Existence and stability of dissipative turbulent solutions to a simple bi-fluid model of compressible fluids

Abstract

Following Abbatiello et al. (DCCDS-A 41(1):1–28, 2020), we introduce dissipative turbulent solutions to a simple model of a mixture of two non interacting compressible fluids filling a bounded domain with general non zero inflow/outflow boundary conditions. We prove existence of such solutions for all adiabatic coefficients $\gamma >1$, their compatibility with classical solutions, the relative energy inequality, and the weak–strong uniqueness principle in this class. The class of dissipative turbulent solutions is so far the largest class of generalized solutions which still enjoys the weak–strong uniqueness property.

Appendix

We recall here some key lemmas which we have used in the proofs and give more details concerning the meaning of the non-homogeneous boundary data.

The first lemma is proved in [2, Lemma 8.1].

Lemma 7.1

Let $Q = (0,T) \times \Omega$, where $\Omega \subset R^d$ is a bounded domain. Suppose that

$$\begin{aligned} r_n \rightarrow r \ \text{ weakly } \text{ in }\ L^p(Q), \ v_n \rightarrow v \ \text{ weakly } \text{ in }\ L^q(Q),\ p> 1, q > 1, \end{aligned}$$

and

$$\begin{aligned} r_n v_n \rightarrow w \ \text{ weakly } \text{ in }\ L^r(Q), \ r > 1. \end{aligned}$$

In addition, let

$$\begin{aligned}&\partial _t r_n = \mathrm{div}_x\mathbf{g}_n + h_n\ \text{ in }\ \mathcal {D}'(Q),\ \Vert \mathbf{g}_n \Vert _{L^s(Q;\, R^d)} {\mathop {\sim }\limits ^{<}}1,\ \\&\quad s> 1,\ h_n \ \text{ precompact } \text{ in }\ W^{-1,z},\ z > 1, \end{aligned}$$

and

$$\begin{aligned} \left\| \nabla _xv_n \right\| _{\mathcal {M}(Q;\, R^d)} {\mathop {\sim }\limits ^{<}}1 \ \text{ uniformly } \text{ for }\ n \rightarrow \infty . \end{aligned}$$

Then

$$\begin{aligned} w = r v \ \text{ a.a. } \text{ in }\ Q. \end{aligned}$$

The second lemma in Lemma 2.11 and Corollary 2.2 in Feireisl [12].

Lemma 7.2

Let $O \subset \mathbb {R}^d$, $d\ge 2$, be a measurable set and $\{ \mathbf{v}_n \}_{n=1}^{\infty }$ a sequence of functions in $L^1(O;\, \mathbb {R}^M)$ such that

$$\begin{aligned} \mathbf{v}_n \rightharpoonup \mathbf{v} \ \text{ in }\ L^1(O;\, \mathbb {R}^M). \end{aligned}$$

Let $\Phi : R^M \rightarrow (-\infty , \infty ]$ be a lower semi-continuous convex function such that $\Phi (\mathbf{v}_n)$ is bounded in $L^1({ O})$.

Then $\Phi (\mathbf{v}):O\mapsto R$ is integrable and

$$\begin{aligned} \int _{O} \Phi (\mathbf{v})\mathrm{d} x\le \liminf _{n\rightarrow \infty } \int _{O} \Phi (\mathbf{v}_n)\mathrm{d} x. \end{aligned}$$

The last lemma we wish to recall deals with the duals of Bochner spaces. Let X be a Banach space. For $1\le p<\infty$, we introduce Bochner-type spaces

$$\begin{aligned}&L^p(I,X)=\{f:I\rightarrow X \text{ measurable, } \Vert f\Vert ^p_{L^p(I;\,X)}:=\int _I\Vert f(t)\Vert ^p_X\mathrm{d}t<\infty \}, \\&L_\mathrm{weak-*}^p(I,X^*)= \{f:I\rightarrow X^* \text{ weakly-* } \text{ measurable, } \\&\Vert f(\cdot )\Vert _{X^*} \text{ measurable, } \Vert f\Vert ^p_{L^p(I;\,X^*)}:=\int _I\Vert f(t)\Vert ^p_{X^*}\mathrm{d}t<\infty \}. \end{aligned}$$

