perp }\right) ^s \,, \end{aligned}$$

where $\gamma := -s - 1 + 2/p >0$. In particular

$$\begin{aligned}{} & {} \left\| D^s W_{\xi ,K,n_0}^p \right\| _{L^p({{\mathbb {T}}}^2)} + \frac{r_{\parallel }}{r_\perp } \Vert D^s W_{\xi ,K,n_0}^c \Vert _{L^p({{\mathbb {T}}}^2)} \le C 2^{- n_0 \gamma } K^{-1/2}(r_{\parallel }r_\perp )^{1/p-1/2} \left( \frac{\lambda }{r_\perp }\right) ^s \,,\qquad \end{aligned}$$

(24)

$$\begin{aligned}{} & {} \Vert D^s Q_{\xi ,K,n_0} \Vert _{L^p({{\mathbb {T}}}^2)} \le C2^{ - n_0 \gamma } \mu ^{-1} K^{-1} (r_{\parallel }r_\perp )^{1/p-1} \left( \frac{\lambda }{r_\perp }\right) ^s. \end{aligned}$$

(25)

Lemma 4.3

(Lorentz estimates) There exists $C=C(|\xi |, \Phi ,\psi )>0$ such that

$$\begin{aligned}{} & {} \Vert D W_{\xi ,K,n_0} \Vert _{L^{1,\infty }} \le C\frac{ \lambda }{K^{1/2}} \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2}\,, \\{} & {} \Vert D Q_{\xi ,K,n_0} \Vert _{L^{1,\infty }} \le C \frac{\lambda }{\mu r_\perp K} \,. \end{aligned}$$

Proof

Observe that

$$\begin{aligned} |D W_{(\xi )}^k|&= \lambda 2^k \left| r_\parallel ^{-1} \frac{\xi }{|\xi |} \otimes \frac{\xi }{|\xi |} (\psi ')^k_{(\xi )}(x,t)\phi ^k_{(\xi )}(x) + r_\perp ^{-1} \frac{\xi }{|\xi |}\otimes \frac{\xi ^\perp }{|\xi |} \psi ^k_{(\xi )}(x,t)(\phi ')^k_{(\xi )}(x)\right| \\&\le \lambda 2^k r_\perp ^{-1} ( |(\psi ')^k_{(\xi )}(x,t)||\phi ^k_{(\xi )}(x)| + |\psi ^k_{(\xi )}(x,t)||(\phi ')^k_{(\xi )}(x)| )\\&= \lambda \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2} \frac{1}{2^{-k}(r_\perp r_\parallel )^{1/2}}( |(\psi ')^k_{(\xi )}(x,t)||\phi ^k_{(\xi )}(x)| + |\psi ^k_{(\xi )}(x,t)||(\phi ')^k_{(\xi )}(x)| )\\&:= \lambda \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2} \Omega _1^k(x,t) , \end{aligned}$$

and similarly

$$\begin{aligned} | D Q_{(\xi )}^k | \le \frac{\lambda }{\mu r_\perp } \Omega _2^k(x,t) \, , \end{aligned}$$

where for $i=1,2$

$$\begin{aligned}{} & {} |\Omega _i^k| \le C 2^{2k}(r_\perp r_\parallel )^{-1}, \quad {{\mathscr {L}}}^{2}(\textrm{supp}(\Omega _i^k)) \le C 2^{-2k} r_\perp r_\parallel ,\nonumber \\{} & {} \textrm{supp}(\Omega _i^k)\cap \textrm{supp}(\Omega _i^{k'})= \emptyset , \, \, \, \text {for}\, k\ne k'. \end{aligned}$$

(26)

Let us now fix $s\ge 1$ and $k_*$ the smallest integer satisfying $k_*\ge n_0+1$ and $C2^{2k_*} \ge s K^{1/2} r_\perp r_\parallel$. It holds

$$\begin{aligned} {{\mathscr {L}}}^{2}\left( \left\{ \frac{1}{K^{1/2}} \sum _{k=n_0+1}^{K + n_0} \Omega _1^k \ge s \right\} \right)&= \sum _{k = n_0+1}^{K + n_0} {{\mathscr {L}}}^{2}(\{\Omega _1^k \ge s K^{1/2}\})\\ {}&\le \sum _{k=k_*}^{K + n_0} {{\mathscr {L}}}^{2}(\{\Omega _1^k \ge s K^{1/2}\}) . \end{aligned}$$

From (26) and the choice of $k_*$ we get

$$\begin{aligned} \sum _{k=k_*}^{K + n_0} {{\mathscr {L}}}^{2}(\{\Omega _k \ge s K^{1/2} \}) \le \sum _{k=k_*}^{K+n_0} C 2^{-2k} r_\perp r_\parallel \le \frac{C}{ s K^{1/2}}\sum _{k\ge k_*} 2^{2k_*-2k} \le \frac{C}{s K^{1/2}}, \end{aligned}$$

hence

$$\begin{aligned} \Vert D W_{\xi ,K,n_0}^p \Vert _{L^{1,\infty }} \le \lambda \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2} \Big \Vert \frac{1}{K^{1/2}} \sum _{k=n_0+1}^{K+n_0} \Omega ^k \Big \Vert _{L^{1,\infty }} \le C^2 \frac{\lambda }{K^{1/2}} \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2}, \end{aligned}$$

