Proof. Multiplying equation (2.1) by $u$, integrating over $\Omega$ and using the Hölder inequality, we get the energy identity

\[ \|\nabla_x u\|^2_{L^2}+(f(u),u)=(g,u)\le C_\kappa\|g\|^q_{L^q}+\frac\kappa2\|u\|_{L^{r+1}}^{r+1}, \]

where $(U,\,V)$ stands for the standard inner product in $[L^2(\Omega )]^3$. Using our assumptions on $f$, we deduce from here that

(2.4)\begin{equation} \|\nabla_x u\|^2_{L^2}+\|u\|_{L^{r+1}}^{r+1}+\|f(u)\|_{L^q}^q\le C\|g\|_{L^q}^q \end{equation}

for some positive constant $C$. To complete the proof of the proposition, we write out equation (2.1) as a linear equation

(2.5)\begin{equation} \Delta_x u-\nabla_x p=\widetilde g:=g+f(u) \end{equation}

and apply the maximal $L^q$-regularity result for solutions of the linear Stokes equation, see e.g. [Reference Galdi11]. Together with the already obtained estimate for the $L^q$-norm of $f(u)$, this give us the desired control of the $W^{2,q}$-norm of $u$ as well as the $L^q$-norm of pressure $p$.

Proof. We divide the proof of this theorem on several steps.

Step 1. Interior regularity. Let $\varphi (x)$ be a sufficiently smooth cut-off function which vanishes near the boundary and equal to one identically outside of the $\varepsilon$-neighbourhood of $\partial \Omega$ and which satisfies the inequality

\[ |\nabla_x\varphi(x)|\le C_\varepsilon\varphi(x)^{1/2},\quad x\in\mathbb{R}^3. \]

Since the domain $\Omega$ is smooth, such a function exists for every $\varepsilon >0$.

Let us multiply equation (2.1) by $\sum \partial _{x_i}(\varphi \partial _{x_i}u)$ and transform the most complicated term containing pressure as follows:

(2.8)\begin{align} \left(\nabla_x p,\sum \partial_{x_i}(\varphi\partial_{x_i}u)\right)& ={-}\left(p, \sum_i\partial_{x_i}(\sum_j\partial_{x_j}\varphi\partial_{x_i}u_j)\right)\nonumber\\ & =\sum_i\left(\partial_{x_i}p,\sum_j\partial_{x_j}\varphi\partial_{x_i}u_j\right)\nonumber\\ & =\sum_i\left(\partial_{x_i}p,\partial_{x_i}(\sum_j \partial_{x_j}\varphi u_j)\right)-\sum_{i,j}\left(\partial_{x_i}p,\partial_{x_i,x_j}\varphi u_j\right)\nonumber\\ & ={-}\left(\Delta_x p,\nabla_x \varphi\cdot u\right)- \sum_{i,j}\left(\partial_{x_i}p,\partial_{x_i,x_j}\varphi u_j\right). \end{align}

The last term in the RHS is easy to estimate using (2.3):

\[ |\sum_{i,j}\left(\partial_{x_i}p,\partial_{x_i,x_j}\varphi u_j\right)|\le \|\nabla_x p\|_{L^q}\|u\|_{L^{r+1}}\le C\|g\|_{L^q}\|g\|_{L^q}^{1/r}=C\|g\|_{L^q}^q. \]

To estimate the first term, we take the divergence from both sides of equation (2.1) and obtain the Poisson equation for pressure:

(2.9)\begin{equation} \Delta_x p={-}\operatorname{div} f(u)-\operatorname{div} g. \end{equation}

Thus, integrating by parts again and using the growth restrictions on $f$ and estimate (2.4), we arrive at

(2.10)\begin{align} |(\Delta_x p,\nabla_x\varphi\cdot u)|& =|(\operatorname{div} f(u)+\operatorname{div} g, \nabla_x\varphi \cdot u)|\nonumber\\ & \le C(|g|,|\nabla_x u|+|u|)+C(|f(u)|,|u|)+(|f(u)|,|\nabla_x\varphi|\,|\nabla_x u|)\nonumber\\ & \le C(1+\|g\|^2_{L^2})+C(|u|^r,|\nabla_x u|\, |\nabla_x \varphi|). \end{align}

Using that $|\nabla _x \varphi |\le C\varphi ^{1/2}$, we transform the last term as follows:

(2.11)\begin{align} (|u|^r,|\nabla_x u|\,|\nabla_x\varphi|)& \le C\left(|u|^{(r-1)/2}\varphi^{1/2}|\nabla_x u|,|u|^{(r+1)/2}\right)\nonumber\\ & \le \beta(|u|^{r-1},\varphi|\nabla_x u|^2)+C_\beta\|u\|_{L^{r+1}}^{r+1}\nonumber\\ & \le \beta(|u|^{r-1},\varphi|\nabla_x u|^2)+C_\beta(1+\|g\|^2_{L^2}), \end{align}

where $\beta >0$ is arbitrary. Combining the obtained estimates, we get

(2.12)\begin{equation} |\left(\nabla_x p,\sum\partial_{x_i}(\varphi\partial_{x_i}u)\right)|\le \beta(\varphi|u|^{r-1},|\nabla_x u|^2)+C_\beta(1+\|g\|^2_{L^2}). \end{equation}

It is now not difficult to complete the interior estimate. Indeed, multiplying equation (2.1) by $\sum _i\partial _{x_i}(\varphi \partial _{x_i}u)$ and arguing in a standard way, with the help of (2.12), we get

(2.13)\begin{align} & (\varphi,|\Delta_x u|^2)+(\varphi f'(u)\nabla_x u,\nabla_x u)\le C\|g\|^2_{L^2}+C\|\nabla_x u\|^2_{L^2}\nonumber\\ & \quad +|\left(\nabla_x p,\sum \partial_{x_i}(\varphi\partial_{x_i}u)\right)|\le C_\beta(1+\|g\|^2_{L^2})+\beta(\varphi|u|^{r-1},|\nabla_x u|^2). \end{align}

Finally, due to our assumptions, $f'(u)\ge \kappa (|u|^{r-1}+1)$, so fixing $\beta <\kappa /2$, we arrive at the desired estimate

(2.14)\begin{equation} (\varphi,|\Delta_x u|^2)+(\varphi |u|^{r-1},|\nabla_x u|^2)\le C(1+\|g\|^2_{L^2}), \end{equation}

which gives the interior $H^2$-regularity.

Step 2. Boundary regularity: tangential directions. We now look at a small neighbourhood of the boundary $\partial \Omega$. We introduce a family of orthogonal vector fields $\tau ^1$, $\tau ^2$ and $n$ in $\bar \Omega$ which are non-degenerate near the boundary and such that $\tau ^i\big |_{\partial \Omega }$ are tangent vector fields and $n\big |_{\partial \Omega }$ is a normal vector field. Note that $n(x)$ is globally defined, but $\tau ^i$ are in general only locally defined, so being pedantic we need to use the cut-off procedure in order to localize them. In order to avoid technicalities, we however will assume that they are globally defined. Let $\tau =(\tau _1,\,\tau _2,\,\tau _3)$ be one of the vectors $\tau ^1$ or $\tau ^2$. Then, we may define the corresponding differential operators

\[ \partial_\tau u:=\sum_{i=1}^3\tau_i\partial_{x_i}u,\quad \partial^*_\tau u={-}\sum_{i=1}^3\partial_{x_i}(\tau_i u). \]

Our idea is to multiply equation (2.1) by $\partial _\tau ^*\partial _\tau u$ and get the $H^1$-norm of $\partial _\tau u$ analogously to Step 1 using the fact that $\partial _\tau u\big |_{\partial \Omega }=0$. As before, we start with the most complicated term related with the pressure:

\[ (\nabla_x p, \partial_\tau^*(\partial_\tau u))=(\partial_\tau\nabla_x p,\partial_\tau u)=(\nabla_x(\partial_\tau p),\partial_\tau u)+([\partial_\tau,\nabla_x]p,\partial_\tau u). \]

Here and below, we denote by $[D_1,\,D_2]$ the commutator of two differential operators $D_1$ and $D_2$, i.e. $[D_1,\,D_2]u:=D_1(D_2u)-D_2(D_1)u$. Let us first estimate the term with the commutator using integration by parts and the fact that $u$ is divergence free:

(2.15)\begin{align} ([\partial_\tau,\nabla_x]p,\partial_\tau u)& =\sum_j(\sum_i\tau_i\partial_{x_i}\partial_{x_j}p- \partial_{x_j}(\tau_i\partial_{x_i}p),\sum_i\tau_i\partial_{x_i}u_j)\nonumber\\ & =\sum_j(\sum_i\partial_{x_j}\tau_i\partial_{x_i}p,\sum_i\tau_i\partial_{x_i}u_j)\nonumber\\ & ={-}\sum_j\left(\sum_k\partial_{x_k}(\tau_k\sum_i\partial_{x_j}\tau_i\partial_{x_i}p),u_j\right) \nonumber\\ & ={-}\left(\sum_{i,j,k}\partial_{x_k}\tau_k\partial_{x_j}\tau_i\partial_{x_i}p,u_j\right)-\left(\sum_{i,j,k}\tau_k\partial_{x_j,x_k}\tau_i\partial_{x_i}p,u_j\right) \nonumber\\ & \quad -\left(\sum_{i,j,k}\tau_k\partial_{x_j}\tau_i\partial_{x_i}(\tau_k\partial_{x_k}p),u_j\right) +\left(\sum_{i,j,k}\partial_{x_i}\tau_k\partial_{x_j}\tau_i\partial_{x_i}p,u_j\right)\nonumber\\ & =(A(x)\nabla_x(\partial_\tau p),u)+(B(x)\nabla_x p, u) \end{align}

for some smooth (and therefore bounded) matrices $A$ and $B$. The remaining term is analogous, but even simpler

\[ (\nabla_x(\partial_\tau p),\partial_\tau u)={-}(\partial_\tau p,[\partial_\tau,\operatorname{div}]u)=(A_1(x)\nabla_x(\partial_\tau p),u)+(B_1(x)\nabla_x p,u) \]

Thus, we have proved the following key relation:

(2.16)\begin{equation} (\nabla_x p,\partial_\tau^*\partial_\tau u)=(\widetilde A(x)\nabla_x (\partial_\tau p),u)+(\widetilde B(x)\nabla_x p,u) \end{equation}

for some smooth matrices $\widetilde A(x)$ and $\widetilde B(x)$. We now turn to the term containing the Laplacian:

(2.17)\begin{align} (\Delta_x u,\partial_\tau^*\partial_\tau u)& =(\partial_\tau\Delta_x u,\partial_\tau u)=(\Delta_x(\partial_\tau u),\partial_\tau u)\nonumber\\ & \quad +([\partial_\tau ,\Delta_x]u,\partial_\tau u)={-}\|\nabla_x(\partial_\tau u)\|^2_{L^2}+([\partial_\tau ,\Delta_x]u,\partial_\tau u). \end{align}

