2.1 General strategy

Our techniques for extending subsolutions and performing convex integration in the nonperiodic setting rely on obtaining compactly supported solutions to the (matrix) divergence equation when the source term is compactly supported. Let us illustrate the key ideas behind our method with the following toy problem:

It is easy to see that $v_0=\nabla \,\Delta ^{-1}\rho $ solves our equation, but in general it is not compactly supported. To fix this, let B be a ball containing the support of $\rho $ , so that $v_0$ is divergence-free outside B. In addition, it follows from the divergence theorem that

$$\begin{align*}0=\int_B \rho=\int_B \operatorname{\mathrm{div}} v_0=\int_{\partial B}v_0\cdot \nu.\end{align*}$$

This ensures that in $\mathbb {R}^3\backslash B$ the divergence-free field $v_0$ can be written as $v_0=\operatorname {\mathrm {curl}} w_0$ for some smooth field $w_0$ . We extend $w_0$ to a smooth field $w\in C^\infty (\mathbb {R}^3,\mathbb {R}^3)$ and we define

Since w extends $w_0$ , we see that v vanishes outside B. Furthermore, $\operatorname {\mathrm {div}} v=\rho $ because $\operatorname {\mathrm {div}}\operatorname {\mathrm {curl}}\equiv 0$ . Therefore, v is the sought field, which is clearly not unique.

Our approach to solving the divergence equation in the matrix case is the same: the potential-theoretic solution of the equation is not compactly supported. However, far from the support of the source our matrix will be the image of certain differential operator $\mathscr {L}$ applied to a smooth potential, which is in the kernel of the divergence. We will extend the potential to the whole space and then subtract it from the potential-theoretic solution, obtaining a compactly supported solution.

Just like in the vector case, we will have to impose certain integrability conditions on the source term for this to be possible. As we will see, these conditions are related to the classical conservation laws of linear and angular momentum in the Euler equations.

A totally different method to construct compactly supported solutions to the divergence equation in the (symmetric) matrix case was developed by Isett and Oh in [Reference Isett and Oh36]. Their theorem is stated in a very different setting and adapting it to what we need would require certain work. On the other hand, it will be relatively easy to deduce our result as a consequence of our analysis of the operator $\mathscr {L}$ introduced below, which is necessary for our result on the extension of subsolutions. Hence, we have preferred to take this path, which we believe is simpler (partly because it has a nice interpretation in terms of the elementary operations of vector calculus).

2.2 Basic tools

The tools that we will need come from the Hodge decomposition theorem for manifolds with boundary. A good reference for this topic is [Reference Schwarz45]. Nevertheless, we do not need the full generality of these results, as we will work in bounded domains of $\mathbb {R}^n$ . Let us summarize the notation and main definitions that we will need.

Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary. We denote by $\Lambda ^k$ the vector space of skew-symmetric k-forms over $\mathbb {R}^n$ , for $0\leq k\leq n$ . In this setting, differential k-forms are maps $\omega \in C^\infty (\overline {\Omega },\Lambda ^k)$ . They form vector spaces in which we have two differential operators: the exterior derivative $d:C^\infty (\overline {\Omega },\Lambda ^k)\to C^\infty (\overline {\Omega },\Lambda ^{k+1})$ and the codifferential $\delta :C^\infty (\overline {\Omega },\Lambda ^{k+1})\to C^\infty (\overline {\Omega },\Lambda ^k)$ . The Euclidean product induces a natural scalar product $(\cdot ,\cdot )$ in $C^\infty (\overline {\Omega },\Lambda ^k)$ . The tangential part of a differential form is , where $j:\partial \Omega \hookrightarrow \overline {\Omega }$ is the natural inclusion, and $j^*$ is the pushforward. We define the Dirichlet harmonic k-forms as:

Finally, to obtain quantitative estimates we will need to work in Hölder spaces. We refer to Appendix A for the definition of these norms. We also recommend to take a look at the appendix to check our convention of Hölder norms when the field is time-dependent.

With this notation, the first basic lemma that we shall use is:

Lemma 2.3. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $\rho \in C^\infty (\overline {\Omega },\Lambda ^k)$ . The boundary value problem

$$\begin{align*}(P)\begin{cases}\delta \omega=\rho,\\d\omega=0,\\ \mathbf{t}\omega=0,\\ ( \omega, \lambda) =0 \quad \forall \lambda\in \mathcal{H}^{k+1}_D(\overline{\Omega})\end{cases}\end{align*}$$

is solvable if and only if

$$\begin{align*}\delta \rho=0\qquad \text{and}\qquad \int_{C_{n-k}}\star \rho=0\quad \forall\; (n-k)\text{-cycle }C_{n-k}.\end{align*}$$

In that case, the solution is unique and we have

$$\begin{align*}\left\|\omega\right\|_{C^{N+1+\alpha}(\Omega)}\leq C\left\|\rho\right\|_{C^{N+\alpha}(\Omega)}\end{align*}$$

for any $N\geq 0$ , $\alpha \in (0,1)$ and certain constants $C\equiv C(N,\alpha ,\Omega )$ .

For a proof, see [Reference Csató, Dacorogna and Kneuss20, Theorem 7.2] and [Reference Schwarz45, Corollary 3.2.4]. In [Reference Schwarz45, Theorem 3.2.5] we can find the general case of a Riemannian manifold with boundary, but it does not include estimates for Hölder norms, only for Sobolev norms. We recall that, as usual, $\star $ is the Hodge star operator acting on differential forms and an $(n-k)$ -cycle is an $(n-k)$ -chain (in the sense of algebraic topology) whose boundary is zero.

We are mainly interested in the problem $\delta \omega =\rho $ , but we have to add the other conditions to select a single solution. This allows us to obtain a time-dependent version of Lemma 2.3:

Lemma 2.4. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $I\subset \mathbb {R}$ be a closed and bounded interval. Let $\rho \in C^\infty (\overline {\Omega }\times I,\Lambda ^k)$ . There exists a differential form $\omega \in C^\infty (\overline {\Omega }\times I,\Lambda ^{k+1})$ solving the boundary value problem $(P)$ at each $t\in I$ if and only if

$$\begin{align*}\delta \rho=0\qquad \text{and}\qquad \int_{C_{n-k}}\star \rho=0\quad \forall\; (n-k)\text{-cycle }C_{n-k}, \forall t\in I.\end{align*}$$

In that case, the solution is unique and we have

$$\begin{align*}\left\|\omega\right\|_{N+1+\alpha}\leq C\left\|\rho\right\|_{N+\alpha}\end{align*}$$

for any $N\geq 0$ , $\alpha \in (0,1)$ and certain constants $C\equiv C(N,\alpha ,\Omega )$ .

Proof. Given a time-dependent differential form, we denote by a subscript the differential form at a given time. By Lemma 2.3, the necessity of the conditions is clear. To prove that they are also sufficient, let us suppose that $\rho _t$ satisfies the conditions of Lemma 2.3 at all times $t\in I$ . Hence, applying Lemma 2.3 at each time, we see that there exists a time-dependent $(k+1)$ -form $\omega $ solving $(P)$ at each $t\in I$ . The question is whether $\omega $ depends smoothly on t.

Since $\partial _t \rho $ also satisfies the hypotheses of Lemma 2.3 at all times $t\in I$ , there exists a $(k+1)$ -form $\widetilde {\omega }$ solving $(P)$ with data $\partial _t \rho $ . For a fixed $t_0\in I$ and $h\neq 0$ small we see that $h^{-1}(\omega _{t_0+h}-\omega _{t_0})-\widetilde {\omega }_{t_0}$ is the unique solution of $(P)$ with data $h^{-1}(\rho _{t_0+h}-\rho _{t_0})-(\partial _t \rho )_{t_0}$ . Therefore,

$$\begin{align*}\left\|\frac{\omega_{t_0+h}-\omega_{t_0}}{h}-\widetilde{\omega}_{t_0}\right\|_{k+1+\alpha}\leq C\left\|\frac{\rho_{t_0+h}-\rho_{t_0}}{h}-(\partial_t \rho)_{t_0}\right\|_{k+\alpha}\longrightarrow 0 \quad \text{as }h\longrightarrow 0.\end{align*}$$

We deduce that $\widetilde {\omega }$ is the partial derivative with respect to time of $\omega $ . Iterating this argument, we conclude that $\omega $ depends smoothly on time. The claimed estimates are easily obtained by taking the supremum on $t\in I$ in the estimates of Lemma 2.3.

Remark 2.5. If $\delta \rho =0$ , the integral of $\star \rho $ on an $(n-k)$ -cycle depends only on the homology class of the cycle. Indeed, if C and $C'$ are two $(n-k)$ -cycles in $\overline {\Omega }$ that are the boundary of an $(n-k+1)$ -chain, by Stokes’ theorem we have

$$\begin{align*}\int_{C'}\star \rho-\int_{C}\star \rho=\int_{\partial \mathcal{N}}\star \rho=\int_{\mathcal{N}}d\star \rho=(-1)^k\int_{\mathcal{N}}\star \delta \rho=0.\end{align*}$$

The machinery of differential geometry is quite powerful, but we are interested in the simpler setting of bounded domains $\Omega \subset \mathbb {R}^n$ . Taking advantage of the canonical basis of Euclidean space, we may forget about differential forms and work with simpler objects. Indeed, there is a natural correspondence between 1-forms and vector fields and between 2-forms and skew-symmetric matrices $\mathcal {A}^n$ :

$$ \begin{align*} &C^\infty\left(\overline{\Omega},\Lambda^1\right)\to C^\infty\left(\overline{\Omega},\mathbb{R}^n\right), \qquad \sum_{i,j=1}^na_{i}\,dx_i \mapsto\;\, \sum_{i=1}^na_{i} e_i, \\ &C^\infty\left(\overline{\Omega},\Lambda^2\right)\to C^\infty\left(\overline{\Omega},\mathcal{A}^n\right), \qquad \frac12\sum_{i,j=1}^na_{ij}\,dx_i\wedge dx_j\;\,\mapsto\;\, \sum_{i,j=1}^na_{ij}e_i\otimes e_j. \end{align*} $$

Here we have used that $dx_i\wedge dx_j=-dx_j\wedge dx_i$ . Using the canonical base of $1$ -forms, the action of the codifferential can be summarized as

$$\begin{align*}\delta(f dx_i)=\partial_if, \delta(f dx_i\wedge dx_j)=-\partial_jf dx_i,\end{align*}$$

where f is any smooth function and $1\leq i<j\leq n$ . One can then check that the following diagram commutes:

This allows us to write everything in terms of matrices and to simplify the notation. Using this correspondence and Remark 2.5, we may formulate a particular case of Lemma 2.4 as follows:

Lemma 2.6. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $I\subset \mathbb {R}$ be a closed and bounded interval. Let $v\in C^\infty (\overline {\Omega }\times I,\mathbb {R}^n)$ . The following are equivalent:

1. 1. there exists $A\in C^\infty (\overline {\Omega }\times I,\mathcal {A}^n)$ such that $\operatorname {\mathrm {div}} A=v$ ,

2. 2. $\operatorname {\mathrm {div}} v=0$ and $\int _\Sigma v\cdot \nu =0$ for any connected component $\Sigma $ of $\partial \Omega $ and any fixed $t\in I$ ,

3. 3. $\int _{C_{n-1}} v\cdot \nu =0$ for any $(n-1)$ -cycle $C_{n-1}$ and any $t\in I$ .

In that case, $A\in C^\infty (\overline {\Omega }\times I,\mathcal {A}^n)$ may be chosen so that

$$\begin{align*}\left\|A\right\|_{N+1+\alpha}\leq C\left\|v\right\|_{N+\alpha}\end{align*}$$

for any $N\geq 0$ , $\alpha \in (0,1)$ and certain constants $C\equiv C(N,\alpha ,\Omega )$ .

The proof is straightforward taking into account that the codifferential $\delta $ becomes the operator $\operatorname {\mathrm {div}}$ and the integral on cycles becomes the flux of the corresponding vector field across a closed surface. By Remark 2.5, the integral only depends on the homology class, so we can choose to integrate on the connected component of $\partial \Omega $ belonging to each class.

2.3 The divergence equation

After collecting some basic tools from Hodge theory in the previous subsection, we will now show how to obtain compactly supported solutions to the divergence equation. We begin by introducing some potential-theoretic solutions, which we will later modify in order to fix the support. Let us consider the following differential operator that maps smooth vector fields (with bounded derivatives) to $C^\infty (\mathbb {R}^n,\mathcal {S}^n)$ :

(2.5)

Here $\Delta ^{-1}$ refers to the potential-theoretic solution of the Poisson equation, that is, the (spatial) convolution of the source term with the fundamental solution of the Laplace equation in $\mathbb R^n$ . We remind the reader that partial derivatives with Latin subscripts denote partial derivatives in the spatial coordinates, whereas temporal partial derivatives are always denoted by $\partial _t$ .

A direct calculation shows that $\operatorname {\mathrm {div}}\mathscr {R}f=f$ . We notice that $\mathscr {R}$ is not trace-free. This is not an issue in our proofs, because our constructions with potentials do not preserve being trace-free, so we will usually absorb the trace into the pressure at the end.

Let us now derive a very useful identity. Let $\Omega \subset \mathbb {R}^n$ be a bounded open set with smooth boundary. Let $v\in C^\infty (\overline {\Omega },\mathbb {R}^n)$ and $S\in C^\infty (\overline {\Omega },\mathcal {S}^n)$ . Integrating by parts, we have

(2.6) $$ \begin{align} \int_\Omega v\cdot \operatorname{\mathrm{div}} S+\int_\Omega(\nabla_{\text{sym}} v):S=\int_{\partial \Omega}v^t S \,\nu, \end{align} $$

where $\nu $ is the unitary normal vector associated to the exterior orientation and the operator $\nabla _{\text {sym}}$ is given by:

$$\begin{align*}\nabla_{\text{sym}}:C^\infty(\Omega,\mathbb{R}^n)\to C^\infty(\Omega,\mathcal{S}^n), v\mapsto \frac{1}{2}(\nabla v+\nabla v^t).\end{align*}$$

Its kernel are the so-called Killing vector fields. It is a finite-dimensional vector space that plays an important role in Riemannian geometry. It is well known (see [Reference Petersen43, page 52]) that in $\mathbb {R}^n$ a basis of this vector space is given by

(2.7)

where

(2.8)

Next, we introduce two vector spaces and a differential operator that will be very important in our construction:

Definition 2.7. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain and let $I\subset \mathbb {R}$ be a closed and bounded interval. We define two vector spaces:

and we consider the differential operator

$$\begin{align*}\mathscr{L}:\mathcal{P}(\overline{\Omega}\times I)\to C^\infty(\overline{\Omega}\times I,\mathbb{R}^{n^2}), \;\; \left[\mathscr{L}(A)\right]_{ij}=\frac{1}{2}\sum_{k,l}\partial_{kl}\left(A^{ik}_{jl}+A^{jk}_{il}\right).\end{align*}$$

This operator already appeared in the context of convex integration in the original article by De Lellis and Székelyhidi [Reference De Lellis and Székelyhidi23], who noticed that the image of the operator $\mathscr {L}$ is contained in the space of divergence-free matrices.