the estimate on $\Vert D W^c_{\xi ,K,n_0} \Vert _{L^{1,\infty }}$ can be obtained following the same strategy. An analogous argument gives

$$\begin{aligned} {{\mathscr {L}}}^{2}\left( \left\{ \frac{1}{K} \sum _{k=n_0+1}^{K + n_0} \Omega _2^k \ge s \right\} \right) \le \frac{C}{s K} , \end{aligned}$$

yielding

$$\begin{aligned} \Vert D Q_{\xi ,K,n_0} \Vert _{L^{1,\infty }} \le C\frac{\lambda }{\mu r_\perp } \Big \Vert \frac{1}{K} \sum _{k=n_0+1}^{K+n_0} \Omega ^k \Big \Vert _{L^{1,\infty }} \le C^2 \frac{ \lambda }{ \mu r_\perp K} \,. \end{aligned}$$

$\square$

Remark 4.4

It is not hard to prove the following extension of Lemma 4.3. For any $q\ge 1$ it holds

$$\begin{aligned}{} & {} \Vert D W_{\xi ,K,n_0} \Vert _{L^{1,q}} \le C\frac{ \lambda }{K^{1/2 - 1/q}} \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2}\,, \\{} & {} \Vert D Q_{\xi ,K,n_0} \Vert _{L^{1,\infty }} \le C \frac{\lambda }{\mu r_\perp K^{1-1/q}} \,. \end{aligned}$$

Lemma 4.5

There exists a smooth $\lambda$-periodic function $A_{\xi , K, n_0}: {{\mathbb {T}}}^2 \rightarrow \textrm{Sym}_2$ such that

$$\begin{aligned}{} & {} \partial _t W_{\xi ,K,n_0} = \mathrm{{div}}\,(A_{\xi ,K,n_0}) \,, \end{aligned}$$

(27)

$$\begin{aligned}{} & {} \Vert A_{\xi ,K,n_0} \Vert _{L^1} \le C(|\xi |,\Phi ,\psi ) \mu K^{1/2} r_\perp ^{3/2} r_{\parallel }^{-1/2} \,. \end{aligned}$$

(28)

Proof

Setting

$$\begin{aligned} A_{(\xi ),k}:= & {} -2^k \left( \frac{r_\perp }{r_\parallel }\right) \mu \left( \left( \frac{\xi }{|\xi |} \otimes \frac{\xi ^{\perp }}{|\xi |} + \frac{\xi ^{\perp }}{|\xi |}\otimes \frac{\xi }{|\xi |} \right) (\psi ')_{(\xi )}^k (\Phi '')_{(\xi )}^k\right. \\{} & {} + \left. \frac{r_\perp }{r_\parallel } \frac{\xi ^\perp }{|\xi |} \otimes \frac{\xi ^\perp }{|\xi |} (\psi '')_{(\xi )}^k (\Phi ')_{(\xi )}^k \right) \,, \\ A_{(\xi ),k}^c:= & {} 2^k \left( \frac{r_\perp }{r_\parallel }\right) ^2 \mu \left( \left( \frac{\xi }{|\xi |} \otimes \frac{\xi ^{\perp }}{|\xi |} + \frac{\xi ^{\perp }}{|\xi |}\otimes \frac{\xi }{|\xi |}\right) (\psi ')_{(\xi )}^k (\Phi ')_{(\xi )}^k\right. \\{} & {} - \left. \frac{r_\perp }{r_\parallel }\frac{\xi ^\perp }{|\xi |} \otimes \frac{\xi ^\perp }{|\xi |} (\psi '')_{(\xi )}^k (\Phi )_{(\xi )}^k \right) \,, \end{aligned}$$

it holds

$$\begin{aligned} \partial _t W_{(\xi )}^k&= 2^{2k} \mu \lambda r_{\parallel }^{-1} \xi ( \psi ')_{(\xi )}^k(x, t) \phi _{(\xi )}^k(x)\\&= -2^k\mu r_\parallel ^{-1} r_\perp \mathrm{{div}}\,\left( \left( \frac{\xi }{|\xi |} \otimes \frac{\xi ^{\perp }}{|\xi |} + \frac{\xi ^{\perp }}{|\xi |}\otimes \frac{\xi }{|\xi |}\right) (\psi ')_{(\xi )}^k (\phi '')_{(\xi )}^k \right. \\&\quad + \left. \frac{\xi ^\perp }{|\xi |} \otimes \frac{\xi ^\perp }{|\xi |} (\psi '')_{(\xi )}^k (\phi ')_{(\xi )}^k \right) \\&=\mathrm{{div}}\,( A_{(\xi ),k} ) , \end{aligned}$$

and similarly

$$\begin{aligned} \partial _t (W_{(\xi )}^k)^c&= \frac{r_\perp }{r_\parallel } 2^{2k} \mu \lambda r_{\parallel }^{-1} \xi ( \psi ')_{(\xi )}^k(x, t) (\Phi '')_{(\xi )}^k(x) =\mathrm{{div}}\,( A_{(\xi ),k}^c ) . \end{aligned}$$

Hence (27) is satisfied. Defining

$$\begin{aligned} A_{\xi ,K,n_0} := \frac{1}{K^{1/2}} \sum _{k=n_0+1}^{K + n_0} (A_{(\xi ),k} + A_{(\xi ),k}^c) \, \end{aligned}$$

and arguing as in Lemma 4.2, we obtain that

$$\begin{aligned} \Vert A_{(\xi ),k} \Vert _{L^1} + \Vert A_{(\xi ),k}^c \Vert _{L^1} \le C(|\xi |,\Phi ,\psi ) \mu K^{1/2} r_{\perp } r_\parallel ^{-1} (r_\perp r_\parallel )^{1/2} , \end{aligned}$$

which yields (28). $\square$

4.6 Combinatorial lemma

The following proposition shows that, up to a suitable (time) translation of each element in the bundle, the building blocks associated to different directions can be taken disjoint.

