June 01, 2026
We study the flow of an isothermal compressible Newtonian fluid around a body that performs a (time-independent) rigid motion. We derive a weak-strong uniqueness principle, and show that in the low Mach number limit, the governing equation is well approximated by the Navier–Stokes equations for incompressible rotating flow. Both results are based on the derivation of a relative energy inequality for weak solutions to this exterior-domain problem.
We consider a rigid body \(\mathcal{S}\) that rotates with a fixed angular velocity \(\boldsymbol{\omega}\in\mathbb{R}^3\) in the three-dimensional space filled with a compressible viscous fluid. In a frame attached to the body, we denote the fluid domain by \(\Omega = \mathbb{R}^3 \setminus \overline{\mathcal{S}}\), which is the complement of the rigid body. The fluid flow in \(\Omega\) is then governed by the Navier–Stokes equations \[\begin{align} \label{NSE} \left\{ \begin{aligned} \partial_t \rho + \relax(\rho (\boldsymbol{u}- \boldsymbol{\omega}\times x)) &= 0 &&in(0,T)\times \Omega,\\ \partial_t (\rho\boldsymbol{u}) + \relax( \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \otimes \boldsymbol{u}) + \rho \boldsymbol{\omega}\times \boldsymbol{u}&= \rho \boldsymbol{f} + \relax\mathbb{S}(\nabla \boldsymbol{u}) - \nabla p(\rho) && in (0,T)\times \Omega,\\ \boldsymbol{u}&= \boldsymbol{\omega}\times x && on(0,T)\times \partial\Omega,\\ \rho(\cdot,x) \to \rho_\infty, \;\boldsymbol{u}(\cdot,x) &\to \boldsymbol{a}_\infty && as|x| \to \infty,\\ \rho(0,\cdot)=\rho_0, \;(\rho \boldsymbol{u})(0,\cdot) &= \rho_0\boldsymbol{u}_0 && in\Omega. \end{aligned} \right. \end{align}\tag{1}\] Here, \(T>0\) is a given time horizon, and \(\rho\colon(0,T)\times\Omega\to[0,\infty)\) and \(\boldsymbol{u}\colon(0,T)\times\Omega\to\mathbb{R}^3\) denote the unknown fluid density and velocity, respectively. We further prescribe an external force \(\boldsymbol{f}\colon(0,T)\times\Omega\to\mathbb{R}^3\), and the density \(\rho_\infty>0\) and the velocity \(\boldsymbol{a}_\infty\in\mathbb{R}^3\) at spatial infinity, where we assume that \(\boldsymbol{a}_\infty\) and \(\boldsymbol{\omega}\) are parallel. The viscous stress tensor \(\mathbb{S}(\nabla\boldsymbol{u})\) satisfies Newton’s rheological law \[\label{eq:stress} \mathbb{S}(\nabla\boldsymbol{u}) = \mu \Big( \nabla\boldsymbol{u}+ \nabla \boldsymbol{u}^T - \frac{2}{3} \relax\boldsymbol{u}\operatorname{Id}\Big) + \eta \relax\boldsymbol{u}\operatorname{Id},\tag{2}\] where \(\mu>0\) and \(\eta \geq 0\) denote shear and bulk viscosity coefficients, respectively. The pressure \(p(\rho)\) satisfies \[\label{assPress} p \in C^1([0,\infty)) \cap C^2((0,\infty)), \quad p(0)=0, \quad p'(\rho)>0 \;(\rho > 0), \quad \liminf_{\rho \to \infty} \frac{p'(\rho)}{\rho^{\gamma-1}}>0\tag{3}\] for some \(\gamma>1\). While the existence of weak solutions requires \(\gamma>\frac{3}{2}\), see also Theorem 3 below, we merely have to assume \(\gamma>\frac{6}{5}\) for the results established in this article, hence including a larger part of the physically relevant regime where \(\gamma \in (1, \frac{5}{3}]\).
The first two lines of 1 describe mass conservation and momentum balance, respectively. We assume that fluid particles are attached to the body, which is reflected by no-slip boundary conditions. The system is complemented by conditions at spatial infinity as well as initial conditions \(\rho_0\) and \(\rho_0\boldsymbol{u}_0\) for the density and the momentum, respectively. For more details on the physical motivation, the derivation of the model, and the transformation into the body frame, we refer the reader to [1]. See also [2] for the case of an incompressible fluid. Additional to the rotation of \(\mathcal{S}\), one could prescribe a time-independent translational motion with velocity \(\boldsymbol{\tau}\in\mathbb{R}^3\) parallel to \(\boldsymbol{\omega}\). By a simple change of frames, this is equivalent to considering 1 with \(\boldsymbol{a}_\infty\) replaced with \(\boldsymbol{a}_\infty-\boldsymbol{\tau}\). Therefore, this case is included in the setting treated here.
Concerning the mathematical theory of compressible fluids, the fundamental results on existence of global-in-time weak solutions in three dimensions were obtained by P.-L. Lions [3] in the barotropic case with \(p(\rho)= a\rho^\gamma\), \(a>0\), \(\gamma\geq\frac{9}{5}\), and by Feireisl, Novotný, Petzeltová [4], who provided an extension to exponents \(\gamma>\frac{3}{2}\). In particular, the latter result includes the physically relevant case \(\gamma=\frac{5}{3}\) of a monoatomic gas. While these works consider the flow in a bounded domain, the results have been generalized to a plethora of other configurations in the recent years. Several results and further references can be found in the monographs [5]–[7], where, among other cases, the flow around a body at rest, the temperature-dependent case, and the flow in moving domains are considered, respectively. Concerning the present case of the flow around a rigid body with prescribed translational and rotational motion, the existence of weak solutions to 1 was derived in [1]. An existence result for a system describing self-propelled motion in an unbounded three-dimensional domain can be found in [8].
The uniqueness of weak solutions to the compressible Navier–Stokes equations is only known under restrictive assumptions, see [9] for instance, but it is still an open question in the general case, even in a bounded domain. However, there are multiple results on weak-strong uniqueness, that is, the property that any weak solution coincides with the (hypothetical) strong solution with the same initial data as long as the latter exists. The first result in this respect is due to Germain [10], where the considered weak solutions satisfied a regularity assumption that could not be expected to hold in general. In [11], [12] this condition was omitted, and the weak-strong uniqueness property was instead established for the class of weak solutions constructed in [3], [4], which satisfy an energy inequality, so-called finite-energy weak solutions. Since then, this result has been generalized in various ways, see e.g. [13], [14] for the temperature-dependent case, or [15], [16] for the case with moving rigid body in a bounded domain filled by a compressible barotropic fluid with homogeneous or inhomogeneous boundary data. For the case of a prescribed moving domain filled by a compressible fluid, the weak-strong uniqueness was shown in [6]. One main result of the present article is to establish a weak-strong uniqueness principle for finite-energy weak solutions to 1 , see Theorem 9 below.
The weak-strong uniqueness results in all these works were derived from a relative energy inequality (REI). In the present case, the relative energy is given by \[\label{eq:relen} E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) = \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + P(\rho | \sigma)\tag{4}\] for suitable functions \((\rho,\boldsymbol{u})\) and \((\sigma, \boldsymbol{V})\), where the absolute and relative internal energy are defined as \[\begin{align} \label{eq:H46def} H(\rho) = \rho \int_1^\rho \frac{p'(z)}{z^2} \,\mathrm{d}z, && P(\rho | \sigma) = H(\rho) - H'(\sigma)(\rho - \sigma) - H(\sigma), \end{align}\tag{5}\] respectively. Due to \(p'>0\), the function \(H\) is strictly convex, and we have \(E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V})=0\) if and only if \((\rho,\boldsymbol{u})=(\sigma,\boldsymbol{V})\). In this way, the relative energy provides a natural distance measure. We derive a REI that describes (an inequality for) the evolution of \(E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V})\) when \((\rho,\boldsymbol{u})\) is a solution to 1 . More precisely, we consider finite-energy weak solutions \((\rho,\boldsymbol{u})\) to 1 and sufficiently smooth comparison functions \((\sigma,\boldsymbol{V})\), see Theorem 4 below.
Besides the weak-strong uniqueness principle, we also use the REI to study the limit of low Mach number, which means that the flow velocity is small compared to the speed of sound, which is the case in many real-world applications. In this setting, the fluid is nearly incompressible, and one would expect convergence towards the Navier–Stokes equations for incompressible fluids, in our framework given by \[\label{NSE-LM-Limit} \left\{\begin{align} \relax\boldsymbol{U}&= 0 && in(0,T)\times \Omega,\\ \rho_\infty \big( \partial_t \boldsymbol{U}+ \relax( (\boldsymbol{U}- \boldsymbol{\omega}\times x) \otimes \boldsymbol{U}) + \boldsymbol{\omega}\times \boldsymbol{U}\big) &= \rho_\infty \boldsymbol{f} + \relax\mathbb{S}(\nabla \boldsymbol{U}) - \nabla \Pi && in(0,T)\times \Omega,\\ \boldsymbol{U}&= \boldsymbol{\omega}\times x && on(0,T)\times \partial\Omega,\\ \boldsymbol{U}(\cdot,x) &\to \boldsymbol{a}_\infty && as|x| \to \infty,\\ \boldsymbol{U}(0,\cdot) &= \boldsymbol{U}_0 && in\Omega, \end{align}\right.\tag{6}\] where \(\boldsymbol{U}\colon(0,T)\times\Omega\to\mathbb{R}^3\) and \(\Pi\colon(0,T)\times\Omega\to\mathbb{R}\) are the velocity and pressure field of the incompressible fluid flow, respectively. A brief sketch of the proof of existence, further properties like asymptotic behavior, and an overview of the state-of-the-art theory on system 6 can be found in the recent monograph [17].
The mathematical study of the low Mach number limit for systems of equations describing a motion of fluids goes back to the seminal work of Klainerman and Majda [18]. In general, studying various types of singular limits allows to eliminate unimportant or unwanted features of the motion as a consequence of scaling and asymptotic analysis. The mathematical analysis of singular limits
in the frame of both strong and weak solutions was carried out in [19], [20] and [5], [21], respectively. We show that in the low Mach number limit, solutions \((\rho,\boldsymbol{u})\)
to 1 can be approximated by \((\rho_\infty,\boldsymbol{U})\), where \((\boldsymbol{U},\Pi)\) solves 6 , see Theorem 10 below. The analogous question in the case of pure rotation and with a gravitational force, partial slip boundary conditions, and additional low Froude number limit (strong
stratification) was studied in [22] using compactness arguments. In contrast to [22], we only consider so-called well-prepared initial data here, which prevents the occurrence of acoustic waves, but we allow the fluid
not to be at rest when \(|x| \to \infty\).
In Section 2 we introduce the notion of weak solutions and we recall the current existence results. We then derive a REI for these weak solutions in Section 3. This result is then
applied to prove a weak-strong uniqueness principle in Section 4, and to study the limit of vanishing Mach number in Section 5.
Throughout the whole article, we make the following assumptions: Let \(\mathcal{S}\subset\mathbb{R}^3\) be a bounded domain and \(\Omega = \mathbb{R}^3 \setminus \overline{\mathcal{S}}\) be an exterior domain with Lipschitz boundary \(\partial\Omega=\partial\mathcal{S}\). Without loss of generality, we assume \(0 \in {\rm int} \, \mathcal{S}\) and \(\overline{\mathcal{S}}\subset B_{1/2}(0)\). We assume that \(\boldsymbol{\omega},\boldsymbol{a}_\infty \in\mathbb{R}^3\) satisfy \(\boldsymbol{\omega}\times \boldsymbol{a}_\infty = 0\), and that \(\rho_\infty>0\). The viscous stress and the pressure are given by 2 and 3 with viscosity coefficients \(\mu>0\) and \(\eta\geq 0\), and with \(\gamma>\frac{6}{5}\). We consider an external force that satisfies \(\boldsymbol{f} \in L^\infty(0,T; [L^1 \cap L^\infty] (\Omega))\). For the initital data \((\rho_0, \boldsymbol{u}_0)\in L^1_{\mathrm{loc}}(\Omega)\times L^1_{\mathrm{loc}}(\Omega;\mathbb{R}^3)\) we assume that \[\rho_0 \geq 0,\qquad E(\rho_0, \boldsymbol{u}_0 | \rho_\infty, \boldsymbol{a}_\infty) \in L^1(\Omega),\] that is, finiteness of the initial (relative) energy.