For our purposes, this operator can be regarded as a matrix analog of the $\operatorname {\mathrm {curl}}$ operator in arbitrary dimension. In order to perform the construction sketched in Subsection 2.1, the next step is to understand how to invert this operator (under the appropriate boundary conditions). The following lemma is the key to our approach to solve the divergence equation:

Lemma 2.8. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $I\subset \mathbb {R}$ be a closed and bounded interval. Then $\mathcal {G}(\overline {\Omega }\times I)$ is the image of the differential operator $\mathscr {L}$ . Furthermore, given $S\in \mathcal {G}(\overline {\Omega }\times I)$ , there exists $A\in \mathcal {P}(\overline {\Omega }\times I)$ such that $S=\mathscr {L}(A)$ and for any $\alpha \in (0,1)$ we have

$$\begin{align*}\left\|A\right\|_{N+2+\alpha}\leq C\left\|S\right\|_{N+\alpha}\end{align*}$$

for all $N\geq 0$ and certain constants $C\equiv C(N,\alpha ,\Omega )$ .

Proof. First, we prove that the image of $\mathscr {L}$ is contained in $\mathcal {G}(\overline {\Omega }\times I)$ . Fix an arbitrary $A\in \mathcal {P}(\overline {\Omega }\times I)$ . It is clear from the definition that $\mathscr {L}(A)$ is symmetric. The fact that it is divergence-free follows from the skew-symmetric properties of A:

$$ \begin{align*} [\operatorname{\mathrm{div}} \mathscr{L}(A)]_i&=\frac{1}{2}\sum_{j,k,l}\partial_{jkl}\left(A^{ik}_{jl}+A^{jk}_{il}\right)=\\ &=\sum_k\frac{1}{2}\,\partial_k\left(\sum_{j,l}\partial_{jl}A^{ik}_{jl}\right)+\sum_l\frac{1}{2}\,\partial_l\left(\sum_{j,k}\partial_{jk}A^{jk}_{il}\right)=0. \end{align*} $$

Next, we fix an arbitrary Killing vector field $\xi $ and a connected component $\Sigma $ of $\partial \Omega $ . We choose a smooth cut-off function $\varphi $ that vanishes in a neighborhood of $\Sigma $ and it is identically 1 in a neighborhood of the other connected components of $\partial \Omega $ . By the choice of $\varphi $ we have

$$\begin{align*}\int_{\partial \Omega} \xi^t\mathscr{L}(\varphi A)\,\nu=\int_{\partial \Omega}\xi^t\mathscr{L}(A)\,\nu-\int_\Sigma \xi^t\mathscr{L}(A)\,\nu.\end{align*}$$

By our previous discussion, $\mathscr {L}(A)$ and $\mathscr {L}(\varphi A)$ are symmetric and divergence-free. In addition, $\xi $ is a Killing vector, so $\nabla _{\text {sym}} w=0$ . Thus, from (2.6) we deduce that the term on the left-hand side of the previous equation vanishes and so does the first term on the right-hand side. Therefore, we see that $\int _\Sigma \xi ^t\mathscr {L}(A)\,n=0$ and, since A, $\xi $ , and $\Sigma $ are arbitrary, we conclude that the image of $\mathscr {L}$ is contained in $\mathcal {G}(\overline {\Omega }\times I)$ .

Now we will prove the other inclusion and the stated estimate. We fix an arbitrary $S\in \mathcal {G}(\overline {\Omega }\times I)$ . By definition, when choosing the canonical basis of $\mathbb {R}^n$ as Killing vectors, we obtain

$$\begin{align*}\int_\Sigma S_{ij}\nu_j=0\qquad \forall \Sigma \text{ connected component of }\partial \Omega\end{align*}$$

for any $i=1, \dots , m$ . If we fix i, we may apply Lemma 2.6 to conclude that there exists $B^i\equiv B^i_{jl}\in C^\infty (\overline {\Omega }\times I,\mathcal {A}^n)$ such that $\partial _l B^i_{jl}=S_{ij}$ .

Next we fix a Killing field $\xi $ of the form $\xi _i=R_{ik}x_k$ , where $R\in \mathcal {A}^n$ . For $j=1, \dots m$ we compute

$$\begin{align*}\partial_l(\xi_i B^i_{jl})=\xi_i\partial_l B^i_{jl}+(\partial_l \xi_i)B^i_{jl}=\xi_i S_{ij}+R_{il} B^i_{jl}.\end{align*}$$

Note that, since $S\in \mathcal {G}(\overline {\Omega }\times I)$ ,

$$\begin{align*}\int_\Sigma \xi_i S_{ij} \nu_j=0 \qquad \forall \Sigma \text{ connected component of }\partial \Omega, \quad \forall t\in I.\end{align*}$$

Regarding the left-hand side term, we define the forms

Since $B^i_{jl}$ is skew-symmetric in the lower indices, we see that $\delta \eta =\omega $ . Using the properties of the Hodge star operator and the codifferential, we have $\star \delta \eta =d\star \eta $ . These forms allow us to rewrite the integral on an $(n-1)$ -cycle at any $t\in I$ as:

$$\begin{align*}\int_{C_{n-1}}\partial_l(\xi_iB^i_{jl})\,\nu_j=\int_{C_{n-1}}\omega(n)\,\tilde{\mu}=\int_{C_{n-1}}\star \,\omega=\int_{C_{n-1}}d(\star\, \eta)=0,\end{align*}$$

where $\tilde {\mu }$ is the measure induced by the standard measure in $\mathbb {R}^n$ . We have used Stokes’ theorem and the fact that $(n-1)$ -cycles have no boundary. We conclude that for any $(n-1)$ -cycle

$$\begin{align*}\int_{C_{n-1}} R_{il}B^i_{jl}\nu_j=0.\end{align*}$$

Choosing $R=e_{i_0}\otimes e_{l_0}-e_{l_0}\otimes e_{i_0}$ , that is, choosing $\xi $ as $\xi _{l_0i_0}$ , we see that for any $i,l=1, \dots m$ we have:

$$\begin{align*}\int_{C_{n-1}} (B^i_{jl}-B^l_{ji})\,\nu_j=0 \qquad \text{for any }(n-1)\text{-cycle }C_{n-1} \text{ and all }t\in I.\end{align*}$$

Applying again Lemma 2.6, we obtain $A^{ik}_{jl}$ skew-symmetric in $j, k$ such that

$$\begin{align*}\partial_k A^{ik}_{jl}=B^i_{jl}-B^l_{ji}.\end{align*}$$

Therefore,

$$ \begin{align*} \frac{1}{2}\partial_{kl}(A^{ik}_{jl}+A^{jk}_{il})&=\frac{1}{2}\partial_l\left[\partial_k A^{ik}_{jl}+\partial_k A^{jk}_{il}\right]=\frac{1}{2}\partial_l\left[(B^i_{jl}-B^l_{ji})+(B^j_{il}-B^l_{ij})\right]\\&=\frac{1}{2}(\partial_l B^i_{jl} +\partial_l B^j_{il})=S_{ij}, \end{align*} $$

where we have used that B is skew-symmetric in the lower indices and the symmetry of S: $\partial _l B^i_{jl}=S_{ij}=S_{ji}=\partial _l B^j_{il}$ . In summary, we have found $A\in C^\infty \left (\overline {\Omega }\times I,\mathbb {R}^{n^4}\right )$ such that:

1. (i) $A^{ik}_{jl}=-A^{ij}_{kl}$ ,

2. (ii) $\partial _k A^{ik}_{jl}=-\partial _k A^{lk}_{ji}$ ,

3. (iii) $\frac {1}{2}\partial _{kl} \left (A^{ik}_{jl}+A^{jk}_{il}\right )=S_{ij}$ .

We define

It is clear that $\widetilde {A}^{ik}_{jl}$ is skew-symmetric in $i,k$ . In addition, it is skew-symmetric in $j,l$ by (i). Furthermore,

$$\begin{align*}\partial_l \widetilde{A}^{ik}_{jl}=\frac{1}{2}\left(\partial_l A^{il}_{jk}-\partial_l A^{kl}_{ji}\right)\stackrel{\text{(ii)}}{=}\partial_l A^{il}_{jk}.\end{align*}$$

Hence, using (iii) we conclude

$$\begin{align*}\frac{1}{2}\partial_{kl} \left(\widetilde{A}^{ik}_{jl}+\widetilde{A}^{jk}_{il}\right)=\frac{1}{2}\partial_{kl} \left(A^{ik}_{jl}+A^{jk}_{il}\right)=S_{ij}.\end{align*}$$

Therefore, $\widetilde {A}\in \mathcal {P}(\overline {\Omega }\times I)$ and $\mathscr {L}(\widetilde {A})=S$ , as we wanted. The estimates for $\widetilde {A}$ follow from applying twice the estimates from Lemma 2.6.

Finally, we are ready to prove the main result of this subsection, which establishes the existence of compactly supported solutions to the divergence equation. In a different setting, a related class of compactly supported solutions to the divergence equation were constructed by Isett and Oh [Reference Isett and Oh36, Theorem 10.1]. Our approach is based on the operator $\mathscr {L}$ , which will be essential for the extension of subsolutions in Lemma 2.15. We observe that the compatibility conditions (2.9) in Lemma 2.9 are precisely the conditions (202) in [Reference Isett and Oh36].

We recall that the Besov norms that we use are defined in Appendix A. We need to work with these norms because they are necessary to derive estimates for the $C^\alpha $ norm of the resulting matrix. This will be essential in the proof of Theorem 1.7.

Lemma 2.9. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $I\subset \mathbb {R}$ be a closed and bounded interval. Let $f\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^n)$ such that $\operatorname {\mathrm {supp}} f(\cdot ,t)\subset \Omega $ for all $t\in I$ . Then, there exists $S\in C^\infty (\mathbb {R}^n\times I,\mathcal {S}^n)$ such that $\operatorname {\mathrm {div}} S=f$ and $\operatorname {\mathrm {supp}} S(\cdot ,t)\subset \Omega $ for all $t\in I$ if and only if

(2.9) $$ \begin{align} \int_\Omega f\cdot \xi=0\qquad \forall \xi\in \ker \nabla_{\text{sym}}, \;\forall t\in I. \end{align} $$

In that case, we may choose S so that for all $N\geq 0$ and any $\alpha \in (0,1)$ we have

$$\begin{align*}\left\|S\right\|_{N+\alpha}\leq C\left\|f\right\|_{B^{N-1+\alpha}_{\infty,\infty}}\end{align*}$$

for certain constants $C=C(\Omega ,N,\alpha )$ .

Proof. First of all, we show that the integrability condition (2.9) is necessary. Let us suppose that such an S exists. We fix a ball $B\supset \overline {\Omega }$ and use the identity (2.6) to obtain

$$\begin{align*}0=\int_{\partial B}\xi^tS \nu=\int_{B}\xi\cdot \operatorname{\mathrm{div}} S=\int_{\Omega}\xi\cdot f \qquad \forall \xi\in\ker\nabla_{\text{sym}},\;\forall t\in I.\end{align*}$$

Let us show that condition (2.9) is also sufficient. The field $S_0\in C^\infty (\mathbb {R}^n\times I,\mathcal {S}^n)$ given by solves the equation $\operatorname {\mathrm {div}} S_0=f$ , where $\mathscr {R}$ was defined in (2.5). It is easy to check that it satisfies the estimates

(2.10) $$ \begin{align} \left\|S_0\right\|_{N+\alpha}\leq C\left\|f\right\|_{B^{N-1+\alpha}_{\infty,\infty}} \end{align} $$

for certain constants $C=C(N,\alpha )$ because $\mathscr {R}$ is an operator of order $-1$ . However, it is not compactly supported, in general. We must modify it far from the support of f.

We begin by studying the boundary conditions. Let $\Sigma _i$ be a connected component of $\partial \Omega $ and let $U_i$ be the domain bounded by it. We claim that

(2.11) $$ \begin{align} \int_{U_i}\xi\cdot f=0 \qquad \forall\xi\in\ker\nabla_{\text{sym}},\;\forall t\in I. \end{align} $$

Indeed, since $\Omega $ is a bounded domain, $U_i$ must be either the complement of the unbounded connected component of $\mathbb {R}^3\backslash \overline {\Omega }$ or one of the bounded connected components of $\mathbb {R}^3\backslash \overline {\Omega }$ (if there are any). In the first case, (2.11) follows from the integrability condition (2.9) because $f(\cdot ,t)\subset \Omega \subset U_i$ . In the second case, (2.11) is trivial because $f(\cdot ,t)$ vanishes on $U_i\subset \mathbb {R}^3\backslash \overline {\Omega }$ . Thus, applying the identity (2.6) to each $U_i$ we obtain

(2.12) $$ \begin{align} \int_{\Sigma_i}\xi^t S_0\nu=0 \qquad \forall\xi\in\ker\nabla_{\text{sym}},\;\forall t\in I. \end{align} $$

Next, note that for sufficiently small $r>0$ the boundary of the open set

has twice as many connected components as $\partial \Omega $ . Furthermore, the boundary of each connected component $G_i$ of G consists of exactly two hypersurfaces, which we denote as $\Sigma _i$ and $\Sigma ^{\prime }_i$ , and we have $\Sigma _i\subset \partial \Omega $ . By further reducing $r>0$ , we may assume that $f(\cdot ,t)$ vanishes on G at all times $t\in I$ . Then, it follows from (2.12) and the identity (2.6) that

$$\begin{align*}\int_{\Sigma^{\prime}_i}\xi^tS\nu=-\int_{\Sigma_i}\xi^tS\nu+\int_{\partial G_i}\xi^tS\nu=-\int_{\Sigma_i}\xi^tS\nu+\int_{G_i}\xi\cdot f=0\end{align*}$$

for any $\xi \in \ker \nabla _{\text {sym}}$ . Next, we fix a ball $B\supset \overline {\Omega }$ and we consider the domain

We see that

$$\begin{align*}\partial U=\partial B\cup\bigcup_i \Sigma_i'.\end{align*}$$

Again, it follows from (2.6) and the integrability condition (2.9) that

$$\begin{align*}\int_{\partial B}\xi^tS_0\nu=\int_B\xi\cdot f=\int_{\Omega}\xi\cdot f=0 \qquad \forall\xi\in\ker\nabla_{\text{sym}},\;\forall t\in I.\end{align*}$$

We conclude that $S_0$ is divergence-free on U and in each connected component $\Sigma $ of $\partial U$ we have

$$\begin{align*}\int_\Sigma \xi^tS_0\nu=0\qquad \forall\xi\in\ker\nabla_{\text{sym}},\;\forall t\in I,\end{align*}$$

that is, $S_0\in \mathcal {G}(\overline {U}\times I)$ . By Lemma 2.8 there exists $A_0\in \mathcal {P}(\overline {U}\times I)$ such that $S_0(x,t)=\mathscr {L}(A_0)(x,t)$ for all $x\in \overline {G}$ and $t\in I$ . Furthermore, for any $N\geq 0$ and $\alpha \in (0,1)$ we have

$$\begin{align*}\left\|A_0\right\|_{N+2+\alpha}\leq C(U,N,\alpha)\left\|S_0\right\|_{N+\alpha}\leq C(U,N,\alpha)\left\|f\right\|_{B^{N-1+\alpha}_{\infty,\infty}}.\end{align*}$$

The constants depend on U, which depends not only on the geometry of $\Omega $ but also on the minimum distance between the support of $f(\cdot ,t)$ and $\partial \Omega $ through the parameter r. However, since U tends to $B\backslash \overline {\Omega }$ as $r\to 0$ , the constants remain uniformly bounded, so they ultimately depend only on $\Omega $ . For this, smoothness of $\partial \Omega $ is essential, as it allows us to choose parametrizations of $\overline {U}$ converging to parametrizations of $\overline {B}\backslash \Omega $ in any Hölder norm as $r\to 0$ .