Proposition 4.6

Let $\xi _1=e_1$, $\xi _2 = e_2$, $\xi _3= e_1+e_2$ and $\xi _4= e_1-e_2$. Then for $n_0= 5K$ the functions in the family $\{ W_{(\xi _{i+1})}^k(x,t+ i \mu ^{-1} 2^{-5K})\}_{k=n_0,\ldots ,n_0+K;\; i=0,1,2,3}$ have all supports mutually disjoint in space-time.

Proof

We apply Lemma 4.7 below to the families $\{ W_{(\xi _2)}^k(x,t+ i \mu ^{-1} 2^{-5K})\}_{k=n_0,...,n_0+K}$ and $\{ W_{(\xi _2)}^k(x,t+ j \mu ^{-1} 2^{-5K})\}_{k=n_0,...,n_0+K}$; up to shifting the time axis, we can assume that $i=0$ and that $j\in \{1,2,3\}$ and conclude the proof.$\square$

Lemma 4.7

Let $\xi _1, \xi _2 \in \{ e_1,e_2, e_1+e_2, e_1-e_2\}$ be two different vector fields. Let us consider two families $\{ W_{(\xi _1)}^k(x,t)\}_{k=n_0,\ldots ,n_0+K}$ and $\{ W_{(\xi _2)}^k(x,t+t_0)\}_{k=n_0,\ldots ,n_0+K}$ for some $t_0 \in [\mu ^{-1} 2^{-7K}, \mu ^{-1} 2^{-7K+2}]$ and for $n_0=5K$. Then the supports of all these functions are disjoint in space-time, namely

$$\begin{aligned} W_{(\xi _1)}^k(x,t) \otimes W_{(\xi _2)}^h(x,t+t_0)=0 \qquad \text{ for } \text{ all } k, h \in \{1,..., K\}. \end{aligned}$$

Proof

The family $\{ W_{(\xi _1)}^k(x,t)\}_{k=n_0,\ldots ,n_0+K}$ is supported by (17) in space in a tube along $\xi _1$ of size $r_\parallel 2^{-n_0}$ and similarly the family $\{ W_{(\xi _2)}^k(x,t+t_0)\}_{k=n_0,\ldots ,n_0+K}$ is supported in the tube along $\xi _2$ of size $r_\parallel 2^{-n_0}$. Since these two thin tubes intersect only in a neighborhood of the origin, we deduce that the supports of $W_{(\xi _1)}^k(x,t)$ and $W_{(\xi _2)}^h(x,t)$, where $h, k \in \{n_0,\ldots , n_0+K\}$, can intersect for some time $t>0$ only if they both belong to $B_{R}(0)$, where $R:= r_\parallel 2^{-n_0 + 1}$.

We claim the following: suppose that for a certain $t>0$ and $k \in \{n_0,\ldots , n_0+K\}$ we have $\textrm{supp}W_{(\xi _1)}^k (\cdot , t) \cap B_{r_\parallel 2^{-n_0+1}} \ne \emptyset$. Then $\textrm{supp}W_{(\xi _2)}^h(\cdot , t+t_0) \cap B_R = \emptyset$ for every $h \in \{n_0,\ldots , n_0+K\}$.

The previous claim excludes the simultaneous presence at any $t>0$ of the support of $W_{(\xi _1)}^k (\cdot , t)$ and the support of $W_{(\xi _2)}^h(\cdot , t+t_0)$ in $B_R(0)$, thereby concluding the proof of the lemma.

We now prove the claim. Let us fix a time t such that $\textrm{supp}W_{(\xi _1)}^k(\cdot ,t) \cap B_{R} \ne \emptyset$. Since $\textrm{supp}W_{(\xi _1)}^k(\cdot ,t)$ is moving at constant speed $\mu 2^k$ along the tube on the torus, there exists ${\bar{t}}$ such that $|t -{\bar{t}} | \le R \mu ^{-1} 2^{-k}$ and $\textrm{supp}W_{(\xi _1)}^k(\cdot ,{\bar{t}}) = \textrm{supp}W_{(\xi _1)}^k(x,0)$. At time ${\bar{t}}$ we have information about the position of $\textrm{supp}W_{(\xi _2)}^h(\cdot ,{\bar{t}}+t_0)$; more precisely, we have that

$$\begin{aligned} \textrm{supp}W_{(\xi _2)}^h(\cdot ,{\bar{t}}+t_0) \subseteq \bigcup _{n\in {\mathbb {N}}} \Big (\textrm{supp}W_{(\xi _2)}^h(\cdot ,t_0) + n \frac{\xi _2}{2^K}\Big ) \end{aligned}$$

(29)

because the ratio between the (constant) velocity of $\textrm{supp}W_{(\xi _1)}^k(\cdot ,t)$ and the velocity of $\textrm{supp}W_{(\xi _2)}^k(\cdot ,t)$ is of the form $2^j$ for some $j \in \{-K,..., K\}$.