We introduce the class of weak solutions to 1 satisfying an energy inequality. Since we study the problem in an unbounded domain, the formulation of the latter requires the consideration of a reference velocity \(\boldsymbol{U}_\infty:\mathbb{R}^3\to\mathbb{R}^3\) satisfying \[\label{Uinfty} \boldsymbol{U}_\infty\in C^\infty(\mathbb{R}^3;\mathbb{R}^3), \qquad \boldsymbol{U}_\infty(x)= \boldsymbol{\omega}\times x \quad \text{for }x\in\mathcal{S}, \qquad \boldsymbol{U}_\infty(x)= \boldsymbol{a}_\infty \quad\text{for }x\in\mathbb{R}^3\setminus B_1.\tag{7}\] Such a vector field can be constructed by standard methods. Observe that \(\relax\boldsymbol{U}_\infty = 0\) in \(\mathbb{R}^3 \setminus (B_1 \setminus \mathcal{S})\). Moreover, we have \(E(\rho, \boldsymbol{u}| \rho_\infty, \boldsymbol{a}_\infty) \in L^1(\Omega)\) if and only if \(E(\rho, \boldsymbol{u}| \rho_\infty, \boldsymbol{U}_\infty) \in L^1(\Omega)\).
Definition 1. We call a couple \((\rho, \boldsymbol{u})\) a finite-energy weak solution to system 1 with initial data \((\rho_0, \boldsymbol{u}_0)\) if it satisfies:
the following regularity assumptions: \[\label{eq:reg46weak} \begin{gather} \rho \geq 0, \quad \rho - \rho_\infty \in L^\infty(0,T; [L^\gamma + L^2] (\Omega)), \\ \rho |\boldsymbol{u}- \boldsymbol{a}_\infty|^2 \in L^\infty(0,T; L^1(\Omega)),\\ \boldsymbol{u}- \boldsymbol{a}_\infty \in L^2(0,T; W^{1,2}(\Omega)), \\ \boldsymbol{u}(t) |_{\partial\Omega} = \boldsymbol{\omega}\times x \text{ for a.e. } t \in (0,T). \end{gather}\tag{8}\]
the weak form of the continuity equation: we have \(\rho \in C_{weak}([0,T]; L^\gamma(K))\) for any compact \(K \subset \overline{\Omega}\), and \[\label{eq:cont46weak} \int_0^T \int_\Omega \rho \partial_t \phi + \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t = - \int_\Omega \rho_0 \,\phi(0, \cdot) \,\mathrm{d}x\tag{9}\] for any \(\phi \in C_c^1([0,T) \times \overline{\Omega})\).
the weak form of the momentum equation: we have \(\rho \boldsymbol{u}\in C_{weak}([0,T]; L^\frac{2\gamma}{\gamma+1}(K))\) for any compact \(K \subset \overline{\Omega}\), and \[\label{eq:mom46weak} \begin{align} \int_0^T \int_\Omega \rho \boldsymbol{u}\cdot \partial_t \psi &+ \rho ((\boldsymbol{u}- \boldsymbol{\omega}\times x) \otimes \boldsymbol{u}) : \nabla \psi - \rho (\boldsymbol{\omega}\times \boldsymbol{u}) \cdot \psi\,\mathrm{d}x \,\mathrm{d}t \\ & + \int_0^T \int_\Omega p(\rho) \relax\psi - \mathbb{S}(\nabla \boldsymbol{u}):\nabla \psi - \rho \boldsymbol{f} \cdot \psi \,\mathrm{d}x \,\mathrm{d}t = - \int_\Omega \rho_0 \boldsymbol{u}_0 \cdot \psi(0, \cdot) \,\mathrm{d}x \end{align}\tag{10}\] for any \(\psi \in C_c^1([0,T) \times \Omega; \mathbb{R}^3)\).
the energy inequality: there exists a reference velocity \(\boldsymbol{U}_\infty:\mathbb{R}^3\to\mathbb{R}^3\) with 7 such that for a.a. \(\tau\in[0,T]\) it holds \[\label{enIneq} \begin{align} &\left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{U}_\infty|^2 + P(\rho | \rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq - \int_0^\tau \int_\Omega p(\rho) \relax\boldsymbol{U}_\infty \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times \boldsymbol{U}_\infty) \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\tag{11}\]
Remark 2. At first glace, the energy inequality 11 only yields a bound on \(\nabla\boldsymbol{u}=\nabla(\boldsymbol{u}-\boldsymbol{a}_\infty)\) in \(L^2((0,T)\times\Omega)\), but not on the function \(\boldsymbol{u}-\boldsymbol{a}_\infty\) itself. However, using the finiteness of the total energy and the Poincaré inequality from [14], one can conclude \(\boldsymbol{u}- \boldsymbol{a}_\infty \in L^2(0,T; W^{1,2}(\Omega))\).
We have the following existence result:
Theorem 3. Let \(\gamma>\frac{3}{2}\). Then there exists a finite-energy weak solution to system 1 in the sense of Definition 1.
Proof. Weak solutions to 1 were constructed in [1] for the case \(\boldsymbol{f}=0\). However, it is evident that the same proof applies in the case \(\boldsymbol{f} \in L^\infty(0,T; [L^1 \cap L^\infty] (\Omega))\). For the validity of the energy inequality 11 , see also [1]. ◻
As mentioned above, although we cannot ensure the existence of solutions if \(\gamma\leq \frac{3}{2}\), all arguments in the subsequent sections will only make use of the weaker assumption \(\gamma>\frac{6}{5}\).
In this section, we derive a REI for weak solutions in the sense of Defintion 1. Recall the definition of the relative energy \(E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V})\) from 4 .
Theorem 4. Let \((\rho,\boldsymbol{u})\) be a finite-energy weak solution to 1 in the sense of Definition 1. Let \(0 < \underline{\sigma} \leq \rho_\infty \leq \overline{\sigma} < \infty\), and let \((\sigma,\boldsymbol{V})\in C^1((0,T)\times\Omega)\) be in the following regularity class: \[\label{regStr} \begin{gather} \sigma - \rho_\infty \in L^\infty(0,T; [L^2 \cap L^{\frac{\gamma}{\gamma-1}}](\Omega)), \qquad \underline{\sigma} \leq \sigma \leq \overline{\sigma}, \\ \nabla \sigma \in L^1(0,T; [L^2 \cap L^4 \cap L^\frac{2\gamma}{\gamma-1}] (\Omega)), \qquad \partial_t \sigma \in L^1(0,T; [L^2 \cap L^{\frac{\gamma}{\gamma-1}}] (\Omega)), \\ (\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x)\cdot\nabla \sigma \in L^1(0,T; [L^1 \cap L^2 \cap L^{\frac{\gamma}{\gamma-1}}] (\Omega)), \\ \boldsymbol{V}- \boldsymbol{a}_\infty \in L^\infty(0,T; [L^2 \cap L^4 \cap L^{\frac{2\gamma}{\gamma-1}}] (\Omega)), \\ \nabla \boldsymbol{V}\in L^1(0,T; [L^\frac{\gamma}{\gamma-1} \cap L^\infty] (\Omega)) \cap L^2((0,T)\times \Omega), \\ (\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x)\cdot\nabla\boldsymbol{V}\in L^1(0,T; [L^2 \cap L^4 \cap L^{\frac{2\gamma}{\gamma-1}}] (\Omega)), \\ \partial_t \boldsymbol{V}\in L^1(0,T; [L^2 \cap L^4 \cap L^{\frac{2\gamma}{\gamma-1}}] (\Omega)). \end{gather}\qquad{(1)}\] Further assume that \(\boldsymbol{V}= \boldsymbol{\omega}\times x\) on \((0,T)\times\partial\Omega\). Then \[\label{REI} \begin{align} &\left[ \int_\Omega E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}- \boldsymbol{V})) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{V}+ \boldsymbol{\omega}\times\boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}- \boldsymbol{f} ) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega ( p(\rho) - p(\sigma)) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega (\rho - \sigma) \partial_t H'(\sigma) + (\rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) - \sigma (\boldsymbol{V}- \boldsymbol{\omega}\times x)) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t \end{align}\qquad{(2)}\] for a.a. \(\tau\in(0,T)\).
Remark 5. To see where the assumptions ?? come from, let us rewrite the REI ?? in the following form, where we also used that \(\boldsymbol{\omega}\times \boldsymbol{a}_\infty = 0\): \[\begin{align} &\left[ \int_\Omega E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}- \boldsymbol{V})) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{a}_\infty - \boldsymbol{u}) \cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot ( \boldsymbol{\omega}\times (\boldsymbol{V}- \boldsymbol{a}_\infty) ) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{a}_\infty - \boldsymbol{u}) \cdot ( \boldsymbol{\omega}\times (\boldsymbol{V}- \boldsymbol{a}_\infty) ) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot (\rho(\boldsymbol{u}- \boldsymbol{a}_\infty) \cdot \nabla) \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{a}_\infty - \boldsymbol{u}) \cdot ((\boldsymbol{u}- \boldsymbol{a}_\infty) \cdot \nabla) \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot ((\boldsymbol{a}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{a}_\infty - \boldsymbol{u}) \cdot ((\boldsymbol{a}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot \boldsymbol{f} \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho (\boldsymbol{a}_\infty - \boldsymbol{u}) \cdot \boldsymbol{f} \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}) : \nabla (\boldsymbol{u}- \boldsymbol{a}_\infty) \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}) : \nabla (\boldsymbol{a}_\infty - \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega ( p(\rho) - p'(\rho_\infty)(\rho - \rho_\infty) - p(\rho_\infty) ) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega p'(\rho_\infty)[ (\rho - \rho_\infty) - (\sigma - \rho_\infty) ] \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega ( p(\sigma) - p'(\rho_\infty)(\sigma-\rho_\infty) - p(\rho_\infty) ) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega (\rho - \rho_\infty) H''(\sigma) \partial_t \sigma \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega (\rho_\infty - \sigma) H''(\sigma) \partial_t \sigma \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}- \boldsymbol{a}_\infty) \cdot H''(\sigma) \nabla \sigma \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega (\rho - \rho_\infty) (\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x) \cdot H''(\sigma) \nabla \sigma \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho_\infty (\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x) \cdot H''(\sigma) \nabla \sigma \,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega \sigma (\boldsymbol{V}- \boldsymbol{a}_\infty) \cdot H''(\sigma) \nabla \sigma \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \sigma (\boldsymbol{a}_\infty - \boldsymbol{\omega}\times x) \cdot H''(\sigma) \nabla \sigma \,\mathrm{d}x \,\mathrm{d}t . \end{align}\] Using the regularity of \((\rho, \boldsymbol{u})\) mentioned in 8 , which also gives us \(\sqrt{\rho}\in L^\infty(L^{2\gamma}+L^4+L^\infty)\) and hence \[\begin{align} \rho (\boldsymbol{u}- \boldsymbol{a}_\infty) = \sqrt{\rho}\sqrt{\rho} (\boldsymbol{u}- \boldsymbol{a}_\infty) \in L^\infty (L^{\frac{2\gamma}{\gamma+1}}+L^{4/3}+L^2), \end{align}\] leads to the regularity from ?? , where we also use the general facts \((L^p \cap L^q)' = L^{p'} + L^{q'}\) and \(L^p \cap L^q \subset L^p \subset L^p + L^q\).
In the case \(\boldsymbol{\omega}= 0\), formula ?? corresponds to the usual REI as can be found, for instance, in [11]. Note also that this inequality is independent of \(\boldsymbol{a}_\infty\) and thus of the choice of \(\boldsymbol{U}_\infty\).