Applying Theorem B.3 and antisymmetrizing, we see that there exists a map $A\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^{n^4})$ such that $A^{ik}_{jl}=-A^{ki}_{jl},\, A^{ik}_{jl}=-A^{ik}_{lj}$ that extends $A_0$ outside $\overline {U}\times I$ . Furthermore, for any $N\geq 0$ and $\alpha \in (0,1)$ , we have

(2.13) $$ \begin{align} \left\|A\right\|_{N+2+\alpha}\leq C(U,N)\left\|A_0\right\|_{N+2+\alpha}\leq C(U,\Omega,N,\alpha)\left\|f\right\|_{B^{N-1+\alpha}_{\infty,\infty}}. \end{align} $$

Again, since U tends to $B\backslash \overline {\Omega }$ as $r\to 0$ in a suitable manner, the constants ultimately depend only on $\Omega $ , N, and $\alpha $ , since they will be uniformly bounded on $r\in (0,1)$ .

Finally, for $x\in B$ and $t\in I$ we define

Since the image of $\mathscr {L}$ is contained in the kernel of the divergence, we see that $\operatorname {\mathrm {div}} S=f$ . By construction A extends $A_0$ , so $\mathscr {L}(A)=\mathscr {L}(A_0)=S_0$ on $\overline {U}\times I$ . Therefore, $S(\cdot ,t)$ vanishes in U, so we may extend it by 0 to $\mathbb {R}^n\times I$ .

In conclusion, we have constructed $S\in C^\infty (\mathbb {R}^n\times I,\mathcal {S}^n)$ such that $\operatorname {\mathrm {div}} S=f$ and $\operatorname {\mathrm {supp}} S(\cdot ,t)\subset \Omega $ for all $t\in I$ . Furthermore, the desired estimate follows from (2.10) and (2.13) because $\mathscr {L}$ is a second-order differential operator.

2.4 Subsolutions and proof of Theorem 1.6

In this subsection we use Lemma 2.8 and Lemma 2.9 to glue and extend subsolutions, which will yield the proof of Theorem 1.6. It should be apparent by now that controlling the $L^2$ -product with the Killing fields is very important in these constructions. It is not difficult to construct $f\in C^\infty _c(\mathbb {R}^n,\mathbb {R}^n)$ with the desired $L^2$ -product with the Killing fields. However, when working with subsolutions we will also need that f be divergence-free. In addition, in our constructions we will work in domains of a certain form. Our approach is based on the following:

Lemma 2.10. Let $\Omega \subset \mathbb {R}^n$ be a bounded domain with smooth boundary and let $I\subset \mathbb {R}$ be a closed and bounded interval. Let $r>0$ and let $L_{ij}\in C^\infty (I)$ for $1\leq i<j\leq n$ . There exists a divergence-free field $w\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^n)$ such that the support of $w(\cdot ,t)$ is contained in $\{x\in \mathbb {R}^n:0<\operatorname {\mathrm {dist}}(x,\Omega )<r\}$ and

$$ \begin{align*} &\int a\cdot w\,dx=0, \qquad \int \xi_{ij}\cdot w\,dx=L_{ij}(t) \end{align*} $$

for all $t\in I$ and $1\leq i<j\leq n$ , where $\xi _{ij}$ is given by (2.8). Furthermore, for any $N\geq 0$ we have

$$ \begin{align*} &\left\|w\right\|_{N}\leq C(N,n)\left\vert\Omega\right\vert{}^{-1}\, r^{-(N+1)}\max_{ij,\, t\in I}\left\vert L_{ij}(t)\right\vert, \\ &\left\|\partial_t w\right\|_{N}\leq C(N,n)\left\vert\Omega\right\vert{}^{-1}\, r^{-(N+1)}\max_{ij,\, t\in I}\left\vert L^{\prime}_{ij}(t)\right\vert. \end{align*} $$

Proof. We will construct our field as $w=\operatorname {\mathrm {div}} A$ for some $A\in C^\infty _c(\mathbb {R}^n\times I,\mathcal {A}^n)$ that we will choose later. Since A is compactly supported, it follows from the divergence theorem that $\int a\cdot w=0$ for any $a\in \mathbb {R}^n$ . Furthermore, for any $1\leq i<j\leq n$ we have

(2.14) $$ \begin{align} \int \xi_{ij}\cdot w=\int (\xi_{ij})_k\partial_l A_{kl}=-\int \partial_l(\xi_{ij})_k A_{kl}=-\int (A_{ji}-A_{ij})=2\int A_{ij} \end{align} $$

because $\partial _l(\xi _{ij})_k=\delta _{il}\delta _{jk}-\delta _{jl}\delta _{ik}$ . Here we have denoted by $(\xi _{ij})_k$ the k-th component of the vector $\xi _{ij}$ and we have used Einstein’s summation convention when summing over k and l. By Lemma B.1 we may choose a nonnegative cutoff function $\varphi \in C^\infty _c(\Omega +B(0,r))$ that is identically 1 in a neighborhood of $\Omega $ and such that

$$\begin{align*}\left\|\varphi\right\|_N\leq C(N,n)\,r^{-N}.\end{align*}$$

We define

Since $\varphi $ is constant in a neighborhood of $\Omega $ and its support is contained in $\Omega +B(0,r)$ , we see that the support of $w(\cdot ,t)$ is contained in $\{x\in \mathbb {R}^n:0<\operatorname {\mathrm {dist}}(x,\Omega )<r\}$ . By construction

$$\begin{align*}2A_{ij}(x,t)=L_{ij}(t)\left(\int\varphi\right)^{-1}\varphi(x),\end{align*}$$

so it follows from Equation (2.14) that $\int \xi _{ij}\cdot w=L_{ij}$ . Finally, the claimed estimates follow at once from the bounds for $\varphi $ and the fact that $\int \varphi \geq \left \vert \Omega \right \vert $ .

Now we have all the ingredients that we need to glue subsolutions in space. The following lemma is the key tool in this section. It will be used not only in the proof of Theorem 1.6, but also in the convex integration scheme. We use skew-symmetric matrices instead of potential vectors because the lemma is stated in any dimension $n\geq 2$ .

Lemma 2.11. Let $T>0$ and let $\Omega _1\Subset \Omega _2\subset \mathbb {R}^n$ be bounded domains with smooth boundary. Let $(v_i,p_i,\mathring {R}_i)\in C^\infty (\Omega _2\times [0,T])$ be subsolutions for $i=1,2$ . Let $r>0$ be sufficiently small. There exists a subsolution $(v,p,\mathring {R})\in C^\infty (\Omega _2\times [0,T])$ such that

(2.15) $$ \begin{align} (v,p,\mathring{R})(x,t)=\begin{cases}(v_1,p_1,\mathring{R}_1)(x,t) \qquad x\in \overline{\Omega}_1, \\ (v_2,p_2,\mathring{R}_2)(x,t)\qquad \operatorname{\mathrm{dist}}(x,\Omega_1)\geq r\end{cases} \end{align} $$

if and only if for each connected component $\Sigma $ of $\partial \Omega _1$ , and all times $t\in [0,T]$ , we have

(2.16) $$ \begin{align} &\int_{\Sigma}v_1\cdot \nu=\int_{\Sigma}v_2\cdot \nu,\\&\int_{\Sigma}\left[(a\cdot x)\partial_t v_1+(a\cdot v_1)v_1+p_1 a-a^t\mathring{R}_1\right]\cdot \nu \nonumber \end{align} $$

(2.17) $$ \begin{align} &\int_{\Sigma}\left[(a\cdot x)\partial_t v_1+(a\cdot v_1)v_1+p_1 a-a^t\mathring{R}_1\right]\cdot \nu \nonumber \\ &\qquad \quad =\int_{\Sigma}\left[(a\cdot x)\partial_t v_2+(a\cdot v_2)v_2+p_2 a-a^t\mathring{R}_2\right]\cdot \nu \qquad \forall a\in\mathbb{R}^n. \end{align} $$

Suppose that, in addition, we have $v_1=\operatorname {\mathrm {div}} A_1$ and $v_2=\operatorname {\mathrm {div}} A_2$ for some potentials $A_i\in C^\infty (\Omega _2\times I,\mathcal {A}^n)$ . Then, there exists $A\in C^\infty (\Omega _2\times I,\mathcal {A}^n)$ such that $v=\operatorname {\mathrm {div}} A$ and $A(x,t)=A_2(x,t)$ if $\operatorname {\mathrm {dist}}(x,\Omega _1)\geq r$ .

Proof. First of all, note that a subsolution $(v_0,p_0,\mathring {R}_0)$ in a bounded domain G with smooth boundary satisfies

(2.18) $$ \begin{align} 0&=\int_G \operatorname{\mathrm{div}} v_0=\int_{\partial G}v_0\cdot\nu, \end{align} $$

(2.19) $$ \begin{align} &0=\int_G a\cdot\left[\partial_tv_0+\operatorname{\mathrm{div}}\left(v_0\otimes v_0+p_0\operatorname{\mathrm{Id}}-\mathring{R}_0\right)\right] \\&=\int_{\partial G}\left[(a\cdot x)\partial_t v_0+(a\cdot v_0)v_0+p_0 a-a^t\mathring{R}_0\right]\cdot \nu \qquad \forall a\in\mathbb{R}^n, \nonumber \end{align} $$

where we have used the divergence theorem, identity (2.6) and the fact that $\operatorname {\mathrm {div}}[(a\cdot x)\partial _tv_0]=a\cdot \partial _t v_0$ because $\partial _t v_0$ is divergence-free.

From these equations we readily deduce that the compatibility conditions (2.16) and (2.17) are automatically satisfied if $\partial \Omega _1$ is connected, as both integrals vanish for each field. In the case $\Omega _2=\mathbb {R}^n$ , we apply Equations (2.18) and (2.19) to the domain bounded by each connected component of $\partial \Omega _1$ . We conclude that both integrals vanish for each field in each connected component of $\partial \Omega _1$ .

Next, we check that the conditions are necessary; we study (2.16) because the expressions are shorter, but the argument for (2.17) is exactly the same. First, if $\partial \Omega _1$ is connected, it readily follows from (2.18) that (2.16) must be satisfied. Hence, we focus on bounded domains $\Omega _1$ whose boundary is not connected. In that case, $\mathbb {R}^n \backslash \Omega _1$ must have at least one bounded connected component. Given a bounded connected component of $\mathbb {R}^n\backslash \Omega _1$ , we define G to be its intersection with $\Omega _2$ . Then, $\partial G$ is composed of a connected component $\Sigma $ of $\partial \Omega _1$ and (possibly) some connected components $\Sigma ^{\prime }_1, \dots , \Sigma ^{\prime }_m$ of $\partial \Omega _2$ . Since v equals $v_1$ on $\Sigma $ and $v_2$ on the other connected components of $\partial G$ , it follows from (2.18) that:

$$\begin{align*}0=\int_{\partial G} v\cdot \nu=\int_{\Sigma}v_1\cdot\nu+\sum_{i=1}^n\int_{\Sigma^{}{\prime}{}_i}v{}_2\cdot\nu.\end{align*}$$

On the other hand, applying (2.18) to $v_2$ on G, we have

$$\begin{align*}-\int_\Sigma v_2\cdot\nu=\sum_{i=1}{}^n\int_{\Sigma^{}{\prime}{}_i}v{}_2\cdot\nu, \end{align*}$$

which, combined with the previous equation, yields

$$\begin{align*}\int_\Sigma v_1\cdot\nu=\int_\Sigma v_2\cdot\nu.\end{align*}$$

Since this applies to any bounded connected component of $\mathbb {R}^n\backslash \Omega _1$ , we can combine it with (2.18) with $G=\Omega _1$ to obtain an analogous identity for the remaining connected component of $\partial \Omega _1$ , that is, the boundary of the unbounded connected component of $\mathbb {R}^n\backslash \Omega _1$ . We conclude (2.16).

Let us now prove that the compatibility conditions (2.16) and (2.17) are also sufficient. Let $r>0$ be small enough so that $\{x\in \Omega _2:\operatorname {\mathrm {dist}}(x,\Omega _1)=r\}$ is diffeomorphic to $\partial \Omega _1$ . We define . Then, the condition (2.16) ensures that there exists $A_{12}\in C^\infty (\overline {U}\times [0,T],\mathcal {A}^n)$ such that $v_2-v_1=\operatorname {\mathrm {div}} A_{12}$ in $U\times [0,T]$ . Indeed, let $U_i$ be a connected component of U and let $\Sigma _i$ and $\Sigma ^{\prime }_i$ be the connected components of $\partial U_i$ , where $\Sigma _i\subset \partial \Omega _1$ . Using (2.16) and the fact that $v_2-v_1$ is divergence-free:

$$\begin{align*}\int_{\Sigma^{\prime}_i}(v_2-v_1)\cdot\nu=\int_{\partial U}(v_2-v_1)\cdot\nu-\int_{\Sigma_i}(v_2-v_1)\cdot\nu=0.\end{align*}$$

Hence, the flux of $v_2-v_1$ through each connected component of U vanishes, so by Lemma 2.6 there exists $A_{12}\in C^\infty (\overline {U}\times [0,T],\mathcal {A}^n)$ such that $v_2-v_1=\operatorname {\mathrm {div}} A_{12}$ .

Next, using Lemma B.1 we choose a cutoff function $\varphi \in C^\infty _c(\Omega _1+B(0,r))$ that equals 1 in a neighborhood of $\Omega _1$ . We define

where so that $\varphi \, v_1+(1-\varphi )v_2+w_c$ is divergence-free. The additional correction $w_L$ is a divergence-free field supported within U that will be defined later. Its purpose is to cancel the angular momentum so that the gluing can be performed in the interior of U. After a tedious computation we obtain

(2.20) $$ \begin{align} \partial_t v+\operatorname{\mathrm{div}} (v\otimes v)+\nabla \widetilde p=\operatorname{\mathrm{div}}\left(\varphi\,\mathring{R}_1+(1-\varphi)\mathring{R}_2+S_1\right)+\partial_t w_L+M\cdot\nabla\varphi, \end{align} $$

where

(2.21)

(2.22)

Let and $\rho =\widetilde {\rho }+\partial _t w_L$ . Our goal is to find $S_2\in C^\infty (\Omega _2\times [0,T],\mathcal {S}^n)$ supported on U for all $t\in [0,T]$ and such that $\operatorname {\mathrm {div}} S_2=\rho $ . Thus, we may set $R=\varphi \,\mathring {R}_1+(1-\varphi )\mathring {R}_2+S_1+S_2$ and absorb the trace into the pressure, obtaining $\mathring {R}$ and the final pressure p. To do so, first we must check that $\rho $ satisfies the compatibility conditions (2.9).