It only remains to estimate the term containing commutator. It is a sum of second and first order differential operators, so integrating by parts again in the second order part, we arrive at

\[ |([\partial_\tau ,\Delta_x]u,\partial_\tau u)|\le \beta\|\nabla_x(\partial_\tau u)\|^2_{L^2}+C_\beta\|\nabla_x u\|^2_{L^2}\le \beta\|\nabla_x(\partial_\tau u)\|^2_{L^2}+C_\beta\|g\|^2_{L^2} \]

and, therefore, we end up with the estimate

(2.18)\begin{equation} -(\Delta_x u,\partial_\tau^*\partial_\tau u)\ge\frac12\|\nabla_x(\partial_\tau u)\|^2_{L^2}-C\|g\|^2_{L^2}. \end{equation}

The non-linear term is estimated in a standard way:

\[ (f(u),\partial_\tau^*\partial_\tau u)=(\partial_\tau f(u),\partial_\tau u)=(f'(u)\partial_\tau u,\partial_\tau u)\ge \kappa(|u|^{r-1},|\partial_\tau u|^2) \]

and combining the obtained estimates together, we arrive at

(2.19)\begin{equation} \|\nabla_x(\partial_\tau u)\|^2_{L^2}+(|u|^{r-1},|\partial_\tau u|^2)\le C(1+\|g\|^2_{L^2})+C|(\widetilde A(x)\nabla_x(\partial_\tau p),u)|. \end{equation}

In order to estimate the term with pressure in the right-hand side, we differentiate equation (2.1) along $\tau$ and write the obtained expression in the following form:

(2.20)\begin{align} \Delta_x(\partial_\tau u)-\nabla_x(\partial_\tau p)& =\{\partial_\tau f(u)\}+\{\partial_\tau g\}\nonumber\\ & \quad +\{[\Delta_x,\partial_\tau] u+\{[\nabla_x,\partial_\tau]p\}= : h_1+h_2+h_3, \nonumber\\ \operatorname{div} (\partial_\tau u)& =[\operatorname{div},\partial_\tau] u:=H,\quad \partial_\tau u\big|_{\partial\Omega}=0. \end{align}

From estimate (2.3), we know that

\[ \|h_3\|_{L^q}\le C\|g\|_{L^q}, \]

where we have used that $[\partial _\tau,\,\Delta _x]$ and $[\partial _\tau,\,\operatorname {div}]$ are second and first order operators respectively. Also $\|H\|_{W^{1,q}}\le C\|g\|_{L^q}$. We now decompose

\[ \partial_\tau u=v_1+v_2+v_3,\quad \partial_\tau p=p_1+p_2+p_3, \]

where

(2.21)\begin{equation} \Delta_x(v_i)-\nabla_x p_i=h_i,\quad \operatorname{div} v_i=H_i, \quad v_i\big|_{\partial\Omega}=0, \end{equation}

where $H_3=H$ and $H_1=H_2=0$. Then, due to the $L^q$ and $H^{-1}$ regularity estimates for the Stokes operator, see e.g. [Reference Galdi11], we have

\[ \|p_2\|_{L^2}\le C\|g\|_{L^2}, \quad \|\nabla_x p_3\|_{L^q}\le C\|g\|_{L^q}, \quad \|\nabla_x p_1\|_{L^q}\le C\|\partial_\tau f(u)\|_{L^q}. \]

Inserting these estimates to (2.19), we arrive at

(2.22)\begin{align} & \|\nabla_x(\partial_\tau u)\|^2_{L^2}+(|u|^{r-1},|\partial_\tau u|^2)\le C(1+\|g\|^2_{L^2})\nonumber\\ & \quad +C\|u\|_{L^{r+1}}\|\partial_\tau f(u)\|_{L^q}\le C(1+\|g\|^2_{L^2})+C\|\partial_\tau f(u)\|_{L^q}^q. \end{align}

It only remains to estimate the last term via the Hölder inequality:

(2.23)\begin{align} \|\partial_\tau f(u)\|_{L^q}^q& \le (|u|^{q(r-1)},|\partial_\tau u|^q)\\ & =\left(\left(|u|^{(r-1)/2}|\partial_\tau u|\right)^{q},|u|^{q(r-1)/2}\right)\nonumber\\ \end{align}

(2.24)\begin{align} & \le (|u|^{r-1},|\partial_\tau u|^2)^{\frac q2}\left(|u|^{r+1},1\right)^{1-\frac q2}\le \beta(|u|^{r-1},|\partial_\tau u|^2)+C_\beta\|u\|_{L^{r+1}}^{r+1}, \end{align}

where we have used that $1-\frac q2=1-\frac {r+1}{2r}=\frac {r-1}{2r}$ and $q(r-1)\frac {2r}{r-1}=2(r+1)$. Inserting the obtained estimate to the right-hand side of (2.22), we end up with the desired estimate for tangential derivatives:

(2.25)\begin{equation} \|\nabla_x(\partial_\tau u)\|^2_{L^2}+(|u|^{r-1},|\partial_\tau u|^2)\le C(1+\|g\|^2_{L^2}). \end{equation}

Step 3. Interpolation and regularity in normal directions. Let $U:=\nabla _x u$. Then, estimates and (2.3) and (2.25) give us

(2.26)\begin{equation} \|U\|_{W^{1,q}}+\|\partial_\tau U\|_{L^2}\le C(1+\|g\|_{L^2}). \end{equation}

Recall that we are now doing estimates in a small neighbourhood of the boundary only (inside of the domain $\Omega$ we already have better control of the $H^2$-norm of $u$ from the proved interior estimate). Thus, using the partition of unity, we may fix a small neighbourhood of containing a piece of the boundary and the straighten it by an appropriate diffeomorphism in such a way that in new coordinates $y=(y_1,\,y_2,\,y_3)$. The direction $y_1$ corresponds to the normal direction $\vec n$ and the variables $y_2$, and $y_3$ correspond to the tangent directions. Then, estimate (2.26) reads

(2.27)\begin{equation} \|\partial_{y_1}U\|_{L^q([0,1]^3)}+\|\partial_{y_2}U\|_{L^2([0,1]^3)}+\|\partial_{y_3}U\|_{L^2([0,1]^3)}\le C(1+\|g\|_{L^2}). \end{equation}

After that we may use the interpolation for anisotropic Sobolev spaces:

(2.28)\begin{align} & W^{1,q}((0,1),L^q(0,1)^2)\cap L^2((0,1),W^{1,2}(0,1)^2) \nonumber\\ & \quad \subset W^{\theta,s}((0,1),W^{1-\theta,s}(0,1)^2)\subset L^{3q}((0,1),L^{3q}(0,1)^2), \end{align}

where

\[ \frac1s=\frac\theta q+\frac{1-\theta}2,\ \frac1{3q}=\frac1s-\frac\theta1=\frac1s-\frac{1-\theta}2, \]

so $\theta =\frac 13$, see e.g. [Reference Rákosnik34] or [Reference Besov, Ilin and Nikolski5]. Thus, reminding that $U=\nabla _x u$,we end up with the desired partial regularity estimate

(2.29)\begin{equation} \|\nabla_x u\|_{L^{3q}}\le C\|u\|_{W^{2,q}}^{1/3}\|\nabla_x(\partial_\tau u)\|^{2/3}_{L^2}\le C(1+\|g\|_{L^2}). \end{equation}

Indeed, inside of the domain $\Omega$, we have better estimate with the exponent $6$ instead of $3q$ due to the proved interior regularity and estimate (2.29) in a small neighbourhood of the boundary follows from the interpolation mentioned above. Thus, estimate (2.6) is proved and we only need to check (2.7). To this end, we first note that, due to the interior estimate (2.14), our assumptions on $f$ and the Sobolev embedding $H^1\subset L^6$, we have

(2.30)\begin{align} \|f(u)\|_{L^{3q}(\Omega_\varepsilon)}^q& \le C\left(1+\|u\|^{r+1}_{L^{3(r+1)}(\Omega_\varepsilon)}\right)\nonumber\\ & \le C_1\left(1+\|u\|_{L^{r+1}(\Omega_\varepsilon)}^{r+1}+\|\nabla_x (u^{(r+1)/2})\|^2_{L^2(\Omega)}\right)\nonumber\\ & \le C_2\left(1+\|g\|^q_{L^q(\Omega)}+(\varphi f'(u)\nabla_x u,\nabla_x u)\right)\le C_3\left(1+\|g\|^2_{L^2(\Omega)}\right). \end{align}

Thus, we only need to look at a small neighbourhood near the boundary. Using the refined Sobolev estimate

\[ \|v\|_{L^6([0,1]^3)}^3\le C\|\partial_{y_1}v\|_{L^2([0,1]^3)}\|\partial_{y_2}v\|_{L^2([0,1]^3)}\|\partial_{y_3}v\|_{L^2([0,1]^3)} \]

which holds for every $v\in H^1_0((0,\,1)^3)$ and taking $v=|u|^{(r+1)/2}$, analogously to (2.30), we get

(2.31)\begin{align} & \|f(u)\|^q_{L^{3q}(\Omega)}\le C\left(1+\|g\|^q_{L^q(\Omega)}\right.\nonumber\\ & \quad + \left. (f'(u)\partial_n u,\partial_n u)^{1/3}(f'(u)\partial_{\tau_1} u,\partial_{\tau_1} u)^{1/3}(f'(u)\partial_{\tau_2} u,\partial_{\tau_2} u)^{1/3}\right). \end{align}

Combining this estimate with (2.25), we end up with the desired estimate (2.7) and finish the proof of the theorem.

Proof. Using estimates (2.3), (2.7) and the interpolation inequality, we infer that

(2.33)\begin{align} \|f(u)\|^2_{L^2}& \le\|f(u)\|_{L^q}^{\frac{3q-2}2}\|f(u)\|_{L^{3q}}^{\frac{3(2-q)}2} \nonumber\\ & \le C(1+\|g\|_{L^q})^{\frac{3q-2}2}\left(1+\|g\|^q_{L^q}+(1+\|g\|_{L^2}^2)^{\frac23}(f'(u)\nabla_x u,\nabla_x u)^{\frac13}\right)^{\frac{3(2-q)}{2q}}. \end{align}

Since $1< q<2$, using the Young inequality, we may rewrite the last inequality in the form

(2.34)\begin{align} \|f(u)\|^2_{L^2}& \le C_1(1+\|g\|_{L^q})^2 \nonumber\\ & \quad +(1+\|g\|_{L^q})^{\frac{3q-2}2} (1+\|g\|_{L^2})^{\frac{2(2-q)}q}(f'(u)\nabla_x u,\nabla_x u)^{\frac{2-q}{2q}}\nonumber\\ & \le C(1+\|g\|^2_{L^2})+ C_\nu(1+\|g\|_{L^q})^q(1+\|g\|_{L^2})^{\frac{4(2-q)}{3q-2}}+\nu(f'(u)\nabla_x u,\nabla_x u), \end{align}

where $\nu >0$ is arbitrary. Treating now the term $f(u)$ in (2.1) as external forces and applying the maximal $L^2$-regularity estimate to the linear Stokes equation, we arrive at

(2.35)\begin{align} & \|u\|^2_{H^2}+\|\nabla_x p\|^2_{L^2}\le C\|g\|^2_{L^2} \nonumber\\ & \quad + C_\nu(1+\|g\|_{L^q})^q(1+\|g\|_{L^2})^{\frac{4(2-q)}{3q-2}}+\nu(f'(u)\nabla_x u,\nabla_x u). \end{align}