To prove Theorem 4, we first consider the case that \((\sigma,\boldsymbol{V})\) differs from the limit values \((\rho_\infty,\boldsymbol{a}_\infty)\) only on a compact set.
Lemma 6. In addition to the assumptions from Theorem 4, assume that that there exists \(R>1\) such that \(\sigma(t,x)= \rho_\infty\) and \(\boldsymbol{V}(t,x) = \boldsymbol{a}_\infty\) for \(t\in(0,T)\) and \(|x|>R\). Then ?? holds for a.a. \(\tau\in(0,T)\).
Proof. We first use the momentum equation 10 tested by \(\boldsymbol{V}- \boldsymbol{U}_\infty\) to see \[\begin{align} &\left[ \int_\Omega \rho \boldsymbol{u}\cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} = \int_0^\tau \int_\Omega \rho \boldsymbol{u}\cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times\boldsymbol{u})\cdot (\boldsymbol{V}-\boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad+ \int_0^\tau \int_\Omega \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \otimes \boldsymbol{u}: \nabla (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad+ \int_0^\tau \int_\Omega p(\rho) \relax(\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Note that this choice of test function is admissible due to the definition of \(\boldsymbol{U}_\infty\) in 7 and the requirement that \(\boldsymbol{V}= \boldsymbol{a}_\infty\) for \(|x| > R\). Similarly, we can use \(\frac{1}{2} |\boldsymbol{V}- \boldsymbol{U}_\infty|^2\) as test function in the continuity equation 9 , which yields \[\begin{align} \left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{V}- \boldsymbol{U}_\infty|^2 \,\mathrm{d}x \right]_{t=0}^{t=\tau} &= \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{U}_\infty) \cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad+ \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) (\boldsymbol{V}- \boldsymbol{U}_\infty) \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] By the identity \[|\boldsymbol{u}- \boldsymbol{V}|^2 = |\boldsymbol{u}- \boldsymbol{U}_\infty|^2 + |\boldsymbol{V}- \boldsymbol{U}_\infty|^2 - 2 (\boldsymbol{u}- \boldsymbol{U}_\infty) \cdot ( \boldsymbol{V}- \boldsymbol{U}_\infty)\] and the energy inequality 11 , we infer \[\begin{align} &\left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + P(\rho | \rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &= \left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{U}_\infty|^2 + P(\rho | \rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{V}- \boldsymbol{U}_\infty|^2 - \rho (\boldsymbol{u}- \boldsymbol{U}_\infty) \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{U}_\infty - \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t\\ &\leq - \int_0^\tau \int_\Omega p(\rho) \relax\boldsymbol{U}_\infty \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times \boldsymbol{U}_\infty) \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{U}_\infty) \cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla (\boldsymbol{V}- \boldsymbol{U}_\infty) \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho \boldsymbol{u}\cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times\boldsymbol{u})\cdot (\boldsymbol{V}-\boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t\\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \otimes \boldsymbol{u}: \nabla (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega p(\rho) \relax(\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \left[ \int_\Omega \rho \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla\boldsymbol{u}) : \nabla (\boldsymbol{U}_\infty - \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &= - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times\boldsymbol{V})\cdot(\boldsymbol{u}-\boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{U}_\infty - \boldsymbol{u}) \cdot (\partial_t \boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) (\boldsymbol{V}- \boldsymbol{U}_\infty) ) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega p(\rho) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \left[ \int_\Omega \rho \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} , \end{align}\] where we additionally used that \[\begin{align} (\boldsymbol{\omega}\times\boldsymbol{u})\cdot (\boldsymbol{V}-\boldsymbol{U}_\infty) - (\boldsymbol{\omega}\times \boldsymbol{U}_\infty) \cdot (\boldsymbol{u}-\boldsymbol{U}_\infty) &= (\boldsymbol{\omega}\times\boldsymbol{u})\cdot (\boldsymbol{V}-\boldsymbol{U}_\infty) + (\boldsymbol{\omega}\times \boldsymbol{u})\cdot\boldsymbol{U}_\infty\\ &= -(\boldsymbol{\omega}\times\boldsymbol{V})\cdot(\boldsymbol{V}-\boldsymbol{U}). \end{align}\] Lastly, we generate the term \(P(\rho | \rho_\infty)\) that shall replace \(P(\rho | \sigma)\). To this end, recall 5 and find \[P(\rho | \sigma) - P(\rho | \rho_\infty) = - \rho (H'(\sigma) - H'(\rho_\infty)) + \sigma H'(\sigma) - H(\sigma) - \rho_\infty H'(\rho_\infty) + H(\rho_\infty).\] With the identity \[\begin{align} &\left[ \int_\Omega \sigma H'(\sigma) - H(\sigma) - \rho_\infty H'(\rho_\infty) + H(\rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} \\ &=\int_0^\tau \frac{\,\mathrm{d}}{\,\mathrm{d}t} \int_\Omega \sigma H'(\sigma) - H(\sigma) - \rho_\infty H'(\rho_\infty) + H(\rho_\infty) \,\mathrm{d}x \,\mathrm{d}t = \int_0^\tau \int_\Omega \sigma\partial_t H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t, \end{align}\] we thus obtain \[\begin{align} \left[ \int_\Omega P(\rho | \sigma) -P(\rho | \rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} &= \left[ \int_\Omega - \rho (H'(\sigma) - H'(\rho_\infty) ) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \sigma \partial_t H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] We further test the continuity equation 9 with \(H'(\sigma) - H'(\rho_\infty)\), which is admissible due to \(\sigma=\rho_\infty\) for \(|x|>R\), to get \[\begin{align} \left[ \int_\Omega \rho (H'(\sigma) - H'(\rho_\infty)) \,\mathrm{d}x \right]_{t=0}^{t=\tau} = \int_0^\tau \int_\Omega \rho \partial_t H'(\sigma) + \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] With these two additional identities, we find \[\begin{align} \label{REI1} \begin{aligned} &\left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + P(\rho | \sigma) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &= \left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + P(\rho | \rho_\infty) \,\mathrm{d}x\right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \left[ \int_\Omega P(\rho | \sigma) - P(\rho | \rho_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} \\ &\leq - \int_0^\tau \int_\Omega \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times\boldsymbol{V})\cdot(\boldsymbol{u}-\boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{U}_\infty - \boldsymbol{u}) \cdot (\partial_t \boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) (\boldsymbol{V}- \boldsymbol{U}_\infty) ) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega p(\rho) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \left[ \int_\Omega \rho \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} \\ &\qquad - \int_0^\tau \int_\Omega \rho \partial_t H'(\sigma) + \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \sigma\partial_t H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{aligned} \end{align}\tag{12}\] Finally, let \(\boldsymbol{U}_\infty^R\) be a function satisfying \[\begin{align} \boldsymbol{U}_\infty^R \in C^\infty(\mathbb{R}^3), \qquad \relax\boldsymbol{U}_\infty^R = 0, \qquad \boldsymbol{U}_\infty^R(x) = \begin{cases} \boldsymbol{\omega}\times x & \text{for }x\in B_R,\\ \boldsymbol{a}_\infty & \text{for }x\in\mathbb{R}^3 \setminus B_{2R}. \end{cases} \end{align}\] Existence of such a function can be shown with the help of the Bogovskiı̆ operator, compare the construction of \(\boldsymbol{V}_\infty\) in 21 below. We then have by Gauß’ theorem \[\begin{align} 0 = \int_0^\tau \int_\Omega \relax(p(\sigma) (\boldsymbol{V}- \boldsymbol{U}_\infty^R)) \,\mathrm{d}x \,\mathrm{d}t = \int_0^\tau \int_\Omega p(\sigma) \relax\boldsymbol{V}+ (\boldsymbol{V}- \boldsymbol{U}_\infty^R) \cdot \nabla p(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] As \(\sigma = \rho_\infty\) outside \(B_R\), we have \(\nabla p(\sigma) = 0\) in \(\mathbb{R}^3 \setminus B_R\), and we find together with \(\nabla p(\sigma) = \sigma \nabla H'(\sigma)\) that \[\begin{align} 0 = \int_0^\tau \int_\Omega p(\sigma) \relax\boldsymbol{V}+ \sigma (\boldsymbol{V}- \boldsymbol{\omega}\times x) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Moreover, using \(\boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty)\) as test function in the continuity equation 9 gives \[\begin{align} &\left[ \int_\Omega \rho \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} \\ &= \int_0^\tau \int_\Omega \rho \boldsymbol{U}_\infty \cdot \partial_t \boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla ( \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) ) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Employing both identities in 12 yields \[\begin{align} &\left[ \int_\Omega \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + P(\rho | \sigma) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}- \boldsymbol{V})) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq - \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \rho (\boldsymbol{\omega}\times\boldsymbol{V})\cdot(\boldsymbol{u}-\boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho \boldsymbol{f} \cdot (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{U}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) (\boldsymbol{V}- \boldsymbol{U}_\infty) ) \,\mathrm{d}x\,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega ( p(\rho) - p(\sigma)) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega (\rho - \sigma) \partial_t H'(\sigma) + (\rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) - \sigma (\boldsymbol{V}- \boldsymbol{\omega}\times x)) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] The first and the fourth integral on the right-hand side add up to \[\begin{align} &- \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}- \boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{U}_\infty \cdot (\boldsymbol{u}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{U}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &= \int_0^\tau \int_\Omega \rho ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{U}_\infty \cdot (\boldsymbol{V}- \boldsymbol{u}) \,\mathrm{d}x \,\mathrm{d}t, \end{align}\] so that we finally find ?? , provided that \((\sigma, \boldsymbol{U}) = (\rho_\infty, \boldsymbol{a}_\infty)\) for \(|x|>R\). ◻
To omit the additional assumptions from Lemma 6, we proceed with an approximation argument.
Proof of Theorem 4. Let \(\phi\in C^\infty(\mathbb{R})\) with \(\phi(s)=1\) for \(s<1\) and \(\phi(s)=0\) for \(s>2\), and define \(\phi_R(x)=\phi(|x|/R)\) for \(R>1\). We approximate \((\sigma, \boldsymbol{V})\) by pairs \((\sigma_R,\boldsymbol{V}_R)\) defined as \[\sigma_R(t,x)= \phi_R(x) \sigma(t,x) + (1-\phi_R(x)) \rho_\infty, \qquad \boldsymbol{V}_R(t,x)=\phi_R(x) \boldsymbol{V}(t,x) + (1-\phi_R(x)) \boldsymbol{a}_\infty\] such that \(\mathop{\mathrm{supp}}(\sigma_R-\rho_\infty, \boldsymbol{V}_R-\boldsymbol{a}_\infty) \subset (0,T)\times B_{2 R}\). By Lemma 6, we then have the REI \[\label{REI46R} \begin{align} &\left[ \int_\Omega E(\rho, \boldsymbol{u}| \sigma_R, \boldsymbol{V}_R) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}_R) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho (\boldsymbol{V}_R - \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{V}_R + \boldsymbol{\omega}\times\boldsymbol{V}_R + ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}_R - \boldsymbol{f} ) \,\mathrm{d}x\,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega ( p(\rho) - p(\sigma_R)) \relax\boldsymbol{V}_R \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega (\rho - \sigma_R) \partial_t H'(\sigma_R) + (\rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) - \sigma_R (\boldsymbol{V}_R - \boldsymbol{\omega}\times x)) \cdot \nabla H'(\sigma_R) \,\mathrm{d}x \,\mathrm{d}t \end{align}\tag{13}\] for a.a. \(\tau\in(0,T)\). To shorten the presentation, we have collected the viscous terms on the left-hand side. For conclusion of 11 , we now pass to the limit \(R\to\infty\) in each term separately.