In the union in the right-hand side of (29), thanks to the upper bound on $t_0$, the choice $n=0$ identifies the ball of the (finite) union at minimal distance from the origin for every k. By the lower bound on $t_0$ and the fact that the minimal velocity is $\mu 2^{n_0}$, we get that this distance is greater than $2^{n_0-7K}$. At time t the distance between $\textrm{supp}W_{(\xi _2)}^h(\cdot , t+t_0)$ and $B_R(0)$ is therefore bigger than

$$\begin{aligned} 2^{n_0-7K} - |t-{\bar{t}}| \mu 2^h -R\ge & {} 2^{n_0-7K} - R2^{h-k} -R \ge 2^{n_0-7K} - R2^{K} -R\\\ge & {} 2^{n_0-7K} - 2^{-n_0+K+1} = 2^{-2K}- 2^{-4K+1}>0. \end{aligned}$$

This concludes the proof of the claim. $\square$

4.7 Proof of Proposition 4.1

Let $n_0=5K$ and $\{ W_{(\xi _{i+1})}^k(x,t+ i \mu ^{-1} 2^{-5K})\}_{k=n_0,\ldots ,n_0+K;\; i=0,1,2,3}$ be as in Proposition 4.6. Since $\textrm{supp}W_{\xi _{i+1}}^k = \textrm{supp}(W_{\xi _{i+1}}^k)^c = \textrm{supp}Q_{\xi _{i+1}}^k$, by translating in time $(W_{\xi _{i+1}}^k)^c$ and $Q_{\xi _{i+1}}^k$ with $t_{k,i}:= i \mu ^{-1} 2^{-5K}$ we deduce that $W_{i+1}^p:= W^p_{\xi _{i+1},K,n_0}$, $W_{i+1}^c:= W^c_{\xi _{i+1},K,n_0}$, $Q_{i+1}:= Q_{\xi _{i+1},K,n_0}$ and $A_{i+1}:=A_{\xi ,K,n_0}$ satisfy (v) in Lemma 4.1. We refer the reader to Lemma 4.5 for the construction of $A_{\xi ,K,n_0}$. Properties (i) and (ii) in Lemma 4.1 are now immediate from (19), (20) and Lemma 4.5. We are left with the proof of (iii) and (iv) in Lemma 4.1. To do so we have to choose appropriately the parameters $\lambda , \mu , K, r_\perp$ and $r_\parallel$. Let $\delta < 1/2$ to be chosen later in terms of ${\varepsilon }>0$, we set

$$\begin{aligned} \lambda = \left( \frac{r_\perp }{r_\parallel }\right) ^{-1/2} \delta ^4 \quad K= \left( \frac{r_\perp }{r_\parallel }\right) ^{-2}\delta ^4 \quad \mu = (r_\perp r_\parallel )^{-1/2}\delta ^{-1} , \end{aligned}$$

leaving $r_\perp \ll r_\parallel \ll 1$ free. From Lemma 4.3, Lemma 4.5, (21), (22) and (23) we deduce

$$\begin{aligned}{} & {} \Vert D(W_i^c + W_i^p) \Vert _{L^{1,\infty }} \le C\frac{ \lambda }{K^{1/2}} \left( \frac{ r_\parallel }{r_\perp } \right) ^{1/2} = C \delta ^2 \,, \\{} & {} \Vert D Q_i \Vert _{L^{1,\infty }} \le C \frac{\lambda }{\mu K r_\perp } = C \delta \frac{r_\perp }{r_\parallel } \le C \delta \,, \\{} & {} \Vert A_i \Vert _{L^1} \le C \mu K^{1/2} (r^{-1}_\parallel r_\perp ) (r_\parallel r_\perp )^{1/2} = C \delta , \\{} & {} \Vert Q_i \Vert _{L^2({{\mathbb {T}}}^2)} \le \frac{C}{\mu (r_\parallel r_\perp )^{1/2}} = C\delta \,, \\{} & {} \Vert W_i^p \Vert _{L^2} + \frac{ r_\parallel }{r_\perp } \Vert W_i^c \Vert _{L^2} \le 1 \,. \end{aligned}$$

Moreover, from (24) and (25) we deduce

$$\begin{aligned} \Vert D^s W_{\xi ,K,n_0}^p \Vert _{L^p} + \frac{r_{\parallel }}{r_\perp } \Vert D^s W_{\xi ,K,n_0}^c \Vert _{L^p} + \Vert D^s Q_{\xi ,K,n_0} \Vert _{L^p} \le C\exp \left\{ -C(p,s) \delta ^4 \left( \frac{r_\perp }{r_\parallel }\right) ^{-2} \right\} , \end{aligned}$$

for any $s\in (0,1)$ and $p\in \left( 1, \frac{2}{1+s}\right)$. The conclusions (iii) and (iv) in Lemma 4.1 follow by choosing first $\delta$ small enough so that $C\delta \le {\varepsilon }$, and after $r_\perp \ll r_\parallel \ll 1$ so that $\frac{r_\perp }{r_\parallel } \le {\varepsilon }$, $\lambda = \delta ^4 r_\parallel ^{1/2} r_\perp ^{-1/2} \ge {\varepsilon }^{-1}$ and $C\exp \left\{ -C(p,s) \delta ^4 \left( \frac{r_\perp }{r_\parallel }\right) ^{-2} \right\} \le {\varepsilon }^{c(p,s)}$.