Moreover, from the equation (2.1) we have

(2.36)\begin{align} \|f(u)\|^2_{L^2}& \le \|u\|^2_{H^2}+\|\nabla_x p\|^2_{L^2}+\|g\|^2_{L^2}\le C\|g\|^2_{L^2}\nonumber\\ & \quad + C_\nu(1+\|g\|_{L^q})^q(1+\|g\|_{L^2})^{\frac{4(2-q)}{3q-2}}+\nu(f'(u)\nabla_x u,\nabla_x u). \end{align}

Finally, in order to estimate the last term in the right-hand side, we multiply equation (2.1) by $\Delta _x u$ and integrate over $x\in \Omega$. This gives

(2.37)\begin{align} \|\Delta_x u\|^2_{L^2}+(f'(u)\nabla_x u,\nabla_x u)& \le(\|g\|_{L^2}+\|\nabla_x p\|_{L^2})\|\Delta_x u\|_{L^2} \nonumber\\ & \le \|g\|^2_{L^2}+\|\Delta_x u\|^2_{L^2}+\|\nabla_x p\|^2_{L^2}\le \|\Delta_x u\|^2_{L^2} \nonumber\\ & \quad +C\|g\|^2_{L^2}+C_\nu(1+\|g\|_{L^q})^q(1+\|g\|_{L^2})^{\frac{4(2-q)}{3q-2}}\nonumber\\ & \quad +\nu(f'(u)\nabla_x u,\nabla_x u). \end{align}

Fixing $\nu >0$ small enough, we deduce that

\[ (f'(u)\nabla_x u,\nabla_x u)\le C\|g\|^2_{L^2}+C(1+\|g\|_{L^q})^q(1+\|g\|_{L^2})^{\frac{4(2-q)}{3q-2}}, \]

which together with (2.35) and (2.36) finish the proof of the corollary.

Proof. We first note that, due to the uniqueness of weak solutions, it is enough to construct a regular solution $u\in H^2(\Omega )$ for any $g\in L^2(\Omega )$ and this will automatically exclude the existence of weak non-smooth solutions. Since the nonlinear localization seems to not work on the level of Galerkin approximations (at least we do not know how to obtain uniform estimates for higher norms), we need to use the alternative methods. For instance, we may use the continuation by the parameter methods based on the Leray–Schauder degree theory. Indeed, let us consider the family of equations of the form (2.1) depending on a parameter $\varepsilon \in [0,\,1]$:

(2.39)\begin{equation} \Delta_x u-\nabla_x p-\varepsilon f(u)=g,\quad u\big|_{\Omega}=0. \end{equation}

Using the Leray–Helmholtz projector $\Pi$ and the invertibility of the Stokes operator $A:=\Pi \Delta$, we may rewrite equation (2.39) in the equivalent form

(2.40)\begin{equation} u=\varepsilon A^{{-}1}\Pi f(u)+A^{{-}1}\Pi g. \end{equation}

Let us assume that $g\in H^1$ and consider equation (2.40) as an equality in $H^3(\Omega )$. Then, due to the embedding $H^3(\Omega )\subset C^1(\Omega )$ and the $H^1\to H^3$ regularity of the solutions of the linear Stokes operator, see [Reference Galdi11], the right-hand side of (2.40) is a compact and continuous operator in $H^3(\Omega )$, so the Leray–Schauder degree theory works and we only need to obtain the uniform with respect to $\varepsilon \in [0,\,1]$ estimates for the $H^3$-solutions of (2.40) or, which is the same, $H^3$-solutions of (2.39).

To get this estimate, we note that now we a priori know that the solution $u=u_\varepsilon$ is in $H^3$ and the corresponding pressure $p_\varepsilon \in H^2$ (by the regularity of the Leray–Helmholtz projector). Since all the above estimates related with nonlinear localizations obviously hold for such solutions, we may apply corollary 2.3 to equation (2.39), which gives us the estimate of the $H^2$-norm of $u$. Moreover, it is not difficult to see that this estimate is unform with respect to $\varepsilon$, so we end up with

\[ \|u_\varepsilon\|_{H^2}\le Q(\|g\|_{L^2}), \]

where the function $Q$ is independent of $\varepsilon$. After that, using the embedding $H^2(\Omega )\subset C(\Omega )$, we get the control of the $H^1$-norm of $f(u)$ and, due to the regularity of the Stokes operator, we finally get the desired uniform a priori estimate in the $H^3$-norm. Thus, we have proved the existence of $H^3$-solution of problem (2.1) under the extra condition that $g\in H^1$. Since the $H^2$-norm estimate of $u$ depends only on the $L^2$-norm of $u$, we may approximate a given external force $g\in L^2$ by a sequence $g_n\in H^1$, construct the associated solution $u_n$ and pass to the limit $n\to \infty$. This will give us the desired $H^2$-solution and finish the proof of the corollary.

Proof. We first derive estimate (2.42) assuming that $u$ is smooth enough, e.g. $u\in H^2$. To this end, we interpret equation (2.41) and (2.1) with the right-hand side $\widetilde g:=g+(u,\,\nabla _x)u$ and apply estimate (2.32) to it. In addition, analyzing the proof of this estimate, we see that the term containing $\|\widetilde g\|_{L^q}$ in the right-hand side of it comes from the corresponding estimate of $\|f(u)\|_{L^q}$ and this control we already proved for (2.41), so we may replace $\widetilde g$ by $g$ in this part of the estimate and arrive at

(2.43)\begin{align} & \|u\|^2_{H^2}+\|\nabla_x p\|_{H^1}^2 \nonumber\\ & \quad \le C(1+\|g\|_{L^q})^q(1+\|\widetilde g\|_{L^2})^s\le Q(\|g\|_{L^2})(1+\|\,|u|\cdot|\nabla_x u|\,\|_{L^2})^s. \end{align}

Using the appropriate interpolation inequality together with (2.4), we estimate

\[ \|\,|u|\cdot|\nabla_x u|\,\|_{L^2}\le \|u\|_{L^\infty}\|u\|_{H^1}\le C\|u\|_{H^1}^{3/2}\|u\|_{H^2}^{1/2}\le C(1+\|g\|_{L^2})^{3q/4})\|u\|_{H^2}^{1/2} \]

and, therefore,

\[ \|u\|^2_{H^2}+\|\nabla_x p\|_{H^1}^2\le Q(\|g\|_{L^2})(1+\|u\|^{s/2}_{H^2}). \]

Since $s=\max \{2,\,\frac {4(2-q)}{3q-2}\}<4$ for $1\le q\le 2$, the last estimate implies (2.42). Thus, the $H^2$ a priori bound for a sufficiently smooth solution of (2.41) is obtained. The existence of such a solution can be verified, e. g. using again the Leray–Schauder degree theory, see the proof of corollary 2.4, and we only need to check that any weak solution is actually smooth.

In contrast to corollary 2.4, we do not expect the uniqueness of a weak solution for problem (2.41) due to the presence of the non-monotone inertial term. In order to overcome this problem, we add an artificial term $Lu$ with big $L$ in both sides of equation (2.41) and consider the auxiliary problem

(2.44)\begin{equation} \Delta_x v-\nabla_x p-f(v)-Lv-(v,\nabla_x v)=g-Lu:=\bar g, \end{equation}

where $u$ is a fixed weak solution of (2.41). Then, since $\bar g\in L^2$ and the new nonlinearity $f(v)+Lv$ satisfies all of the assumptions posed on $f$, this problem possesses a smooth solution $v\in H^2$. On the other hand, $u$ is still a weak solution of this problem and we only need to check that $u=v$. This will be true if $L$ is large enough to compensate the non-monotonicity of the inertial term. Indeed, let $w=u-v$. Then, this function solves

\[ \Delta_x w-\nabla_x \bar p-[f(w+v)-f(v)]-[(u,\nabla_x)w-(w,\nabla_x)v]-Lw=0. \]

Multiplying this equation by $w$, integrating by $x\in \Omega$ and using the monotonicity of $f$ together with the cancellation $((u,\,\nabla _x)w,\,w)=0$, we arrive at

(2.45)\begin{align} \|\nabla_x w\|^2+L\|w\|^2& \le |(w,\nabla_x)v,w)|\le \|\nabla_x v\|_{L^2}\|w\|_{L^4}^2 \nonumber\\ & \le C_1\|v\|_{H^1}\|w\|^{1/2}_{L^2}\|w\|_{H^1}^{3/2}\le C_2\|v\|^4_{H^1}\|w\|^2_{L^2}+\|\nabla_x w\|^2_{L^2} \end{align}

and we see that the uniqueness holds if $L>C_2\|v\|^4_{H^1}$. This finishes the proof of the corollary.

Proof. We multiply equation (3.1) by $u$ and integrate over $x\in \Omega$ to get

(3.6)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\|u(t)\|^2_{L^2}+\|\nabla_x u(t)\|^2_{L^2}+(f(u(t)),u(t))=(g(t),u(t)). \end{equation}

Then, using assumption (3.4) together with the Hölder and Poincare inequalities, we arrive at

(3.7)\begin{align} & \frac{{\rm d}}{{\rm d}t}\|u(t)\|^2_{L^2}+\alpha\|u(t)\|^2_{L^2} \nonumber\\ & \quad +\alpha(\|u(t)\|^2_{H^1}+ \|u(t)\|_{L^{r+1}}^{r+1}+\|f(u(t))\|_{L^q}^q)\le C(1+\|g\|_{L^q}^q), \end{align}

where $C$ and $\alpha$ are positive constants which are independent of $t$ and $u$. Integrating this inequality, we have

(3.8)\begin{align} & \|u(t)\|^2_{L^2} \nonumber\\ & \quad +\alpha\int_0^te^{-\alpha(t-s)}\left(\|u(s)\|^2_{H^1}+ \|u(s)\|_{L^{r+1}}^{r+1}+\|f(u(s))\|_{L^q}^q\right)\,{\rm d}s \nonumber\\ & \le \|u(0)\|^2_{L^2}e^{-\alpha t}+C\left(\alpha^{{-}1}+\int_0^te^{-\alpha(t-s)}\|g(s)\|_{L^q}^q\,{\rm d}s\right). \end{align}

Thus, the basic energy estimate is proved (note that the assumption $r\ge 3$ is nowhere used here). To complete estimate (3.5), we need to rewrite equation (3.1) in the form of a linear non-stationary Stokes equation:

(3.9)\begin{equation} \partial_t u-\Delta_x u+\nabla_x p=g(t)-f(u(t))-(u(t),\nabla_x)u(t):=g_u(t) \end{equation}

and to apply the standard $L^q$-maximal regularity estimate to this equation. This gives

(3.10)\begin{align} & \|u(t)\|^q_{W^{2(1-1/q),q}}\nonumber\\ & \qquad +\int_0^te^{-\alpha(t-s)}\left(\|\partial_t u(s)\|^q_{L^q}+\|u(s)\|^q_{W^{2,q}}+\|\nabla_x p(s)\|^q_{W^{1,q}}\right)\,{\rm d}s\nonumber\\ & \quad \le C\|u(0)\|^q_{W^{2(1-1/q),2}}e^{-\alpha t}+C\int_0^te^{-\alpha(t-s)}\|g_u(s)\|^q_{L^q}\,{\rm d}s, \end{align}

where $C$ and $\alpha$ are some constants and $|\alpha |$ is small enough, see e.g. [Reference Solonnikov35]. So, it only remains to estimate the norm of $g_u(s)$ in the right-hand side. Moreover, the term containing $f(u)$ is already estimated and we only need to estimate the inertial term. To this end, we need the assumption $r\ge 3$ which allows us to use the Hölder inequality in the form

(3.11)\begin{align} \|(u,\nabla_x)u\|_{L^q}^q& \le \|u\|_{L^{\frac{2q}{2-q}}}^q\|\nabla_x u\|_{L^2}^q \nonumber\\ & \le\|\nabla_x u\|^2_{L^2}+\|u\|_{L^{\frac{2q}{2-q}}}^{\frac{2q}{2-q}}\le C(1+\|\nabla_x u\|^2_{L^2}+\|u\|_{L^{r+1}}^{r+1}), \end{align}

since $\frac {2q}{2-q}=\frac {2(r+1)}{r-1}\le r+1$ if $r\ge 3$. Thus, the inertial term is also controlled by (3.8) and this estimate together with (3.10) give the desired estimate (3.5) and finish the proof of the proposition.