They are Banach spaces with corresponding norms $\Vert f\Vert ^p_{L^p(I;\,X)}$ resp;\, $\Vert f\Vert ^p_{L^p(I;\,X^*)}$. It is not true in general that $(L^p(I,X))^*$ can be identified with $L^{p'}(I,X^*)$, $1/p+1/p'=1$. However, the following lemma holds, cf. Pedregal [25]:

Lemma 7.3

Let X be a separable Banach space. Then

$$\begin{aligned}(L^p(I,X))^*= L^{p'}_\mathrm{weak-*}(I,X^*)\end{aligned}$$

under the duality mapping

$$\begin{aligned}<f,g>=\int _I<f(t),g(t)>_{X^*,X}\mathrm{d}t. \end{aligned}$$

The particular case we are interested in this paper deal with

$$\begin{aligned} X=C(\overline{\Omega }),\;\,{\text{ whence }}\;\, X^*=\mathcal{M}(\overline{\Omega }), \end{aligned}$$

cf. Rudin [26, Theorem 2.14].

1.1 Non-homogeneous boundary data

Let us explain in details how to understand non-homogeneous Dirichlet boundary conditions. We will follow Lions [21], page 209. We will explain the meaning $\mathbf{u}=\mathbf{u}_B$ on $\partial \Omega \times (0,T)$ and $r = r_B$ on $\{(x,t)\in \partial \Omega , \mathbf{u}_B\cdot \mathbf{n} <0\}$. We assume that the boundary data $(r_B,\mathbf{u}_B)$ satisfy

$$\begin{aligned}&\mathbf{u}_B \in L^2(0,T;\,W^{1,2}(\Omega )) \cap L^{\infty }(0,T;\,L^{\frac{2\gamma }{\gamma - 1}}(\Omega )) \\&\text{ div } \mathbf{u}_B \in L^1(0,T;\,L^{\infty }(\Omega )), \mathbf{D}\mathbf{u}_B \in L^1(0,T;\,L^{\infty }(\Omega ))\end{aligned}$$

($\mathbf{D}\mathbf{u}_B$ we denote the symmetrical part of the stress tensor)

$$\begin{aligned} r_B \in L^{\gamma }(\partial \Omega \times (0,T);\, (\mathbf{u}_B \cdot \mathbf{n})^{-}) dS \otimes dt\end{aligned}$$

which means $\int _0^T dt \int _{\partial \Omega } r_B ^{\gamma }(\mathbf{u}_B \cdot \mathbf{n})^{-} dS \otimes dt < \infty$. Since $\mathbf{u}_B$ is trace of $\mathbf{u}$ then $\mathbf{u}- \mathbf{u}_B \in L^2(0,T;\,W^{1,2}_0(\Omega ))$.

With the density the situation is more complicated since there is no trace of r.

How to understand such case?

Taking the test function $\phi$ in the weak formulation of the continuity equation, $\phi$ smooth in $(0,T)\times \overline{\Omega }$, then together with regularity properties :

$$\begin{aligned} r\in L^{\infty }(0,T;\,L^{\gamma }), \sqrt{r}\mathbf{u}\in L^{\infty }(0,T;\, L^2(\Omega ))\end{aligned}$$

and

$$\begin{aligned}\mathbf{u}\in L^{2}(0,T;\,W^{1,2}(\Omega )), \end{aligned}$$

$r (\mathbf{u}_B \cdot \mathbf{n})$ makes sense on $\partial \Omega \times (0,T)$ as an element of the dual space of the space consisting of traces on $\partial \Omega \times (0,T)$ of the following functional space $\mathcal W$,

$$\begin{aligned} \begin{array}{l} \mathcal{W}= \{\phi ;\, \phi \in C([0,T];\,L^{\frac{\gamma }{\gamma - 1}}(\Omega )), \frac{\partial \phi }{\partial t} \in L^1(0,T;\,L^{\frac{\gamma }{\gamma - 1}}(\Omega )),\\ \phi \in L^1(0,T;\,W^{1,\frac{2\gamma }{\gamma -1}})(\Omega ))+L^2(0,T;\,W^{1,r}(\Omega ))\} \end{array} \end{aligned}$$

(7.1)

where $1/r=5/6-1/\gamma$,$\gamma > 6/5$. This implies that the boundary condition on r means

$$\begin{aligned}r(\mathbf{u}_B\cdot \mathbf{n})^{-} = r_B (\mathbf{u}_B\cdot \mathbf{n})^{-}.\end{aligned}$$