A new construction of weak solutions to compressible Navier–Stokes equations

Abstract

We prove the existence of the weak solutions to the compressible Navier–Stokes system with barotropic pressure $p(\varrho )=\varrho ^\gamma$ for $\gamma \ge 9/5$ in three space dimension. The novelty of the paper is the approximation scheme that instead of the classical regularization of the continuity equation (based on the viscosity approximation $\varepsilon \Delta \varrho$) uses more direct truncation and regularisation of nonlinear terms and the pressure. This scheme is compatible with the Bresch–Jabin compactness criterion for the density. We revisit this criterion and prove, in full rigour, that it can be applied in our approximation at any level.

A Appendix

1.1 A.1 Proof of Lemma 2

First note that from (18) we have that

$$\begin{aligned} \overline{K}_h(z) \le |\ln h|^{-1}\left( 2^{d} + \frac{1}{(|z|+h)^d}\right) \le C \overline{K}_h(z). \end{aligned}$$

To show (24) we first write

$$\begin{aligned}{} & {} \int _{{\mathbb {T}}^d} \overline{K}_{h_1}(x-y) \overline{K}_{h_2}(y-z)\, \textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{{\mathbb {T}}^d} \left( 2^{d} + \frac{1}{(|x-y|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y-z|+h_2)^d} \right) \,\textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{{\mathbb {T}}^d} \left( 2^{d} + \frac{1}{(|(x-z)-y|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \,\textrm{d} {y}=I. \end{aligned}$$

Now we split $I=I_1+I_2+I_3$ by considering ${\mathbb {T}}^d=\Omega _1+\Omega _2+\Omega _3$, where $\Omega _1=\{y:|y|\ge 2|x-z|\}$, $\Omega _2=\{y:|y|\le \frac{1}{2}|x-z|\}$, $\Omega _3=\{y: \frac{1}{2}|x-z|\le |y|\le 2|x-z|\}$, respectively.

Since on the set $\Omega _1$ we have $|(x-z)-y|\ge |y|-|x-z|\ge |x-z|$ we can estimate

$$\begin{aligned} I_1= & {} |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _1} \left( 2^{d} + \frac{1}{(|(x-z)-y|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \,\textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _1} \left( 2^{d} + \frac{1}{(|x-z|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \, \textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{{\mathbb {T}}^d} \left( 2^{d} + \frac{1}{(|x-z|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \, \textrm{d} {y}\\{} & {} \le 2\overline{K}_{h_1}(x-z) \end{aligned}$$

Similarly, on the set $\Omega _2$ we have $|(x-z)-y|\ge |x-z|-|y|\ge |x-z|-|y|\ge \frac{1}{2} |x-z|$, and so

$$\begin{aligned} I_2= & {} |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _2} \left( 2^{d} + \frac{1}{(|(x-z)-y|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \,\textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _1} \left( 2^{d} + \frac{1}{(\frac{1}{2}|x-z|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \, \textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _1} \left( 2^{d} + \frac{2^d}{(|x-z|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \, \textrm{d} {y}\\{} & {} \le 2^{d+1} \overline{K}_{h_1}(x-z). \end{aligned}$$

Finally, for the set $\Omega _3$ we have in particular that $|y|\ge \frac{1}{2}|x-z|$

$$\begin{aligned} I_3= & {} |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{\Omega _3} \left( 2^{d} + \frac{1}{(|(x-z)-y|+h_1)^d}\right) \left( 2^{d}+\frac{1}{ (|y|+h_2)^d} \right) \,\textrm{d} {y}\\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{{\mathbb {T}}^d }\left( 2^{d} + \frac{1}{(|(x-z)-y|+h_1)^d}\right) \left( 2^{d}+\frac{2^d}{ (|x-z|+h_2)^d} \right) \,\textrm{d} {y}\\{} & {} \{ \text{ changing } y-(x-z) \rightarrow y\} \\{} & {} \le |\ln h_1|^{-1} |\ln h_2|^{-1} \int _{{\mathbb {T}}^d} \left( 2^{d} + \frac{1}{(|y|+h_1)^d}\right) \left( 2^{d}+\frac{2^d}{ (|x-z|+h_2)^d} \right) dy\\{} & {} \le 2^{d+1} \overline{K}_{h_2}(x-z). \end{aligned}$$

Summing up $I_1+I_2+I_3$ we verify (24). $\square$

1.2 A.2 Proof of Lemma 11

In what follows we treat the function defined over the torus as a periodic function and $x,y \in [-\pi ,\pi ]^d$. This way we avoid questions with definition of function u. We take $u \in C^1$, and then conclude the result for Lebesgue points of u. We have

$$\begin{aligned} u(x)-u(y)=u(x)-u(z)+u(z)-u(y)= \int _{l(x,z)} \partial _{(x,z)}u(s)dl_s + \int _{l(z,y)}\partial _{(z,y)}u(s) dl_s, \end{aligned}$$

(103)

where l(x, z) is the line connecting x to z, $\partial _{(x,z)}$ is the directional derivative in direction of $\mathbf {[x,z]}$.

Next we integrate (103) over $B(\frac{x+y}{2},\frac{|x-y|}{2})$ w.r.t. z. Note that

$$\begin{aligned} B\left( \frac{x+y}{2},\frac{|x-y|}{2} \right) \subset B(x,|x-y|)\cap B(y,|x-y|) \end{aligned}$$

and thus we obtain

$$\begin{aligned}{} & {} \omega _d 2^{-d} |x-y|^d |u(x)-u(y)| \\{} & {} \le \int _{B(x,|x-y|)} \textrm{d}z \int _{l(x,z)} |\nabla u(s)| \textrm{d}l_s + \int _{B(y,|x-y|)} \textrm{d}z \int _{l(z,y)} |\nabla u(s)| \textrm{d}l_s \\{} & {} \le \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(x+\omega \tau )| \\{} & {} \quad + \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(y+\omega \tau )| \\{} & {} \le \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^r d\tau |\nabla u(x+\omega \tau )| \\{} & {} \quad + \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^r d\tau |\nabla u(y+\omega \tau )| \\{} & {} \le \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(x+\omega \tau )|\\{} & {} \quad + \int _0^{|x-y|} r^{d-1}dr \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(y+\omega \tau )| \\{} & {} \le c|x-y|^d \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(x+\omega \tau )|\frac{\tau ^{d-1}}{\tau ^{d-1}}\\{} & {} \quad + c|x-y|^d \int _{S^{d-1}} d\omega \int _0^{|x-y|} d\tau |\nabla u(y+\omega \tau )| \frac{\tau ^{d-1}}{\tau ^{d-1}} \\{} & {} \le c|x-y|^d \left( \int _{B(0,|x-y|)}\frac{|\nabla u(x+z)|}{|z|^{d-1}} \textrm{d}z + \int _{B(0,|x-y|)}\frac{|\nabla u(y+z)|}{|z|^{d-1}} \textrm{d}z \right) , \end{aligned}$$

where the last inequality follows by substituting $z=\tau \omega$ and by noticing that $|z|=\tau$ and $\omega _d$ stands for the volume of unit sphere in $\mathbb {R}^d$. $\square$