Proof. Indeed, the existence of a solution follows in a standard way from estimate (3.8) using, e.g. Galerkin approximations. The uniqueness of a solution is known for $r>3$ only. First of all, we use this assumption in order to check that the inertial term is in $L^q$, see (3.11). After that, we justify the multiplication of equation (3.1) by $u$ as well as the multiplication of for difference of two weak solutions $u$ and $v$ by $u-v$ and this gives us the following identity for the difference $w:=u-v$:

(3.12)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\|w\|^2_{L^2}+\|\nabla_x w\|^2_{L^2}+([f(w+v)-f(v)],w)+ ((w,\nabla_x)v,w)=0, \end{equation}

see [Reference Gallbally and Zelik10, Reference Kalantarov and Zelik16] for the details. Then, using the integration by parts, we estimate the inertial term as follows

\[ |((w,\nabla_x)v,w)|\le (|w|\cdot |v|,|\nabla_x w|)\le \|\nabla_x w\|^2+(|w|^2,|v|^2). \]

Moreover, using assumption (3.4) on the nonlinearity, we arrive at

\[ (f(u)-f(v),w)\ge \kappa'(|u|^{r-1}+|v|^{r-1},v^2)\ge (|v|^2,|w|^2)-C_{r,\kappa}\|w\|^2_{L^2} \]

for some positive constants $\kappa '$ and $C_{r,\kappa }$ (here we have used that $r>3$ again). Inserting these estimates into (3.12), we have

\[ \frac12\frac{{\rm d}}{{\rm d}t}\|w\|^2_{L^2}\le C_{r,\kappa}\|w\|^2_{L^2} \]

and the Gronwall inequality gives us that $w=0$ which proves the uniqueness. Finally, if $u_0\in \mathcal {H}\cap W^{2(1-1/q),q}(\Omega )$, we may apply the maximal $L^q$-regularity estimate (3.10) to the linear Stokes equation (3.9) and verify that the unique weak solution (3.9) satisfies (3.5). This finishes the proof of the proposition.

Proof. Of course, the analogue of the smoothing property (3.14) holds for all $t>0$ with the function $Q$ depending also on $1/t$ and we state the smoothing property for $t\ge 1$ just in order to avoid extra technicalities.

We first assume that $u$ is a sufficiently smooth solution of (3.1), for instance,

(3.15)\begin{equation} u\in C_{loc}([0,\infty),H^2(\Omega))\cap L^2_{loc}([0,\infty),H^3(\Omega)), \end{equation}

and derive the desired estimates for it. We divide the proof on several steps.

Step 1. Interior estimates. This step is almost identical to Step1 in the proof of theorem 2.2. Indeed, let $\varphi$ be the same as in that step. Multiplying equation (3.1) by $-\sum _{i}\partial _{x_i}(\varphi \partial _{x_i}u)$, we arrive at

(3.16)\begin{align} & \frac{{\rm d}}{{\rm d}t}(\varphi,|\nabla_x u|^2)+(\varphi,|\Delta_x u|^2)+\kappa(|u|^{r-1}\varphi,|\nabla_x u|^2) \nonumber\\ & \quad \le C(1+\|g(t)\|^2_{L^2}+\|\nabla_x u\|^2_{L^2}+\|u\|^{r+1}_{L^{r+1}}+\|\nabla_x p\|^q_{L^q}) \nonumber\\ & \qquad +2(\Delta_x p,\nabla_x\varphi\cdot u)-2((u,\nabla_x)u,\operatorname{div}(\varphi\nabla_x u)). \end{align}

Thus, we just have an extra term in the right-hand side of (3.16) which is related with the inertial term which should be properly estimated and also we now have

\[ \Delta_x p={-}\operatorname{div} f(u)-\operatorname{div} g-\operatorname{div}(u,\nabla_x)u \]

with the extra term related with the divergence of the inertial term (in comparison with (2.9)). Both of the extra terms are not difficult to control. Indeed,

(3.17)\begin{align} |((u,\nabla_x)u,\operatorname{div}(\varphi\nabla_x u))|& \le C(|u|\cdot |\nabla_x u|,\varphi|\Delta_x u|)\nonumber\\ & \quad +C(|u|,\varphi^{1/2}|\nabla_x u|^2)|\le \frac14(\varphi,|\Delta_x u|^2)\nonumber\\ & \quad +\frac14\kappa(\varphi |u|^{r-1},|\nabla_x u|^2)+C\|\nabla_x u\|^2_{L^2}. \end{align}

The extra term in the expression for the Laplacian of pressure $p$, after inserting it into the right-hand side of (3.16) gives us the term

\[ |(\operatorname{div}(u,\nabla_x)u,\nabla_x\varphi\cdot u)|=|((u,\nabla_x) u, \nabla_x (\nabla_x\varphi\cdot u))|, \]

which can be estimated exactly as in (3.17). Combining all of the estimates together, we arrive at

(3.18)\begin{align} & \frac{{\rm d}}{{\rm d}t}(\varphi,|\nabla_x u|^2)+\frac12(\varphi,|\Delta_x u|^2)+\frac\kappa2(|u|^{r-1}\varphi,|\nabla_x u|^2) \nonumber\\ & \quad \le C(1+\|g(t)\|^2_{L^2}+\|\nabla_x u\|^2_{L^2}+\|u\|^{r+1}_{L^{r+1}}+\|\nabla_x p\|^q_{L^q}). \end{align}

Finally, integrating this relation in time and using (3.5), we arrive at the desired interior estimate

(3.19)\begin{align} & (\varphi,|\nabla_x u(t)|^2) \nonumber\\ & \qquad +\alpha\int_0^t e^{-\alpha(t-s)}\left((\varphi,|\Delta_x u(s)|^2)+(|u|^{r-1}\varphi,|\nabla_x u|^2)\right)\,{\rm d}s \nonumber\\ & \quad \le C\|u_0\|^2_{\mathcal{V}}e^{-\alpha t}+C\left(1+\int_0^te^{-\alpha(t-s)}\|g(s)\|^2_{L^2}\,{\rm d}s\right). \end{align}

for some positive constants $C$ and $\alpha$.

Step 2. Boundary regularity: tangential directions. Analogously to § 2, we multiply equation (3.1) by $\partial ^*_\tau \partial _\tau u$ and integrate over $x\in \Omega$. Then, in comparison with (2.19), we will have an extra good term $1/2 {{\rm d}}/{{\rm d}t}\|\partial _\tau u\|^2_{L^2}$ as well as the term $((u,\,\nabla _x)u,\,\partial _\tau ^*\partial _\tau u)$ related with the extra inertial term, which can be estimated as follows:

(3.20)\begin{align} & |((u,\nabla_x)u,\partial_\tau^*\partial_\tau u)|\le||((u,\nabla_x)\partial_\tau u, \partial_\tau u)| \nonumber\\ & \quad +C(|u|\cdot|\partial_\tau u|,|\nabla_x u|)\le |((\partial_\tau u,\nabla_x) u,\partial_\tau u)|+C(|u|^2,|\partial_\tau u|^2)+C\|\nabla_x u\|^2_{L^2}. \end{align}

Furthermore, integrating by parts, we get

(3.21)\begin{align} |((\partial_\tau u,\nabla_x) u,\partial_\tau u)|& \le C|(|\partial_\tau u|\cdot |u|,|\nabla_x(\partial_\tau u)|)|\nonumber\\ & \le \varepsilon\|\nabla_x(\partial_\tau u)\|^2+C_\varepsilon(|u|^2,|\partial_\tau u|^2). \end{align}

Thus, using again that $r>3$, we arrive at

\[ |((u,\nabla_x)u,\partial_\tau^*\partial_\tau u)|\le \varepsilon\left(\|\nabla_x(\partial_\tau u)\|^2_{L^2}+(|u|^{r-1},|\partial_\tau u|^2)\right)+C_\varepsilon\|\nabla_x u\|^2_{L^2}, \]

where $\varepsilon >0$ is arbitrary, and, therefore, this extra term is under the control and, arguing as in the stationary case, we arrive at

(3.22)\begin{align} & \frac{{\rm d}}{{\rm d}t}\|\partial_\tau u\|^2_{L^2}+\alpha\left(\|\nabla_x(\partial_\tau u)\|^2_{L^2}+(|u|^{r-1},|\partial_\tau u|^2)+\|\partial_\tau f(u)\|^q_{L^q}\right) \nonumber\\ & \quad \le C(1+\|g(t)\|_{L^2}^2)+ C\|\nabla_x u(t)\|^2_{L^2}+C\|u(t)\|^{r+1}_{L^{r+1}}+C|(A(x)\nabla_x(\partial_\tau p),u)|. \end{align}

Integrating this inequality in time and using (3.8), after the standard transformations, we arrive at

(3.23)\begin{align} & \sup_{s\in[0,t]}\{e^{-\beta(t-s)}\|\partial_\tau u(s)\|^2\}+\kappa\int_0^te^{-\beta(t-s)}\left(\|\nabla_x(\partial_\tau u(s))\|^2_{L^2}\right. \nonumber\\ & \quad \left.+(|u(s)|^{r-1},|\partial_\tau u(s)|^2)+\|\partial_\tau f(u(s))\|^q_{L^q}\right){\rm d}s\le C\|u_0\|^2_{\mathcal{V}}e^{-\beta t}+C + \nonumber\\ & \quad +C\int_0^te^{-\beta(t-s)}\|g(s)\|^2_{L^2}\,{\rm d}s+C\int_0^te^{-\beta(t-s)}|(A(x)\nabla_x(\partial_\tau p(s)),u(s))|\,{\rm d}s, \end{align}

where $\kappa$, $\beta$ and $C$ are some positive constants, which are independent of $t$ and $u$. Thus, we only need to estimate the term, containing pressure. To this end, we introduce a function $G=G(t)$ as a solution of the linear Stokes equation:

\[ \partial_t G-\Delta_x G+\nabla_x p_G=g(t),\quad G\big|_{t=0}=u_0,\quad \operatorname{div} G=0,\quad G\big|_{\partial\Omega}=0. \]