First of all, we have \[\begin{align} &E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) - E(\rho, \boldsymbol{u}| \sigma_R, \boldsymbol{V}_R) \\ &=\frac{1}{2}\rho\big[-2(\boldsymbol{u}-\boldsymbol{a}_\infty)\cdot(\boldsymbol{V}-\boldsymbol{a}_\infty) +|\boldsymbol{V}-\boldsymbol{a}_\infty|^2 +2(\boldsymbol{u}-\boldsymbol{a}_\infty)\cdot(\boldsymbol{V}_R-\boldsymbol{a}_\infty) -|\boldsymbol{V}_R-\boldsymbol{a}_\infty|^2\big] \\ &\qquad +\int_0^1 \frac{\mathrm d}{\mathrm d\theta}P(\rho\mid (1-\theta)\sigma_R+\theta \sigma)\,\mathrm{d}\theta \\ &=\frac{1}{2}\rho\big[2(\phi_R-1)(\boldsymbol{u}-\boldsymbol{a}_\infty)\cdot(\boldsymbol{V}-\boldsymbol{a}_\infty) +(1-\phi_R^2)|\boldsymbol{V}-\boldsymbol{a}_\infty|^2\big] \\ &\qquad -(1-\phi_R)(\sigma -\rho_\infty)\int_0^1 H''\big(\sigma - (1-\theta)(1-\phi_R)(\sigma-\rho_\infty)\big) \big[\rho-\sigma - (1-\theta)(1-\phi_R)(\sigma-\rho_\infty) \big]\,\mathrm{d}\theta, \end{align}\] where we used 5 . Since \(\sqrt{\rho}|\boldsymbol{u}-\boldsymbol{a}_\infty|\in L^\infty(0,T;L^2(\Omega))\) and \(\rho = (\rho - \rho_\infty) + \rho_\infty \in L^\infty(0,T; [L^2 + L^\gamma + L^\infty] (\Omega))\), see Definition 1, the integrability properties of \(\boldsymbol{V}-\boldsymbol{a}_\infty\) stated in ?? imply that the first term on the right-hand converges to \(0\) in \(L^\infty(0,T;L^1(\Omega))\) as \(R\to\infty\), which is due to the fact that \(\phi_R, \phi_R^2 \stackrel{*}{\rightharpoonup} 1\) in \(L^\infty(\Omega)\). For the second term, we use the identity \[\begin{align} &(1-\phi_R)(\sigma-\rho_\infty)\big[\rho-\sigma - (1-\theta)(1-\phi_R)(\sigma-\rho_\infty) \big] \\ &\quad =(1-\phi_R)(\sigma-\rho_\infty)(\rho-\rho_\infty)-(1-\phi_R)\big[1- (1-\theta)(1-\phi_R)\big](\sigma-\rho_\infty)^2 \end{align}\] and conclude the convergence to \(0\) in a similar way, using that convex combinations of \(\sigma\) and \(\rho_\infty\) are bounded from above and below. In summary, we see \(E(\rho, \boldsymbol{u}| \sigma_R, \boldsymbol{V}_R)\to E(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V})\) in \(L^\infty(0,T;L^1(\Omega))\) as \(R\to\infty\).
For the diffusive term, note that \[\begin{align} &\mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V})-\mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}_R) =\mathbb{S}(\nabla\boldsymbol{u}): \big[ (1-\phi_R)\nabla\boldsymbol{V}- (\boldsymbol{V}-\boldsymbol{a}_\infty)\otimes\nabla\phi_R)\big]. \end{align}\] Due to the convergences \(\phi_R\stackrel{*}{\rightharpoonup} 1\) in \(L^\infty(\Omega)\) and \(\nabla\phi_R\rightharpoonup 0\) in \(L^3(\Omega)\) as \(R\to\infty\), which one readily verifies, and due to ?? , which also implies \(\boldsymbol{V}-\boldsymbol{a}_\infty\in L^2(0,T;L^6(\Omega))\) by Sobolev inequality, we conclude the convergence \(\mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V}_R)\to\mathbb{S}(\nabla \boldsymbol{u}) : \nabla (\boldsymbol{u}- \boldsymbol{V})\) in \(L^1((0,T)\times\Omega)\) as \(R\to\infty\).
To derive convergence of the first term on the right-hand side of 13 , we observe that \[\begin{align} &\rho (\boldsymbol{V}- \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{V}+ \boldsymbol{\omega}\times\boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}- \boldsymbol{f} ) \\ &\qquad -\rho (\boldsymbol{V}_R - \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{V}_R + \boldsymbol{\omega}\times\boldsymbol{V}_R + ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}_R - \boldsymbol{f} ) \\ &=(1-\phi_R)\rho(\boldsymbol{V}-\boldsymbol{a}_\infty)\cdot ( \partial_t \boldsymbol{V}+ \boldsymbol{\omega}\times\boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}- \boldsymbol{f} ) \\ &\qquad + (1-\phi_R)\rho (\boldsymbol{V}- \boldsymbol{u}) \cdot (\partial_t \boldsymbol{V}+ \boldsymbol{\omega}\times(\boldsymbol{V}-a_\infty) + ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}) \\ &\qquad+\rho (\boldsymbol{V}- \boldsymbol{u})\cdot (\boldsymbol{V}-\boldsymbol{a}_\infty) ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla\phi_R). \end{align}\] In virtue of the integrability assumptions from 8 and ?? and the aforementioned convergence properties of \(\phi_R\) and \(\nabla\phi_R\), one concludes that all terms on the right-hand side converge to \(0\) in \(L^1((0,T)\times\Omega)\) as \(R\to\infty\).
For the remaining terms in 13 , we can use similar arguments as before. In particular, for the last term, we use the representation \[\begin{align} &(\rho - \sigma) \partial_t H'(\sigma) + (\rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) - \sigma (\boldsymbol{V}- \boldsymbol{\omega}\times x)) \cdot \nabla H'(\sigma) \\ & =H''(\sigma)\big[(\rho{-}\rho_\infty)-(\sigma{-}\rho_\infty)\big]\partial_t \sigma +H''(\sigma)\big[\sqrt{\rho} \sqrt{\rho}(\boldsymbol{u}-\boldsymbol{a}_\infty) -\sigma (\boldsymbol{V}-\boldsymbol{a}_\infty) + (\rho{-}\sigma)(\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x)\big] \cdot\nabla \sigma. \end{align}\] Since \(\rho-\rho_\infty\in L^\infty(0,T; [L^2 + L^\gamma](\Omega))\), and \(\sqrt{\rho}(\boldsymbol{u}-\boldsymbol{a}_\infty)\in L^\infty(0,T;L^2(\Omega))\), this explains the regularity assumptions on \(\partial_t \sigma\) and \(\nabla \sigma\) in ?? . Finally, passing to the limit \(R\to\infty\) in 13 leads to ?? and completes the proof. ◻
Remark 7. In the proof of Lemma 6 we did not make use of any structural assumptions on the pressure \(p\), so that the statement keeps valid for any \(p\in C^0([0,\infty)) \cap C^1((0,\infty))\). For the approximation argument to obtain Theorem 4, we use the growth conditions from 3 , but the restriction on \(\gamma\) could be relaxed to \(\gamma>1\).
In this section, we show that a weak solution to 1 coincides with a strong one emanating from the same initial data, at least as long as the latter exists. To this end, we make use of the following Korn-type inequality.
Lemma 8. Let \(p\in[1,\infty)\). For any \(\boldsymbol{u}\in L^p(\mathbb{R}^3;\mathbb{R}^3)\) with \(\nabla \boldsymbol{u}\in L^2(\mathbb{R}^3;\mathbb{R}^{3\times 3})\) it holds \[\begin{align} \sqrt{2} \|\nabla \boldsymbol{u}\|_{L^2(\mathbb{R}^3)} \leq \Big\| \nabla \boldsymbol{u}+ \nabla^T \boldsymbol{u}- \frac{2}{3} \relax\boldsymbol{u}\operatorname{Id}\Big\|_{L^2(\mathbb{R}^3)}. \end{align}\]
Proof. First, let \(\boldsymbol{u}\in C_c^\infty(\mathbb{R}^3;\mathbb{R}^3)\). Then integration by parts yields \[\Big\| \nabla \boldsymbol{u}+ \nabla^T \boldsymbol{u}- \frac{2}{3} \relax\boldsymbol{u}\operatorname{Id}\Big\|_{L^2(\mathbb{R}^3)}^2 =\int_{\mathbb{R}^3} 2|\nabla \boldsymbol{u}|^2 + 2\nabla \boldsymbol{u}:\nabla\boldsymbol{u}^T - \frac{4}{3}(\relax\boldsymbol{u})^2 \,\mathrm{d}x =\int_{\mathbb{R}^3} 2|\nabla \boldsymbol{u}|^2 +\frac{2}{3}(\relax\boldsymbol{u})^2 \,\mathrm{d}x.\] Since \((\relax\boldsymbol{u})^2\geq 0\), this gives the asserted inequality for \(\boldsymbol{u}\in C_c^\infty(\mathbb{R}^3;\mathbb{R}^3)\). The general case follows by a standard approximation argument. ◻
Using the REI established in Theorem 4, we derive the following weak-strong uniqueness result. Our argument follows the presentation in [23] and [24].
Theorem 9. Let \((\rho, \boldsymbol{u})\) be a finite-energy weak solution to system 1 with initial data \((\rho_0, \boldsymbol{u}_0)\). Let further \((\sigma, \boldsymbol{V}) = (r, \boldsymbol{v})\) belong to the regularity class specified in ?? and satisfy \[\label{eq:wsu46regularity} \relax\mathbb{S}(\nabla \boldsymbol{v})\in L^\infty(0,T; [L^3 \cap L^{\frac{6\gamma}{5\gamma-6}}] (\Omega)), \qquad \nabla\boldsymbol{v}\in L^\infty(0,T;L^\infty(\Omega)).\qquad{(3)}\] Moreover, assume that \((r, \boldsymbol{v})\) satisfies the system 1 with the same initial data as \((\rho, \boldsymbol{u})\). Then \(\rho = r\) and \(\boldsymbol{u}= \boldsymbol{v}\) a.e. in \((0,T) \times \Omega\).