1.3 A.3 Proof of Lemma 12

Our first goal is to control the time regularisation

$$\begin{aligned} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho (t,x) (\textbf{u}(t,x) -\pi _{\eta _t} *\textbf{u}(t,x))^2 \ \textrm{d} {x}\,\textrm{d} t , \end{aligned}$$

where $\pi _{\eta _t}$ denotes the time mollifier. To proceed, we rewrite this term as follows

$$\begin{aligned}{} & {} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho (t,x) (\textbf{u}(t,x) -\pi _{\eta _t}*\textbf{u}(t,x))^2 \ \textrm{d} {x}\,\textrm{d} t \\{} & {} =\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho (t,x) \left( \textbf{u}(t,x) -\int _0^T\pi _{\eta _t}(t-s) \textbf{u}(s,x)\, ds \right) ^2 \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T\varrho (t,x) (\textbf{u}(t,x) - \textbf{u}(s,x))^2 \pi _{\eta _t} (t-s)\, ds \ \textrm{d} {x}\,\textrm{d} t , \end{aligned}$$

where we used the Jensen’s inequality. Thus we have

$$\begin{aligned}{} & {} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho (t,x) (\textbf{u}(t,x) -\pi _{\eta _t}*\textbf{u}(t,x))^2 \ \textrm{d} {x}\,\textrm{d} t \\{} & {} {\le }\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T\varrho (t,x) (\textbf{u}(t,x) - \textbf{u}(s,x)) \pi _{\eta _t}(t-s) (\textbf{u}(t,x) - \textbf{u}(s,x))\, ds \ \textrm{d} {x}\,\textrm{d} t \\{} & {} =\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) \textbf{u}(t,x) - \varrho (s,x) \textbf{u}(s,x)) \pi _{\eta _t}(t-s) (\textbf{u}(t,x) - \textbf{u}(s,x)) \, ds \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \quad {-} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) -\varrho (s,x)) \textbf{u}(s,x) \pi _{\eta _t}(t-s) ((\textbf{u}(t,x) - \textbf{u}(s,x))\, ds \ \textrm{d} {x}\,\textrm{d} t \\{} & {} =J_1 +J_2. \end{aligned}$$

To estimate $J_1$ we first note that

$$\begin{aligned} \partial _{t}(\varrho \textbf{u})\in L^{q}_t W^{-1,q}_x \text { for } q>1, \end{aligned}$$

where q being restricted by the integrability of the pressure mostly, and $q\ge \frac{10}{9}$ when $\gamma \ge \frac{9}{5}$. Therefore, we introduce additional regularisation in space for the term $\textbf{u}(t,x)-\textbf{u}(s,x)$ in $J_1$ and $J_2$, denoted by ${\eta _x}$. From the compact embedding of $W^{1,2}_x \subset L^q_x \text { with } q<6$, it is clear that there exists $\theta >0$ such that

$$\begin{aligned} \Vert \textbf{u}- \pi _{\eta _x}*\textbf{u}\Vert _{L^2_t L^q_x}\le C {\eta _x}^\theta \ \hbox { for all}\ q< 6. \end{aligned}$$

Moreover, we also have

$$\begin{aligned} \Vert \pi _{\eta _x}*\left( \textbf{u}(t,x)-\textbf{u}(s,x) \right) \Vert _{W^{1, r}_x}\le {\eta _x}^{-\alpha }\ \hbox { for some} \alpha >0 \hbox {and} r<\infty . \end{aligned}$$

Therefore, $J_1$ can be estimated as follows

$$\begin{aligned}{} & {} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) \textbf{u}(t,x) - \varrho (s,x) \textbf{u}(s,x)) \pi _{\eta _t}(t-s) (\textbf{u}(t,x) - \textbf{u}(s,x)) \, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} =\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) \textbf{u}(t,x) - \varrho (s,x) \textbf{u}(s,x)) \\{} & {} \hspace{3cm} \pi _{\eta _t}(t-s) [\textbf{u}(t,x) -\pi _{\eta _x}*\textbf{u}(t,x)-( \textbf{u}(s,x)- \pi _{\eta _x}*\textbf{u}(s,x))] \, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} +\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) \textbf{u}(t,x) - \varrho (s,x) \textbf{u}(s,x)) \pi _{\eta _t}(t-s) \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x)) \, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta +\int _0^T\!\!\!\int _0^T\pi _{\eta _t}(t-s)\int _{\mathbb {T}^d} \!\!\left( \int _t^s\partial _\tau (\varrho \textbf{u})\, d\tau \, \right) \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x)) \, \ \textrm{d} {x} \,\textrm{d}s\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta +\int _0^T\!\!\!\int _0^T\pi _{\eta _t}(t-s)\left( \int _t^s\Vert \partial _\tau (\varrho \textbf{u})\Vert _{W^{-1,q}_x}\, d\tau \right) \Vert \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x))\Vert _{W^{1,r}_x} \, \,\textrm{d}s\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta \\{} & {} + \left( \int _0^T\!\!\!\int _0^T\pi _{\eta _t}(t-s)\left( \int _t^s\Vert \partial _\tau (\varrho \textbf{u})\Vert _{W^{-1,q} } d\tau \right) ^2 \,\textrm{d}s\,\textrm{d} t \right) ^{\frac{1}{2}}\\{} & {} \quad \quad \left( \int _0^T\!\!\!\int _0^T\pi _{\eta _t}(t-s)\Vert \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x))\Vert ^2_{W^{1,r}} \, \,\textrm{d}s\,\textrm{d} t \right) ^{\frac{1}{2}}\\{} & {} \le C{\eta _x}^\theta +C{\eta _x}^{-\alpha }{\eta _t}^{\beta }. \end{aligned}$$