Then, using the $L^2$-maximal regularity estimate for the linear Stokes equation, see [Reference Solonnikov35], we end up with

(3.24)\begin{align} & \|G(t)\|^2_{L^2} \nonumber\\ & \qquad +\int_0^te^{-\beta_1(t-s)}\left(\|\partial_t G(s)\|^2_{L^2}+\|G(s)\|^2_{H^2}+\|\nabla_x p_G(s)\|^2_{L^2}\right)\,{\rm d}s\nonumber\\ & \quad \le C\|u_0\|_{\mathcal{V}}^2e^{-\beta_1(t-s)}+C\int_0^te^{-\beta_1(t-s)}\|g(s)\|^2_{L^2}\,{\rm d}s \end{align}

for some positive constants $\beta _1>\beta$, and $C$ (and we also have the $L^q$-version of this estimate). We also introduce a new function $v:=u-G$ which also solves the linear Stokes equation:

(3.25)\begin{equation} \partial_t v-\Delta_x v+\nabla_x p_v={-}f(u)-(u,\nabla_x)u, \quad v\big|_{t=0}=0, \quad \operatorname{div} v=0. \end{equation}

Differentiating this equation with respect to $\tau$ and denoting $w:=\partial _\tau v$ and $p_w:=\partial _\tau p_v$, we arrive at

(3.26)\begin{align} \partial_t w-\Delta_x w+\nabla_x p_w& =\partial_\tau f(u)-\partial_\tau (u,\nabla_x)u-[\Delta_x,\partial_\tau]v \nonumber\\ & \quad +[\nabla_x,\partial_\tau]p_v,\ \ w\big|_{\partial\Omega}=0,\quad w\big|_{t=0}=0,\nonumber\\ & \qquad\operatorname{div} w=[\operatorname{div},\partial_\tau]v:=H(t). \end{align}

Our plan is to apply the $L^q$-maximal regularity estimate to this linear non-homogeneous Stokes equation. Indeed, from estimates (3.5) and (3.24), we only know that

(3.27)\begin{align} \|H(t)\|_{W^{1,q}}+\|\partial_t H(t)\|_{W^{{-}1,q}}& \le C(\|u(t)\|_{W^{2,q}}+\|\partial_t u(t)\|_{L^q}\nonumber\\ & \quad +\|\partial_t G(t)\|_{L^q}+\|G(t)\|_{W^{2,q}}) \end{align}

and the $L^q$-norm in time from the left-hand side is under the control. However, this is not enough in general to get the maximal $L^q$-regularity estimate (in general, we need $\partial _t H$ to belong to $L^q(0,\,t;L^q)$, see e. g. [Reference Filinov and Shilkin9] for the counterexample). Fortunately, the function $H(t)$ in (3.26) has a special structure which allows us to overcome this problem. Namely, it is not difficult to see that

\[ H(t)=\operatorname{div} W(t)+h(t), \quad W_i:={-}v\cdot \nabla_x \tau_i, \quad h:=v\cdot \nabla_x\operatorname{div}\tau. \]

Important is that $W\big |_{\partial \Omega }=0$. Therefore, we may subtract the function $W$ from the solution $w$ and get a new linear Stokes problem for the function $\bar w:=w-W$:

(3.28)\begin{align} \partial_t \bar w-\Delta_x \bar w+\nabla_x p_w& =\partial_\tau f(u)-\partial_\tau (u,\nabla_x)u-[\Delta_x,\partial_\tau]v \nonumber\\ & \quad +[\nabla_x,\partial_\tau]p_v-g_W:=g_{\bar w},\nonumber\\ & \qquad w\big|_{\partial\Omega}=0, \quad w\big|_{t=0}=0, \quad \operatorname{div} w=h(t), \end{align}

where $g_W:=\partial _t W-\Delta _x W$. Since both $W$ and $h$ are proportional to $v$, we have the control of the $L^q$-norms of $G_W$ and $\partial _t h$ from (3.5). Moreover, since

\[ \|[\Delta_x,\partial_\tau]v\|_{L^q}\le C\|v\|_{W^{2,q}}, \quad \|[\nabla_x,\partial_\tau] p_v\|_{L^q}\le C\|p_v\|_{W^{1,q}} \]

and $p=p_G+p_v$, all terms with commutators are also controlled by (3.5). We actually need not to estimate $\partial _\tau f(u)$ since this term is presented in the left-hand side of (3.23) and will be finally cancelled out. However, we still need to estimate the most complicated term related with the inertial term in (3.28), but we prefer to postpone this estimate and first complete the exclusion of pressure. To this end, we apply the $L^q$-regularity estimate to problem (3.28), see [Reference Filinov and Shilkin9] and get

(3.29)\begin{align} & \int_0^te^{-\beta_1(t-s)}\|\nabla_x p_w(s)\|_{L^q}^q\,{\rm d}s\nonumber\\ & \quad \le C\int_0^t e^{-\beta_1(t-s)}\left(\|g_{\bar w}(s)\|^q_{L^q}+\|h(s)\|^q_{W^{1,q}}+\|\partial_t h(s)\|^q_{L^q}\right)\,{\rm d}s\nonumber\\ & \quad \le C\int_0^t e^{-\beta_1(t-s)}\left(\|\partial_\tau f(u(s))\|^q_{L^q}+\|\partial_\tau(u(s),\nabla_x)u(s))\|^q_{L^q}\right)\,{\rm d}s\nonumber\\ & \qquad + C\int_0^te^{-\beta_1(t-s)}\left(\| v(s)\|^q_{W^{2,q}}+\|\partial_t v(s)\|^q_{L^q}+\|\nabla_x p_v(s)\|_{L^q}^q\right)\,{\rm d}s. \end{align}

Using the obvious estimate

(3.30)\begin{equation} \int_0^te^{-\beta_1(t-s)}\left(\int_0^se^{-\alpha(s-\tau)}|U(\tau)|\,{\rm d}\tau\right)\,{\rm d}s\le C^*\!\!\int_0^te^{-\beta_1(t-s)}|U(s)|\,{\rm d}s, \end{equation}

where $\alpha >\beta _1>0$ and the constant $C^*$ depends only on $\alpha$ and $\beta$, together with estimate (3.5) and the $L^q$-version of estimate (3.24), we arrive at

(3.31)\begin{align} & \int_0^te^{-\beta_1(t-s)}\|\nabla_x p_w(s)\|_{L^q}^q\,{\rm d}s \nonumber\\ & \quad \le C\int_0^t e^{-\beta_1(t-s)}\left(\|\partial_\tau f(u(s))\|^q_{L^q}+\|\partial_\tau(u(s),\nabla_x)u(s))\|^q_{L^q}\right)\,{\rm d}s \nonumber\\ & \qquad + C\|u_0\|_{\mathcal{V}}e^{-\beta_1 t}+C\left(1+\int_0^te^{-\beta_1(t-s)}\| g(s)\|^q_{L^q}\,{\rm d}s\right). \end{align}

We are now ready to return to estimation of the last term in (3.23). Namely,

(3.32)\begin{align} |(A(x)\nabla_x(\partial_\tau p),u)|& \le |(\nabla_x(\partial _\tau p_G),A^*(x)u)|+|(A(x)\nabla_x p_w,u)|| \nonumber\\ & \le C\left(\|u\|^2_{H^1}+\|\nabla_x p_G\|_{L^2}^2\right)+\nu\|\nabla_x p_w\|^q_{L^q}+C_\nu\|u\|^{r+1}_{L^{r+1}}, \end{align}

where $\nu >0$ is arbitrary. Using the obtained estimates (3.31) and (3.24) together with (3.30), we exclude the pressure from (3.23) and get

(3.33)\begin{align} & \sup_{s\in[0,t]}\big\{e^{-\beta(t-s)}\|\partial_\tau u(s)\|^2\big\}+\int_0^te^{-\beta(t-s)}\left(\|\nabla_x(\partial_\tau u(s))\|^2_{L^2}\right. \nonumber\\ & \quad \left.+(|u(s)|^{r-1},|\partial_\tau u(s)|^2)+\|\partial_\tau f(u(s))\|^q_{L^q}\right){\rm d}s\le C_\nu\|u_0\|^2_{\mathcal{V}}e^{-\beta t}+C_\nu \nonumber\\ & \quad +C_\nu\int_0^te^{-\beta(t-s)}\|g(s)\|^2_{L^2}\,{\rm d}s+ \nu \int_0^te^{-\beta(t-s)}\|\partial_\tau((u(s),\nabla_x)u(s))\|_{L^q}^q\,{\rm d}s. \end{align}

Moreover, the last term in this estimate can be further simplified. Namely,

(3.34)\begin{equation} \partial_\tau(u,\nabla_x)u=(\partial_\tau u,\nabla_x)u+(u,\nabla_x)\partial_\tau u+(u,[\nabla_x,\partial_\tau])u \end{equation}

and

\[ \|(u,\nabla_x)\partial_\tau u\|_{L^q}^q\le \|\nabla_x(\partial_\tau u)\|_{L^2}^q\|u\|_{L^{\frac{2q}{2-q}}}^q\le \nu\|\nabla_x(\partial_\tau u)\|^2_{L^2}+C_\nu\|u\|_{L^{\frac{2q}{2-q}}}^{\frac{2q}{2-q}}. \]

Since $\frac {2q}{q-2}=\frac {2(r+1)}{r-1}< r+1$ if $r>3$, this term is under the control. The third term in the right-hand side of (3.34) can be estimated analogously using the fact that the commutator $[\nabla _x,\,\partial _\tau ]$ is a first order differential operator. So, we only need to estimate the first term. We will do this with the help of (2.29), the Hölder inequality and the fact that $\frac {3q}2\le 2$, namely,

(3.35)\begin{align} & \|\,|\partial_\tau u|^q\cdot|\nabla_x u|^q\|_{L^1}\le \|\partial_\tau u\|^q_{L^{3q/2}}\|u\|_{W^{1,3q}}^q \nonumber\\ & \quad \le C\|\partial_\tau u\|_{L^2}^q\|\nabla_x(\partial_\tau u)\|_{L^2}^{2q/3}\|\Delta_x u\|_{W^{2,q}}^{q/3}\le \|\nabla_x(\partial_\tau u)\|^2_{L^2}\nonumber\\ & \qquad +C\|\partial_\tau u\|_{L^2}^{\frac{3q}{3-q}}\|u\|^{\frac q{3-q}}_{W^{2,q}}\le \|\nabla_x(\partial_\tau u)\|^2_{L^2}+C\|\partial_\tau u\|_{L^2}^{\frac{5q-4}{3-q}}\|\partial_\tau u\|_{L^2}^{\frac{2(2-q)}{3-q}}\|u\|_{W^{2,q}}^{\frac q{3-q}}. \end{align}

Crucial for us is the fact that $\frac {5q-4}{3-q}<2$ for $q<\frac {10}7$ and therefore, due to our assumptions, $q<\frac 43<\frac {10}7$, so the number $m:=2- \frac {5q-4}{3-q}$ is always positive. In addition, the first term in the right-hand side of (3.35) is not dangerous since it is absorbed by the corresponding term in the left hand side of (3.33), so we only need to estimate the integral

\[ I:=\int_0^te^{-\beta(t-s)}\|\partial_\tau u(s)\|_{L^2}^{\frac{5q-4}{3-q}}\|\nabla_x u(s)\|_{L^2}^{\frac{2(2-q)}{3-q}}\|u(s)\|_{W^{2,q}}^{\frac q{3-q}}\,{\rm d}s. \]