5 Definition of the Perturbations

Let us begin by observing that there exist $\Gamma _i \in C^\infty (\textrm{Sym}_2, \mathbb {R})$, $i=1,\ldots , 4$ such that

$$\begin{aligned} R = \sum _{i=1}^4 \Gamma _i(R)^2 e_i \otimes e_i, \quad \text {for any}\, R\in \textrm{Sym}_2\,\text { such that}\, |R-I|<1/8, \end{aligned}$$

where $e_1:=(1,0)$, $e_2:=(0,1)$, $e_3:= (1/\sqrt{2}, 1/\sqrt{2})$ and $e_4:= (1/\sqrt{2},-1/\sqrt{2})$.

We can define, for instance,

$$\begin{aligned}{} & {} \Gamma _1(R)^2 := R_{1,1}-R_{1,2}-\frac{1}{2} , \quad \Gamma _2(R)^2 := R_{2,2} - R_{1,2} - \frac{1}{2} , \\{} & {} \Gamma _3(R)^2 := 2R_{1,2} + \frac{1}{2} , \quad \Gamma _4(R)^2 := \frac{1}{2} . \end{aligned}$$

It is immediate to show the identity $R= \sum _{i=1}^4 \Gamma _i(R)^2 e_i \otimes e_i$. Moreover, using that $|R-I|<1/8$, we deduce

$$\begin{aligned} \Gamma _1(R)^2= & {} \frac{1}{2} + (R_{1,1}-1) - R_{1,2} \ge \frac{1}{2} - |R_{1,1}-1| - |R_{1,2}| \ge \frac{1}{4} \,, \\ \Gamma _2(R)^2= & {} \frac{1}{2} + (R_{2,2}-1) - R_{1,2} \ge \frac{1}{2} - |R_{2,2}-1| - |R_{1,2}| \ge \frac{1}{4} \,, \\ \Gamma _3(R)^2\ge & {} \frac{1}{2} - 2| R_{1,2}| \ge 1/4 \,, \end{aligned}$$

which implies that $\Gamma _i$ are smooth functions.

We define

$$\begin{aligned} a_{i}(x,t) := (10\chi (t)(|R_0(x,t)| + \Vert R_0 \Vert _{L^1}))^{1/2} \, \Gamma _i \Big (I - \frac{10^{-1}}{|R_0(x,t)| + \Vert R_0 \Vert _{L^\infty (L^1)}} R(x,t) \Big ) , \end{aligned}$$

where $\chi \in C^\infty ({\mathbb {R}})$ satisfies $0\le \chi \le 1$, $\chi = 0$ on $[0,t_0/2]$, and $\chi = 1$ on $[t_0, \infty )$. Our choice leads to

$$\begin{aligned} \sum _{i=1}^4 a_i(x,t)^2 \frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|} = -R_0(x,t) + \chi (t) 10(|R_0(x,t)| + \Vert R_0 \Vert _{L^\infty (L^1)}) I , \end{aligned}$$

where $\xi _1=(1,0)$, $\xi _2=(0,2)$, $\xi _3=(1,1)$ and $\xi _4=(1,-1)$. The latter implies that

$$\begin{aligned} -\mathrm{{div}}\,(R_0) = \mathrm{{div}}\,\Big ( \sum _{i=1}^4 a_i(x,t)^2 \frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|} \Big ) + \nabla P , \end{aligned}$$

(30)

for some smooth pressure function P.

We observe that the coefficient $a_i$ is a “slow function”, namely its derivatives are estimated only in terms of the smoothness of $R_0$

$$\begin{aligned}{} & {} \Vert \partial _t^M \nabla ^N a_i\Vert _{L^\infty } \le C(t_0, \Vert R_0 \Vert _{C^{N+M}}, N, M) \,, \\{} & {} \Vert a_i \Vert _{L^\infty (L^2)} \le 5 \Vert R_0 \Vert _{L^\infty (L^1)}^{1/2} \,. \end{aligned}$$

For ${\varepsilon }>0$ to be chosen later, we consider the functions $W_i^p$, $W_i^c$, $Q_i$, $A_i$ from Proposition 4.1. We define the new velocity field as the sum of the previous one, a principal perturbation, a divergence corrector and a temporal corrector

$$\begin{aligned} u_{1} := u_0 + u_{1}^{(p)} + u_{1}^{(c)} + u_{1}^{(t)} , \end{aligned}$$

where

$$\begin{aligned} u_{1}^{(p)} = \sum _{i=1}^4a_i(W_{i}^p + W_{i}^c) , \quad u_{1}^{(c)} = - \sum _{i=1}^4 \mathcal {R}\left( \nabla a_i \cdot (W_{i}^p + W_{i}^c)\right) , \quad u_{1}^{(t)} = -\mathbb {P}(\sum _{i=1}^4 a_i^2 Q_{i}) , \end{aligned}$$

where $\mathbb {P} = \nabla ^\perp \Delta ^{-1}\mathrm{{div}}\,: C^\infty ({{\mathbb {T}}}^2;{\mathbb {R}}^2) \rightarrow C^\infty ({{\mathbb {T}}}^2;{\mathbb {R}}^2)$ is the Leray projector.