Now, we want to estimate the term $J_2$ in a similar way. Since $\varrho \in L^\infty _t L^{\gamma }_x$ and $\textbf{u}\in L^2_t W^{1,2}_x$, so the continuity equation gives $\partial _{t}\varrho \in L^2_t W^{-1, \frac{6\gamma }{\gamma +6}}_x$, with $\frac{6\gamma }{\gamma +6} \ge \frac{18}{13}$ when $\gamma \ge \frac{9}{5}$. Again using the identity

$$\begin{aligned}&\textbf{u}(t,x) - \textbf{u}(s,x)= (\textbf{u}(t,x) -\pi _{\eta _x}*\textbf{u}(t,x)-( \textbf{u}(s,x)- \pi _{\eta _x}*\textbf{u}(s,x)) ) \\ {}&\quad + \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x)) \end{aligned}$$

for the term $J_2$, we obtain

$$\begin{aligned}{} & {} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) -\varrho (s,x)) \textbf{u}(s,x) \pi _{\eta _t}(t-s) (\textbf{u}(t,x) - \textbf{u}(s,x))\, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta +\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) -\varrho (s,x))\pi _{\eta _x}*\textbf{u}(s,x) \pi _{\eta _t}(t-s) (\textbf{u}(t,x) - \textbf{u}(s,x))\, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta +\int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T (\varrho (t,x) -\varrho (s,x))\pi _{\eta _x}*\textbf{u}(s,x) \pi _{\eta _t}(t-s) \pi _{\eta _x}*(\textbf{u}(t,x)\\{} & {} \quad - \textbf{u}(s,x))\, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta + \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \int _0^T\left( \int _t^s\partial _\tau \varrho \ d\tau \right) \pi _{\eta _x}*\textbf{u}(s,x) \pi _{\eta _t}(t-s) \pi _{\eta _x}*(\textbf{u}(t,x) - \textbf{u}(s,x))\, \textrm{d}s \ \textrm{d} {x}\,\textrm{d} t \\{} & {} \le C{\eta _x}^\theta + \int _0^T \!\!\!\int _0^T\pi _{\eta _t}(t-s)\left( \int _t^s\Vert \partial _\tau \varrho \Vert _{W^{-1,\frac{18}{13}}_x}\ d\tau \right) \Vert \pi _{\eta _x}*\textbf{u}(s,x) \pi _{\eta _x}*(\textbf{u}(t,x)\\{} & {} \quad - \textbf{u}(s,x))\Vert _{W^{1,\frac{18}{5}}_x}\, \textrm{d}s \ \textrm{d} t \\{} & {} \le C{\eta _x}^\theta +\Vert \partial _{t}\varrho \Vert _{L^2_t W^{-1,\frac{18}{13}}_x} \int _0^T \!\!\!\int _0^T\pi _{\eta _t}(t-s) (t-s)^{1/2} \Vert \pi _{\eta _x}*\textbf{u}(s,x) \pi _{\eta _x}*(\textbf{u}(t,x) \\{} & {} \quad - \textbf{u}(s,x))\Vert _{W^{1,\frac{18}{5}}_x}\, \textrm{d}s \ \textrm{d} t \\{} & {} \le C{\eta _x}^\theta +C \int _0^T \!\!\!\int _0^T\pi _{\eta _t}(t-s) (t-s)^{1/2}\left( \Vert \pi _{\eta _x}*\textbf{u}(t,x)\Vert ^2_{W^{1,\frac{18}{5}}_x} + \Vert \pi _{\eta _x}*\textbf{u}(s,x)\Vert ^2_{W^{1,\frac{18}{5}}_x} \right) \textrm{d}s \ \textrm{d} t \\{} & {} \quad \le C{\eta _x}^\theta +C{\eta _x}^{-\alpha }{\eta _t}^\beta , \end{aligned}$$

where we use the fact that there exists $\alpha >0$ such that $\Vert \pi _{\eta _x}*\textbf{u}(s,x)\Vert _{L^2_t W^{1,\frac{18}{5}}_x} \le {\eta _x}^{-\alpha }.$

Choosing ${\eta _t}$ sufficiently small w.r.t. ${\eta _x}$, for example ${\eta _t}={\eta _x}^{\frac{\alpha }{\beta }+\frac{\vartheta }{\beta }}$, we finally estimate the sum of $J_1$ and $J_2$ as

$$\begin{aligned}{} & {} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho (\textbf{u}-\pi _\eta *\textbf{u})^2 \ \textrm{d} {x}\,\textrm{d} t \le C{\eta _t}^{\theta '}, \end{aligned}$$

for some constant $\theta '>0$. From the above relation, we want to deduce the following:

$$\begin{aligned} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho |\textbf{u}-\pi _{\eta _t}*\textbf{u}| \ \textrm{d} {x}\,\textrm{d} t \le C{\eta _t}^{\theta '}. \end{aligned}$$

To do this, we consider two cases:

$$\begin{aligned} |\textbf{u}-\pi _{\eta _t}*\textbf{u}(t,x)| \le {\eta _t}^{\frac{\theta '}{2}} \text{ and } |\textbf{u}(t,x) -\pi _{\eta _t}*\textbf{u}(t,x)| > {\eta _t}^{\frac{\theta '}{2}}. \end{aligned}$$

Then a straightforward computation yields

$$\begin{aligned} \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho |\textbf{u}-\pi _{\eta _t}*\textbf{u}| \ \textrm{d} {x}\,\textrm{d} t \le \int _0^T\!\!\!\! \int _{\mathbb {T}^d} \varrho \left( {\eta _t}^{\frac{\theta '}{2}}+ {\eta _t}^{-\frac{\theta '}{2}} |\textbf{u}-\pi _{\eta _t}*\textbf{u}|^2 \right) \ \textrm{d} {x}\,\textrm{d} t \le C {\eta _t}^{\frac{\theta '}{2}}. \end{aligned}$$

Now, taking $\eta ={\eta _t}$ and $\vartheta =\frac{\theta '}{2}$ we conclude the proof. $\square$

1.4 A.4 Proof of Lemma 13

At first we use Hölder inequality to obtain

$$\begin{aligned}{} & {} \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \Vert D_{|z|}u(\cdot ) - D_{|z|}u(\cdot + z)\Vert _{L^2} \\{} & {} \le \left( \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \right) ^{1/2} \left( \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \Vert D_{|z|}u(\cdot ) - D_{|z|}u(\cdot + z)\Vert _{L^2}^2\right) ^{1/2} \end{aligned}$$

The first component is proportional to $|\ln h|^{1/2}$, and so we only need to show that the latter term is bounded by $\Vert u\Vert _{W^{1,2}({\mathbb {T}}^d)}$.