Proof. We use the momentum equation satisfied by \((r, \boldsymbol{v})\), specifically, \[\begin{align} \partial_t \boldsymbol{v}+ \boldsymbol{\omega}\times\boldsymbol{v}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{v}- \boldsymbol{f} = \frac{1}{r} \relax\mathbb{S}(\nabla \boldsymbol{v}) + ((\boldsymbol{u}- \boldsymbol{v}) \cdot \nabla) \boldsymbol{v}- \frac{1}{r} \nabla p(r). \end{align}\] Moreover, the continuity equation for \((r, \boldsymbol{v})\) implies \[\begin{align} (r-\rho) (\partial_t H'(r) + \nabla H'(r) \cdot (\boldsymbol{v}- \boldsymbol{\omega}\times x)) &= (r-\rho)H''(r) (\partial_t r + \nabla r \cdot (\boldsymbol{v}- \boldsymbol{\omega}\times x)) \\ &= (r-\rho) H''(r) (-r\relax(\boldsymbol{v}- \boldsymbol{\omega}\times x)) \\ &= (\rho - r) p'(r)\relax\boldsymbol{v}. \end{align}\] Then the REI ?? , together with \(r^{-1} \nabla p(r) = \nabla H'(r)\), yields \[\label{mezi1} \begin{align} &\left[ \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}- \boldsymbol{v})) : \nabla (\boldsymbol{u}- \boldsymbol{v}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho (\boldsymbol{v}- \boldsymbol{u}) \cdot \Bigg[ \frac{1}{r}\relax\mathbb{S}(\nabla \boldsymbol{v})+((\boldsymbol{u}- \boldsymbol{v}) \cdot \nabla) \boldsymbol{v}-\frac{1}{r}\nabla p(r)\Bigg]\,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{v}):\nabla(\boldsymbol{u}-\boldsymbol{v}) \,\mathrm{d}x \,\mathrm{d}t - \int_0^\tau \int_\Omega (p(\rho)-p(r))\relax\boldsymbol{v}\,\mathrm{d}x \,\mathrm{d}t\\ &\quad - \int_0^\tau \int_\Omega (\rho-r)p'(r)\relax\boldsymbol{v}-\rho(\boldsymbol{u}-\boldsymbol{v})\cdot\nabla H'(r)\,\mathrm{d}x \,\mathrm{d}t \\ &= \int_0^\tau \int_\Omega \rho (\boldsymbol{v}- \boldsymbol{u}) \cdot ((\boldsymbol{u}- \boldsymbol{v}) \cdot \nabla) \boldsymbol{v}\,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega \Big( \frac{\rho}{r} - 1 \Big) (\boldsymbol{v}- \boldsymbol{u}) \cdot \relax\mathbb{S}(\nabla \boldsymbol{v}) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \big(p(\rho) - p'(r) (\rho - r) - p(r)\big)\relax\boldsymbol{v}\,\mathrm{d}x \,\mathrm{d}t, \end{align}\tag{14}\] where we integrated by parts in the second term of the first inequality and used that \((\boldsymbol{u}- \boldsymbol{v}) |_{\partial\Omega} = 0\). To treat the viscous terms on the left-hand side, we observe that \[\begin{align} \int_\Omega \mathbb{S}(\nabla(\boldsymbol{u}- \boldsymbol{v})) : \nabla(\boldsymbol{u}- \boldsymbol{v}) \,\mathrm{d}x &\geq \frac{\mu}{2} \Big\| \nabla (\boldsymbol{u}- \boldsymbol{v}) + \nabla^T(\boldsymbol{u}- \boldsymbol{v}) - \frac{2}{3} \relax(\boldsymbol{u}- \boldsymbol{v}) \operatorname{Id}\Big\|_{L^2(\Omega)}^2 \\ &\geq \mu \|\nabla(\boldsymbol{u}- \boldsymbol{v})\|_{L^2(\Omega)}^2, \end{align}\] where we prolonged \(\boldsymbol{u}\) and \(\boldsymbol{v}\) by \(\boldsymbol{\omega}\times x\) outside of \(\Omega\), and used the Korn-type inequality from Lemma 8. The first integral on the right-hand side of 14 is easily estimated as \[\begin{align} \int_0^\tau \int_\Omega \rho (\boldsymbol{v}- \boldsymbol{u}) \cdot((\boldsymbol{u}- \boldsymbol{v}) \cdot \nabla) \boldsymbol{v}\,\mathrm{d}x \,\mathrm{d}t &\leq \|\nabla \boldsymbol{v}\|_{L^\infty(0,\tau; L^\infty(\Omega))} \int_0^\tau \int_\Omega \rho |\boldsymbol{u}- \boldsymbol{v}|^2 \,\mathrm{d}x \,\mathrm{d}t \\ &\leq 2\|\nabla \boldsymbol{v}\|_{L^\infty(0,\tau; L^\infty(\Omega))} \int_0^\tau \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Introducing the essential and residual part of a function \(f\) as \[\begin{align} [f]_{\mathrm{ess}} = \chi_{\frac{1}{2} \leq \rho/r \leq 2} f, && [f]_{\mathrm{res}} = f - [f]_{\mathrm{ess}}, \end{align}\] we find that the relative internal energy is coercive with respect to \(\rho\) in the sense that \[\label{eq:coerciveP} [\rho]_{\mathrm{res}}^\gamma + [1]_{\mathrm{res}} + [\rho - r]_{\mathrm{ess}}^2 \leq C P(\rho | r),\tag{15}\] where \(C>0\) only depends on \(\underline r\) and \(\overline{r}\) if \(\underline r\leq r\leq \overline{r}\), see [25] or [14]. For the second integral on the right-hand side of 14 , we now calculate for the residual part that \[\begin{align} &\int_\Omega \Big[ \frac{\rho}{r} - 1 \Big]_{\mathrm{res}} (\boldsymbol{v}- \boldsymbol{u}) \cdot \relax\mathbb{S}(\nabla \boldsymbol{v}) \,\mathrm{d}x \\ &\quad\leq \big(\|[\rho/r]_{\mathrm{res}}\|_{L^\gamma(\Omega)} + \| [1]_{\mathrm{res}} \|_{L^\gamma(\Omega)} \big)\|\boldsymbol{v}- \boldsymbol{u}\|_{L^6(\Omega)} \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^\frac{6\gamma}{5\gamma-6}(\Omega)} \\ &\quad\leq C \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^\frac{6\gamma}{5\gamma-6}(\Omega)} ^2 (\underline r^{-2}\| [\rho]_{\mathrm{res}} \|_{L^\gamma(\Omega)}^2 + \| [1]_{\mathrm{res}} \|_{L^\gamma(\Omega)}^2) + \delta \|\nabla(\boldsymbol{v}- \boldsymbol{u})\|_{L^2(\Omega)}^2, \end{align}\] where \(\delta>0\) is arbitrary, and where we used the Sobolev inequality. In view of the coercivity estimate 15 , we find for \(\frac{6}{5}\leq\gamma \leq 2\) that \[\begin{align} \|[\rho]_{\mathrm{res}}\|_{L^\gamma(\Omega)}^2 + \|[1]_{\mathrm{res}}\|_{L^\gamma(\Omega)}^2 &\leq C \Big( \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x \Big)^\frac{2}{\gamma} \leq C \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x, \end{align}\] since also \(E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \in L^\infty(0,T; L^1(\Omega))\). If \(\gamma > 2\) and \(\relax\mathbb{S}(\nabla \boldsymbol{v}) \in L^2(0,T; L^3(\Omega))\), we may estimate \[\begin{align} &\int_\Omega \Big[ \frac{\rho}{r} - 1 \Big]_{\mathrm{res}} (\boldsymbol{v}- \boldsymbol{u}) \cdot \relax\mathbb{S}(\nabla \boldsymbol{v}) \,\mathrm{d}x \\ &\quad \leq \big(\|[\rho/r]_{\mathrm{res}}\|_{L^2(\Omega)} + \| [1]_{\mathrm{res}} \|_{L^2(\Omega)} \big) \|\boldsymbol{v}- \boldsymbol{u}\|_{L^6(\Omega)} \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^3(\Omega)} \\ &\quad \leq C \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^3(\Omega)}^2 (\underline{r}^{-2}\| [\rho]_{\mathrm{res}} \|_{L^2(\Omega)}^2 + \| [1]_{\mathrm{res}} \|_{L^2(\Omega)}^2) + \delta \|\nabla(\boldsymbol{v}- \boldsymbol{u})\|_{L^2(\Omega)}^2 \end{align}\] by the Sobolev inequality. By \(\gamma > 2\) and Young’s inequality, we have \[\begin{align} [\rho]_{\mathrm{res}}^2 + [1]_{\mathrm{res}}^2\lesssim [\rho]_{\mathrm{res}}^\gamma + [1]_{\mathrm{res}} \lesssim E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \end{align}\] due to 15 . Consequently, for any \(\gamma \geq \frac{6}{5}\), we arrive at \[\begin{align} \label{eq:reldiff46res} \int_\Omega \Big[ \frac{\rho}{r} - 1 \Big]_{\mathrm{res}} (\boldsymbol{v}- \boldsymbol{u}) \cdot \relax\mathbb{S}(\nabla \boldsymbol{v}) \,\mathrm{d}x \leq \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^p(\Omega)} ^2 \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x + \delta \|\nabla(\boldsymbol{v}- \boldsymbol{u})\|_{L^2(\Omega)}^2 \end{align}\tag{16}\] with \(p=\max\{\frac{6\gamma}{5\gamma-6},3\}\). An analogous argument for the essential part leads to \[\label{eq:reldiff46ess} \begin{align} &\int_\Omega \Big[ \frac{\rho}{r} - 1 \Big]_{\mathrm{ess}} (\boldsymbol{v}- \boldsymbol{u}) \cdot \relax\mathbb{S}(\nabla \boldsymbol{v}) \,\mathrm{d}x \leq \underline{r}^{-1} \|[\rho - r]_{\mathrm{ess}}\|_{L^2(\Omega)} \|\boldsymbol{u}- \boldsymbol{v}\|_{L^6(\Omega)} \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^3(\Omega)}\\ &\qquad\leq C \|\relax\mathbb{S}(\nabla \boldsymbol{v})\|_{L^3(\Omega)}^2 \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x + \delta \|\nabla(\boldsymbol{u}- \boldsymbol{v})\|_{L^2(\Omega)}^2. \end{align}\tag{17}\] The very last pressure integral in 14 is handled similarly. Here we use Taylor’s theorem for the essential part, exploiting \(p\in C^2((0,\infty))\), and Young’s inequality for the residual part, and find (compare [14]) \[\begin{align} &\int_\Omega \relax\boldsymbol{v}\big(p(\rho) - p'(r)(\rho - r) - p(r)\big) \,\mathrm{d}x \\ &\quad \leq C \|\relax\boldsymbol{v}\|_{L^\infty(\Omega)} \big(\|[\rho - r]_{\mathrm{ess}}\|_{L^2(\Omega)}^2 + \|[p(\rho)]_{\mathrm{res}}\|_{L^1(\Omega)} + \|[1]_{\mathrm{res}}\|_{L^1(\Omega)}\big) \\ &\quad \leq C \|\relax\boldsymbol{v}\|_{L^\infty(\Omega)} \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x, \end{align}\] where we used \(p(\rho) \lesssim 1 + \rho^\gamma\) and 15 .
Gathering all estimates, choosing \(\delta>0\) small enough to absorb the dissipative terms to the left-hand side of 14 , and employing ?? , we are left with the inequality \[\begin{align} \left[ \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \|\nabla( \boldsymbol{u}- \boldsymbol{v}) \|_{L^2((0,T) \times \Omega)}^2 \leq C \int_0^\tau \int_\Omega E(\rho, \boldsymbol{u}| r, \boldsymbol{v}) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Applying Grönwall’s inequality completes the proof. ◻
In this section, we consider system 1 , but with a rescaled pressure of the form \[\label{NSE-LM} \left\{\begin{align} \partial_t \rho + \relax(\rho (\boldsymbol{u}- \boldsymbol{\omega}\times x)) &= 0 && in(0,T)\times \Omega,\\ \partial_t (\rho\boldsymbol{u}) + \relax( \rho (\boldsymbol{u}- \boldsymbol{\omega}\times x) \otimes \boldsymbol{u}) + \rho \boldsymbol{\omega}\times \boldsymbol{u}&= \rho \boldsymbol{f} + \relax\mathbb{S}(\nabla \boldsymbol{u}) - \frac{1}{\varepsilon^2} \nabla p(\rho) && in (0,T)\times \Omega,\\ \boldsymbol{u}&= \boldsymbol{\omega}\times x && on(0,T)\times \partial\Omega,\\ \rho(\cdot,x) \to \rho_\infty,\;\boldsymbol{u}(\cdot,x) &\to \boldsymbol{a}_\infty && as|x| \to \infty,\\ \rho(0,\cdot)=\rho_0, \;(\rho \boldsymbol{u})(0,\cdot) &= \boldsymbol{\rho}_0\boldsymbol{u}_0 && in\Omega. \end{align}\right.\tag{18}\] The additional factor \(\varepsilon^{-2}\) in front of the pressure plays the role of a low Mach number. In turn, we obtain the rescaled relative energy \[\begin{align} E_\varepsilon(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) = \frac{1}{2} \rho |\boldsymbol{u}- \boldsymbol{V}|^2 + \frac{1}{\varepsilon^2} P(\rho | \sigma) \end{align}\] and find the relative energy inequality of the form \[\label{REI-LM} \begin{align} &\left[ \int_\Omega E_\varepsilon(\rho, \boldsymbol{u}| \sigma, \boldsymbol{V}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}- \boldsymbol{V})) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho (\boldsymbol{V}- \boldsymbol{u}) \cdot ( \partial_t \boldsymbol{V}+ \boldsymbol{\omega}\times\boldsymbol{V}+ ((\boldsymbol{u}-\boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}- \boldsymbol{f} ) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}) : \nabla (\boldsymbol{u}- \boldsymbol{V}) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \frac{1}{\varepsilon^2} \int_0^\tau \int_\Omega ( p(\rho) - p(\sigma)) \relax\boldsymbol{V}\,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \frac{1}{\varepsilon^2} \int_0^\tau \int_\Omega (\rho - \sigma) \partial_t H'(\sigma) + (\rho (\boldsymbol{u}-\boldsymbol{\omega}\times x) - \sigma (\boldsymbol{V}- \boldsymbol{\omega}\times x)) \cdot \nabla H'(\sigma) \,\mathrm{d}x \,\mathrm{d}t, \end{align}\tag{19}\] which follows from Theorem 4. We shall use 19 to quantify the convergence of solutions to 18 (for \(\varepsilon>0\)) as \(\varepsilon\to 0\). This limit passage formally leads to the incompressible Navier–Stokes equations 6 . We specify this low Mach number limit, where the convergence is stated in terms of the relative energy.