Note that $\widetilde {\rho }=\operatorname {\mathrm {div}}(\varphi M)$ because $\operatorname {\mathrm {div}} M=0$ , since $(v_i,p_i,\mathring {R}_i)$ are subsolutions. Hence, by the divergence theorem for any $a\in \mathbb {R}^n$ we have

(2.23) $$ \begin{align} \int_U a\cdot \widetilde{\rho}=\int_{\partial U}a^t(\varphi M)\nu=\int_{\partial \Omega_1} a^tM\nu. \end{align} $$

Note that

(2.24) $$ \begin{align} \int_{\partial\Omega_1}a^t(\partial_tA_{12})\nu&=-\int_{\partial\Omega_1}\nu^t(\partial_t A_{12})\nabla(a\cdot x)=\int_{\partial\Omega_1}(a\cdot x)\operatorname{\mathrm{div}}(\partial_t A_{12})\cdot \nu \nonumber \\&=\int_{\partial\Omega_1}(a\cdot x)(\partial_t v_1-\partial_t v_2)\cdot \nu. \end{align} $$

Therefore, combining Equations (2.17), (2.23) and (2.24) we conclude that $\int _U a\cdot \widetilde {\rho }=0$ for all $a\in \mathbb {R}^n$ . Since $\partial _t w_L$ is divergence-free and its support is contained in U, the same holds for $\partial _t w_L$ , so $\int _U a\cdot \rho =0$ for all $a\in \mathbb {R}^n$ .

Next, we study the product with nonconstant Killing fields. For each pair $1\leq i<j\leq n$ we define

where $\xi _{ij}$ are the elements of the basis of Killing fields defined in (2.8). It will be useful later on to write the coefficients as:

(2.25) $$ \begin{align} l_{ij}=\int_U\xi_{ji}\cdot \operatorname{\mathrm{div}}(\varphi M)=\int_{\partial\Omega_1}\xi_{ij}^tM\nu, \end{align} $$

where we have used the fact that Killing fields are divergence-free as well as the values of $\varphi $ on $\partial \Omega _1$ and $\partial \Omega _2$ . We define

(2.26)

We then define the correction $w_L$ to be the divergence-free field obtained by applying Lemma 2.10 to the domain $\Omega _1$ with coefficients $L_{ij}\in C^\infty ([0,T])$ . Thus, we have

$$\begin{align*}\int \xi_{ij}\cdot\partial_tw_L=\frac{d}{dt}\int\xi_{ij}\cdot w_L=L_{ij}'=-l_{ij}.\end{align*}$$

We conclude that $\int _U\xi \cdot \rho =0$ for any Killing field $\xi $ , as we wanted. Therefore, by Lemma 2.9 there exists $S_2\in C^\infty (\Omega _2\times [0,T],\mathcal {S}^n)$ supported on U for all $t\in [0,T]$ and such that $\operatorname {\mathrm {div}} S_2=\rho $ . We define the final pressure and the Reynolds stress as

It follows from Equation (2.20) that the resulting triplet $(v,p,\mathring {R})$ is a subsolution and it satisfies (2.15) because $S_1$ and $S_2$ are supported in U for all $t\in [0,T]$ .

Finally, let us consider that the velocity fields are given by $v_i=\operatorname {\mathrm {div}} A_i$ . In that case, we may simply take $A_{12}=A_2-A_1$ instead of constructing a suitable potential using Lemma 2.6. We see that

$$\begin{align*}v-w_L=\varphi v_1+(1-\varphi)v_2+(A_2-A_1)\cdot\nabla \varphi=\operatorname{\mathrm{div}}(\varphi A_1+(1-\varphi)A_2).\end{align*}$$

Inspecting Lemma 2.10 leads us to define

(2.27)

Hence, we have $v=\operatorname {\mathrm {div}} A$ and we see that A equals $A_2$ in $\{\operatorname {\mathrm {dist}}(x,\Omega _1)\geq r\}$ because $\varphi $ vanishes in a neighborhood of this set.

In the convex integration scheme we will need estimates of the glued subsolution. For the sake of clarity, we keep them separate in a different lemma:

Lemma 2.14. Let $\alpha \in (0,1)$ . In the conditions of Lemma 2.11 and using the notation of its proof, the new subsolution satisfies:

(2.28) $$ \begin{align} \left\|v-(\varphi v_1+(1-\varphi)v_2)\right\|_N&\lesssim Tr^{-(N+1)}\left\|M\right\|_{0; U}+\sum_{k=0}^Nr^{-(k+1)}\left\|A_{12}\right\|_{N-k; U}, \end{align} $$

(2.29) $$ \begin{align} \left\|\partial_t(v-\varphi v_1-(1-\varphi)v_2)\right\|_N&\lesssim r^{-(N+1)}\left\|M\right\|_{0; U}+\sum_{k=0}^Nr^{-(k+1)}\left\|\partial_t A_{12}\right\|_{N-k; U}, \end{align} $$

(2.30) $$ \begin{align} \|\mathring{R}-\varphi\,\mathring{R}_1-(1-\varphi)\mathring{R}_2\|_0&\lesssim r^{-\alpha}\left\|M\right\|_{0; U}+\left\|v_1-v_2\right\|_{0; U}^2\\& +(\left\|v_1\right\|_{0; U}+\left\|v_2\right\|_{0; U}+\left\|w\right\|_{0; U})\left\|w\right\|_{0; U}, \nonumber \end{align} $$

In addition, if $v_1=\operatorname {\mathrm {div}} A_1$ and $v_2=\operatorname {\mathrm {div}} A_2$ , the potential A satisfies

(2.31) $$ \begin{align} \left\|A-(\varphi A_1+(1-\varphi)A_2)\right\|_N&\lesssim Tr^{-N}\left\|M\right\|_{0; U}, \end{align} $$

(2.32) $$ \begin{align} \left\|\partial_t(A-\varphi A_1-(1-\varphi)A_2)\right\|_N&\lesssim r^{-N}\left\|M\right\|_{0; U}. \end{align} $$

The implicit constants in these inequalities depend on $\Omega _1$ , N, and $\alpha $ .

Proof. We begin by estimating $w_c=A_{12}\cdot \nabla \varphi $ . Since $\varphi $ satisfies $\left \|\varphi \right \|_N\lesssim r^{-N}$ and it is independent of time, it is clear that

$$\begin{align*}\left\|w_c\right\|_N\lesssim\sum_{k=0}^Nr^{-(k+1)}\left\|A_{12}\right\|_{N-k; U}, \qquad \left\|\partial_t w_c\right\|_N\lesssim\sum_{k=0}^Nr^{-(k+1)}\left\|\partial_t A_{12}\right\|_{N-k; U}\end{align*}$$

because the support of $\nabla \varphi $ is contained in U. Regarding $w_L$ , it follows from (2.25) that $\left \vert l_{ij}\right \vert \lesssim \left \|M\right \|_{0; U}$ , so $\left \vert L_{ij}\right \vert \lesssim T\left \|M\right \|_{0; U}$ . Hence, by Lemma 2.10 we have the bounds

$$\begin{align*}\left\|w_L\right\|_N\lesssim Tr^{-(N+1)}\left\|M\right\|_{0; U}, \qquad \left\|\partial_t w_L\right\|_N\lesssim r^{-(N+1)}\left\|M\right\|_{0; U}.\end{align*}$$

The claimed estimates for $v-(\varphi v_1+(1-\varphi )v_2)=w_c+w_L$ follow at once. Let us now focus on the Reynolds stress. Using the assumption $\left \|w\right \|_0\leq \left \|v_1\right \|_0+\left \|v_2\right \|_0$ , we deduce from the definition (2.21) that

$$\begin{align*}\left\|S_1\right\|_0\lesssim \left\|v_1-v_2\right\|_{0; U}^2+(\left\|v_1\right\|_{0; U}+\left\|v_2\right\|_0)\left\|w\right\|_{0; U}.\end{align*}$$

Concerning $S_2$ , let us first estimate $\rho $ :

$$\begin{align*}\left\|\rho\right\|_0\leq \left\|M\cdot\nabla\varphi\right\|_{0; U}+\left\|\partial_tw_L\right\|_0\lesssim r^{-1}\left\|M\right\|_{0; U}.\end{align*}$$

Since the support of $\rho (\cdot ,t)$ is contained in $\{x\in \mathbb {R}^n:0<\operatorname {\mathrm {dist}}(x,\Omega _1)<r\}$ , we may apply Lemma B.4, obtaining

$$\begin{align*}\left\|\rho\right\|_{B^{-1+\alpha}_{\infty,\infty}}\lesssim r^{1-\alpha}\left\|\rho\right\|_0\lesssim r^{-\alpha}\left\|M\right\|_{0; U}.\end{align*}$$

Hence, it follows from the estimates in Lemma 2.9 that

$$\begin{align*}\left\|S_2\right\|_0\lesssim \left\|\rho\right\|_{B^{-1+\alpha}_{\infty,\infty}}\lesssim r^{-\alpha}\left\|M\right\|_{0; U}.\end{align*}$$

Since

$$\begin{align*}\mathring{R}-(\varphi\,\mathring{R}_1+(1-\varphi)\mathring{R}_2)=S_1+S_2-\frac{1}{n}\operatorname{\mathrm{tr}}(S_1+S_2)\operatorname{\mathrm{Id}},\end{align*}$$

the claimed bound follows.

Finally, let us estimate A in the case that the velocities are given by $v_i=\operatorname {\mathrm {div}} A_2$ . By (2.27) we have

$$\begin{align*}A-(\varphi A_1-(1-\varphi)A_2)=\sum_{1\leq i<j\leq n}L_{ij}(t)\,(e_i\otimes e_j-e_j\otimes e_i)\left(2\int \varphi\right)^{-1}\varphi(x).\end{align*}$$

The claimed bounds then follow from the estimates derived in Lemma 2.10.

Lemma 2.11 is almost what we want, but it can be made a bit sharper. In particular, in Theorem 1.6 we do not want to assume that the subsolution $(v_0,p_0,\mathring {R}_0)$ is defined in a neighborhood of $\overline {\Omega }_0$ . Fortunately, it turns out that all subsolutions can be extended, at least a little bit. Our operator $\mathscr {L}$ is essential for this:

Proof. We begin by constructing the velocity field v. We choose $\rho \in C^\infty _c(\mathbb {R}^n\times I,\mathbb {R})$ such that $\operatorname {\mathrm {supp}} \rho (\cdot ,t)$ is contained in $\mathbb {R}^n\backslash \overline {\Omega }$ for all $t\in I$ and such that

$$\begin{align*}\int_{G}\rho(x,t)\,dx=\int_{\partial G} v_0\cdot \nu\qquad \forall t\in I\end{align*}$$

for each bounded connected component G of $\mathbb {R}^n\backslash \overline {\Omega }_0$ , whose boundary we have oriented with the outer normal with respect to G. This can be done if $\Omega $ is a sufficiently small neighborhood of $\overline {\Omega }_0$ so that the intersection of G with $\mathbb {R}^n\backslash \overline {\Omega }$ is nonempty.

Next, let so that $\widetilde {v}\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^n)$ and $\operatorname {\mathrm {div}} \widetilde {v}=\rho $ for all $t\in I$ . By the divergence theorem we have

$$\begin{align*}\int_{\partial G} \widetilde{v}\cdot \nu=\int_G \rho=\int_{\partial G} v_0\cdot \nu\end{align*}$$

in each bounded connected component G of $\mathbb {R}^n\backslash \overline {\Omega }_0$ and for all $t\in I_0$ . In addition, $\widetilde {v}-v_0$ is divergence-free in $\Omega _0\times I$ because $\rho $ vanishes in this set by construction. In particular, by the divergence theorem we have $\int _{\partial \Omega _0}(\widetilde {v}-v_0)\cdot \nu =0$ , from which we conclude

$$\begin{align*}\int_\Sigma (\widetilde{v}-v_0)\cdot \nu=0 \qquad \forall t\in I_0\end{align*}$$

for all connected components $\Sigma $ of $\partial \Omega _0$ . Therefore, by Lemma 2.6 there exists $A\in C^\infty (\overline {\Omega }_0\times I,\mathcal {A}^n)$ such that $\operatorname {\mathrm {div}} A=\widetilde {v}-v_0$ in $\Omega _0\times I$ . We choose a smooth extension $\widetilde {A}\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^{n\times n})$ and then we take the skew-symmetric part, so that $\widetilde {A}\in C^\infty (\mathbb {R}^n\times \mathbb {R},\mathcal {A}^n)$ . We define:

Since the support of $\rho (\cdot ,t)$ is contained in $\mathbb {R}^n\backslash \overline {\Omega }$ and the second term is divergence-free, v is divergence-free in $\Omega \times I$ . In addition, our choice of $\widetilde {A}$ ensures that the restriction of v to $\overline {\Omega }_0\times I$ is $v_0$ , as we wanted.

Next, we will extend in a similar manner. First, we choose $f\in C^\infty _c(\mathbb {R}^n\times I,\mathbb {R}^n)$ such that $\operatorname {\mathrm {supp}} f(\cdot ,t)$ is contained in $\mathbb {R}^n\backslash \overline {\Omega }$ for all $t\in I$ and such that

(2.33) $$ \begin{align} \int_{G_i} \xi\cdot f=\int_{G_i} \xi\cdot\partial_t v+\int_{\partial G_i} \xi^tU_0\nu\qquad \forall \xi\in \ker \nabla_{\text{sym}},\;\forall t\in I \end{align} $$

for all bounded connected components $G_i$ of $\mathbb {R}^n\backslash \overline {\Omega }_0$ , whose boundary we have oriented with the outer normal with respect to $G_i$ . To find such an f, we fix a nonnegative radial function $\psi \in C^\infty _c(B(0,1))$ and a ball $\overline {B}(x_i,r_i)\subset G_i$ . Due to symmetry, for any two elements $w_j\neq w_k$ of the basis $\mathcal {B}$ defined in (2.7) we have

$$\begin{align*}\int \psi\left(r_i^{-1}(x-x_i)\right)\,w_j(x)\cdot w_k(x)\,dx=0.\end{align*}$$

Therefore, it suffices to choose

for the appropriate coefficients $c_{ij}$ . Then, noticing that $\partial _t v$ is a smooth vector field with bounded derivatives, we define so that $\widetilde {S}\in C^\infty (\mathbb {R}^n\times I,\mathcal {S}^n)$ , and

(2.34) $$ \begin{align} \operatorname{\mathrm{div}} \widetilde{S}=-\partial_t v+f. \end{align} $$

Since v extends $v_0$ and $(v_0,p_0,\mathring {R}_0)$ is a subsolution in $\Omega _0\times I$ , we have

(2.35) $$ \begin{align} \operatorname{\mathrm{div}}(\widetilde{S}-S_0)(x,t)=0\qquad \forall(x,t)\in \Omega_0\times I \end{align} $$

because f vanishes in that set. In addition, using (2.6) it follows from (2.34) and (2.33) that

$$\begin{align*}\int_{\partial G} \xi^t (\widetilde{S}-S_0)\nu=0 \qquad \forall \xi\in\ker \nabla_{\text{sym}}\; \forall t\in I\end{align*}$$

for all bounded connected components G of $\mathbb {R}^m\backslash \overline {\Omega }_0$ . Due to (2.6) and (2.35), this integral also vanishes for the remaining connected component of $\partial \Omega _0$ , that is, the connected component that separates $\Omega _0$ and the unbounded connected component of $\mathbb {R}^n\backslash \Omega _0$ .