To avoid technical problems with properties of functions on torus we extend u onto the whole $\mathbb {R}^d$ in such a way that u is treated as a periodic function and then it is localized in such a way that it is u for $x \in [-3\pi ,3\pi ]^d$ and zero for $\mathbb {R}^d{\setminus } [-4\pi ,4\pi ]^d$ (as the torus is ${\mathbb {T}}^d=[-\pi ,\pi ]^d$) and we name it Eu. Then it is clear that $\Vert Eu\Vert _{W^{1,2}(\mathbb {R}^d)} \le C\Vert u\Vert _{W^{1,2}({\mathbb {T}}^d)}$ and

$$\begin{aligned} \Vert D_{|z|} u(\cdot )-D_{|z|}u(\cdot +z)\Vert _{L^2({\mathbb {T}}^d)} \le \Vert D_{|z|}Eu(\cdot )-D_{|z|}Eu(\cdot +z)\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Using the above inequality, we have

$$\begin{aligned}{} & {} \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \Vert D_{|z|}u(\cdot ) - D_{|z|}u(\cdot + z)\Vert _{L^2({\mathbb {T}}^d)}^2\\{} & {} \le \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \int _{\mathbb {R}^d}\left| \frac{1}{|z|}\left( \int _{B(0,|z|)} \frac{|\nabla Eu(x+w)|}{|w|^{d-1} } \textrm{d}w- \int _{B(0,|z|)} \frac{|\nabla Eu(x+z+w)|}{|w|^{d-1} } \textrm{d}w \right) \right| ^2\textrm{d} {x}\\{} & {} = \int _{{\mathbb {T}}^d} \frac{dz}{(|z|+h)^d} \int _{\mathbb {R}^d} \left| \frac{1}{|z|} \int _{B(0,|z|)} \frac{\textrm{d}w}{|w|^{d-1}} e^{ {iw \cdot \xi }} (1-e^{{iz\cdot \xi }})\widehat{|\nabla Eu|}(\xi ) \right| ^2 \textrm{d} \xi \\{} & {} =\int _{\mathbb {R}^d} \textrm{d}\xi \widehat{|\nabla Eu|}^2(\xi ) \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \frac{|1-e^{iz\cdot \xi }|^2}{|z|^2} \left| \int _{B(0,|z|)} \frac{\textrm{d}w}{|w|^{d-1}} e^{iw\cdot \xi } \right| ^2. \end{aligned}$$

We will next show that the integral

$$\begin{aligned}I={} & {} \int _{{\mathbb {T}}^d} \frac{\textrm{d}z}{(|z|+h)^d} \frac{|1-e^{iz\cdot \xi }|^2}{|z|^2} \left| \int _{B(0,|z|)} \frac{\textrm{d}w}{|w|^{d-1}} e^{iw\cdot \xi } \right| ^2 \\ {}{} & {} \le \int _{\mathbb {R}^d} \frac{\textrm{d}z}{|z|^d} \frac{|1-e^{iz\cdot \xi }|^2}{|z|^2} \left| \int _{B(0,|z|)} \frac{\textrm{d}w}{|w|^{d-1}} e^{iw\cdot \xi } \right| ^2 \end{aligned}$$

is independent of $\xi$ and is bounded by a constant.

We first denote $\xi =|\xi |\omega _\xi$, and then by changing the variables for $t=|\xi | z$ and $v=|\xi |w$ we get the estimate

$$\begin{aligned} I\le \int _{\mathbb {R}^d} \frac{\textrm{d}t}{|t|^d} \frac{|1-e^{i\omega _\xi \cdot t}|^2}{|t|^2}\left| \int _{B(0,|t|)} \frac{\textrm{d}v}{|v|^{d-1}} e^{i\omega _\xi \cdot v} \right| ^2. \end{aligned}$$

(104)

Because both integrals are spherically symmetric, we can fix $\omega _\xi$ to be a unit vector, say $\omega _\xi ={\hat{e}}_1$. So the integral I is independent of $\xi$, and therefore in order to end the proof one should only show that it is bounded. We denote $v=(v_1,v')$, and for now we focus on the second integral only, i.e. $\int _{B(0,|t|)} \frac{dv}{|v|^{d-1}} e^{iv_1}$.

Case 1. For t small we can estimate $|e^{iv_1}|\le 1$ and the integral is smaller than Ct.

Case 2. For $t>>1$ we split the integral as follows

$$\begin{aligned} \int _{B(0,|t|)} \frac{\textrm{d}v}{|v|^{d-1}} e^{iv_1}= 2\int _0^{|t|} \textrm{d}v_1 \int _{{ B'(0,\sqrt{|t|^2-v_1^2})}} e^{iv_1} \frac{\textrm{d}v'}{(\sqrt{v_1^2+|v'|^2})^{(d-1)/2}}, \end{aligned}$$

(105)

where $B'$ denotes a ball in $d-1$ dimensions. We first write $e^{iv_1}=\cos (v_1)+i\sin (v_1)$. We split the real part of the integral $\int _0^{|t|}$ into the integrals on the intervals $(0,\frac{\pi }{2})$, $(-\frac{\pi }{2}+k\pi ,-\frac{\pi }{2}+ (k+1)\pi )$, $(-\frac{\pi }{2}+(k_{|t|}+1)\pi ,|t|)$ for $k=1,\ldots ,k_{|t|}$, $k_{|t|}=\left\lfloor \frac{t-\pi /2}{\pi }\right\rfloor$. The idea is just to show that the above division gives rise to the alternating series, just to show that the integral is uniformly bounded in the meaning of Riemann’s integral i.e.

$$\begin{aligned} \frac{1}{2}\int _{B(0,|t|)} \frac{\textrm{d}v}{|v|^{d-1}} e^{iv_1}= & {} \int _0^{\pi /2} \Gamma \, \textrm{d}v_1+ \sum _{k=1}^{k_{|t|}} (-1)^k\int _{-\pi /2+k\pi }^{-\pi /2+(k+1)\pi } \Gamma \, \textrm{d}v_1\\{} & {} \quad + (-1)^{k_{|t|}+1}\int _{-\frac{\pi }{2}+(k_{|t|}+1)\pi }^{|t|} \Gamma \, \textrm{d}v_1, \end{aligned}$$

where we denoted

$$\begin{aligned} \Gamma = |\cos {v_1}| \int _{B'(0,\sqrt{|t|^2-v_1^2})} \frac{\textrm{d}v'}{(\sqrt{v_1^2+|v'|^2})^{(d-1)/2}}. \end{aligned}$$

Note that the irrational part of the integral, related to $i\sin v_1$, is odd, so it sums up to zero.