Theorem 10. For \(\varepsilon\in (0, 1)\) let \((\rho_\varepsilon, \boldsymbol{u}_\varepsilon)\) be a finite energy weak solution to system 18 in the sense of Definition 1, emanating from the initial data \((\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0})\). Let \((\boldsymbol{U},\Pi)\) be such that the pair \((\sigma,\boldsymbol{V})=(\rho_\infty, \boldsymbol{U})\) lies in the regularity class specified in ?? , with the additional properties \[\begin{align} \label{eq:singLim46assstrong} \begin{aligned} \relax\mathbb{S}(\nabla \boldsymbol{U}) &\in L^2(0,T; [L^2\cap L^3 \cap L^\frac{6\gamma}{5\gamma-6}] (\Omega)), & \partial_t \boldsymbol{U}&\in L^\infty(0,T; L^2(\Omega;\mathbb{R}^3)), \\ \nabla \Pi &\in L^2(0,T; [L^2 \cap L^3 \cap L^\frac{6\gamma}{5\gamma-6}] (\Omega)), & \Pi &\in L^2((0,T) \times \Omega), \end{aligned} \end{align}\qquad{(4)}\] and such that system 6 is satisfied for some initial data \(\boldsymbol{U}_0\) with \(\boldsymbol{U}_0-\boldsymbol{a}_\infty\in L^2(\Omega;\mathbb{R}^3)\). If \[\begin{align} \label{ass:initDat} \lim_{\varepsilon\to 0} \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x = 0, \end{align}\qquad{(5)}\] then \[\label{eq:limit46lm} \lim\limits_{\varepsilon\to 0} \sup_{\tau \in [0, T]} \int_\Omega E_\varepsilon(\rho_{\varepsilon}, \boldsymbol{u}_{\varepsilon} | \rho_\infty, \boldsymbol{U})(\tau) \,\mathrm{d}x = 0.\qquad{(6)}\]
The existence of local-in-time strong solutions \((\boldsymbol{U},\Pi)\) to 6 was shown in [26] for \(\boldsymbol{a}_\infty=0\), and global-in-time strong solutions were found in [26] for small initial data and small rotation. We emphasize that the regularity of the strong solutions in [26] differs from ours. For extensions of this existence result, allowing even for time-dependence of \(\boldsymbol{a}_\infty\) and \(\boldsymbol{\omega}\), see the monograph [17], the recent article [27], and references therein.
Under additional assumptions on the pressure \(\Pi\), we can show the following quantified version of Theorem 10:
Theorem 11. Under the assumptions of Theorem 10, assume additionally that \[\begin{align} \label{eq:sigLimRate46pressurereg} \begin{aligned} \partial_t\Pi,\, (\boldsymbol{a}_\infty-\boldsymbol{\omega}\times x)\cdot \nabla\Pi \in L^1(0,T; [L^1 \cap L^2 \cap L^{\frac{\gamma}{\gamma-1}}] (\Omega)), && \Pi(0,\cdot) \in [L^1 \cap L^2 \cap L^{\frac{\gamma}{\gamma-1}}] (\Omega). \end{aligned} \end{align}\qquad{(7)}\] Then \[\begin{align} &\sup_{\tau \in [0,T]} \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U})(\tau) \,\mathrm{d}x + \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{U})\|_{L^2(0,T; L^2(\Omega))}^2 \\ &\leq C \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x + C(\varepsilon+ \varepsilon^\frac{2}{\gamma}), \end{align}\] where the constant \(C>0\) is independent of \(\varepsilon\).
Remark 12. Thanks to the coercivity properties of the relative energy, we also have \[\begin{align} &\|[\boldsymbol{u}_\varepsilon- \boldsymbol{U}]_{\mathrm{ess},\varepsilon}\|_{L^\infty(0,T; L^2(\Omega))}^2 + \frac{1}{\varepsilon^2} \|[\rho_\varepsilon- \rho_\infty]_{\mathrm{ess}, \varepsilon}\|_{L^\infty(0,T; L^2(\Omega))}^2 + \frac{1}{\varepsilon^2} \|[\rho_\varepsilon]_{\mathrm{res}, \varepsilon}\|_{L^\infty(0,T; L^\gamma(\Omega))}^\gamma \\ &\leq C \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x + C(\varepsilon+ \varepsilon^\frac{2}{\gamma}), \end{align}\] where we define the \(\varepsilon\)-dependent essential and residual part of a function \(f\) as \[\label{eq:ess46res46eps} [f]_{{\mathrm{ess},\varepsilon}} = \chi_{\frac{1}{2} \leq \rho_\varepsilon/ \rho_\infty \leq 2} f, \qquad [f]_{{\mathrm{res},\varepsilon}} = f - [f]_{{\mathrm{ess},\varepsilon}}.\tag{20}\]
The following two subsections are devoted to the proofs of Theorems 10 and 11.
We split the proof of Theorem 10 into several lemmas. First, we use the REI 19 to derive suitable bounds on \((\rho_\varepsilon, \boldsymbol{u}_\varepsilon)\).
Lemma 13. The weak solutions \((\rho_\varepsilon,\boldsymbol{u}_\varepsilon)\) satisfy the \(\varepsilon\)-uniform bounds \[\begin{align} \|[\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T; L^2(\Omega))}^2 + \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^\gamma(\Omega))}^\gamma + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^1(\Omega))} &\lesssim \varepsilon^2, \\ \|[\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))}^2 + \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty)\|_{L^2(0,T; L^2(\Omega))}^2 &\lesssim 1, \end{align}\] where the essential and residual parts are defined as in 20 .
Proof. Since the limiting system 6 is incompressible, we wish to take an incompressible test function that is not “too far away” from \(\boldsymbol{u}_\varepsilon\) in the REI ?? . To this end, we first recall the notion of the Bogovskiı̆ operator. Let \(q\in(1,\infty)\). By \(L_0^q(B_1 \setminus \mathcal{S})\) we denote the set of all functions \(f \in L^q(B_1 \setminus \mathcal{S})\) with \(\int_{B_1 \setminus \mathcal{S}} f \,\mathrm{d}x = 0\). Since \(B_1 \setminus \mathcal{S}\) is a bounded Lipschitz domain, there exists a bounded linear operator \(\mathcal{B}: (L_0^q \cap C^\infty)(B_1 \setminus \mathcal{S}) \to C_0^\infty(B_1 \setminus \mathcal{S}; \mathbb{R}^3)\) such that \[\begin{align} \relax\mathcal{B}(f) &= f \;\text{ in } \;B_1 \setminus \mathcal{S}, \\ \|\mathcal{B}(f)\|_{W^{1,q}(B_1 \setminus \mathcal{S})} &\leq C \|f\|_{L^q(B_1 \setminus \mathcal{S})} \end{align}\] for any \(f \in (L_0^q \cap C^\infty)(B_1 \setminus \mathcal{S})\), see e.g. [28]. Recalling the definition of \(\boldsymbol{U}_\infty\) from 7 , we immediately see that \[\begin{align} \int_{B_1} \relax\boldsymbol{U}_\infty \,\mathrm{d}x &= \int_{\partial B_1} \boldsymbol{U}_\infty \cdot \boldsymbol{n} \,\mathrm{d}S = \int_{\partial B_1} \boldsymbol{a}_\infty \cdot \boldsymbol{n} \,\mathrm{d}S = 0, \\ \int_{\mathcal{S}} \relax\boldsymbol{U}_\infty \,\mathrm{d}x &= \int_{\mathcal{S}} \relax(\boldsymbol{\omega}\times x) \,\mathrm{d}x = 0, \end{align}\] and hence \(\relax\boldsymbol{U}_\infty \in (L_0^q \cap C^\infty)(B_1 \setminus \mathcal{S})\) for any \(q\in(1,\infty)\). In turn, we define \[\label{eq:V46def} \boldsymbol{V}_\infty = \boldsymbol{U}_\infty - \mathcal{B}(\relax\boldsymbol{U}_\infty) \in C^\infty(\mathbb{R}^3; \mathbb{R}^3).\tag{21}\] Note especially that we have \(\boldsymbol{V}_\infty |_{\partial \mathcal{S}} = \boldsymbol{U}_\infty |_{\partial \mathcal{S}} = \boldsymbol{\omega}\times x\) and \({\rm supp} (\boldsymbol{V}_\infty - \boldsymbol{a}_\infty) \subset \overline{B}_1\) due to \(\mathcal{B}(f)|_{\partial(B_1 \setminus \mathcal{S})} = 0\) for any \(f \in L_0^q(B_1 \setminus \mathcal{S})\). Moreover, \(\relax\boldsymbol{V}_\infty = 0\) in \(\mathbb{R}^3\) and \[\begin{align} \label{estV} \|\boldsymbol{V}_\infty - \boldsymbol{a}_\infty\|_{W^{1,q}(\mathbb{R}^3)} \leq C \Big( \|\boldsymbol{U}_\infty - \boldsymbol{a}_\infty\|_{W^{1,q}(B_1)} + \|\relax\boldsymbol{U}_\infty \|_{L^q(B_1)} \Big) \leq C \|\boldsymbol{U}_\infty - \boldsymbol{a}_\infty\|_{W^{1,q}(B_1)}. \end{align}\tag{22}\] By ?? , we further have \[\begin{align} \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x \lesssim 1 . \end{align}\] In particular, since \(\boldsymbol{U}_0-\boldsymbol{a}_\infty\in L^2(\Omega)\), this implies \[\begin{align} \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x \lesssim 1. \end{align}\]
Now we use the REI 19 with test functions \((\rho_\infty, \boldsymbol{V}_\infty)\). In view of \(\partial_t H'(\rho_\infty) = 0\) and \(\nabla H'(\rho_\infty) = 0\), we find \[\label{REI2} \begin{align} &\left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x\right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot (\boldsymbol{\omega}\times \boldsymbol{V}_\infty + ((\boldsymbol{V}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla) \boldsymbol{V}_\infty - \boldsymbol{f} ) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad + \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot ((\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\cdot \nabla) \boldsymbol{V}_\infty \,\mathrm{d}x \,\mathrm{d}t \\ &\quad - \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}_\infty) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\tag{23}\] Note that the choice \((\sigma,\boldsymbol{V})=(\rho_\infty, \boldsymbol{V}_\infty)\) satisfies ?? in Theorem 4. In particular, we have \(\nabla \boldsymbol{V}_\infty = 0\) outside the bounded domain \(B_1\), and \(\nabla \boldsymbol{V}_\infty \in (L^1\cap L^\infty)(\mathbb{R}^3)\). Having this in mind, we estimate \[\begin{align} \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot ((\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\cdot \nabla) \boldsymbol{V}_\infty \,\mathrm{d}x \,\mathrm{d}t &\leq \|\nabla \boldsymbol{V}_\infty\|_{L^\infty(\mathbb{R}^3)} \int_0^\tau \int_\Omega \rho_\varepsilon|\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty|^2 \,\mathrm{d}x \,\mathrm{d}t \\ &\lesssim \int_0^\tau \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t, \end{align}\] and for any \(\delta>0\), \[\begin{align} \int_0^\tau \int_\Omega \mathbb{S}(\nabla \boldsymbol{V}_\infty) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t &= \int_0^\tau \int_{B_1} \mathbb{S}(\nabla \boldsymbol{V}_\infty) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq C_\delta \|\nabla \boldsymbol{V}_\infty\|_{L^2(B_1)}^2 + \delta \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\|_{L^2((0,\tau) \times \Omega)}^2. \end{align}\] For the first integral on the right-hand side of 23 we abbreviate \(\boldsymbol{b} = \boldsymbol{\omega}\times \boldsymbol{V}_\infty + ((\boldsymbol{V}_\infty - \boldsymbol{\omega}\times x)\cdot \nabla)\boldsymbol{V}_\infty) - \boldsymbol{f}\) to see \[\begin{align} \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot \boldsymbol{b} \,\mathrm{d}x \,\mathrm{d}t = \int_0^\tau \int_\Omega [\rho_\varepsilon]_{\mathrm{ess}, \varepsilon} (\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot \boldsymbol{b} \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega [\rho_\varepsilon]_{\mathrm{res}, \varepsilon} (\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot \boldsymbol{b} \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Similarly to the estimate 15 , we see that the relative energy is coercive in the sense that \[\begin{align} \label{eq:coercivity46eps} \frac{1}{\varepsilon^2} \Big( [\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}^\gamma + [1]_{{\mathrm{res},\varepsilon}} + [\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}^2 \Big) + [\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty]_{{\mathrm{ess},\varepsilon}}^2 \leq C E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \end{align}\tag{24}\] with a constant \(C>0\) only depending on \(\rho_\infty\), and thus independent of \(\varepsilon\). Note that \([\rho_\varepsilon]_{{\mathrm{ess},\varepsilon}} \leq 2\rho_\infty\) and \(\boldsymbol{b} \in L^\infty(0,T; [L^1 \cap L^\infty] (\Omega))\) since \(\boldsymbol{f} \in L^\infty(0,T; [L^1 \cap L^\infty] (\Omega))\). Hence, we find \[\begin{align} \int_0^\tau \int_\Omega [\rho_\varepsilon]_{{\mathrm{ess},\varepsilon}} (\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot \boldsymbol{b} \,\mathrm{d}x \,\mathrm{d}t &\lesssim \|\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty\|_{L^2(0,\tau; L^6(\Omega))} \|\boldsymbol{b}\|_{L^2(0,\tau; L^\frac{6}{5}(\Omega))} \\ &\lesssim 1 + \delta \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\|_{L^2((0,\tau) \times \Omega)}^2 \end{align}\] for any \(\delta>0\). For the residual part, we use Hölder’s and Young’s inequalities multiple times to conclude \[\begin{align} \int_\Omega [\rho_\varepsilon]_{{\mathrm{res},\varepsilon}} (\boldsymbol{V}_\infty - \boldsymbol{u}_\varepsilon) \cdot \boldsymbol{b} \,\mathrm{d}x &\leq \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}^{1/2} \|_{L^2(\Omega)} \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}^{1/2} (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty) \|_{L^2(\Omega)} \|\boldsymbol{b}\|_{L^\infty(\Omega)} \\ &\lesssim \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\|_{L^1(\Omega)} + \|\rho_\varepsilon|\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty|^2\|_{L^1(\Omega)} \\ &\lesssim \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}^\gamma\|_{L^1(\Omega)} + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^1(\Omega)} + \|\rho_\varepsilon|\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty|^2\|_{L^1(\Omega)} \\ &\lesssim \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x \end{align}\] due to 24 and \(\varepsilon<1\). Choosing \(\delta>0\) small enough and using the Korn-type inequality from Lemma 8, we are left with \[\begin{align} \left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega |\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)|^2 \,\mathrm{d}x \,\mathrm{d}t \lesssim 1 + \int_0^\tau \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] By Grönwall’s inequality, this yields \[\begin{align} \sup_{t \in [0,T]} \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{V}_\infty) \,\mathrm{d}x + \int_0^T \int_\Omega |\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)|^2 \,\mathrm{d}x \,\mathrm{d}t \lesssim 1. \end{align}\] The coercivity of the relative energy from 24 now enforces \[\begin{align} \|[\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T; L^2(\Omega))}^2 + \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^\gamma(\Omega))}^\gamma + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^1(\Omega))} &\lesssim \varepsilon^2, \\ \|[\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))}^2 + \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\|_{L^2(0,T; L^2(\Omega))}^2 &\lesssim 1. \end{align}\] Returning to \(\boldsymbol{U}_\infty\), we use 22 to see that \[\begin{align} \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty)\|_{L^2(0,T; L^2(\Omega))} \leq \|\nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty)\|_{L^2(0,T; L^2(\Omega))} + \|\nabla(\boldsymbol{V}_\infty - \boldsymbol{U}_\infty)\|_{L^2(0,T; L^2(\Omega))} &\lesssim 1, \\ \|[\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))} \leq \|[\boldsymbol{u}_\varepsilon- \boldsymbol{V}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))} + \|[\boldsymbol{V}_\infty - \boldsymbol{U}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))} &\lesssim 1. \end{align}\] In conclusion, we obtain the asserted bounds. ◻
From the uniform estimates obtained in Lemma 13, we can extract convergent subsequences.
Lemma 14. There exists a (not relabeled) subsequence of \((\rho_\varepsilon,\boldsymbol{u}_\varepsilon)\) such that \[\begin{align} {[\rho_\varepsilon- \rho_\infty]}_{{\mathrm{ess},\varepsilon}} &\to 0 &&\text{ strongly in } L^\infty(0,T; L^2(\Omega)),\\ [\rho_\varepsilon]_{{\mathrm{res},\varepsilon}} &\to 0 &&\text{ strongly in } L^\infty(0,T; L^\gamma(\Omega)), \\ [1]_{{\mathrm{res},\varepsilon}} &\to 0 &&\text{ strongly in } L^\infty(0,T; L^p(\Omega)),\;p\in[1,\infty), \\ [\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty]_{{\mathrm{ess},\varepsilon}} &\rightharpoonup^\ast \boldsymbol{u}- \boldsymbol{U}_\infty &&\text{ weakly-\ast in } L^\infty(0,T; L^2(\Omega)), \\ \nabla(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) &\rightharpoonup\nabla(\boldsymbol{u}- \boldsymbol{U}_\infty)&& \text{ weakly in } L^2(0,T; L^2(\Omega)), \end{align}\] for some \(\boldsymbol{u}\) with \(\boldsymbol{u}-\boldsymbol{U}_\infty\in L^2(0,T;W^{1,2}(\Omega))\cap L^\infty(0,T;L^2(\Omega))\) and \(\relax\boldsymbol{u}=0\).
Proof. If we take into account that \[\|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^p(\Omega))}^p = \|[1]_{{\mathrm{res},\varepsilon}}^p\|_{L^\infty(0,T; L^1(\Omega))} = \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^1(\Omega))}\] for any \(1 \leq p < \infty\), the existence of a convergent subsequence in the asserted topologies follows directly from Lemma 13. It only remains to show that \(\boldsymbol{u}\) is divergence free. To this end, we take the limit in the weak form of the continuity equation. For any \(\phi \in C_c^1((0,T) \times \Omega)\) we have \[\begin{align} \int_0^T \int_\Omega (\rho_\varepsilon- \rho_\infty) \partial_t \phi + \rho_\varepsilon(\boldsymbol{u}_\varepsilon- \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t = 0. \end{align}\] For the first part, we see that \[\begin{align} \int_0^T \int_\Omega (\rho_\varepsilon- \rho_\infty) \partial_t \phi \,\mathrm{d}x \,\mathrm{d}t \lesssim \|[\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T; L^2(\Omega))} + \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^\gamma(\Omega))} + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^1(\Omega))} \end{align}\] and the right-hand side vanishes as \(\varepsilon\to 0\) by the convergences obtained before. For the second term, we rewrite \[\begin{align} \int_0^T \int_\Omega \rho_\varepsilon(\boldsymbol{u}_\varepsilon- \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t = \int_0^T \int_\Omega \rho_\varepsilon(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t + \int_0^T \int_\Omega \rho_\varepsilon(\boldsymbol{U}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Using the convergences of \([\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\), \([\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\), and \([1]_{{\mathrm{res},\varepsilon}}\), the last integral converges to \[\begin{align} \lim_{\varepsilon\to 0} \int_0^T \int_\Omega \rho_\varepsilon(\boldsymbol{U}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t = \int_0^T \int_\Omega \rho_\infty (\boldsymbol{U}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] For the first integral, we have \[\begin{align} &\int_0^T \int_\Omega \rho_\varepsilon(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t \\ &= \int_0^T \int_\Omega (\rho_\varepsilon- \rho_\infty) (\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t + \rho_\infty \int_0^T \int_\Omega (\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t \\ &= \int_0^T \int_\Omega (\rho_\varepsilon- \rho_\infty) (\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t - \rho_\infty \int_0^T \int_\Omega \relax(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \phi \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] We obtain \[\begin{align} \int_0^T\int_\Omega [\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}} (\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x\,\mathrm{d}t &\lesssim \int_0^T\|[\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^2(\Omega)} \|[\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^2(\Omega)} \,\mathrm{d}t \\ &\lesssim \|[\rho_\varepsilon- \rho_\infty]_{{\mathrm{ess},\varepsilon}}\|_{L^\infty(0,T;L^2(\Omega))} \to 0, \\ \int_0^T \int_\Omega [\rho_\varepsilon- \rho_\infty]_{{\mathrm{res},\varepsilon}} (\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t &\lesssim \int_0^T \|\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty\|_{L^6(\Omega)} \Big( \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}\|_{L^\frac{6}{5}(\Omega)} + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\frac{6}{5}(\Omega)} \Big)\,\mathrm{d}t \\ &\lesssim \|[\rho_\varepsilon]_{{\mathrm{res},\varepsilon}}^\gamma\|_{L^\infty(0,T;L^1(\Omega))} + \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T;L^\frac{6}{5}(\Omega))} \to 0, \\ \int_0^T \int_\Omega \relax(\boldsymbol{u}_\varepsilon- \boldsymbol{U}_\infty) \phi \,\mathrm{d}x \,\mathrm{d}t &\to \int_0^T \int_\Omega \relax(\boldsymbol{u}- \boldsymbol{U}_\infty) \phi \,\mathrm{d}x \,\mathrm{d}t. \end{align}\] Gathering the terms above, we infer \[\begin{align} 0 &= \lim_{\varepsilon\to0} \int_0^T \int_\Omega (\rho_\varepsilon- \rho_\infty) \partial_t \phi + \rho_\varepsilon(\boldsymbol{u}_\varepsilon- \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t \\ &= -\int_0^T \int_\Omega \rho_\infty \relax(\boldsymbol{u}- \boldsymbol{U}_\infty) \phi \,\mathrm{d}x \,\mathrm{d}t + \int_0^T \int_\Omega \rho_\infty(\boldsymbol{U}_\infty - \boldsymbol{\omega}\times x) \cdot \nabla \phi \,\mathrm{d}x \,\mathrm{d}t \\ &= -\int_0^T \int_\Omega \rho_\infty \phi \relax\boldsymbol{u}\,\mathrm{d}x \,\mathrm{d}t \end{align}\] for any \(\phi \in C_c^1((0,T) \times \Omega)\), leading to \(\relax\boldsymbol{u}= 0\). ◻
To conclude the proof of the theorem, we invoke the scaled REI 19 again, now with the strong solution \(\boldsymbol{U}\) of 6 .