We refer the reader to Remark 3.2 for the definition of $\mathcal {R}$. From now on, in order to simplify our notation, for any function space X and any map f which depends on t and x, we will write $\Vert f \Vert _X$ meaning $\Vert f \Vert _{L^\infty (X)}$.

5.1 Estimate on $\Vert u_{1} - u_0 \Vert _{L^2}$ and on $\Vert u_{1} - u_0 \Vert _{L^1}$

By the triangular inequality,

$$\begin{aligned} \Vert u_{1} - u_0 \Vert _{L^2} \le \Vert u_{1}^{(p)} \Vert _{L^2} + \Vert u_{1}^{(c)} \Vert _{L^2} + \Vert u_{1}^{(t)} \Vert _{L^2} \end{aligned}$$

and we estimate the right-hand side separately as

$$\begin{aligned} \Vert u_{1}^{(p)} \Vert _{L^2}&\le \sum _{i=1}^4 \Vert a_i (W_{i}^p + W_{i}^c) \Vert _{L^2}\\&\le \sum _{i=1}^4\left( \Vert a_i \Vert _{L^2} \Vert W_{i}^p + W_{i}^c\Vert _{L^2} + C \frac{ \Vert a_i \Vert _{C_1} \Vert W_{i}^p + W_{i}^c\Vert _{L^2}}{\lambda ^{1/2}} \right) \\&\le \Vert R_0 \Vert _{L^1} + {\varepsilon }^{1/2}C(t_0, \Vert R_0 \Vert _{C_1}) , \end{aligned}$$

where in the second line we used the improved Holder inequality (8) and (iii) in Proposition 4.1.

From Remark 3.2 we deduce

$$\begin{aligned} \Vert u_{1}^{(c)} \Vert _{L^2} \le C{\varepsilon }\sum _{i=1}^4 \Vert a_i \Vert _{C_2} \Vert W_{i}^p + W_{i}^c\Vert _{L^2} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^2}) . \end{aligned}$$

Finally we employ (iv) in Proposition 4.1 to get

$$\begin{aligned} \Vert u_{1}^{(t)} \Vert _{L^2} \le \sum _{i=1}^4 \Vert a_i \Vert _{L^\infty } \Vert Q_{i} \Vert _{L^2} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{L^\infty }) . \end{aligned}$$

Analogously

$$\begin{aligned} \Vert u_{1} - u_0 \Vert _{L^1 }&\le \sum _{i=1}^4 \left( \Vert u_1^{(p)} \Vert _{L^1} + \Vert u^{(c)}_1 \Vert _{L^1} + \Vert u^{(t)}_1 \Vert _{L^1}\right) \nonumber \\&\le C\sum _{i=1}^4 (\Vert a_i \Vert _{L^\infty } \Vert W_{i}^p + W_{i}^c\Vert _{L^1} + \Vert u^{(c)}_1 \Vert _{L^2} + \Vert u^{(t)}_1 \Vert _{L^2} )\nonumber \\&\le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^2}) . \end{aligned}$$

(31)

5.2 Estimate on $\Vert {{\,\textrm{curl}\,}}(u_{1} - u_0 )\Vert _{L^{1,\infty }}$ and $\Vert D^s (u_1 - u_0)\Vert _{L^p}$

By triangular inequality,

$$\begin{aligned}&\Vert {{\,\textrm{curl}\,}}(u_{1} - u_0) \Vert _{L^{1,\infty }} \\&\quad \le C\sum _{i=1}^4 \left( \Vert D( a_i(W_{i}^p + W_{i}^c)) \Vert _{L^{1,\infty }} + \Vert D\mathcal {R}(\nabla a_i \cdot (W_{i}^p + W_{i}^c) \Vert _{L^{1,\infty }} + \Vert \text {curl}\, \mathbb {P}(a_i Q_{i}) \Vert _{L^{1,\infty }}\right) , \end{aligned}$$

we estimate the right-hand side separately as

$$\begin{aligned} \Vert D( a_i(W_{i}^p + W_{i}^c)) \Vert _{L^{1,\infty }}{} & {} \le \Vert a_i \Vert _{C_1} \Vert (W_{i}^p + W_i^c) \Vert _{L^1} + \Vert a_i\Vert _{L^\infty } \Vert D (W_{i}^p + W_i^c) \Vert _{L^{1,\infty }}\\{} & {} \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^1}) \,,\\ \Vert \text {curl} \, \mathbb {P}(a_i Q_{i}) \Vert _{L^{1,\infty }}{} & {} = \Vert \text {curl}(a_i Q_{i}) \Vert _{L^{1,\infty }} \le C \Vert a_i \Vert _{C_1} \Vert Q_{i} \Vert _{L^1} + \Vert a_i\Vert _{L^\infty } \Vert D Q_{i} \Vert _{L^{1,\infty }}\\{} & {} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^1})\,, \end{aligned}$$

where we employed (iv) in Proposition 4.1. Using that $D\mathcal {R}$ is a Calderon-Zygmund operator we deduce

$$\begin{aligned} \Vert D\mathcal {R}(\nabla a_i \cdot (W_{i}^p + W_{i}^c)) \Vert _{L^{1,\infty }} \le C \Vert \nabla a_i \cdot (W_{i}^p + W_{i}^c) \Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^1})\, . \end{aligned}$$