In the above we have used that $\cos t = -\cos (\pi +t)$. We see that the integrals of $|\Gamma |$ are decreasing with k, and so the series is bounded by the first term, i.e. we have

$$\begin{aligned} \frac{1}{2}\left| \int _{B(0,|t|)} \frac{\textrm{d}v}{|v|^{d-1}} e^{iv_1}\right| \le \int _0^{\pi /2} \Gamma \, \textrm{d}v_1+ \int _{\pi /2}^{3\pi /2} \Gamma \, \textrm{d}v_1. \end{aligned}$$

We now want to estimate these integrals, for this we include the irrational part again, and estimate (105) on the intervals $(0,\frac{\pi }{2})$, $(\frac{\pi }{2},\frac{3\pi }{2})$, assuming that $t > 2\pi$, we can write that

$$\begin{aligned}{} & {} \int _0^{\pi /2} \textrm{d}v_1 \int _{B'(0,\sqrt{|t|^2-v_1^2})} \frac{\textrm{d} v'}{|v|^{d-1}} \le \int _0^{\pi /2} \textrm{d}v_1 \int _{B'(0,|t|)} \frac{\textrm{d} v'}{|v|^{d-1}}\\{} & {} \le C \int _0^{\pi /2} \textrm{d}v_1 \int _{B'(0,|t|)} \frac{ \textrm{d} v' }{ (|v_1|+|v'|)^{d-1}} \le C \int _0^{\pi /2} \textrm{d}v_1 \int _0^{|t|} \frac{ \textrm{d} |v'|\, |v'|^{d-2} }{ (|v_1|+|v'|)^{d-1}} \\{} & {} \le C \int _0^{\pi /2} \textrm{d}v_1 \int _0^{|t|} \frac{ \textrm{d} |v'|}{|v_1|+|v'|}= C\int _0^{\pi /2} \textrm{d}v_1 (\ln ( |t|+|v_1|) - \ln |v_1|) \le C (\ln |t| +1). \end{aligned}$$

For the second part, the same way we obtain

$$\begin{aligned} \int _{\pi /2}^{3\pi /2} \textrm{d}v_1 \int _{B'(0,|t|)} \frac{\textrm{d}v'}{|v|^{d-1}} \le C\int _{\pi /2}^{3\pi /2} \textrm{d}v_1 \int _0^{|t|} \frac{\textrm{d}|v'|}{\frac{\pi }{2} + |v'|} \le C(\ln |t| +1). \end{aligned}$$

Since $t>2\pi$, we have $1+\ln |t| \le C \ln |t|$, so, summing up cases 1 and 2, we get that

$$\begin{aligned} \left| \int _{B(0,|t|)} \frac{\textrm{d}v}{|v|^{d-1}} e^{iv_1} \right| \le C\min \{ t, \ln |t|\}. \end{aligned}$$

At last, coming back to (104), we obtain

$$\begin{aligned}I\le C\int _{\mathbb {R}^d} \frac{\textrm{d}t}{|t|^d} \frac{|1-e^{i\omega _\xi t}|^2}{|t|^2}(\min \{ t, \ln |t|\})^2 \le C\left( \int _{B(0,1)} \frac{\textrm{d}t}{|t|^{d-2}} +\int _{B(0,1)^c} \frac{\textrm{d}t}{|t|^d} \frac{\ln | t|^2}{|t|^2} \right) \le C, \end{aligned}$$

where we used that for |t| small, i.e. for $t\in B(0,1)$ we can estimate $|1-e^{i\omega _\xi t}|\le |t|$. With this estimate in hand, we complete the proof. $\square$

1.5 A.5 Proof of Lemma 15

Before we begin the proof, we introduce the following notation: For $0<a,b \in \mathbb {R}$, the notation $a\approx b$ means that there exists a $\mu >0$ such that

$$\begin{aligned} \mu a \le b \le \frac{1}{\mu } a. \end{aligned}$$

From Parseval’s theorem for Fourier transform, we have

$$\begin{aligned} \int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} \Big ((- \Delta _x)^{-1}{\text {div}}_x \partial _{t}\pmb {\varphi } \Big ) W(t,x) \textrm{d} {x}\; \textrm{d} t&= \int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} {\mathcal {F}}_{\xi _0,\xi } {\Big ((- \Delta _x)^{-1}{\text {div}}_x \partial _{t}\pmb {\varphi } \Big )} \overline{{\mathcal {F}}_{\xi _0,\xi } ({W} )}\text {d}\xi \; \text {d}\xi _0\\&=- \int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} \;\xi _0 \vert \xi \vert ^{-2} \; \xi \cdot \; {\mathcal {F}}_{\xi _0,\xi }({\pmb {\varphi }}) \; \overline{{\mathcal {F}}_{\xi _0,\xi } (W)} \text {d}\xi \; \text {d}\xi _0. \end{aligned}$$

Using ${\mathcal {F}}_{\xi _0,\xi }(\pmb {\varphi })= |\xi | {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})$ and $0<\beta <1$, we obtain

$$\begin{aligned}&\int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} \;\xi _0 \vert \xi \vert ^{-2} \; \xi \cdot \; {\mathcal {F}}_{\xi _0,\xi }({\pmb {\varphi }}) \; \overline{{\mathcal {F}}_{\xi _0,\xi } (W)} \text {d}\xi \; \text {d}\xi _0\\&= \int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} \;\left( \xi _0 |\xi _0|^{(\beta -1)} \vert \xi \vert ^{-\beta } \xi \cdot {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})\right) \;\left( \vert \xi _0\vert ^{1-\beta } \vert \xi \vert ^{-(1-\beta )} \overline{{\mathcal {F}}_{\xi _0,\xi } (W)} \right) \text {d}\xi \; \text {d}\xi _0\\&= \int _{{\mathbb {R}}} \int _{{\mathbb {R}}^d} \; {\mathcal {F}}^{-1}_{t,x} \left( \xi _0 |\xi _0|^{\beta -1} \vert \xi \vert ^{-\beta } \xi \cdot {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})\right) \overline{{\mathcal {F}}^{-1}_{t,x} } \left( \vert \xi _0\vert ^{1-\beta } \vert \xi \vert ^{-(1-\beta )} \overline{{\mathcal {F}}_{\xi _0,\xi } (W)} \right) \textrm{d} {x}\; \textrm{d} t \\&\le C \left\| {\mathcal {F}}^{-1}_{t,x} \left( |\xi _0|^\beta \vert \xi \vert ^{1-\beta } {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})\right) \right\| _{L^p_{t,x}} \left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi _0\vert ^{1-\beta } \vert \xi \vert ^{-(1-\beta )} {\mathcal {F}}_{\xi _0,\xi } (W) \right) \right\| _{L^{p^{\prime }}_{t,x}}, \end{aligned}$$

with $\frac{1}{p}+ \frac{1}{p^\prime } =1$. Now our idea is to use the following observation

$$\begin{aligned} \left\| {\mathcal {F}}^{-1}_{t,x} \left( |\xi _0|^\beta \vert \xi \vert ^{1-\beta } {\mathcal {F}}_{\xi _0,\xi }(\pmb {g})\right) \right\| _{L^p_{t,x}} \approx \Vert \pmb {g} \Vert _{{\dot{W}}^{\beta ,p}(\mathbb {R}; {\dot{W}}^{1-\beta ,p}(\mathbb {R}^d))}, \end{aligned}$$

which means that

$$\begin{aligned} {\mathcal {F}}^{-1}_{t}\left( |\xi _0|^\beta {\mathcal {F}}_{\xi _0}(\pmb {g})\right) \in L^p(\mathbb {R};W^{1-\beta ,p}(\mathbb {R}^d)). \end{aligned}$$

Then we want to interpolate $\Vert \pmb {g} \Vert _{{\dot{W}}^\beta _p(\mathbb {R}; {\dot{W}}^{1-\beta ,p}(\mathbb {R}^d))}$ with $\Vert \pmb {g} \Vert _{L_{q}(\mathbb {R}^{d+1})}$ and $\Vert \partial _t \pmb {g} \Vert _{L^{q^\prime }(\mathbb {R}; W^{-1,q^\prime }(\mathbb {R}^{d+1}))}$ for some q and $q^\prime$.