We conclude that $\widetilde {S}-S_0$ is in $\mathcal {G}(\overline {\Omega }_0\times I)$ . Therefore, by Lemma 2.8 there exists $\widetilde {E}\in \mathcal {P}(\overline {\Omega }_0\times I)$ such that $\mathscr {L}(\widetilde {E})=\widetilde {S}-S_0$ in $\Omega _0\times I$ . We choose a smooth extension $E\in C^\infty (\mathbb {R}^n\times I,\mathbb {R}^{n^4})$ and then we make the appropriate antisymmetrization. We define

By construction of E, we see that S extends $S_0$ . Additionally, we have

$$\begin{align*}\operatorname{\mathrm{div}} S=\operatorname{\mathrm{div}} \widetilde{S}=-\partial_t v+f\end{align*}$$

because the image of $\mathscr {L}$ is contained in the kernel of the divergence. We define

Since $f(\cdot ,t)$ vanishes in $\Omega $ for all $t\in I$ , we conclude that $(v,p,\mathring {R})$ is a subsolution in $\Omega \times I$ that extends $(v_0,p_0,\mathring {R}_0)$ .

Combining Lemma 2.11 and Lemma 2.15 we can finally prove Theorem 1.6:

3.1 The iterative process

Let us assume all along this subsection that the subsolution $(v_0,p_0)(\cdot ,t)$ is compactly supported for each time $t\in [0,T]$ . We will construct the desired weak solution of the Euler equations as the limit of a sequence of subsolutions, that is, at a given step $q\geq 0$ we have $(v_q,p_q,\mathring {R}_q)\in C^\infty (\mathbb {R}^3\times [0,T])$ solving the Euler-Reynolds system:

(3.1) $$ \begin{align} \begin{cases} \partial_t v_q+\operatorname{\mathrm{div}}(v_q\otimes v_q)+\nabla p_q=\operatorname{\mathrm{div}}\mathring{R}_q,\\ \operatorname{\mathrm{div}} v_q=0, \end{cases} \end{align} $$

to which we add the constraint that

(3.2) $$ \begin{align} \operatorname{\mathrm{tr}}\mathring{R}_q=0. \end{align} $$

The matrix $\mathring {R}_q$ measures the deviation from being a solution of the Euler equations. The goal of the process is to make $\mathring {R}_q$ vanish at the limit $q\to +\infty $ , so that the limit field is a weak solution of the Euler equations.

Assume we are given the initial subsolution $(v_0,p_0,\mathring {R}_0)\in C^\infty (\mathbb {R}^3\times [0,T])$ . Let us then show how to construct the rest of the terms iteratively. To construct the subsolution at step q from the one in step $q-1$ , we will add an oscillatory perturbation with frequency $\lambda _q$ . Meanwhile, the size of the Reynolds stress will be measured by an amplitude $\delta _q$ . These parameters are given by

(3.3) $$ \begin{align} \lambda_q&=2\pi \lceil a^{b^q}\rceil, \end{align} $$

(3.4) $$ \begin{align} \delta_q&=\lambda_q^{-2\beta}, \end{align} $$

where $\lceil x\rceil $ denotes the ceiling, that is, the smallest integer $n\geq x$ . The parameters $a,b>1$ are very large and very close to 1, respectively. They will be chosen depending on the exponent $0<\beta <1/3$ that appears in Theorem 1.7, on $\Omega $ and on the initial subsolution. We introduce another parameter $\alpha>0$ that will be very small. The necessary size of all of the parameters will be discovered in the proof.

Throughout the process we will also try to achieve a given energy profile $e\in C^\infty ([0,T])$ , which must satisfy the inequality (1.3). We will also assume

(3.5) $$ \begin{align} \sup_{t\in[0,T]}\left\vert\frac{d}{dt}e(t)\right\vert\leq 1. \end{align} $$

We will see that this can be assumed without losing generality.

Unlike the construction on the torus in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7], it is essential that we only perturb the field in the region where the Reynolds stress is nonzero. Hence, we have to pay special attention to the support of the fields.

Since the map $(v_0,p_0,\mathring {R}_0)(\cdot ,t)$ is assumed to be compactly supported at each time $t\in [0,T]$ , we shall see that with a suitable rescaling we may assume that its support and $\Omega $ are contained in $(0,1)^3$ . This is useful because sometimes it will be convenient to consider that we are working with periodic boundary conditions (that is, in $\mathbb {T}^3$ ) to reuse the results in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7]. On the other hand, $\mathring {R}_0(\cdot ,t)$ is supported in a potentially smaller domain $\overline {\Omega }$ . In our construction we must ensure that we do not perturb the subsolution outside of this set.

It will be convenient to do an additional rescaling in our problem. In the rescaled problem the initial subsolution will depend on a, but we assume that nevertheless there exists a sequence $\{y_N\}_{N=0}^\infty $ independent of the parameters such that

(3.6) $$ \begin{align} \left\|v_0\right\|_N+\left\|\partial_t v_0\right\|_N&\leq y_N, \end{align} $$

(3.7) $$ \begin{align} \left\|p_0\right\|_N&\leq y_N, \end{align} $$

(3.8) $$ \begin{align} \|\mathring{R}_0\|_N+\|\partial_t \mathring{R}_0\|_N&\leq y_N. \end{align} $$

Since the initial Reynolds stress $\mathring {R}_0$ and its derivatives vanish at $\partial \Omega \times [0,T]$ , for any $k\in \mathbb {N}$ there exists a constant $C_k$ such that for any $x\in \Omega $ we have

(3.9) $$ \begin{align} \vert\mathring{R}_0(x,t)\vert\leq C_k\operatorname{\mathrm{dist}}(x,\partial \Omega)^k. \end{align} $$

The constants $C_k$ are independent of a by (3.8). We define

(3.10)

Hence, we have

(3.11) $$ \begin{align} \vert\mathring{R}_0(x,t)\vert\leq \frac{1}{4}\delta_{q+2}\lambda_{q+1}^{-6\alpha} \qquad \forall x\in \Omega,\; \operatorname{\mathrm{dist}}(x,\partial\Omega)\leq d_q. \end{align} $$

At step q the perturbation will be localized in a central region

(3.12)

so that $(v_q,p_q,\mathring {R}_q)$ equals the initial subsolution in $(\mathbb {R}^3\backslash A_q)\times [0,T]$ . Note that $d_q\to 0$ as $q\to \infty $ because so does $\delta _{q+2}$ . Therefore, in the limit the perturbation covers all of the region where $\mathring {R}_0$ is nonzero. However, the velocity is not modified outside this set.

As we have mentioned, the error introduced in the gluing step of [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] is spread throughout the whole space. To avoid this, we introduce an additional gluing in space, which we will explain in more detail in the following sections.

The complete list of inductive estimates is the following:

(3.13) $$ \begin{align} (v_q,p_q,\mathring{R}_q)&=(v_0,p_0,\mathring{R}_0) \qquad \text{outside } A_q\times[0,T], \end{align} $$

(3.14) $$ \begin{align} \|\mathring{R}_q\|_0&\leq \delta_{q+1}\lambda_q^{-6\alpha}, \end{align} $$

(3.15) $$ \begin{align} \left\|v_q\right\|_1&\leq M\delta_q^{1/2}\lambda_q, \end{align} $$

(3.16) $$ \begin{align} \left\|v_q\right\|_0&\leq 1-\delta_q^{1/2}, \end{align} $$

(3.17) $$ \begin{align} \delta_{q+1}\lambda_q^{-\alpha}&\leq e(t)-\int_\Omega\left\vert v_q\right\vert{}^2dx\leq \delta_{q+1}, \end{align} $$

where M is a geometric constant that depends on $\Omega $ and is fixed throughout the iterative process.

Remark 3.1. If $\Omega $ has several connected components $\Omega ^j$ , we may fix an energy profile $e^j$ in each of them. In that case, (3.17) would have to be replaced by

$$\begin{align*}\delta_{q+1}\lambda_q^{-\alpha}\leq e^j(t)-\int_{\Omega^j}\left\vert v_q\right\vert{}^2dx\leq \delta_{q+1}.\end{align*}$$

Since the construction does not differ much, for simplicity we will assume that $\Omega $ is connected.

The following proposition is the key result to prove Theorem 1.7, because it establishes the existence of the iterative scheme in the convex integration process.

Proposition 3.2. Let $T>0$ and let $\Omega \subset (0,1)^3\subset \mathbb {R}^3$ be an open set with smooth boundary and with a finite number of connected components. Let $(v_0,p_0,\mathring {R}_0)\in C^\infty (\mathbb {R}^3\times [0,T])$ be a subsolution whose support is contained in $(0,1)^3\times [0,T]$ and such that $\operatorname {\mathrm {supp}}\mathring {R}_0\subset \overline {\Omega }\times [0,T]$ . Furthermore, assume that (3.6) $-$ (3.8) are satisfied for some sequence of positive numbers $\{y_N\}_{N=0}^\infty $ . There exists a constant M depending only on $\Omega $ with the following property: Assume $0<\beta <1/3$ and

(3.18) $$ \begin{align} 1<b<\min\left\{\frac{1-\beta}{2\beta},\;\frac{11}{10}\right\}. \end{align} $$

Then there exists an $\alpha _0$ depending on $\beta $ and b such that for any $0<\alpha <\alpha _0$ there is an $a_0$ depending on $\beta $ , b, $\alpha $ , $\Omega $ , and $\{y_N\}_{N=0}^\infty $ such that for any $a\geq a_0$ the following holds: Given a strictly positive energy profile satisfying (3.5) and a subsolution $(v_q,p_q,\mathring {R}_q)$ satisfying (3.13) $-$ (3.17), there exists a subsolution $(v_{q+1},p_{q+1},\mathring {R}_{q+1})$ satisfying the same equations (3.13) $-$ (3.17) with q replaced by $q+1$ . Furthermore, we have the estimate

(3.19) $$ \begin{align} \left\|v_{q+1}-v_q\right\|_0+\frac{1}{\lambda_{q+1}}\left\|v_{q+1}-v_q\right\|_1\leq M\delta_{q+1}^{1/2}. \end{align} $$

We wish to iterate this result to construct a sequence of subsolutions whose limit will be the desired weak solution. However, in order to start the process, the first term in the sequence must satisfy the inductive hypotheses (3.13) $-$ (3.17). Since we do not assume any bounds on $(v_0,p_0,\mathring {R}_0)$ , these hypotheses will not be satisfied by the initial subsolution, in general. Although a time dilation would almost solve the problem, we need the following lemma to fully prepare the initial subsolution:

Lemma 3.3. Let $T>0$ and let $\Omega \subset (0,1)^3\subset \mathbb {R}^3$ be an open set with smooth boundary and with a finite number of connected components. Let $(v_0,p_0,\mathring {R}_0)\in C^\infty (\mathbb {R}^3\times [0,T])$ be a subsolution whose support is contained in $(0,1)^3\times [0,T]$ and such that $\operatorname {\mathrm {supp}}\mathring {R}_0\subset \overline {\Omega }\times [0,T]$ . Let $\lambda>0$ be a sufficiently large constant. There exists a subsolution $(v,p,\mathring {R})\in C^\infty (\mathbb {R}^3\times [0,T])$ such that for any $N\geq 0$ we have

$$\begin{align*}\left\|v\right\|_N\lesssim \lambda^N, \|\mathring{R}\|_0\leq\lambda^{-1/2} \end{align*}$$

where the implicit constants are independent of $\lambda $ . In addition, the energy satisfies

(3.20) $$ \begin{align} \int_\Omega \left\vert v_0\right\vert{}^2dx<\int_\Omega \left\vert v\right\vert{}^2dx<\int_\Omega \left\vert v_0\right\vert{}^2dx+6\|\mathring{R}_0\|_0\left\vert\Omega\right\vert. \end{align} $$

Furthermore, $(v,p,\mathring {R})$ equals the initial subsolution outside the set

While necessary, this result is nothing new and one could easily obtain it by combining [Reference De Lellis and Székelyhidi24] and Lemma 2.9, or using the ideas in [Reference Isett and Oh36]. For completeness, we sketch its proof at the end of Section 7, considering a simplified version of the preceding construction.

3.2 Proof of Theorem 1.7

We first prove the theorem under the assumption that the subsolution $(v_0,p_0,\mathring {R}_0)(\cdot ,t)$ is compactly supported for each time $t\in [0,T]$ . This assumption will be relaxed in next subsection to a condition on the support of $\mathring {R}_0$ . We fix $0<\beta <1/3$ , we choose b satisfying (3.18) and $\alpha $ smaller than the threshold given by Proposition 3.2. Next, we use the scale invariance of the Euler equations and subsolutions

$$\begin{align*}v_0(x,t)\mapsto v_0(\rho x,\rho t), \quad p_0(x,t)\mapsto p_0(\rho x,\rho t), \quad \mathring{R}_0(x,t)\mapsto \mathring{R}_0(\rho x,\rho t)\end{align*}$$

to assume that $\Omega \subset (0,1)^3$ and $\operatorname {\mathrm {supp}} (v_0,p_0,\mathring {R}_0)\subset (0,1)^3\times [0,T]$ . The desired energy profile must also be modified: $e(t) \mapsto \rho ^{-3}e(t)$ . Note that this preserves (1.3). For convenience, we denote the rescaled subsolution like the original. It then suffices to construct the desired solution in this case and then reverse the change of variables.

Next, we use Lemma 3.3 with $\lambda =\lambda _1^{12\alpha }$ to obtain a subsolution $(v_1,p_1,\mathring {R}_1)$ that equals the initial subsolution outside the set

$$\begin{align*}\{x\in \Omega:d(x,\partial \Omega)>\lambda_1^{-\alpha}\}\times[0,T]\end{align*}$$

and satisfies the estimates

$$\begin{align*}\left\|v_1\right\|_N\leq c_N\lambda_1^{12N\alpha}, \|\mathring{R}_1\|_0\leq \lambda_1^{-6\alpha},\end{align*}$$

where the constants $c_N$ are independent of $\lambda _1$ but they will depend on $\Omega $ and the initial subsolution. In addition, by (1.3) we have

$$\begin{align*}e(t)>\int_\Omega \left\vert v_0\right\vert{}^2dx+6\|\mathring{R}_0\|_0\left\vert\Omega\right\vert>\int_\Omega\left\vert v_1\right\vert{}^2dx.\end{align*}$$

Next, we use another scale invariance of the Euler equations (and the definition of subsolution):

$$\begin{align*}v(x,t)\mapsto \Gamma v(x,\Gamma t), \quad p(x,t)\mapsto \Gamma^2 p( x,\Gamma t), \quad \mathring{R}(x,t)\mapsto \Gamma^2\mathring{R}( x,\Gamma t).\end{align*}$$

We choose

and we begin to work in this rescaled setting, which we will indicate with a superscript r. We are thus working in the interval $[0,\widetilde {T}]$ , where , and we try to prescribe the energy profile . By construction of $\Gamma $ we have

$$\begin{align*}\sup_t\left(\tilde{e}(t)-\int_\Omega \left\vert v_1^r(x,t)\right\vert{}^2dx\right)\leq \delta_2\end{align*}$$

and

$$\begin{align*}\inf_t\left(\tilde{e}(t)-\int_\Omega \left\vert v_1^r(x,t)\right\vert{}^2dx\right)= \Gamma^2\inf_t\left(e(t)-\int_\Omega \left\vert v_1(x,t)\right\vert{}^2dx\right).\end{align*}$$

It follows from the definition of $\Gamma $ that if a is sufficiently large we have

$$\begin{align*}\lambda_1^{\alpha}\left(\frac{\Gamma^2}{\delta_2}\right)\inf_t\left(e(t)-\int_\Omega \left\vert v_0(x,t)\right\vert{}^2dx\right)\geq 1.\end{align*}$$

so (3.17) holds. On the other hand,

$$\begin{align*}\sup_t\left\vert\tilde{e}'(t)\right\vert\leq \Gamma^{3/2}\sup_t\left\vert e'(t)\right\vert\leq 1\end{align*}$$

because $\Gamma $ becomes arbitrarily small by increasing a.