Let us now fix $s\in (0,1)$ and $p\in (1,\frac{2}{1+s})$. Arguing as above employing (15) we get

$$\begin{aligned} \Vert D^s (u_{1} - u_0) \Vert _{L^p}&\le C\sum _{i=1}^4 \big ( \Vert D^s (a_i(W_{i}^p + W_{i}^c)) \Vert _{L^p} + \Vert D^s\mathcal {R}(\nabla a_i \cdot (W_{i}^p + W_{i}^c) \Vert _{L^p}\\&\quad + \Vert D^s\mathbb {P}(a_i Q_{i}) \Vert _{L^p}\big )\\&\le {\varepsilon }^{c(p,s)} C(t_0, \Vert R_0 \Vert _{C^1}) . \end{aligned}$$

6 New Error

We define $R_1$ in such a way that

$$\begin{aligned} \partial _t u_1 + \mathrm{{div}}\,( u_1 \otimes u_1) + \nabla p_1 = \mathrm{{div}}\,( R_1) , \end{aligned}$$

which, by subtracting the equation for $u_0$, is equivalent to

$$\begin{aligned} \mathrm{{div}}\,(R_1)&= \mathrm{{div}}\,( u_0 \otimes (u_1-u_0) + (u_1-u_0) \otimes u_0 + (u_1-u_0) \otimes (u_1-u_0) + R_0) \nonumber \\&\quad + \partial _t (u_1 - u_0) + \nabla (p_1 - p_0) . \end{aligned}$$

(32)

We are going to define

$$\begin{aligned} R_1:= R_1^{(l)} + R_1^{(t)} + R_1^{(q)} \,, \end{aligned}$$

where the various addends are defined in the following paragraphs, and show that

$$\begin{aligned} \Vert R^{(l)}_1\Vert _{L^1} + \Vert R^{(t)} \Vert _{L^1} + \Vert R^{(q)} \Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^3}) . \end{aligned}$$

The proof of Proposition 2.1 will follow by choosing ${\varepsilon }$ small enough.

6.1 Linear error

Let us set

$$\begin{aligned} R^{(l)}_1 := u_0 \otimes (u_1-u_0) + (u_1-u_0) \otimes u_0 , \end{aligned}$$

(33)

thanks to (31) it holds

$$\begin{aligned} \Vert R_1^{(l)}\Vert _{L^1} \le 2\Vert u_0\Vert _{L^\infty } \Vert u_1-u_0\Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^2}) \,. \end{aligned}$$

6.2 Temporal error

Let us set

$$\begin{aligned} R^{(t)}_1&:= \mathcal {R}( \partial _t a_i \cdot (W_{i}^p + W_{i}^c)) + a_i A_i - \mathcal {R}( \nabla a_i \cdot A_{i} )\\&\quad + \mathcal {R}_0 \mathcal {R}( \partial _t(\nabla a_i) \cdot (W_{i}^p + W_{i}^c)) + \mathcal {R}_0 \mathcal {R}(\nabla a_i \cdot \mathrm{{div}}\,(A_i)) -\mathcal {R}_0\mathbb {P}\left( \sum _{i=1}^4 \partial _t a_i^2 Q_{i}\right) . \end{aligned}$$

Using that

$$\begin{aligned} \partial _t u_1^{(t)}&= -\mathbb {P}(\sum _{i=1}^4 \partial _t a_i^2 Q_{i}) - \mathbb {P}(\sum _{i=1}^4 a_i^2 \mathrm{{div}}\,(W_i^p \otimes W^p_i))\\&= -\mathbb {P}(\sum _{i=1}^4 \partial _t a_i^2 Q_{i}) - \sum _{i=1}^4 a_i^2 \mathrm{{div}}\,( W_i^p \otimes W^p_i) - \nabla P , \end{aligned}$$

for some pressure term P, it is immediate to verify the identity

(34)

Since ${\mathcal {R}}$ and $\mathcal {R}_0$ send $L^1$ to $L^1$ (cf. Lemma 3.1 and Remark 3.2), we have that

$$\begin{aligned}&\Vert \mathcal {R}( \partial _t a_i \cdot (W_{i}^p + W_{i}^c)) \Vert _{L^1} + \Vert \mathcal {R}_0\mathcal {R}( \partial _t\nabla a_i \cdot (W_{i}^p + W_{i}^c))\Vert _{L^1} \\&\qquad \le 2\Vert a\Vert _{C^2} \Vert W_{i}^p + W_{i}^c \Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^2}) .\\&\Vert \mathcal {R}_0\mathbb {P}\left( \sum _{i=1}^4 \partial _t a_i^2 Q_{i}\right) \Vert _{L^1} \le \sum _{i=1}^4 \Vert \partial _t a_i^2 Q_{i} \Vert _{L^2} \le \sum _{i=1}^4 \Vert \partial _t a_i^2 \Vert _{L^\infty }\Vert Q_{i} \Vert _{L^2} \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^1}) . \end{aligned}$$