It is well known that for the homogeneous Triebel-Lizorkin space ${\dot{F}}^s_{p,2} (\mathbb {R}^d)$, we have

$$\begin{aligned} {\dot{W}}^{s,p}(\mathbb {R}^d)={\dot{F}}^s_{p,2} (\mathbb {R}^d) \text { for } \; s\in \mathbb {R}\text { and }1<p<\infty \end{aligned}$$

(106)

with equivalent norms. Therefore, using Duoandikoetxea [8, Theorem 8.7], we obtain

$$\begin{aligned} \left\| {\mathcal {F}}^{-1}_{t,x} \left( |\xi _0|^\beta \vert \xi \vert ^{1-\beta } {\mathcal {F}}_{\xi _0,\xi }(\pmb {g})\right) \right\| _{L^p_{t,x}} \approx \left\| \left( \sum _{k,l} 2^{2k(1-\beta )} 2^{2l\beta } {\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}} \right\| _{L^p_{t,x}}, \end{aligned}$$

where

$$\begin{aligned} {\pmb {g}}_{k,l}= {\dot{\Delta }}_k^x {\dot{\Delta }}_l^t {\mathcal {F}}_{\xi _0,\xi } ({\pmb {g}}), \end{aligned}$$

with ${\dot{\Delta }}_k^x$ and ${\dot{\Delta }}_l^t$ are the homogeneous dyadic blocks in space variable and time variable, respectively. Next, using Holder inequality, we obtain

$$\begin{aligned} \left\| \left( \sum _{k,l} 2^{2k(1-\beta )} 2^{2l\beta } {\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}} \right\| _{L^p_{t,x}}&\le \left\| \left( \sum _{k,l} 2^{2k}{\pmb {g}}_{k,l}^2\right) ^{\frac{1-\beta }{2}} \left( \sum _{k,l} 2^{2l} {\pmb {g}}_{k,l}^2\right) ^{\frac{\beta }{2}} \right\| _{L^p_{t,x}}\\&\le \left\| \left( \sum _{k,l} 2^{2k}{\pmb {g}}_{k,l}^2\right) ^{\frac{1-\beta }{2}}\right\| _{L^{r_1}_{t,x}} \left\| \left( \sum _{k,l} 2^{2l} {\pmb {g}}_{k,l}^2\right) ^{\frac{\beta }{2}}\right\| _{L^{r_1^\prime }_{t,x}}, \end{aligned}$$

with $\frac{1}{r_1} + \frac{1}{r_1^\prime } =\frac{1}{p}$. Considering $r= \frac{r_1}{p}$ and $r^\prime = \frac{r_1^\prime }{p}$, we introduce

$$\begin{aligned} q=(1-\beta )p r \text { and } q^\prime = \beta p r^\prime \end{aligned}$$

to deduce

$$\begin{aligned} \left\| \left( \sum _{k,l} 2^{2k(1-\beta )} 2^{2l\beta } {\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}} \right\| _{L^p_{t,x}} \le \left\| \left( \sum _{k,l} 2^{2k}{\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}}\right\| _{L^{q}_{t,x}}^{1-\beta } \left\| \left( \sum _{k,l} 2^{2l} {\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}}\right\| _{L^{q^\prime }_{t,x}}^{\beta } \end{aligned}$$

with the following relation between $p, q, q^\prime$ and $\beta$:

$$\begin{aligned} 1=p \left( \frac{1-\beta }{q} + \frac{\beta }{q^\prime } \right) . \end{aligned}$$

Again, using the equivalence of Sobolev and Triebel-Lizorkin space (106), we have

$$\begin{aligned} \left\| \left( \sum _{k,l} 2^{2k}{\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}}\right\| _{L^{q}_{t,x}} \approx \left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi \vert \; {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})\right) \right\| _{L^q_{t,x}} = \left\| {\mathcal {F}}^{-1}_{t,x} \left( \; {\mathcal {F}}_{\xi _0,\xi }({\pmb {\varphi }})\right) \right\| _{L^q_{t,x}} = \Vert \pmb {\varphi } \Vert _{L^q_{t,x}} \end{aligned}$$

and

$$\begin{aligned} \left\| \left( \sum _{k,l} 2^{2l} {\pmb {g}}_{k,l}^2\right) ^{\frac{1}{2}}\right\| _{L^{q^\prime }_{t,x}}\approx&\left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi _0 \vert \; {\mathcal {F}}_{\xi _0,\xi }({\pmb {g}})\right) \right\| _{L^{q^\prime }_{t,x}}\\ {}&=\left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi _0 \vert \; \vert \xi \vert ^{-1} {\mathcal {F}}_{\xi _0,\xi }({\pmb {\varphi }})\right) \right\| _{L^{q^\prime }_{t,x}}= \left\| \partial _t \pmb {\varphi } \right\| _{L^{q^\prime }_tW^{-1,q^\prime }_x}. \end{aligned}$$

The above two estimates yield

$$\begin{aligned} \left\| {\mathcal {F}}^{-1}_{t,x} \left( |\xi _0|^\beta \vert \xi \vert ^{1-\beta } {\mathcal {F}}_{\xi _0,\xi }(\pmb {g})\right) \right\| _{L^p_{t,x}} \le \Vert \pmb {\varphi } \Vert _{L^{q}_{t,x}}^{1-\beta } \Vert \partial _t \pmb {\varphi } \Vert _{L^{q^\prime }_tW^{-1,q^\prime }_x}^{\beta }. \end{aligned}$$

(107)

For the term $\left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi _0\vert ^{1-\beta } \vert \xi \vert ^{-(1-\beta )} {\mathcal {F}}_{\xi _0,\xi } (W) \right) \right\| _{L^{p^{\prime }}_{t,x}}$, we proceed analogously and obtain

$$\begin{aligned} \left\| {\mathcal {F}}^{-1}_{t,x} \left( \vert \xi _0\vert ^{1-\beta } \vert \xi \vert ^{-(1-\beta )} {\mathcal {F}}_{\xi _0,\xi } (W) \right) \right\| _{L^{p^{\prime }}_{t,x}} \le \Vert W \Vert _{L^{{\bar{q}}}_{t,x}}^{1-\alpha } \Vert \partial _t W \Vert _{L^{{\bar{q}}^{\prime }}_tW^{-1,{\bar{q}}^\prime }_x}^{\alpha } \end{aligned}$$