Next, we observe that $(v_0^r,p_0^r,\mathring {R}{}_0^r)$ still satisfies (3.6) $-$ (3.8) with the same sequence $\{y_N\}_{N=0}^\infty $ . Regarding $(v_1^r,p_1^r,\mathring {R}{}_1^r)$ , it follows from the definition of the rescaling that

$$\begin{align*}\|\mathring{R}{}_1^r\|_0\leq \delta_2\lambda_1^{-6\alpha}.\end{align*}$$

On the other hand, since the constants $c_N$ are independent of $\lambda _1$ , for sufficiently large a we have

$$ \begin{align*} \left\|v_1^r\right\|_0&\leq\delta_{2}^{1/2}\left\|v_1\right\|_0\leq \delta_{2}^{1/2}c_0\leq 1-\delta_1^{1/2},\\ \left\|v_1^r\right\|_1&\leq\delta_{2}^{1/2}\left\|v_1\right\|_1\leq \delta_{2}^{1/2}c_1\lambda_1^{12\alpha}\leq M\delta_1^{1/2}\lambda_1. \end{align*} $$

Finally, $(v_1^r,p_1^r,\mathring {R}{}_1^r)=(v_0^r,p_0^r,\mathring {R}{}_0^r)$ outside

$$\begin{align*}\{x\in \Omega:d(x,\partial \Omega)>\lambda_1^{-\alpha}\}\times[0,\widetilde{T}].\end{align*}$$

Let us consider the sets $A_q$ defined in (3.12). We see that $(v_1^r,p_1^r,\mathring {R}{}_1^r)=(v_0^r,p_0^r,\mathring {R}{}_0^r)$ outside $A_1\times [0,\widetilde {T}]$ for sufficiently small $\alpha $ and sufficiently large a. Remember that $\delta _q$ and $\lambda _q$ depend on a through the expressions (3.4) and (3.3), respectively.

From now on we assume that we are working in this rescaled problem and we omit the superscript r. Once we obtain the desired weak solution in this setting, to obtain the solution to the original problem it suffices to undo the scaling. By the previous discussion, the energy profile satisfies the inductive hypotheses (3.5) and the subsolution $(v_1,p_1,\mathring {R}_1)$ satisfies (3.13) $-$ (3.17). In addition, the initial subsolution $(v_0,p_0,\mathring {R}_0)$ satisfies (3.6) $-$ (3.8) for some sequence $\{y_N\}_{N=0}^\infty $ that does not depend on a. Therefore, we may apply Proposition 3.2 iteratively, obtaining a sequence of compactly supported smooth subsolutions $\{(v_q,p_q,\mathring {R}_q)\}_{q=1}^\infty $ .

It follows from (3.19) that $v_q$ converges uniformly to some continuous map v. On the other hand, note that the pressure $p_q$ is the only compactly supported solution of

Therefore, $p_q$ also converges to some pressure $p\in L^r(\mathbb {R}^3)$ for any $1\leq r<\infty $ . Since $\mathring {R}_q$ converges uniformly to $0$ , we conclude that the pair $(v,p)$ is a weak solution of the Euler equations.

Furthermore, using (3.19) we obtain

$$ \begin{align*} \sum_{q=1}^\infty \left\|v_{q+1}-v_q\right\|_{\beta'}&\leq \sum_{q=1}^\infty C(\beta',\beta) \left\|v_{q+1}-v_q\right\|_{0}^{1-\beta'}\left\|v_{q+1}-v_q\right\|_{\beta}^{\beta'}\\ &\leq C(\beta',\beta) \sum_{q=1}^\infty (M\delta_{q+1}^{1/2})^{1-\beta'}\left(M\delta_{q+1}^{1/2}\lambda_q \right)^{\beta'}\\ &\leq M \,C(\beta',\beta) \sum_{q=1}^\infty \lambda_q^{\beta'-\beta}, \end{align*} $$

so $\{v_q\}_{q=1}^\infty $ is uniformly bounded in $C^0_t C^{\beta '}_x$ for all $\beta '<\beta $ . To recover the time regularity, see [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7].

Next, note that $(v_q,p_q,\mathring {R}_q)=(v_0,p_0,0)$ in $(\mathbb {R}^3\backslash \Omega )\times [0,T]$ for all q by (3.13) and the definition of $A_q$ . Hence, we have $(v,p)=(v_0,p_0)$ in $(\mathbb {R}^3\backslash \Omega )\times [0,T]$ .

Finally, it follows from (3.17) and the fact that $\delta _{q+1}\to 0$ as $q\to \infty $ that $\int _\Omega \left \vert v(x,t)\right \vert {}^2dx=e(t)$ , as we wanted. This completes the proof of the theorem for the case that the initial subsolution is compactly supported for all time.

3.3 Dropping the compact support condition

Once we have proved Theorem 1.7 for the case that $(v_0,p_0,\mathring {R}_0)(\cdot ,t)$ is compactly supported, it is easy to relax this condition and show that it suffices that $\mathring {R}_0(\cdot ,t)$ is compactly supported.

Indeed, let us choose a bounded domain with smooth boundary $U\Supset \Omega $ . As we mentioned in Lemma 2.11 and Remark 2.12, any subsolution defined in all $\mathbb {R}^3$ automatically satisfies the conditions in Theorem 1.6. Hence, there exists a subsolution $(\widetilde {v}_0,\widetilde {p}_0,\mathring {\widetilde {R}}_0)\in C^\infty (\mathbb {R}^3\times [0,T])$ that extends $(v_0,p_0,\mathring {R}_0)$ outside $\overline {U}$ and that is compactly supported in each time slice. Since it is an extension, we see that the Reynolds stress vanishes in $\overline {U}\backslash \Omega $ .

We can now apply Theorem 1.7 in the case that the subsolution is compactly supported for each time slice, thus obtaining a weak solution $(v,p)$ with the appropriate regularity and such that $(v,p)=(v_0,p_0)$ in $(\overline {U}\backslash \Omega )\times [0,T]$ . Since $\Omega \Subset U$ , we may glue this region back into $(v_0,p_0)$ , obtaining a weak solution that equals $(v_0,p_0)$ outside $\Omega \times [0,T]$ .

Finally, since $\Omega $ and $\mathbb {R}^3\backslash \overline {U}$ are disjoint, when we apply Theorem 1.7 we may fix the energy profile in each region independently. Therefore, we can prescribe the energy profile in $\Omega $ of the final weak solution to be any smooth function satisfying condition (1.3).

4.1 Stages of the proof

1. 1. Preparing the subsolution. We mollify our subsolution $(v_q,p_q,\mathring {R}_q)$ to avoid the loss of derivatives problem, obtaining a new subsolution $(v_\ell ,p_\ell ,\mathring {R}_\ell )$ . It is convenient to glue it in space to the original subsolution far from the turbulent zone.

2. 2. Gluing in space. We pick a collection of times $\{t_i\}\subset [0,T]$ and we consider the solutions $(v_i,p_i)$ of the Euler equations with initial data $\widetilde {v}_\ell (t_i)$ . We glue them in space to $(\widetilde {v}_\ell ,\widetilde {p}_\ell ,\mathring {\widetilde {R}}_\ell )$ , obtaining new subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ . The error $\mathring {\widetilde {R}}_i$ is small and these subsolutions equal $(v_0,p_0,\mathring {R}_0)$ near $\partial \Omega $ .

3. 3. Gluing in time. We glue together the subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ , obtaining a new subsolution $(\overline {v}_q,\overline {p}_q,\mathring {\overline {R}}_q)$ in which most of the error is concentrated in temporally disjoint regions. The error remains localized within $\Omega $ owing to the fact that the differences between the subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ vanish near $\partial \Omega $ .

4. 4. Perturbation. We add a highly oscillatory perturbation to reduce the error. In fact, we add many corrections, each of them reducing the error in one of the temporally disjoint regions. These perturbations do not interact with each other, which yields the optimal estimates for the new subsolution $(v_{q+1},p_{q+1},\mathring {R}_{q+1})$ .

Throughout the iterative process we will use the notation $x\lesssim y$ to denote $x\leq C y$ for a sufficiently large constant $C>0$ that is independent of a, b, and q. However, the constant is allowed to depend on $\alpha $ , $\beta $ , $\Omega $ , and $\{y_N\}_{N=0}^\infty $ and it may change from line to line.

4.2 Preparing the subsolution

The first step consists in mollifying the field in order to avoid the loss of derivatives problem, which is typical of convex integration. The problem is the following: to control a Hölder norm of $(v_{q+1},p_{q+1},\mathring {R}_{q+1})$ we need estimates of higher-order Hölder norms of $(v_q,p_q,\mathring {R}_q)$ . As the iterative process goes on, we need to estimate higher and higher Hölder norms of the initial terms of the sequence to control just the first few Hölder norms of the subsolution. However, if we mollify the subsolution $(v_q,p_q,\mathring {R}_q)$ , we can control all the derivatives in terms of the first few Hölder norms and the mollification parameter.

Note that this process changes the subsolution in the whole space, yet mollification is only strictly necessary in $A_q\times [0,T]$ , as $(v_q,p_q,\mathring {R}_q)$ equals $(v_0,p_0,\mathring {R}_0)$ outside of this set. Furthermore, it will be convenient for later estimates that the resulting subsolution equals the initial subsolution far from the turbulent zone, as this is a property that we want to impose unto $(v_{q+1},p_{q+1},\mathring {R}_{q+1})$ . Hence, our approach consists in gluing the mollified subsolution to the initial subsolution, which is not quite demanding because both subsolutions are very close far from the turbulent zone.

We begin by fixing a mollification kernel in space $\psi \in C^\infty _c(\mathbb {R}^3)$ and we introduce the mollification parameter

(4.1)

Since $(\delta _{q+1}/\delta _q)^{1/2}=\lambda _q^{-\beta (b-1)}$ and our assumption (3.18) implies that $\beta (b-1)<1/2$ , we see that we may choose $\alpha $ sufficiently small and a sufficiently large so as to have

(4.2) $$ \begin{align} \lambda_q^{-3/2}\leq \ell\leq \lambda_q^{-1}. \end{align} $$

Note that our definition of $\ell $ differs from the one in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] by a factor of $\lambda _q^{-3\alpha /2}$ . The coefficients of $\alpha $ used in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] are fine-tuned to their proof. Since we do some things differently, it is not surprising that we have to change some of these coefficients. In general the factor $\lambda _q^{-3\alpha /2}$ is quite harmless, as only simple relationships like (4.2) are used throughout most of the paper, and these are the same here and in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7]. The actual value of $\ell $ is only used at the very end, when fine relationships between the parameters are needed to estimate $\mathring {R}_{q+1}$ . We will study these situations when they arise, but in any case we will see that the extra factor is essentially irrelevant. Indeed, $\alpha $ is assumed to be so small that in those inequalities the term containing $\alpha $ is negligible. Our definition of $\ell $ leads to a different coefficient multiplying $\alpha $ , but this only changes how small $\alpha $ has to be chosen, so it is not important.

After this brief digression, we define

where the convolution with $\psi _\ell $ is in space only and $f\mathring {\otimes }g$ denotes the traceless part of the tensor $f\otimes g$ . It is easy to check that the triplet $(v_\ell ,p_\ell ,\mathring {R}_\ell )$ is a subsolution and by [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7, Proposition 2.2] we have the following estimates:

$$ \begin{align*} \left\|v_\ell-v_q\right\|_0&\lesssim \delta_{q+1}^{1/2}\lambda_q^{-\alpha}, \\ \left\|v_\ell\right\|_{N+1}&\lesssim \delta_q^{1/2}\lambda_q\,\ell^{-N} \forall N\geq 0, \\ \|\mathring{R}_\ell\|_{N+\alpha}&\lesssim \delta_{q+1}\ell^{-N+3\alpha} \quad \forall N\geq 0,\\ \left\vert\int_\Omega \left\vert v_q\right\vert{}^2-\left\vert v_\ell\right\vert{}^2\,dx\right\vert&\lesssim \delta_{q+1}\ell^\alpha. \end{align*} $$

Note that we have an extra factor $\ell ^{2\alpha }$ in the estimate for the Reynolds stress in comparison to [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7]. This comes from the extra factor $\lambda _q^{-3\alpha }$ in $\|\mathring {R}_q\|_0$ and from our definition of $\ell $ . They cause an extra factor $\lambda _q^{-3\alpha }$ to appear in the estimate, which yields an extra factor $\ell ^{2\alpha }$ by (4.2).

Obtaining an extra factor $\ell ^{2\alpha }$ (actually $\ell ^\alpha $ would suffice) is our reason for modifying the inductive estimate of the Reynolds stress and the definition of $\ell $ . We will use this extra factor to compensate a suboptimal estimate that we will be forced to use in Section 5.