From (iv) in Proposition 4.1 we get

$$\begin{aligned} \Vert a_i A_i\Vert _{L^1} + \Vert \mathcal {R}( \nabla a_i \cdot A_{i} ) \Vert _{L^1} \le 2\Vert a_i \Vert _{C^1} \Vert A_i \Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^1}) . \end{aligned}$$

By employing (14) we bound

$$\begin{aligned} \Vert \mathcal {R}_0\mathcal {R}(\nabla a_i \cdot \mathrm{{div}}\,(A_i)) \Vert _{L^1} \le C\Vert a_i \Vert _{C^3} \Vert A_i \Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^3}) \, . \end{aligned}$$

6.3 Quadratic error terms

Let us set

$$\begin{aligned} R_1^{(q)} = (u_1 - u_0)\otimes (u_1 - u_0) - \sum _{i=1}^4 a_i^2 W_{i}^p\otimes W_{i}^p + \sum _{i=1}^4\mathcal {R}\Big ( \nabla a_i^2 \cdot \Big (W_i^p \otimes W^p_i - \int _{{{\mathbb {T}}}^2} W^p_i \otimes W_i^p\Big ) \Big ) , \end{aligned}$$

and show that (32) holds. In view of (33), (30) and (34) it amounts to check that

$$\begin{aligned} \mathrm{{div}}\,(R_1^{(q)})&= \mathrm{{div}}\,\Big ( (u_1-u_0) \otimes (u_1-u_0) - \sum _{i=1}^4 a_i^2 \Big (\frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|} \Big )\Big ) \\&\quad - \sum _{i=1}^4 a_i^2 \mathrm{{div}}\,( W_i^p \otimes W^p_i) + \nabla (p_1 - p_2) . \end{aligned}$$

The latter easily follows by noticing that, as a consequence of (ii) in Proposition 4.1, one has

$$\begin{aligned}&\sum _{i=1}^4 \nabla a_i^2 \cdot \Big (W_i^p \otimes W^p_i - \int _{{{\mathbb {T}}}^2} W^p_i \otimes W_i^p \Big ) = \sum _{i=1}^4 \nabla a_i^2 \cdot \Big (W_i^p \otimes W^p_i - \int _{{{\mathbb {T}}}^2} W^p_i \otimes W_i^p \Big )\\&\quad = \sum _{i=1}^4 \mathrm{{div}}\,\Big ( a_i^2 \Big (W_i^p \otimes W^p_i - \frac{\xi _i}{|\xi _i|} \otimes \frac{\xi _i}{|\xi _i|} \Big )\Big ) - \sum _{i=1}^4 \mathrm{{div}}\,(a_i^2 W_i^p \otimes W^p_i) . \end{aligned}$$

Let us finally prove that $\Vert R_1^{(q)}\Vert _{L^1} \le {\varepsilon }C(t_0, \Vert R_0\Vert _{C^2})$. We begin by observing that

$$\begin{aligned}&(u_1 - u_0) \otimes (u_1-u_0) - \sum _{i=1}^4 a_i^2 W_{i}^p\otimes W_{i}^p\\&\quad = \sum _{i=1}^4 ( a_i^2 W_{i}^p\otimes W_{i}^c + a_i^2W_{i}^c\otimes W_{i}^p + a_i^2 W_{i}^c\otimes W_{i}^c) \\&\qquad + ( u_{1}^{(c)} + u_{1}^{(t)}) \otimes ( u_1-u_0)+ ( u_1-u_0) \otimes ( u_{1}^{(c)} + u_{1}^{(t)}) , \end{aligned}$$

From (iv) in Proposition 4.1, the estimates in Sect. 5.1 on $\Vert u_{1}^{(c)}\Vert _{L^2},\Vert u_{1}^{(t)} \Vert _{L^2}, \Vert u_1-u_0\Vert _{L^2}$ and Lemma 3.1 we deduce

$$\begin{aligned}{} & {} \Vert a_i^2 W_{i}^p\otimes W_{i}^c + a_i^2W_{i}^c\otimes W_{i}^p + a_i^2 W_{i}^c\otimes W_{i}^c \Vert _{L^1} \le \Vert a_i \Vert _{L^\infty }( 2 \Vert W_i^p\Vert _{L^2}\Vert W_i^c\Vert _{L^2} + \Vert W_i^c\Vert _{L^2}^2) \\{} & {} \qquad \le {\varepsilon }C(t_0, \Vert R_0\Vert _{L^\infty }) \,, \\{} & {} \Vert ( u_{1}^{(c)} + u_{1}^{(t)}) \otimes ( u_1-u_0)+ ( u_1-u_0) \otimes ( u_{1}^{(c)} + u_{1}^{(t)})\Vert _{L^1} \le 2 \Vert u_{1}^{(c)} + u_{1}^{(t)} \Vert _{L^2} \Vert u_1-u_0\Vert _{L^2} \\{} & {} \qquad \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C_2}), \\{} & {} \Big \Vert \mathcal {R}\left( \nabla a_i^2 \cdot (W_i^p \otimes W^p_i - \int _{{{\mathbb {T}}}^2} W^p_i \otimes W_i^p) \right) \Big \Vert _{L^1} \le C {\varepsilon }\Vert \nabla a_1\Vert _{C^1} \Vert W_i^p \otimes W^p_i\Vert _{L^1} \\{} & {} \qquad \le {\varepsilon }C(t_0, \Vert R_0 \Vert _{C^2}) \,. \end{aligned}$$