(108)

where $1-\beta =\alpha$ and the relation between $p^\prime , {\bar{q}}, {{\bar{q}}}^\prime$ and $\alpha$ is given by

$$\begin{aligned} 1=p^\prime \left( \frac{1-\alpha }{{\bar{q}}} + \frac{\alpha }{{\bar{q}}^\prime } \right) . \end{aligned}$$

Moreover, we have

$$\begin{aligned} 1-\frac{1}{q^\prime }-\frac{1}{{\bar{q}}}=\alpha \left( \frac{1}{q} + \frac{1}{{\bar{q}}^\prime } -\frac{1}{q^\prime }-\frac{1}{{\bar{q}}} \right) . \end{aligned}$$

The assumption (80) implies $0<\alpha <1$. Summing up the estimates (107) and (108), we get our desired result. $\square$

1.6 A.6 Existence of solutions to parabolic systems

Lemma 16

Let $p>1$. Let a be sufficiently smooth and satisfies

$$\begin{aligned}{} & {} 0< a_* \le a(t,x) \le a^* < \infty ,\; \nabla a \in L^\infty (0,T;L^\infty ({\mathbb {T}}^d)) \\ {}{} & {} \text { and }a \in C^\alpha (0,T;C_b({\mathbb {T}}^d)) \text { for }\alpha >0 \end{aligned}$$

Moreover, we assume

$$\begin{aligned} \textbf{u}_0 \in W^{1-2/p,p}({\mathbb {T}}^d),\; {\textbf{F}} \in L^p(0,T;L^p({\mathbb {T}}^d)\text { and } {\textbf{G}}\in L^p(0,T;L^p({\mathbb {T}}^d)) \end{aligned}$$

Then the solution $\textbf{u}$ of

$$\begin{aligned} a\partial _{t}\textbf{u}-{\text {div}}\textbf{S}(\textbf{u}) = {\text {div}}{\textbf{F}} +{\textbf{G}}, \qquad \textbf{u}|_{t=0}=\textbf{u}_0 \end{aligned}$$

(109)

satisfies the following bound

$$\begin{aligned} \Vert \partial _{t}\textbf{u}\Vert _{L^p_t W^{-1,p}_x} + \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}} \le C(T)(\Vert {\textbf{F}}\Vert _{L^p_t L^p_x}+\Vert {\textbf{G}}\Vert _{L^p_t L^p_x}+ \Vert \textbf{u}_0\Vert _{W^{1-2/p,p}_x}). \end{aligned}$$

(110)

Proof

The result belongs to the classical theory of parabolic systems. Since such results are consequences of several results from [7] and [12] we here give a sketch of the proof to satisfy readers. The first step which is required by the structure of regularity is reformulation of (109) in the following form:

$$\begin{aligned}{} & {} \partial _{t}\textbf{u}-b{\text {div}}\textbf{S}(\textbf{u}) = {\text {div}}\tilde{{\textbf{F}}} +\tilde{{\textbf{G}}}, \qquad u|_{t=0}=u_0 \text{ with } b=1/a,\; \tilde{{\textbf{F}}}= \frac{{\textbf{F}}}{b} \text { and } \\{} & {} \tilde{{\textbf{G}}}={\textbf{G}} - {\textbf{F}} \nabla \left( \frac{1}{b} \right) \end{aligned}$$

The proof we perform for a simplified version, to avoid unnecessary technicalities. Therefore, instead of ${\text {div}}{\textbf{S}}(\textbf{u})$, we consider

$$\begin{aligned} \partial _{t}\textbf{u}- b \Delta \textbf{u}= \nabla f, \end{aligned}$$

(111)

with $f \in L^p(0,T;L^p({\mathbb {T}}^d))$.

Step 1. The basic result is: if ${{\bar{b}}} \in [b_*,b^*]$ it is a constant then the solution to

$$\begin{aligned} \partial _{t}\textbf{u}- {{\bar{b}}} \Delta \textbf{u}= \nabla f \end{aligned}$$

fulfills

$$\begin{aligned} \Vert \partial _{t}\textbf{u}\Vert _{L^p_t W^{-1,p}_x} + \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}} \le C(b_*,b^*)(\Vert f\Vert _{L^p_t L^{p}_x}+ \Vert \textbf{u}_0\Vert _{W^{1-2/p,p}_x}). \end{aligned}$$

Step 2. In this part we fix the time in the coefficient b, i.e. we consider a time-independent in (110). Let $\pi _k$ be a partition of unity. We are required to

$$\begin{aligned} \mathrm{diam \;\; supp\;} \pi _k \le \lambda . \end{aligned}$$

Let k be fixed, and let us set $\pi =\pi _k$. We consider

$$\begin{aligned} \partial _{t}(\pi \textbf{u}) - b \Delta (\pi \textbf{u}) = \nabla (\pi f) - \nabla \pi f + b \nabla \pi \nabla \textbf{u}+ b \Delta \pi \textbf{u}. \end{aligned}$$

Let ${{\bar{b}}} = b(x_k)$ such that $\pi _k(x_k)=1$, then

$$\begin{aligned} \partial _{t}(\pi \textbf{u}) - {{\bar{b}}} \Delta (\pi \textbf{u}) = \nabla (\pi f) - \nabla \pi f + b \nabla \pi \nabla \textbf{u}+ b \Delta \pi \textbf{u}+(b-{{\bar{b}}}) \Delta (\pi \textbf{u}). \end{aligned}$$

We just deal with the key elements of the r.h.s.. First, we take $f\nabla \pi$, which is of higher order but with bad dependence form $\lambda$. This part is formally related with the following problem

$$\begin{aligned} \partial _{t}{\textbf{w}}-{{\bar{b}}}\Delta {\textbf{w}} = f \nabla \pi , \qquad {\textbf{w}}|_{t=0} =0. \end{aligned}$$

Then we have

$$\begin{aligned} \sup _t \Vert {\textbf{w}}\Vert _{W^{2-2/p,p}_x} + \Vert \partial _{t}{\textbf{w}},\nabla ^2 {\textbf{w}}\Vert _{L^p_{t,x}} \le C\Vert f\nabla \pi \Vert _{L^p_{t,x}} \end{aligned}$$

with constant independent of T. Then we get (for $p>2$) that

$$\begin{aligned}{} & {} \Vert \nabla {\textbf{w}}\Vert _{L^p_{t,x}} \le CT^{1/p}\sup _t \Vert {\textbf{w}}\Vert _{W^{1-2/p,p}_x} \le CT^{1/p} \Vert f\nabla \pi \Vert _{L^p_{t,x}} \\ {}{} & {} \le CT^{1/p}\lambda ^{-1} \Vert f\Vert _{L^p((0,T) \times supp\,\pi ))}. \end{aligned}$$