To this end, we use that the exponents $\frac {2(2-q)}{3-q}$ and $\frac q{3-q}$ are such that, due to the Young inequality

\[ \|\nabla_x u(s)\|_{L^2}^{\frac{2(2-q)}{3-q}}\|u(s)\|_{W^{2,q}}^{\frac q{3-q}}\le \|\nabla_x u(s)\|^2_{L^2}+\|u(s)\|^{q}_{W^{2,q}} \]

and, therefore,

(3.36)\begin{align} I& \le C\left(\sup_{s\in[0,t]}\big\{e^{-\beta(t-s)}\|\partial_\tau u(s)\|^2_{L^2}\big\}\right)^{1-m/2} \nonumber\\ & \quad \times \int_0^te^{-\beta(t-s)/2}\left(\|\nabla_x u(s)\|^2_{L^2}+\|u(s)\|^q_{W^{2,q}}\right)\,{\rm d}s\nonumber\\ & \le \nu\sup_{s\in[0,t]}\big\{e^{-\beta(t-s)}\|\partial_\tau u(s)\|^2_{L^2}\big\}\nonumber\\ & \quad+C_\nu\left(\int_0^te^{-\beta(t-s)/2}\left(\|\nabla_x u(s)\|^2_{L^2}+\|u(s)\|^q_{W^{2,q}}\right)\,{\rm d}s\right)^{\frac2m}. \end{align}

Inserting this estimate to the right-hand side of (3.33) and using estimate (3.5) with $\alpha =\beta /2$, we finally end up with the following estimate:

(3.37)\begin{align} & \sup_{s\in[0,t]}\big\{e^{-\beta(t-s)}\|\partial_\tau u(s)\|^2\big\}+\int_0^te^{-\beta(t-s)}\left(\|\nabla_x(\partial_\tau u(s))\|^2_{L^2}\right. \nonumber\\ & \qquad \left.+(|u(s)|^{r-1},|\partial_\tau u(s)|^2)+\|\partial_\tau f(u(s))\|^q_{L^q}\right){\rm d}s\nonumber\\ & \le C\left(\|u_0\|^2_{\mathcal{V}}e^{-\beta t/2}+1 +\int_0^te^{-\beta(t-s)/2}\|g(s)\|^2_{L^2}\,{\rm d}s\right)^{\widetilde m}, \end{align}

where $\widetilde m:=\max \{1,\,2/m\}$. This finishes the boundary regularity estimate in tangential directions.

Step 3. Key interpolation estimate. Namely, we start with the Gagliardo–Nirenberg inequality:

\[ \|u\|_{L^\infty}\le C\|u\|_{L^{r+1}}^{1/4}\|u\|_{W^{1,3q}}^{3/4}, \]

since $\frac 34-3\left (\frac 1{4(r+1)}+\frac 1{4q}\right )=0$, see e.g. [Reference Machihara and Ozawa25] or [Reference Adams and Fournier2]. This inequality, together with (2.29), give us

(3.38)\begin{align} \|u\|_{L^\infty}^2& \le C\|u\|_{L^{r+1}}^{1/2}\|u\|_{W^{2,q}}^{1/2}\|\nabla_x(\partial_\tau u)\|_{L^2} \nonumber\\ & \le C(\|u\|_{L^{r+1}}^{r+1}+\|u\|^q_{W^{2,q}}+\|\nabla_x(\partial_\tau u)\|^2_{L^2}), \end{align}

where we have used that $1=\frac 1{2(r+1)}+\frac 1{2q}+\frac 12$. Using estimates (3.5) and (3.37), we get the desired estimate

(3.39)\begin{align} & \int_0^te^{-\beta(t-s)}\|u(s)\|^2_{L^\infty}\,{\rm d}s \nonumber\\ & \quad \le C\left(\|u_0\|_{\mathcal{V}}^2e^{-\beta t/2}+\int_0^te^{-\beta(t-s)/2}\|g(s)\|^2_{L^2}\,{\rm d}s\right)^{\widetilde m}. \end{align}

Step 4. $\Phi$-energy estimate. Till that moment, we have nowhere used our extra assumption that the nonlinearity $f$ is gradient, but it is essentially used at this step. Indeed, we now multiply equation (3.1) by $\partial _t u$ and integrate over $x\in \Omega$. This gives

(3.40)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\frac12\|\nabla_x u\|^2_{L^2}+(F(u),1) + L \| u \|_{L^2}^2 \right) \nonumber\\ & \quad +\frac12\|\partial_t u\|^2_{L^2}={-}\frac12\|\partial_t u\|^2_{L^2} \nonumber\\ & \quad- ((u,\nabla_x) u,\partial_t u) +(g + 2Lu,\partial_t u)\le C(\|g\|^2_{L^2}+\|u\|_{L^2}^2)\nonumber\\ & \quad+ C \|u\|_{L^\infty}^2\left(\frac12\|\nabla_x u\|^2+(F(u),1) + L \|u \|_{L^2}^2\right), \end{align}

where $L$ is such that $F(u)+L|u|^2\ge 0$ (it exists due to assumption (3.4)). Note that assumption (3.4) implies also that

(3.41)\begin{equation} \kappa_2|u|^{r+1}-C_2\le F(u)\le \kappa_1|u|^{r+1}+C_1 \end{equation}

for some positive constants $\kappa _i$ and $C_i$. Therefore, estimate (3.39) allows us to apply the Gronwall inequality to (3.40) and to get the following control:

(3.42)\begin{align} & \|u(t)\|_{H^1}^2+\|u(t)\|^{r+1}_{L^{r+1}}\le Ce^{C\int_0^t\|u(s)\|_{L^\infty}^2\,{\rm d}s} \nonumber\\ & \quad \times\left(\|u(0)\|_{H^1}^2+\|u(0)\|^{r+1}_{L^{r+1}}+ \int_0^t\left(\|u(s)\|^2_{L^2}+\|g(s)\|^2_{L^2}\right)\,{\rm d}s\right). \end{align}

We will use this estimate for $0\le t\le 1$ only since it is growing in time even if $g(t)$ is bounded and, therefore, is not convenient for study the attractors. Combining (3.42) with (3.39) and (3.5), we arrive at the following estimate for $t\in [0,\,1]$:

(3.43)\begin{equation} \|u(t)\|_{H^1}^2\!+\!\|u(t)\|^{r+1}_{L^{r+1}}\le Q\!\left(\|u(0)\|_{H^1}^2\!+\!\|u(0)\|^{r+1}_{L^{r+1}}+\!\!\int_0^t\|g(s)\|^2_{L^2}\,{\rm d}s\right) \end{equation}

for some monotone increasing function $Q$. For $t\ge 1$, we will use the following smoothing estimate:

(3.44)\begin{equation} \|u(1)\|_{H^1}^2\!+\!\|u(1)\|^{r+1}_{L^{r+1}}\le Q\!\left(\|u(0)\|_{L^2}^2+\!\!\int_0^t\|g(s)\|^2_{L^2}\,{\rm d}s\right) \end{equation}

for some new monotone function $Q$. This estimate is a standard corollary of (3.43) and (3.5). Indeed, from (3.5), we know that

\[ \int_0^1\left(\|u(t)\|_{H^1}^2\!+\!\|u(t)\|^{r+1}_{L^{r+1}}\right)\,{\rm d}t\le C\left(\|u_0\|_{L^2}^2+1+\int_0^1\|g(t)\|^2_{L^2}\,{\rm d}t\right). \]

Therefore, there exists $t_0\in [0,\,1]$ (depending on $u$) such that

\[ \|u(t_0)\|_{H^1}^2\!+\!\|u(t_0)\|^{r+1}_{L^{r+1}}\, \le C\left(\|u_0\|_{L^2}^2+1+\int_0^1\|g(t)\|^2_{L^2}\,{\rm d}t\right). \]

Applying after that estimate (3.43) on the time interval $t\in [t_0,\,1]$, we arrive at (3.44). In turn, combining estimates (3.43) (on interval $t\in [0,\,1]$) with estimate (3.44) (which will give us the estimate of the $H^1\cap L^{r+1}$-norm of $u(t)$ through the $L^2$-norm of $u(t-1)$, $t\ge 1$) together with the dissipative estimate (3.5), we end up with the desired estimates (3.13) and (3.14). The estimate for the $L^2$-norm of $\partial _t u$ in them follows immediately by integrating (3.40) in time.

Step 5. $\Phi$-regularity of solutions. We recall that all previous estimates were derived assuming that $u(t)$ is a sufficiently smooth solution of (3.5), for instance, satisfying (3.15), will be enough to justify all of them (here we used that $H^2\subset C$ in 3D, so all terms related with the nonlinearity are under the control). To get such a regular solution, we approximate the external force $g$ and the initial data $u_0$ by the sequences $g_n$ and $u_0^n$ of smooth functions. Then, as proved in [Reference Kalantarov and Zelik16], there exist an $L^\infty _{loc}(\mathbb {R}_+,\,H^2)$ smooth solution $u_n(t)$ of problem (3.5) where the initial data $u_0$ and the external force $g$ are replaced by $u_0^n$ and $g_n$ respectively. Moreover, since $f\in C^1$, it is easy to see that $\partial _t f(u_n)\in L^2_{loc}(\mathbb {R}_+,\,L^2)$ and, therefore, the standard regularity result for the linear Stokes equation gives us that (3.15) are satisfied. For this reason, the solutions $u_n$ satisfy estimates (3.13) and (3.14) uniformly with respect to $n$. Passing to the limit $n\to \infty$, we see that the limit unique solution $u(t)$ of problem (3.1) also satisfies these estimates. Thus, theorem 3.3 is proved.

Proof. We rewrite equation (3.1) in the form of a stationary problem (2.1) at every fixed time $t$, namely,

(3.46)\begin{align} -\Delta_x u+\nabla_x p+f(u)+Lu& =\widetilde g(t):= \nonumber\\ & =g(t)-\partial_t u(t)-(u(t),\nabla_x)u(t)+L u(t),\nonumber\\ & \qquad \operatorname{div} u=0,\ u\big|_{\partial\Omega}=0. \end{align}

Moreover, due to the proved theorem, the $L^2_{loc}(\mathbb {R}_+,\,L^2)$-norm of $\widetilde g$ is under the control. For this reason, we may apply estimates (2.3) and (2.6) to this equation (recall that the modified non-linearity $\widetilde f(u):=f(u)+Lu$ satisfies (2.2). Then, integrating estimates (2.3) (in a power $2/q$) and (2.6), we arrive at

(3.47)\begin{align} & \int_{t-1}^t\|u(s)\|^2_{W^{2,q}}+\|u(s)\|^2_{W^{1,3q}}\,{\rm d}s \nonumber\\ & \quad \le Q\left(\|u_0\|_{\Phi}^2e^{-\beta t/2}+\int_0^te^{-\beta(t-s)/2}\|g(s)\|^2_{L^2}\,{\rm d}s\right). \end{align}

Thus, we have got the part of the desired estimate related with the $W^{1,3q}$-norm of $u$. In order to get the remaining part, we improve estimate (3.38) using that we now have estimate for the $L^2$-norm in time for $\|u(t)\|_{W^{2,q}}$ and also the $L^\infty$ norm in time for $\|u(t)\|_{L^{r+1}}$:

(3.48)\begin{align} \|u\|_{L^\infty}^{8/3}& \le C\|u\|_{L^{r+1}}^{2/3}\|u\|_{W^{2,q}}^{2/3}\|\nabla_x(\partial_\tau u)\|_{L^2}^{4/3} \nonumber\\ & \le C\|u\|_{L^{r+1}}^{2/3}(\|u\|^2_{W^{2,q}}+\|\nabla_x(\partial_\tau u)\|^2_{L^2}), \end{align}

This estimate, together with (3.13) and (3.47) completes the desired estimate (3.45) and finishes the proof of the corollary.