Proof of Theorem 10. We use \((\sigma,\boldsymbol{V})=(\rho_\infty, \boldsymbol{U})\) in 19 and utilize the identities \(\relax\boldsymbol{U}= 0\), \(\partial_t H'(\rho_\infty) = 0\) and \(\nabla H'(\rho_\infty) = 0\), together with the solution property of \((\boldsymbol{U}, \Pi)\), to find that \[\label{REI-LM-1} \begin{align} &\left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big) (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot ( \relax\mathbb{S}(\nabla \boldsymbol{U}) - \nabla \Pi) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot ((\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \cdot \nabla) \boldsymbol{U}\,\mathrm{d}x \,\mathrm{d}t. \end{align}\tag{25}\] Here we used integration by parts on the term including \(\mathbb{S}(\nabla\boldsymbol{U})\). We decompose the second term on the right-hand side as \[\begin{align} \int_0^\tau \int_\Omega (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t = \int_0^\tau \int_\Omega [\boldsymbol{U}- \boldsymbol{u}_\varepsilon]_{{\mathrm{ess},\varepsilon}} \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t + \int_0^\tau \int_\Omega [\boldsymbol{U}- \boldsymbol{u}_\varepsilon]_{{\mathrm{res},\varepsilon}} \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t =: I_1 + I_2. \end{align}\]
The integral \(I_2\) vanishes as \(\varepsilon\to 0\) due to \[\begin{align} I_2 &\leq \|\boldsymbol{U}- \boldsymbol{u}_\varepsilon\|_{L^2(0,T; L^6(\Omega))} \|\nabla \Pi\|_{L^2(0,T; L^2(\Omega))} \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^3(\Omega))} \\ &\leq \|\nabla(\boldsymbol{U}- \boldsymbol{u}_\varepsilon)\|_{L^2((0,T) \times \Omega)} \|\nabla \Pi\|_{L^2(0,T; L^2(\Omega))} \|[1]_{{\mathrm{res},\varepsilon}}\|_{L^\infty(0,T; L^3(\Omega))} \to 0 \end{align}\] as \(\varepsilon\to 0\) since \(\nabla(\boldsymbol{U}- \boldsymbol{u}_\varepsilon)\) is uniformly bounded in \(L^2((0,T) \times \Omega)\) and \([1]_{{\mathrm{res},\varepsilon}} \to 0\) in \(L^\infty(0,T; L^p(\Omega))\) for any \(1 \leq p < \infty\) by Lemma 14. For \(I_1\), we use that \([\boldsymbol{U}- \boldsymbol{u}_\varepsilon]_{{\mathrm{ess},\varepsilon}} \rightharpoonup^\ast \boldsymbol{U}- \boldsymbol{u}\) in \(L^\infty(0,T; L^2(\Omega))\) to deduce \[\begin{align} I_1=\int_0^\tau \int_\Omega [\boldsymbol{U}- \boldsymbol{u}_\varepsilon]_{{\mathrm{ess},\varepsilon}} \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \to \int_0^\tau \int_\Omega (\boldsymbol{U}- \boldsymbol{u}) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t = 0, \end{align}\] which follows from \((\boldsymbol{U}- \boldsymbol{u})|_{\partial\Omega} = 0\) as well as \(\relax\boldsymbol{U}= \relax\boldsymbol{u}= 0\). The last term in 25 is clearly controlled by the relative energy \(E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U})\) since \(\nabla\boldsymbol{U}\in L^1(0,T;L^\infty(\Omega))\). Lastly, splitting the first term on the right-hand side of 25 into its essential and residual part, we can derive estimates analogous to 16 and 17 . Due to the integrability assumptions from ?? , we finally find \[\begin{align} &\left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\lesssim \int_0^\tau \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t + o_{\varepsilon}, \end{align}\] where \(o_\varepsilon\geq 0\) is a scalar with \(o_\varepsilon\to 0\) as \(\varepsilon\to0\). Grönwall’s inequality then yields \[\begin{align} &\sup_{t \in [0, T]} \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x + \int_0^T \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\lesssim \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x + o_{\varepsilon}. \end{align}\] Due to assumption ?? , we now conclude the limit ?? as \(\varepsilon\to 0\) along the subsequence chosen in Lemma 14. However, since this limit is independent of the chosen subsequence, the sequence itself converges, which finishes the proof of Theorem 10. ◻
Lastly, we derive the convergence rates as given in Theorem 11. To this end, we recall the uniform bounds obtained in Lemma 13 and the inequality 25 , which we rewrite in the form \[\label{REI-LM-R1} \begin{align} &\left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big) (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \relax\mathbb{S}(\nabla \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad - \int_0^\tau \int_\Omega \frac{\rho_\varepsilon}{\rho_\infty} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \rho_\varepsilon(\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot ((\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \cdot \nabla) \boldsymbol{U}\,\mathrm{d}x \,\mathrm{d}t. \end{align}\tag{26}\]
Proof of Theorem 11. As before, the last term on the right-hand side of 26 is estimated by the relative energy itself. The first term we split into its essential and residual part to see \[\begin{align} &\int_0^\tau \int_\Omega \Big[ \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big]_{\mathrm{ess}, \varepsilon} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \relax\mathbb{S}(\nabla \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad \leq C \|[\rho_\varepsilon- \rho_\infty]_{\mathrm{ess}, \varepsilon} \|_{L^\infty(0,T; L^2(\Omega))} \|\boldsymbol{U}- \boldsymbol{u}_\varepsilon\|_{L^2(0,T; L^6(\Omega))} \|\relax\mathbb{S}(\nabla \boldsymbol{U})\|_{L^2(0,T; L^3(\Omega))} \leq C \varepsilon, \\ &\int_0^\tau \int_\Omega \Big[ \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big]_{\mathrm{res}, \varepsilon} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \relax\mathbb{S}(\nabla \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\quad \leq C \|[\rho_\varepsilon- \rho_\infty]_{\mathrm{res}, \varepsilon} \|_{L^\infty(0,T; L^\gamma(\Omega))} \|\boldsymbol{U}- \boldsymbol{u}_\varepsilon\|_{L^2(0,T; L^6(\Omega))} \|\relax\mathbb{S}(\nabla \boldsymbol{U})\|_{L^2(0,T; L^\frac{6\gamma}{5\gamma-6}(\Omega))} \leq C \varepsilon^\frac{2}{\gamma}, \end{align}\] where we used Sobolev embeddings and the uniform bounds obtained in Lemma 13. For the remaining pressure part, we make use of the function \(\boldsymbol{V}_\infty\) defined in 21 and infer \[\begin{align} &\int_0^\tau \int_\Omega \frac{\rho_\varepsilon}{\rho_\infty} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\quad = \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1\Big) (\boldsymbol{U}-\boldsymbol{V}_\infty) \cdot \nabla \Pi - \frac{\rho_\varepsilon}{\rho_\infty} (\boldsymbol{u}_\varepsilon-\boldsymbol{V}_\infty) \cdot \nabla \Pi + (\boldsymbol{U}-\boldsymbol{V}_\infty) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t\\ &\quad = \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1\Big) (\boldsymbol{U}-\boldsymbol{V}_\infty) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t - \frac{1}{\rho_\infty} \int_0^\tau \int_\Omega\rho_\varepsilon(\boldsymbol{u}_\varepsilon-\boldsymbol{V}_\infty) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t, \end{align}\] where we used that the integral over \((\boldsymbol{U}-\boldsymbol{V}_\infty) \cdot \nabla \Pi\) vanishes by the divergence theorem due to \(\relax\boldsymbol{U}=\relax\boldsymbol{V}_\infty=0\) in \(\Omega\) and \(\boldsymbol{U}=\boldsymbol{V}_\infty=\boldsymbol{\omega}\times x\) on \(\partial\Omega\). For the second integral, we deduce from the continuity equation 9 that \[\begin{align} - \int_0^\tau \int_\Omega\rho_\varepsilon(\boldsymbol{u}_\varepsilon-\boldsymbol{V}_\infty) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t &= \int_0^\tau \int_\Omega (\rho_\varepsilon- \rho_\infty) \partial_t \Pi + \rho_\varepsilon(\boldsymbol{V}_\infty-\boldsymbol{\omega}\times x) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad -\left[ \int_\Omega (\rho_{\varepsilon} - \rho_\infty) \Pi\,\mathrm{d}x \right]_{t=0}^{t=\tau}, \end{align}\] which follows by an approximation that requires to introduce the additional terms related to \(\rho_\infty\), which can be carried out similarly to the approximation argument in the proof of Theorem 4. Note that \(\Pi\) is an admissible test function due to the regularity assumed in ?? . Moreover, the divergence theorem yields \[\int_\Omega(\boldsymbol{V}_\infty-\boldsymbol{\omega}\times x) \cdot \nabla \Pi \,\mathrm{d}x = 0 .\] Hence, we may write \[\begin{align} \int_0^\tau \int_\Omega \frac{\rho_\varepsilon}{\rho_\infty} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t &= \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1\Big) (\boldsymbol{U}- \boldsymbol{V}_\infty) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big) (\boldsymbol{V}_\infty - \omega \times x) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \\ &\qquad + \int_0^\tau \int_\Omega \Big( \frac{\rho_\varepsilon}{\rho_\infty} - 1 \Big) \partial_t \Pi \,\mathrm{d}x \,\mathrm{d}t -\left[\int_\Omega \Big(\frac{\rho_{\varepsilon}}{\rho_\infty} - 1 \Big) \Pi \,\mathrm{d}x\right]_{t=0}^{t=\tau}. \end{align}\] Splitting as before in essential and residual part, and using that \[\begin{align} \left[\int_\Omega \Big(\frac{\rho_{\varepsilon}}{\rho_\infty} - 1 \Big) \Pi \,\mathrm{d}x\right]_{t=0}^{t=\tau} \leq C \|\rho_\varepsilon- \rho_\infty\|_{L^\infty(0,T; [L^2+L^\gamma] (\Omega))} \|\Pi\|_{L^\infty(0,T; [L^2 \cap L^\frac{\gamma}{\gamma-1}] (\Omega))} \leq C (\varepsilon+ \varepsilon^\frac{2}{\gamma}) \end{align}\] due to ?? , we thus infer \[\begin{align} \int_0^\tau \int_\Omega \frac{\rho_\varepsilon}{\rho_\infty} (\boldsymbol{U}- \boldsymbol{u}_\varepsilon) \cdot \nabla \Pi \,\mathrm{d}x \,\mathrm{d}t \leq C(\varepsilon+ \varepsilon^\frac{2}{\gamma}). \end{align}\] Back to 26 , we find \[\begin{align} &\left[ \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \right]_{t=0}^{t=\tau} + \int_0^\tau \int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq C(\varepsilon+ \varepsilon^\frac{2}{\gamma}) + C \int_0^\tau \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t, \end{align}\] so that Grönwall’s inequality finally yields \[\begin{align} &\sup_{\tau \in [0,T]} \int_\Omega E_\varepsilon(\rho_\varepsilon, \boldsymbol{u}_\varepsilon| \rho_\infty, \boldsymbol{U})(\tau) \,\mathrm{d}x + \int_0^T\int_\Omega \mathbb{S}(\nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U})) : \nabla (\boldsymbol{u}_\varepsilon- \boldsymbol{U}) \,\mathrm{d}x \,\mathrm{d}t \\ &\leq C \Big( \int_\Omega E_\varepsilon(\rho_{\varepsilon, 0}, \boldsymbol{u}_{\varepsilon, 0} | \rho_\infty, \boldsymbol{U}_0) \,\mathrm{d}x + \varepsilon+ \varepsilon^\frac{2}{\gamma} \Big). \end{align}\] We finish the proof of Theorem 11 by applying Korn’s inequality from Lemma 8. ◻
T. Eiter’s research has been funded by Deutsche Forschungsgemeinschaft (DFG) through grant CRC 1114 “Scaling Cascades in Complex Systems”, Project Number 235221301, Project YIP. F. Oschmann has been supported by the Czech Science Foundation (GAČR) project 22-01591S, and the Czech Academy of Sciences project L100192351. Š. Nečasová was supported by Premium Academia of Š.N. The Institute of Mathematics, CAS is supported by RVO:67985840.