The second term is

$$\begin{aligned} \Vert (b-{{\bar{b}}}) \Delta (\pi \textbf{u})\Vert _{L^p_t W^{-1,p}_x}&\le \Vert {\text {div}}((b-{{\bar{b}}}) \nabla (\pi \textbf{u})\Vert _{L^p_t W^{-1,p}_x}+ +\Vert \nabla b \nabla (\pi \textbf{u})\Vert _{L^p_t W^{-1,p}_x} \\&\le C \Vert b-{{\bar{b}}}\Vert _{L^\infty (supp\;\pi )}\Vert \nabla (\pi \textbf{u})\Vert _{L^p((0,T) \times supp\,\pi ))}\\&\quad + CT^{1/p}\Vert \nabla b\Vert _{L^\infty }\Vert \nabla (\pi \textbf{u})\Vert _{L^p_{t,x}}\\&\le C(\lambda + T^{1/p}) \Vert \nabla b\Vert _{L^\infty } \Vert \nabla (\pi \textbf{u})\Vert _{L^p_{t,x}}. \end{aligned}$$

Thus all together we obtain

$$\begin{aligned} \Vert \nabla (\pi \textbf{u})\Vert _{L^p_{t,x}} \le C&\big ( (1+T^{1/p}\lambda ^{-1})\Vert f\Vert _{L^p((0,T)\times (supp\,\pi ))} \\&+ (\lambda + T^{1/p}) \Vert \nabla (\pi \textbf{u})\Vert _{L^p_{t,x}} + T^{1/p}\lambda ^{-1} \Vert \nabla \textbf{u}\Vert _{L^p(0,T;L^p(supp\,\pi ))}\big ). \end{aligned}$$

Now as $\lambda + T^{1/p}$ is small, we deduce

$$\begin{aligned} \Vert \nabla (\pi \textbf{u})\Vert _{L^p_{t,x}}^p \le C ( (1+T^{1/p}\lambda ^{-1})^p\Vert f\Vert _{L^p(0,T;L^p(supp\,\pi ))}^p + (T^{1/p}\lambda ^{-1})^p \Vert \nabla \textbf{u}\Vert _{L^p(0,T;L^p(supp\,\pi ))}^p). \end{aligned}$$

Let N be the cover number of the union of supp $\pi _k$, then we can estimate

$$\begin{aligned}&\Vert \nabla \textbf{u}\Vert ^p_{L^p_{t,x}} \le N^{p-1} \sum _k \Vert \nabla (\pi _k \textbf{u})\Vert _{L^p_{t,x})}^p \\&\le CN^{p-1} \sum _{k} \left[ (1+T^{1/p}\lambda ^{-1})^p\Vert f\Vert _{L^p(0,T;L^p(supp\,\pi _k))}^p + (T^{1/p}\lambda ^{-1})^p \Vert \nabla \textbf{u}\Vert _{L^p(0,T;L^p(supp\,\pi _k))}^p \right] \\&\le CN^{2p-1} \Vert f\Vert ^p_{t,x} + CN^{2p-1}\epsilon \Vert \nabla \textbf{u}\Vert ^p_{L^p_{t,x}} \end{aligned}$$

as $T^{1/p}\lambda ^{-1} <1$ and $(T^{1/p}\lambda ^{-1})^p \le \epsilon$. For $\epsilon$ small the last term can be absorbed by the l.h.s. and we get

$$\begin{aligned} \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}}\le C\Vert f\Vert _{L^p_{t,x}}. \end{aligned}$$

Hence for b time-independent we get

$$\begin{aligned} \Vert \partial _{t}\textbf{u}\Vert _{L^p_t W^{-1,p}_x} + \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}} \le C(b_*,b^*,\Vert \nabla b\Vert _{L^\infty })(\Vert f\Vert _{L^p_t L^{p}_x}+ \Vert \textbf{u}_0\Vert _{W^{1-2/p,p}_x}).\nonumber \\ \end{aligned}$$

(112)

Now we need to make b time dependent.

Step 3. As b is time dependent we can consider equation (111) in the form

$$\begin{aligned} \partial _{t}\textbf{u}-b(0,x)\Delta \textbf{u}= (b(t,x)-b(0,x)) \Delta \textbf{u}+ \nabla f, \qquad \textbf{u}|_{t=0}=\textbf{u}_0. \end{aligned}$$

Then we restate the first term of the l.h.s. as follows

$$\begin{aligned} (b(t,x)-b(0,x)) \Delta \textbf{u}= {\text {div}}((b(x,t)-b(x,0)) \nabla \textbf{u}+ \nabla b \nabla \textbf{u}. \end{aligned}$$

So from Step 2 we know that this term generates the following impact on the estimate

$$\begin{aligned} \Vert b(t,x)-b(0,x)\Vert _{L^\infty _x} \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}} + CT^{1/p}\Vert \nabla b\Vert _{L^\infty _{t,x}}\Vert \nabla \textbf{u}\Vert _{L^p_{t,x}}\\ \le C(\lambda ^\alpha + T^{1/p})(\Vert b\Vert _{C^\alpha _t L^\infty _x} + \Vert \nabla b\Vert _{L^\infty _{t,x}})\Vert \nabla \textbf{u}\Vert _{L^p_{t,x}}, \end{aligned}$$

where we used the fact that $b\in C^{\alpha }(0,T;C_b)$. Therefore, for a short time we get (112) in the form

$$\begin{aligned} \Vert \partial _{t}\textbf{u}\Vert _{L^p_t W^{-1,p}_x} + \Vert \nabla \textbf{u}\Vert _{L^p_{t,x}} \le C(b_*,b^*,\Vert \nabla b\Vert _{L^\infty _{t,x}},\Vert b\Vert _{C^\alpha _t L^\infty _x} )(\Vert f\Vert _{L^p_t L^{p}_x}+ \Vert \textbf{u}_0\Vert _{W^{1-2/p,p}_x}). \end{aligned}$$

The above inequality is valid for time interval $[0,T_*]$ for small $T_*$, then from the trace theorem we control the $W^{1-2/p,p}_x$ norm of $\textbf{u}|_{t=T_*}$ in terms of the l.h.s. of above and $\textbf{u}_0$. From that, we can obtain the estimate for $[T_*,2T_*]$ with the same constant C considering $\textbf{u}|_{t={T_*}}$ instead of $\textbf{u}_0$. Thus the inequality can be continued step by step with an increase of the constant in (110) with factor $e^{CT}$. The proof of the Lemma is concluded. $\square$