Proof. Note first of all that, without loss of generality, we may assume that $f'(u)\ge 0$, so $F(u)$ is convex. We use the following identities, which can be verified by straightforward computations:

(4.2)\begin{align} & \quad -(\Delta_x u(t),u(t+h)-u(t))+\frac12\|\nabla_x u(t+h)-\nabla_x u(t)\|^2_{L^2} \nonumber\\ & =\frac12\left(\|\nabla_x u(t+h)\|^2_{L^2}-\|\nabla_x u(t)\|^2_{L^2}\right)\nonumber\\ & ={-}(\Delta_x u(t+h),u(t+h)- u(t))-\frac12\|\nabla_x u(t+h)-\nabla_x u(t)\|^2_{L^2} \end{align}

and

(4.3)\begin{align} & (f(u(t+h)),u(t+h)-u(t))) \nonumber\\ & \qquad +\int_0^1(f'(su(t+h)+(1-s)u(t))(u(t+h)-u(t),u(t+h)-u(t))\,{\rm d}s\nonumber\\ & \quad =(F(u(t+h)),1)-(F(u(t)),1)\nonumber\\ & \quad =(f(u(t)),u(t+h)-u(t)))\nonumber\\ & \qquad -\int_0^1(f'(su(t+h)+(1-s)u(t))(u(t+h)-u(t),u(t+h)-u(t))\,{\rm d}s. \end{align}

Note that all terms in (4.2) and (4.3) make sense as functions from $L^1_{loc}(\mathbb {R}_+)$ and these identities and the regularity of a solution $u(t)$ is enough to justify them. Taking a sum of these two identities and using that $f'(u)\ge 0$, we end up with the following two-sided inequality:

(4.4)\begin{align} & (-\Delta_x u(t+h)+f(u(t+h)),u(t+h)-u(t)) \nonumber\\ & \le E(u(t+h))-E(u(t))\le (-\Delta_x u(t)+f(u(t)),u(t+h)-u(t)). \end{align}

Inserting the expression for $-\Delta _x u+f(u)$ from (3.1) to (4.4) and integrating over $t\in [S,\,T]$, we arrive at

(4.5)\begin{align} & \frac1h\int_S^{T}(H_u(t+h),u(t+h)-u(t))\,{\rm d}t\nonumber\\ & \quad \le \frac1h\int_T^{T+h}\!\!E(u(t))\,{\rm d}t-\frac1h\int_S^{S+h}\!\!E(u(t))\,{\rm d}t\le \frac1h\int_S^T\!(H_u(t),u(t+h)-u(t))\,{\rm d}t. \end{align}

We know that $H_u\in L^2_{loc}(\mathbb {R}_+,\,L^2)$, $E(u)\in L^1_{loc}(\mathbb {R}_+)$ and $u\in W^{1,2}(\mathbb {R}_+,\,L^2)$, therefore, we may pass to the limit $h\to 0$ in (4.5) and get that, for almost all $S,\,T\in \mathbb {R}_+$, $S\le T$, we have the integral identity

(4.6)\begin{equation} E(u(T))-E(u(S))=\int_S^T(H_u(t),\partial_t u(t))\,{\rm d}t. \end{equation}

To finish the proof of the theorem, we need to remove the condition that (4.6) holds for almost all $S\le T$ only. To this end, let us assume that the energy inequality

(4.7)\begin{equation} E(u(T))-E(u(S))\le\int_S^T(H_u(t),\partial_t u(t))\,{\rm d}t. \end{equation}

is already verified for all $0\le S< T$. We recall that $u(t)$ is weakly continuous as a function of time with values in $\Phi$, therefore the values of $E(u(t))$ are well-defined for all $t\in \mathbb {R}_+$. Moreover, since the function $u\to E(u)$ is convex, we have

(4.8)\begin{equation} E(u(t))\le \liminf_{s\to t}E(u(s)) \end{equation}

for all $t\in \mathbb {R}_+$. In particular, this property, together with (4.7) imply that the function $t\to E(u(t))$ is continuous from the right for all $t\in \mathbb {R}_+$. In turn, this ensures us that the energy equality (4.6) holds for all $0\le S< T$. Indeed, since (4.6) holds almost everywhere, we may find sequences $S_n\to S$, $S_n\ge S$, and $T_n\to T$, $T_n\ge T$ such that (4.6) holds for $S_n$ and $T_n$ for all $n$. Passing to the limit $n\to \infty$, we see that it holds for $S$ and $T$ as well. This finishes the proof of the theorem since the energy equality (4.6), which holds for all $S$ and $T$ is equivalent to the absolute continuity of the function $t\to E(u(t))$ and equality (4.1).

Thus, we only need to verify the energy inequality (4.7). Moreover, due to the uniqueness of a solution for problem (3.1), it is sufficient to verify it for $S=0$ only (the general case will follow just by replacing the initial time $t=0$ by $t=S$). To do this, we approximate the initial data $u_0\in \Phi$ and the external force $g\in L^2_{loc}(\mathbb {R}_+,\,L^2)$ by smooth functions $u_0^n$ and $g_n$ respectively. Let $u_n(t)$ be the corresponding solutions of (3.1). Then, analogously to the end of the proof of theorem 3.3, on the one hand, the solutions $u_n(t)\in C_{loc}(\mathbb {R}_+,\,H^2)$ and therefore, the energy equality (4.6) holds for them for every $T\ge 0$, i.e.

(4.9)\begin{equation} E(u_n(T))-E(u_n(0))=\int_0^T(H_{u_n}(t),\partial_t {u_n}(t))\,{\rm d}t. \end{equation}

On the other hand, the solutions $u_n$ satisfy the key estimates obtained in § 3 uniformly with respect to $n$. These estimates guarantee that, in particular, $u_n(t)\to u(t)$ weakly in $\Phi$ for all $t\ge 0$ and, by the choice of the initial data, we have the strong convergence at $t=0$. Therefore, without loss of generality we have

\[ E(u(0))=\lim_{n\to\infty}E(u_n(0)),\quad E(u(t))\le\lim_{n\to\infty}E(u_n(t)) \]

for all $t\in \mathbb {R}$. In order to pass to the limit in the term containing $H_{u_n}$, we note that we also know that $\partial _t u_n\to \partial _t u$ weakly in $L^2_{loc}(\mathbb {R}_+,\,L^2)$ and, therefore,

\[ \int_S^T\|\partial_t u(t)\|^2_{L^2}\,{\rm d}t\le \lim_{n\to\infty} \int_S^T\|\partial_t u_{n}(t)\|^2_{L^2}\,{\rm d}t. \]

Taking into account that $g_n\to g$ strongly in $L^2_{loc}(\mathbb {R}_+,\,L^2)$, it only remains to verify that

(4.10)\begin{equation} (u_n,\nabla_x)u_n\to (u,\nabla_x)u\ \text{strongly in}\ L^2_{loc}(\mathbb{R}_+,L^2). \end{equation}

Indeed, due to (3.45), the functions $u_n$ are uniformly bounded in the space $L^2_{loc}(\mathbb {R}_+,\,W^{1,3q})$ and their time derivatives are uniformly bounded in the space $L^2_{loc}(\mathbb {R}_+,\,L^2)$ (due to theorem 3.3). Therefore, without loss of generality, we have the strong convergence of $u_n$ in $L^2_{loc}(\mathbb {R}_+,\,W^{1-\varepsilon,3q})$ for all $\varepsilon >0$. Since $W^{1-\varepsilon,3q}\subset L^\infty$ if $\varepsilon >0$ is small enough, we have verified that $u_n\to u$ strongly in $L^2_{loc}(\mathbb {R}_+,\,L^\infty )$. Using also that $u_n$ is bounded in $L^{8/3}_{loc}(\mathbb {R}_+,\,L^\infty )$ (again by (3.45)), we conclude that

(4.11)\begin{equation} u_n\to u,\ \text{strongly in}\ L^{r_1}_{loc}(\mathbb{R}_+,L^\infty) \end{equation}

for all $2< r_1<\frac 83$. We now turn to the sequence $\nabla _x u_n$. According to estimate (3.5) and the compactness lemma, we conclude that $\nabla _x u_n\to \nabla _x u$ strongly in $L^q_{loc}(\mathbb {R}_+,\,W^{1-\varepsilon,q})$ for all $\varepsilon >0$. Combining this result with the uniform boundedness of $\nabla _x u_n$ in $L^2_{loc}(\mathbb {R}_+,\,L^{3q})$ (due to (3.45)) and using that $3q>2$, we end up with the strong convergence

(4.12)\begin{equation} \nabla_x u_n\to \nabla_x u,\ \text{strongly in}\ L^{r_2}_{loc}(\mathbb{R}_+,L^2) \end{equation}

for all $1< r_2<2$. Moreover, combining this result with the uniform boundedness of the sequence $\nabla _x u_n$ in $L^\infty (\mathbb {R}_+,\,L^2)$, we see that the convergence (4.12) holds for all $1\le r_2<\infty$. Fixing finally the exponents $r_1$ and $r_2$ in such a way that $\frac 1{r_1}+\frac 1{r_2}=2$ and using the convergences (4.11) and (4.12), we end up with the desired convergence (4.10). This convergence allows us to pass to the limit $n\to \infty$ in the energy equality (4.9) and get the desired inequality (4.7). Thus, the theorem is proved.

Proof. This fact is a standard corollary of the uniform convexity of the function $u\to E(u)$ as the function from $\Phi$ to $\mathbb {R}$ (without loss of generality, we assume that $f'(u)\ge \kappa |u|^{r-1}$ which gives us the desired uniform convexity) and the proved energy equality. Namely, let $t_n\to t$, $t_n\ge 0$, be an arbitrary sequence of times. Then, due to the weak continuity of $u(t)$ and the energy equality, we have

(4.13)\begin{equation} E(u(t))=\lim_{n\to\infty}E(u(t_n)) \quad \text{and} \quad u(t_n)\rightharpoondown u(t) \text{in}\ \Phi. \end{equation}

Since $u\to E(u)$ is uniformly convex, the above convergence imply that $u(t_n)\to u(t)$ strongly in $\Phi$. Although this fact seems to be well-known, see e.g. [Reference Zalinescu40, Reference Zhikov and Pastukhova43], for the convenience of the reader, we present its proof below.