Once we have mollified the subsolution, we will glue it to the initial subsolution far from the turbulent zone. It will be convenient to divide $A_{q+1}\backslash A_q$ into several pieces because we will have to do several constructions in this region. We define

(4.3) $$ \begin{align} \sigma = \frac{1}{5}(d_q-d_{q+1}), \end{align} $$

where $d_q$ was defined in (3.10). Using the elementary inequalities

(4.4) $$ \begin{align} 2\pi\leq \frac{\lambda_q}{a^{b^q}}\leq 4\pi \end{align} $$

we deduce $\delta _{q+2}\gtrsim \lambda _q^{-2\beta b^2}$ . By (3.18) we have $b^2<5/4$ , so for sufficiently small $\alpha $ we have

$$\begin{align*}d_q\gtrsim\lambda_q^{-1/11}.\end{align*}$$

Therefore,

(4.5) $$ \begin{align} \sigma^{-1}\lesssim \lambda_q^{1/11}. \end{align} $$

In particular, $\sigma \gg \ell $ . For $1\leq j\leq 5$ we define

By hypothesis, $(v_q,p_q,\mathring {R}_q)$ equals $(v_0,p_0,\mathring {R}_0)$ outside $A_q\times [0,T]$ . Hence, it follows from (3.6) $-$ (3.8) that

(4.6) $$ \begin{align} \left\|v_\ell-v_0\right\|_{N;B_1}+\left\|\partial_t v_\ell-\partial_t v_0\right\|_{N;B_1}&\lesssim \ell^2, \end{align} $$

(4.7) $$ \begin{align} \left\|p_\ell-p_0\right\|_{N;B_1}&\lesssim \ell^2, \end{align} $$

(4.8) $$ \begin{align} \|\mathring{R}_\ell-\mathring{R}_0\|_{N;B_1}&\lesssim \ell^2. \end{align} $$

Thus, both subsolutions are very close in this region, which makes gluing them much easier. Taking into account (4.6), it follows from Lemma 2.6 that there exists a potential $A\in C^\infty (B_1\times [0,T],\mathcal {A}^3)$ such that $\operatorname {\mathrm {div}} A=v_\ell -v_0$ in $B_1\times [0,T]$ and

$$\begin{align*}\left\|A\right\|_{N+1+\alpha;B_1}+\left\|\partial_t A\right\|_{N+1+\alpha;B_1}\lesssim \ell^2\qquad \forall N\geq 0.\end{align*}$$

Therefore, the matrices $S_1$ and M that appear in Lemma 2.11 satisfy the estimates $\left \|S_1\right \|_{N+\alpha }+\left \|M\right \|_{N+\alpha }\lesssim \ell ^2$ . We introduce this estimates into Lemma 2.11 to perform a gluing in the region $B_1\backslash B_2$ . Since $\sigma \gg \ell $ we may essentially ignore any factor coming from the derivatives of the cutoff by absorbing it into the $\ell $ factor. Carrying out the rest of the construction of Lemma 2.11, we conclude that there exists a smooth subsolution such that

(4.9) $$ \begin{align} (\widetilde{v}_\ell,\widetilde{p}_\ell,\mathring{\widetilde{R}}_\ell)(x,t)=\begin{cases} (v_\ell,p_\ell,\mathring{R}_\ell)(x,t) \qquad \text{if } x\in A_q+B(0,\sigma), \\ (v_0,p_0,\mathring{R}_0)(x,t) \text{if } x\in B_2\end{cases} \end{align} $$

and satisfies the estimates

(4.10) $$ \begin{align} \left\|\widetilde{v}_\ell-v_q\right\|_0&\lesssim \delta_{q+1}^{1/2}\lambda_q^{-\alpha}, \end{align} $$

(4.11) $$ \begin{align} \left\|\widetilde{v}_\ell\right\|_{N+1}&\lesssim \delta_q^{1/2}\lambda_q\,\ell^{-N} \forall N\geq 0, \end{align} $$

(4.12) $$ \begin{align} \|\mathring{\widetilde{R}}_\ell\|_{N+\alpha}&\lesssim \delta_{q+1}\ell^{-N+3\alpha} \quad \forall N\geq 0, \end{align} $$

(4.13) $$ \begin{align} \left\vert\int_\Omega \left\vert\widetilde{v}_q\right\vert{}^2-\left\vert v_\ell\right\vert{}^2\,dx\right\vert&\lesssim \delta_{q+1}\ell^\alpha. \end{align} $$

That is, we have obtained a new subsolution satisfying the same estimates as $(v_\ell ,p_\ell ,\mathring {R}_\ell )$ but with the additional property of being equal to the initial subsolution far from the turbulent zone.

Although many constructions in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] work in $\mathbb {R}^3$ with very little or no modification, it is more convenient to work with periodic fields so that we may use results from [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] directly. Note that the subsolution that we have just obtained is supported in $(0,1)^3\times [0,T]$ because it equals the initial subsolution far form the turbulent zone. This allows us to consider its periodic extension to , which we denote the same. From now on, we consider that we are working in this setting.

4.3 Overview of the gluing in space

The key idea introduced by Isett in [Reference Isett35] is to glue in time exact solutions of the Euler equations to obtain a new subsolution $(\overline {v}_q,\overline {p}_q,\mathring {\overline {R}}_q)$ such that $\overline {v}_q$ is close to $v_q$ but $\mathring {\overline {R}}_q$ is supported in a series of disjoint temporal regions of the appropriate length. This allows the use of Mikado flows, leading to the optimal regularity $C^\beta $ for any $\beta <1/3$ (as in Onsager’s conjecture).

More specifically, we define the length

(4.14) $$ \begin{align} \tau_q=\frac{\ell^{2\alpha}}{\delta_q^{1/2}\lambda_q} \end{align} $$

and we consider the smooth solutions of the Euler equations

(4.15) $$ \begin{align} \begin{cases} \partial_t v_i+\operatorname{\mathrm{div}}(v_i\otimes v_i)+\nabla p_i=0,\\ \operatorname{\mathrm{div}} v_i=0,\\ v_i(\cdot,t_i)= \widetilde{v}_\ell(\cdot,t_i), \end{cases} \end{align} $$

where . We will see that they are defined in the time interval $[t_i-\tau _q,t_i+\tau _q]$ . The pressure is recovered as the unique solution to the equation $-\Delta p_i=\operatorname {\mathrm {tr}}(\nabla v_i\nabla v_i)$ with the normalization

(4.16) $$ \begin{align} \int_{\mathbb{T}^3}p_i(x,t)\,dx=\int_{\mathbb{T}^3}\widetilde{p}_\ell(x,t)\,dx. \end{align} $$

We will see that these solutions remain sufficiently close to $\widetilde {v}_\ell $ in their respective intervals. Following Isett’s ideas, we would like to glue them in time to obtain a subsolution in the whole interval $[0,T]$ . The velocity field would remain sufficiently close to $\widetilde {v}_\ell $ while the Reynolds stress would be localized to the intersection of the consecutive time intervals.

However, even if $\mathring {R}_\ell $ is well localized, the solutions $v_i$ will immediately differ from $\widetilde {v}_\ell $ in the whole space. Furthermore, different solutions $v_i$ , $v_{i+1}$ will also differ in the whole space during the intersection of their temporal domains. If we tried to apply Isett’s procedure to them we would obtain a Reynolds stress that spreads throughout the whole space. This is not suitable for our purposes, so we must modify Isett’s approach.

What we will do is to glue in space the exact solutions $v_i$ to $\widetilde {v}_\ell $ in the region where $\mathring {\widetilde {R}}_\ell $ is small, obtaining subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ . The Reynolds stress will no longer be 0, but it will be so small that we may ignore it in the current iteration. Since these subsolutions will coincide far from the turbulent zone, we will be able to glue them in time while keeping the Reynolds stress localized.

The actual process is not so simple because the difference between the exact solutions $(v_i,p_i)$ and the subsolution $(\widetilde {v}_\ell ,\widetilde {p}_\ell ,\mathring {\widetilde {R}}_\ell )$ is too big, leading to an unacceptably large error if we try to glue them. Our approach consists in producing a series of intermediate subsolutions that act as a sort of interpolation between them. Instead of a single gluing we perform a big number of them, going from $v_i$ to $\widetilde {v}_\ell $ far from the turbulent zone. The difference between two consecutive intermediate subsolutions will be very small so that the error introduced in each of these middle gluings is small.

At the end of this process we will obtain subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ such that

(4.17) $$ \begin{align} (\widetilde{v}_i,\widetilde{p}_i,\mathring{\widetilde{R}}_i)&=(v_0,p_0,\mathring{\widetilde{R}}_0) \text{ in } B_3\times[t_i-\tau_q,t_i+\tau_q], \end{align} $$

(4.18) $$ \begin{align} \widetilde{v}_i(t_i,\cdot)&=\widetilde{v}_\ell(t_i,\cdot). \end{align} $$

In addition, for $\left \vert t-t_i\right \vert \leq \tau _q$ and any $N\geq 0$ we will have the following estimates:

(4.19) $$ \begin{align} \|\mathring{\widetilde{R}}_i\|_0&\leq \frac{1}{2}\delta_{q+2}\lambda_{q+1}^{-6\alpha}, \end{align} $$

(4.20) $$ \begin{align} \left\|\widetilde{v}_i-\widetilde{v}_\ell\right\|_{N+\alpha}&\lesssim \tau_q\delta_{q+1}\ell^{-N-1+\alpha}, \end{align} $$

(4.21) $$ \begin{align} \left\|D_{t,\ell}(\widetilde{v}_i-\widetilde{v}_\ell)\right\|_{N+\alpha}&\lesssim \delta_{q+1}\ell^{-N-1+\alpha}, \end{align} $$

where we write

for the transport derivative. Furthermore, there exist smooth vector potentials $\widetilde {z}_i$ defined in $[t_i-\tau _q,t_i+\tau _q]$ such that $\widetilde {v}_i=\operatorname {\mathrm {curl}} \widetilde {z}_i$ and at the intersection of two intervals we have:

(4.22) $$ \begin{align} \left\|\widetilde{z}_i-\widetilde{z}_{i+1}\right\|_{N+\alpha}&\lesssim \tau_q\delta_{q+1}\ell^{-N+\alpha}, \end{align} $$

(4.23) $$ \begin{align} \|D_{t,\ell}(\widetilde{z}_i-\widetilde{z}_{i+1})\|_{N+\alpha}&\lesssim\delta_{q+1}\ell^{-N+\alpha}. \end{align} $$

These estimates are completely analogous to the ones in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7, Proposition 3.3, Proposition 3.4] but we have the additional benefit of the fields being equal to the initial subsolution far from the turbulent zone. We do have to pay a price because now we have subsolutions instead of solutions of the Euler equations. Nevertheless, the errors $\mathring {\widetilde {R}}_i$ are so small that we may ignore them until the $(q+1)$ -th iteration.

All of the steps summarized here will be discussed in full detail in Section 5, where we derive the claimed estimates.

4.4 Overview of the gluing in time

Once we have our subsolutions $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ we will glue them in time. The goal is to obtain a subsolution defined in all $[0,T]$ that remains close to $\widetilde {v}_\ell $ but in which most of the Reynolds stress is localized to temporally disjoint regions of the appropriate length. This will allow us to correct the error in each region separately using Mikado flows. Let

Note that $\{I_j,J_i\}$ is a pairwise disjoint decomposition of $[0,T]$ . We choose a smooth partition of unity $\{\chi _i\}$ such that:

• • $\sum _i \chi _i=1$ .

• • $\operatorname {\mathrm {supp}} \chi _i\cap \operatorname {\mathrm {supp}} \chi _{i+2}=\varnothing $ . Furthermore,

  (4.24) $$ \begin{align} \begin{aligned} \operatorname{\mathrm{supp}} \chi_i\subset \left(t_i-\frac{2}{3}\tau_q,t_i+\frac{2}{3}\tau_q\right),\\ \chi_i(t)=1 \qquad \forall t\in J_i. \end{aligned} \end{align} $$

• • For any i and $N\geq 0$ we have

  (4.25) $$ \begin{align} \left\|\partial_t^N\chi_i\right\|_0\lesssim \tau_q^{-N}. \end{align} $$

We define

It is clear that $\overline {v}_q$ is divergence-free and it equals $v_0$ in $B_3\times [0,T]$ because of (4.17). In addition, it inherits the estimates of $\widetilde {v}_i$ :

Proposition 4.1. The velocity field $\overline {v}_q$ satisfies the following estimates:

(4.26) $$ \begin{align} \left\|\overline{v}_q-\widetilde{v}_\ell\right\|_\alpha &\lesssim \delta_{q+1}^{1/2}\ell^{\alpha}, \end{align} $$

(4.27) $$ \begin{align} \left\|\overline{v}_q-\widetilde{v}_\ell\right\|_{N+\alpha} &\lesssim \tau_q\delta_{q+1}\ell^{-1-N+\alpha}, \end{align} $$

(4.28) $$ \begin{align} \left\|\overline{v}_q\right\|_{1+N}&\lesssim \delta_q^{1/2}\lambda_q\ell^{-N}, \end{align} $$

(4.29) $$ \begin{align} \left\vert\int_\Omega \left\vert\overline{v}_q\right\vert{}^2-\left\vert\widetilde{v}_\ell\right\vert{}^2\right\vert&\lesssim \delta_{q+1}\ell^{\alpha}, \end{align} $$

for all $N\geq 0$ .

The proof is exactly the same as the proof of [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7, Proposition 4.3, Proposition 4.5] because our fields $\widetilde {v}_i$ satisfy completely analogous estimates to the solutions $v_i$ in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7].

We conclude that the new velocity $\overline {v}_q$ equals $v_0$ far from the turbulent zone, it is close to $\widetilde {v}_\ell $ and satisfies suitable bounds. Nevertheless, we must check if it leads to a subsolution. If $t\in J_i$ , then in a neighborhood of t we have $\chi _i=1$ while the rest of the cutoffs vanish. Thus, for all $t\in J_i$ we have

$$\begin{align*}\overline{v}_q= \widetilde{v}_i, \qquad \overline{p}^{(1)}_q= \widetilde{p}_i, \qquad \overline{\mathcal R}^{(1)}_q=\mathring{\widetilde{R}}_i.\end{align*}$$

Since $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ is a subsolution, we have

$$\begin{align*}\partial_t\overline{v}_q+\operatorname{\mathrm{div}}(\overline{v}_q\otimes\overline{v}_q)+\nabla \overline{p}^{(1)}_q=\operatorname{\mathrm{div}}\overline{\mathcal R}^{(1)}_q\end{align*}$$

On the other hand, if $t\in I_i$ , then $\chi _j=0$ for $j\neq i,i+1$ and $\chi _i+\chi _{i+1}=1$ . Hence, on $I_i$ we have

$$\begin{align*}\overline{v}_q=\chi_i\widetilde{v}_i+\chi_{i+1}\widetilde{v}_{i+1}, \qquad \overline{p}^{(1)}_q=\chi_i\widetilde{p}_i+\chi_{i+1}\widetilde{p}_{i+1}, \qquad \overline{\mathcal R}^{(1)}_q=\chi_i\mathring{\widetilde{R}}_i+\chi_{i+1}\mathring{\widetilde{R}}_{i+1}.\end{align*}$$

After a tedious computation we obtain

(4.30) $$ \begin{align} \partial_t\overline{v}_q&+\operatorname{\mathrm{div}}(\overline{v}_q\otimes\overline{v}_q)+\nabla \overline{p}^{(1)}_q-\operatorname{\mathrm{div}}\overline{\mathcal R}^{(1)}_q= \\ &=\partial_t\chi_i(\widetilde{v}_i-\widetilde{v}_{i+1})-\chi_i(1-\chi_i)\operatorname{\mathrm{div}}((\widetilde{v}_i-\widetilde{v}_{i+1})\otimes (\widetilde{v}_i-\widetilde{v}_{i+1})), \nonumber \end{align} $$

where we have used the fact that $(\widetilde {v}_i,\widetilde {p}_i,\mathring {\widetilde {R}}_i)$ are subsolutions.

In conclusion, for $t\in J_i$ the triplet $(\overline {v}_q,\overline {p}_q^{(1)},\overline {\mathcal R}^{(1)}_q)$ is trivially a subsolution, whereas for $t\in I_i$ it suffices to express the right-hand side of Equation (4.30) as the divergence of a symmetric matrix. However, we must do this carefully because it is very important to keep under control the spatial support of the Reynolds stress. Indeed, since $\overline {\mathcal R}^{(1)}_q$ equals $\mathring {R}_0$ outside $A_{q+1}\times [0,T]$ , any perturbation in the Reynolds stress must be contained within $A_{q+1}\times [0,T]$ in order to satisfy the inductive property (3.13). Fortunately, we will be able to achieve this because the right-hand side of Equation (4.30) is supported in $A_q+B(0,3\sigma )$ due to (4.17). We see that performing the gluing in space in the previous stage is crucial.

The logical approach would be to apply Lemma 2.9 to the right-hand side of (4.30) and define the new Reynolds stress as the sum of the obtained matrix and $\overline {\mathcal R}^{(1)}_q$ . Unfortunately, we cannot simply do that because we cannot obtain the necessary estimates for the transport derivative from Lemma 2.9.