For the quadratic part of the functional $E(u)$, we will use the following obvious identity:

\[ \|\nabla_x v\|^2_{L^2}-\|\nabla_x u\|^2_{L^2}=2(\nabla_x u,\nabla_x v-\nabla_x u)+\|\nabla_x u-\nabla_x v\|^2_{L^2}. \]

The analogue of this identity for the nonlinearity $F$ follows from the Taylor expansions near the point $u$, namely,

\[ F(v)-F(u)=f(u).(v-u)+\frac12 \int_0^1(1-s)f'(sv+(1-s)u)\,{\rm d}s(v-u).(v-u). \]

It is not difficult to verify that

\[ \int_0^1|su+(1-s)v|^{r-1}\,{\rm d}s\ge\alpha(|u|^{r-1}+|v|^{r-1})\ge\alpha_1|u-v|^{r-1} \]

for some positive $\alpha$ and $\alpha _1$. Therefore, putting $u=u(x)$, $v=v(x)$ in the above Taylor's expansion of the nonlinearity $F$ and integrating it with respect to $x$, we get the inequality

\[ (F(v),1)-(F(u),1)\ge (f(u),v-u)+\frac{\kappa\alpha_1}2\|u-v\|_{L^{r+1}}^{r+1}, \]

which holds for every $u,\,v\in L^{r+1}(\Omega )$. Combining this inequality with the identity for the quadratic part of $E_u$ mentioned above, we end up with the key inequality

(4.14)\begin{align} E(v)-E(u)& \ge (\nabla_x u,\nabla_x v-\nabla_x u)+(f(u),v-u) \nonumber\\ & \quad +\frac12\|\nabla_x u-\nabla_x v\|^2_{L^2}+\frac{\kappa\alpha_2}2\|u-v\|_{L^{r+1}}^{r+1}\nonumber\\ & \ge (\nabla_x u,\nabla_x v-\nabla_x u)+(f(u),v-u)+\alpha_3\|u-v\|^2_\Phi, \end{align}

where $\alpha _3$ is a positive constant. This inequality gives us the uniform convexity of $E(u)$ as well as the desired continuity of $u(t)$. Indeed, let us take $u=u(t)$ and $v=u(t_n)$ in this inequality. Then, we have

(4.15)\begin{align} & \|u(t_n)-u(t)\|_\Phi^2\le\alpha_3^{{-}1}\left(E(u(t_n))-E(u(t))\right. \nonumber\\ & \quad -\left.(\nabla_x u(t),\nabla_x u(t_n)-\nabla_x u(t))-(f(u(t)),u(t_n)-u(t))\right). \end{align}

It only remains to note that the convergence (4.13) implies that the right hand side of (4.15) tends to zero as $n\to \infty$, therefore, $u(t_n)\to u(t)$ strongly in $\Phi$. This finishes the proof of the corollary.

Proof. The statement of the theorem is an almost immediate corollary of theorem 3.3. Indeed, since

\[ \|\xi\|_{L^2_b(\mathbb{R},L^2(\Omega))}\le \|g\|_{L^2_b(\mathbb{R},L^2(\Omega))},\ \ \forall\xi\in\mathcal{H}(g), \]

estimate (3.13) implies that

(5.10)\begin{equation} \|\mathcal{S}_\xi(t)u_0\|_\Phi\le Q(\|u_0\|_\Phi)e^{-\beta t}+Q(\|g\|_{L^2_b}) \end{equation}

for some monotone increasing function $Q$ and a positive constant $\beta$ which are independent of $u_0\in \Phi$ and $t\in \mathbb {R}_+$. Estimate (5.10) guarantees that the set

(5.11)\begin{equation} \mathcal{B}:=\{u_0\in\Phi,\ \|u_0\|_\Phi\le 2Q(\|g\|_{L^2_b})\} \end{equation}

is a bounded uniformly attracting set for the cocycle $\mathcal {S}_\xi (t)$. Therefore, the existence of a weak uniform attractor $\mathcal {A}_{un}^w$ follows from proposition 5.3. The weak continuity of the maps $(u_0,\,\xi )\to \mathcal {S}_\xi (t)u_0$ for every fixed $t$ can be checked in a standard way (we leave the rigorous proof of this fact to the reader). This gives the representation formula (5.8) and finishes the proof of the theorem.

Proof. According to proposition 5.3, we only need to find a (pre)compact uniformly absorbing set for the cocycle $\mathcal {S}_\xi (t)$. We claim that the set

(5.14)\begin{equation} \mathcal{B}_1:=\cup_{\xi\in\mathcal{H}(g)}\mathcal{S}_\xi(1)\mathcal{B}, \end{equation}

where $\mathcal {B}$ is defined via (5.11) is a desired absorbing set. Indeed, due to estimate (5.10) this set is absorbing and bounded, so we only need to check its compactness. Let $v_n\in \mathcal {B}_1$ is an arbitrary sequence. Then there exists a sequence of the initial data $u_0^n\in \mathcal {B}$ and a sequence of right-hand sides $\xi _n\in \mathcal {H}(g)$ such that $v_n=u_n(1)$, where $u_n(t):=\mathcal {S}_{\xi _n}u_0^n$ is the corresponding sequence of solutions of problem (5.4). Since $\mathcal {B}$ and $\mathcal {H}(g)$ are compact in a weak topology, we may assume without loss of generality that

\[ u_0^n\rightharpoondown u_0 \ \text{ and }\ \xi_n\rightharpoondown\xi. \]

Moreover, since the maps $(u_0,\,\xi )\to \mathcal {S}_\xi (t)u_0$ are continuous in a weak topology for every fixed $t\ge 0$, we conclude that $u_n(t)\rightharpoondown u(t)$ in $\Phi$ for every fixed $t$, where $u(t):=\mathcal {S}_\xi (t)u_0$. In particular, $v_n\rightharpoondown u(1)$. Therefore, we only need to verify that this convergent is actually strong.

Arguing as in the proof of corollary 4.2, we see that we only need to verify that $E(u_n(1))\to E(u(1))$. In turn, in order to verify this convergence, we will use the energy identity (4.1) together with the trick suggested in [Reference Zelik41]. In order to avoid technicalities, we first give the proof for the particular case where $g$ is normal and after that indicate briefly the changes which should be made to cover the general case where $g$ is only weakly normal.

Namely, we multiply equation (5.4) for solutions $u_n$ by $Nu_n$, where $N$ is a big positive constant, integrate over $x\in \Omega$ and take a sum with (4.1) for $u_n$. This gives the following identity:

(5.15)\begin{align} & \frac{{\rm d}}{{\rm d}t}E(u_n(t))+2NE(u_n(t))\nonumber\\ & \quad+\|\partial_t u_n(t)\|^2_{L^2}+N(f(u_n).u_n-2F(u_n),1)\nonumber\\ & \quad+N(u_n(t),\partial_t u_n(t))= (\xi_n,\partial_t u_n+Nu_n)-((u_n,\nabla_x)u_n,\partial_t u_n). \end{align}

Multiplying this identity by $t$ and integrating over $t\in [0,\,1]$, we arrive at

(5.16)\begin{align} & E(u_n(1))+\int_0^1te^{{-}2N(1-t)}\|\partial_t u_n(t)\|^2_{L^2}\,{\rm d}t\nonumber\\ & \qquad+N\int_0^1te^{{-}2N(1-t)}(f(u_n(t)).u_n(t)-2F(u_n(t)),1)\,{\rm d}t\nonumber\\ & \quad=\int_0^1e^{{-}2N(1-t)}E(u_n(t))\,{\rm d}t+2\int_0^1te^{{-}2N(1-t)}((u_n(t),\nabla_x)u_n(t),\partial_t u_n(t))\,{\rm d}t\nonumber\\ & \qquad+N\int_0^1e^{{-}2N(1-t)}t(\xi_n(t),u_n(t))\,{\rm d}t+\int_0^1te^{{-}2N(1-t)}(\xi_n(t),\partial_t u_n(t))\,{\rm d}t\nonumber\\ & \quad-N\int_0^1te^{{-}2N(1-t)}(\partial_t u_n(t),u_n(t))\,{\rm d}t. \end{align}

We want to pass to the limit $n\to \infty$ in this identity. To this end, we note that by convexity arguments

\[ \int_0^1te^{{-}2N(1-t)}\|\partial_t u(t)\|^2_{L^2}\,{\rm d}t\le \liminf_{n\to\infty}\int_0^1te^{{-}2N(1-t)}\|\partial_t u_n(t)\|^2_{L^2}\,{\rm d}t. \]

To pass to the limit in the other terms, we recall that, due to estimates (3.5) and (3.45), the sequence $u_n$ is bounded in the space

\[ L^{8/3}(0,1;L^\infty(\Omega))\cap L^2(0,1;W^{1.3q}(\Omega))\cap H^1(0,1;L^2(\Omega))\cap L^\infty(0,1;L^{r+1}(\Omega)) \]

and therefore, since this space is compactly embedded to $L^2(0,\,1;W^{1,2}(\Omega ))\cap L^{r+1}(0,\,1;L^{r+1}(\Omega ))$, we have the strong convergence $u_n\to u$ in this space. This allows us to pass to the limit in the last term in the left-hand side of (5.16) as well as in the first and the last terms in the right-hand side there. Moreover, arguing analogously to the proof of theorem 4.1 (see (4.10)), we prove that $(u_n,\,\nabla _x)u_n\to (u,\,\nabla _x)u$ strongly in $L^2(0,\,1;L^2(\Omega ))$ and this allows us to pass to the limit in the second term in the right-hand side of (5.16). The third term is obvious since $u_n\to u$ strongly in $L^2(0,\,1;L^2(\Omega ))$ and the only problem is the last term in the right-hand side. In contrast to the other terms, we have here only weak convergence $\xi _n\to \xi$ and $\partial _t u_n\to \partial _t u$ in $L^2(0,\,1;L^2(\Omega ))$ and cannot use the convexity arguments. Instead, we prove that this term is small when $N$ is large uniformly with respect to $n$ and this will be enough for our purposes.

Lemma 5.8 Under the above assumptions, the following convergence holds:

(5.17)\begin{equation} \lim_{N\to\infty}\sup_{n\in\mathbb{N}}\int_0^1te^{{-}2N(1-t)}|(\xi_n(t),\partial_t u_n(t))|\,{\rm d}t=0.\end{equation}

Proof of the lemma Indeed, since $\partial _t u_n$ are uniformly bounded in the space $L^2(0,\,1;L^2(\Omega ))$, due to Cauchy-Schwarz inequality, we have

\[ \int_0^1te^{{-}2N(1-t)}|(\xi_n(t),\partial_t u_n(t))|\,{\rm d}t\le C\left(\int_0^1te^{{-}2N(1-t)}\|\xi_n(t)\|^2_{L^2}\,{\rm d}t\right)^{1/2}. \]

It only remains to recall that assumption (5.13) implies that

(5.18)\begin{equation} \lim_{N\to\infty}\sup_{t\in\mathbb{R}}\sup_{\xi\in\mathcal{H}(g)}\int_{-\infty}^te^{{-}2N(t-s)}\|\xi(s)\|^2_{L^2}\,{\rm d}s=0, \end{equation}

see [Reference Zelik41, Reference Zelik42] for the details. This finishes the proof of the lemma.