Nevertheless, this difficulty can be solved. In Section 6 we will find smooth symmetric matrices $\overline {\mathcal R}^{(2)}_q,\overline {\mathcal R}^{(3)}_q$ solving the equation

$$\begin{align*}\operatorname{\mathrm{div}}\left(\overline{\mathcal R}^{(2)}_q+\overline{\mathcal R}^{(3)}_q\right)=\partial_t\chi_i(\widetilde{v}_i-\widetilde{v}_{i+1})\end{align*}$$

for $t\in I_i$ . We set $\overline {\mathcal R}^{(2)}_q,\overline {\mathcal R}^{(3)}_q=0$ for $t\notin \bigcup _{i}I_i$ . Since the source term is supported in $A_q+B(0,3\sigma )$ due to (4.17), we will be able to choose $\overline {\mathcal R}^{(2)}_q,\overline {\mathcal R}^{(3)}_q$ supported in $A_q+B(0,4\sigma )$ . The motivation for each matrix is the following:

• • $\overline {\mathcal R}^{(2)}_q$ solves the equation except for a small error. We have good bounds for the derivative of the material derivative.

• • $\overline {\mathcal R}^{(3)}_q$ corrects the errors introduced when fixing the support of $\overline {\mathcal R}^{(2)}_q$ . We do not have good bounds for its material derivative, but its $C^0$ -norm is very small.

Using these auxiliary matrices, we define

(4.31)

(4.32)

(4.33)

(4.34)

By construction of $\overline {\mathcal R}^{(2)}_q$ and $\overline {\mathcal R}^{(3)}_q$ , we see that $(\overline {v}_q,\overline {p}_q,\mathring {\,\overline {\!{R}}}_q)$ is a smooth subsolution. Furthermore, we have

$$\begin{align*}(\overline{v}_q,\overline{p}_q,\mathring{\,\overline{\!{R}}}_q)=(v_0,p_0,\mathring{R}_0) \text{ in } B_4\times[0,T]\end{align*}$$

because of (4.17) and the fact that $\overline {\mathcal R}^{(2)}_q, \overline {\mathcal R}^{(3)}_q$ are supported in $A_q+B(0,4\sigma )$ . We emphasize again the importance of performing a gluing in space to use $\widetilde {v}_i$ instead of the solutions $v_i$ . Otherwise, we would have no control on the Reynolds stress because $v_i-v_{i+1}$ will in general be spread throughout the whole space.

We summarize the facts about the new Reynolds stress that we will prove in Section 6:

Proposition 4.2. The smooth symmetric matrices $\mathring {\,\overline {\!{R}}}_q\strut {}^{(1)},\mathring {\,\overline {\!{R}}}_q\strut {}^{(2)}$ satisfy

(4.35) $$ \begin{align} \|\mathring{\,\overline{\!{R}}}_q\strut{}^{(1)}\|_0&\leq \frac{3}{4}\delta_{q+2}\lambda_{q+1}^{-3\alpha}, \end{align} $$

(4.36) $$ \begin{align} \operatorname{\mathrm{supp}} \mathring{\,\overline{\!{R}}}_q\strut{}^{(2)}&\subset [A_q+B(0,4\sigma)]\times\bigcup_i I_i, \end{align} $$

(4.37) $$ \begin{align} \|\mathring{\,\overline{\!{R}}}_q\strut{}^{(2)}\|_N&\lesssim \delta_{q+1}\ell^{-N+\alpha}, \end{align} $$

(4.38) $$ \begin{align} \|(\partial_t+\overline{v}_q\cdot\nabla)\mathring{\,\overline{\!{R}}}_q\strut{}^{(2)}\|_N&\lesssim\delta_{q+1}\delta_q^{1/2}\lambda_q\ell^{-N-\alpha}. \end{align} $$

Comparing (4.37) with the analogous estimate in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7], we see that we estimate the $C^N$ -norm instead of the $C^{N+\alpha }$ -norm. This difference is immaterial because they merely use the $C^{N+\alpha }$ -norm to estimate the $C^N$ -norm, which leads to the same bound as (4.37).

In conclusion, $\mathring {\,\overline {\!{R}}}_q\strut {}^{(1)}$ is so small that we may ignore it for the present iteration, whereas $\mathring {\,\overline {\!{R}}}_q\strut {}^{(2)}$ is big but it is supported in temporally disjoint regions and it satisfies good estimates. The next stage of the process is aimed at correcting $\mathring {\,\overline {\!{R}}}_q\strut {}^{(2)}$ by means of highly oscillatory perturbations.

All of the steps summarized here will be discussed in full detail in Section 6, where we derive the claimed estimates.

4.5 Overview of the perturbation step

We have localized most of the Reynolds stress, that is, $\mathring {\,\overline {\!{R}}}_q\strut {}^{(2)}$ , to small disjoint temporal regions but to reduce it we must resort to convex integration.

First of all, we fix a cutoff $\phi _q\in C^\infty (\mathbb {R}^3,[0,1])$ that equals $1$ in $A_q+\overline {B}(0,4\sigma )$ and whose support is contained in $A_q+B(0,5\sigma )$ . In particular, $\phi _q\equiv 1$ on the support of $\mathring {\,\overline {\!{R}}}_q\strut {}^{(2)}(\cdot ,t)$ for all $t\in [0,T]$ .

We follow the construction of [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] with Mikado flows, but there are some differences:

• • We control the support of the perturbation by introducing the cutoff $\phi _q$ .

• • To obtain the desired energy, we must use a slightly different normalization coefficient for the perturbation to account for the presence of the cutoff $\phi _q$ when integrating.

• • We construct the new Reynolds stress using Lemma 2.9 so that we have control on its support. To apply this lemma we must introduce a minor correction $w_L$ to ensure that the perturbation has vanishing angular momentum.

Since $\left \|\phi _q\right \|_N$ are much smaller than the $C^N$ -norms of the other maps involved, the presence of the cutoff does not affect the estimates. In addition, $w_L$ will be negligible because its size is determined by an integral quantity, which is very small for a highly oscillating perturbation.

The form of the perturbation $\widetilde {w}_{q+1}= v_{q+1}-v_q$ is $\widetilde {w}_{q+1}=w_0+w_c+w_L$ , where $w_0$ is the main perturbation term and it is used to cancel the Reynolds stress, $w_c$ is a small correction to ensure that the perturbation is divergence-free and $w_L$ is a tiny correction that ensures that the perturbation has vanishing angular momentum. This term is not present in [Reference Buckmaster, De Lellis, Székelyhidi and Vicol7] because they do not control the support of the Reynolds stress, so it suffices to use .

At the end of the process we will obtain a new subsolution $(v_{q+1},p_{q+1},\mathring {R}_{q+1})$ that equals $(v_0,p_0,\mathring {R}_0)$ outside $A_{q+1}\times [0,T]$ and satisfying the following estimates:

(4.39) $$ \begin{align} \left\|v_{q+1}\overline{v}_q\right\|_0+\frac{1}{\lambda_{q+1}}\left\|v_{q+1}\overline{v}_q\right\|_1&\leq \frac{3}{4}M\delta_{q+1}^{1/2}, \end{align} $$

(4.40) $$ \begin{align} \|\mathring{R}_{q+1}\|_0&\leq \delta_{q+2}\lambda_{q+1}^{-6\alpha}, \end{align} $$

(4.41) $$ \begin{align} \delta_{q+2}\lambda_{q+1}^{-7\alpha}\leq e(t)-\int_\Omega\left\vert v_{q+1}(x,t)\right\vert{}^2dx&\leq \delta_{q+1}. \end{align} $$

4.6 Proof of Proposition 3.2

Finally, we are ready to complete the proof of the proposition. The estimate (3.19) is a consequence of (4.10), (4.11), (4.26), (4.28) and (4.39):

$$\begin{align*}\left\|v_{q+1}-v_q\right\|_0+\lambda_{q+1}^{-1}\left\|v_{q+1}-v_q\right\|_1\leq \frac{3}{4}M\delta_{q+1}^{1/2}+C\delta_{q+1}^{1/2}\ell^\alpha+C\delta_q^{1/2}\lambda_q\lambda_{q+1}^{-1},\end{align*}$$

where the constant C depends on $\alpha $ , $\beta $ , $\Omega $ and $(v_0,p_0,\mathring {R}_0)$ , but not on a, b or q. Thus, for any fixed b (3.19) holds for sufficiently large a. Regarding (3.15), we use the inequality at level q to get

$$\begin{align*}\left\|v_{q+1}\right\|_1\leq M\delta_q^{1/2}\lambda_q+\frac{3}{4}M\delta_{q+1}^{1/2}+C\delta_{q+1}^{1/2}\ell^\alpha\lambda_{q+1}+C\delta_q^{1/2}\lambda_q.\end{align*}$$

Hence, if we choose a large enough we obtain (3.15). Finally, (3.16) follows from

$$\begin{align*}\left\|v_{q+1}\right\|_0\leq \left\|v_q\right\|_0+\left\|v_{q+1}-v_q\right\|_0\leq 1-\delta_q^{1/2}+M\delta_{q+1}^{1/2}.\end{align*}$$

The inductive hypotheses (3.13), (3.14) and (3.17) were obtained in the perturbation step, and we are done.

5.1 Interpolating sequence

We begin by describing the intermediate subsolutions that we will use. First of all, we recall the following local existence result. It is standard, but we provide a proof for the sake of completeness.

Proposition 5.1. For any $\alpha>0$ there exists a constant $c(\alpha )>0$ with the following property. Given any $C^\infty $ initial data $u_0\in H^3(\mathbb {R}^3)$ , any $C^\infty $ force $f\in L^1_{\mathrm {loc}}(\mathbb {R},H^3(\mathbb {R}^3))$ , let us fix a constant $T>0$ such that

(5.1) $$ \begin{align} T\left\|u_0\right\|_{1+\alpha}+T^2\left\|f\right\|_{1+\alpha}\leq c(\alpha). \end{align} $$

Then there exists a unique solution $u\in C^\infty (\mathbb {R}^3\times [-T,T])\cap C([-T,T],H^3(\mathbb {R}^3))$ to the Euler equation

$$\begin{align*}\begin{cases} \partial_t u+\operatorname{\mathrm{div}}(u\otimes u)+\nabla p=f,\qquad \operatorname{\mathrm{div}} u=0,\\ u(\cdot,0)=u_0. \end{cases}\end{align*}$$

Moreover, u obeys the bounds

$$\begin{align*}\left\|u\right\|_{N+\alpha}\lesssim \left\|u_0\right\|_{N+\alpha}+T\left\|f\right\|_{N+\alpha}\end{align*}$$

for all $N\geq 1$ , where the implicit constant depends on N and $\alpha>0$ .

Proof. It is classical [Reference Swann48, Reference Kato37, Reference Temam50] that the 3d Euler equation is locally well-posed on $H^3(\mathbb {R}^3)$ , and that the solution, which is defined a priori for some time $T=T(\|u_0\|_{H^3(\mathbb {R}^3)})>0$ , can be continued (and stay smooth, provided that $u_0\in C^\infty $ ) as long as the norm $\|u(t)\|_{H^3(\mathbb {R}^3)}$ remains bounded. Furthermore, the weak Beale–Kato–Majda criterion shows [Reference Beale, Kato and Majda3] that this norm is controlled by $\|\nabla u\|_{L^1L^\infty }$ . More precisely, one has

$$ \begin{align*} \|u(t)\|_{H^3(\mathbb{R}^3)} &\leq \|u_0\|_{H^3(\mathbb{R}^3)}+C \int_{-|t|}^{|t|}\|f(\tau)\|_{H^3(\mathbb{R}^3)}\, d\tau\\ &\quad + C \int_{-|t|}^{|t|}\|\nabla u(\tau)\|_{L^\infty(\mathbb{R}^3)}\,\|f(\tau)\|_{H^3(\mathbb{R}^3)}\,e^{C \int_{-|\tau|}^{|\tau|}\|f(\tau')\|_{H^3(\mathbb{R}^3)}\, d\tau'}\, d\tau\,. \end{align*} $$

Therefore, we only need to provide a uniform a priori estimate for $\|\nabla u\|_{L^\infty }$ for all times $t< T$ , where the maximum value $T>0$ is yet to be determined. To this end, note that for any multi-index $\theta $ with $\left \vert \theta \right \vert =N$ we have

$$\begin{align*}\partial_t\partial^\theta u+u\cdot\nabla \partial^\theta u+[\partial^\theta, u\cdot \nabla]u+\nabla\partial^\theta p=\partial^\theta f.\end{align*}$$

Since the pressure satisfies the equation $-\Delta p=\operatorname {\mathrm {tr}}(\nabla u\nabla u)-\operatorname {\mathrm {div}} f$ , it follows that

$$\begin{align*}\left\|\nabla \partial^\theta p\right\|_\alpha\lesssim \left\|\operatorname{\mathrm{tr}}(\nabla u\nabla u)\right\|_{N-1+\alpha}+\left\|f\right\|_{N+\alpha}\lesssim \left\|u\right\|_{1+\alpha}\left\|u\right\|_{N+\alpha}+\left\|f\right\|_{N+\alpha}.\end{align*}$$

Hence, we have

$$\begin{align*}\left\|(\partial_t\partial^\theta+u\cdot\nabla \partial^\theta )u\right\|_\alpha\lesssim \left\|u\right\|_{1+\alpha}\left\|u\right\|_{N+\alpha}+\left\|f\right\|_{N+\alpha}.\end{align*}$$

Thus, by Lemma B.2:

(5.2) $$ \begin{align} \left\|u(\cdot,t)\right\|_{N+\alpha}\lesssim \left\|u_0\right\|_{N+\alpha}+T\left\|f\right\|_{N+\alpha}+\int_0^t\left\|u(\cdot,s)\right\|_{1+\alpha}\left\|u(\cdot,s)\right\|_{N+\alpha}ds. \end{align} $$

Specializing to the case $N=1$ , we note that for all $|t|<T$ one has

(5.3) $$ \begin{align} \left\|u(\cdot,t)\right\|_{1+\alpha}\leq \tilde{c}(\alpha)\left(\left\|u_0\right\|_{1+\alpha}+T\left\|f\right\|_{1+\alpha}\right)+\tilde{c}(\alpha)\int_0^t\left\|u(\cdot,s)\right\|_{1+\alpha}^2ds, \end{align} $$

for some constant $\tilde {c}(\alpha )>0$ . We define the constant $c(\alpha )$ that appears in the statement of the proposition as and then assume that $T>0$ satisfies (5.1). Let $y_0$ be the first term on the right-hand side of (5.3) and let $y(t)$ be the solution to the ODE $y'= \tilde {c}(\alpha ) y^2$ with $y(0)=y_0$ . One finds

$$\begin{align*}\left\|u\right\|_{1+\alpha}\leq y(t)= \frac{y_0}{1-\tilde{c}(\alpha)y_0t}. \end{align*}$$

Since $y(t)\lesssim 1$ for all $|t|<T$ , by our choice of T, we conclude that the solution is well defined for $|t|< T$ . Furthermore, inserting this estimate in (5.2) and applying Grönwall’s inequality, we obtain the desired Hölder bounds for $N>1$ .

To construct the desired sequence of subsolutions, we define

(5.4)

(5.5)

For $i\geq 0$ and $0\leq k\leq m$ we consider the smooth solutions of the forced Euler equations