1 Introduction and Main Result

Consider the inhomogeneous Navier–Stokes equations

$$\begin{aligned} \begin{array}{rcl} \rho _t+(u\cdot \nabla )\rho =0\,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \rho u_t-\mu \Delta u + \rho (u \cdot \nabla ) u+\nabla P = 0 \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,u = 0 \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ \rho (0,x)=\rho _0,\quad u(0,x)=u_0 &{}\text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(1.1)

where \(0<T\le \infty \), \(n\ge 2\), \(\mu \) is the dynamic viscosity, and \(\rho \), u, P are density, velocity and pressure of an incompressible flow, respectively. The system (1.1) describes the motion of viscous incompressible flows with variable density and its global well-posedness has been widely studied, (cf. [1,2,3, 7,8,9,10,11,12,13,14,15,16,17, 19] and the references therein).

Writing \(a:=\frac{{\bar{\rho }}-\rho }{\rho }\) with a suitable constant \({\bar{\rho }}>0\), the system (1.1) is reduced to the following equivalent system:

$$\begin{aligned} \begin{array}{rcl} a_t+(u\cdot \nabla )a=0\,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ u_t+(u \cdot \nabla ) u+(1+a)(-\nu \Delta u +\nabla p) = 0 \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,u = 0 \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ a(0,x)=a_0,\quad u(0,x)=u_0 &{}\text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(1.2)

where and in what follows

$$\begin{aligned} \nu \equiv \frac{\mu }{{\bar{\rho }}}\quad \text {and}\quad p\equiv \frac{P}{{\bar{\rho }}}. \end{aligned}$$

The equations of (1.2) are invariant under the scaling

$$\begin{aligned} a_\lambda (t,x)= & {} a(\lambda ^2 t, \lambda x), u_\lambda (t,x)=\lambda u(\lambda ^2 t, \lambda x), \,\, p_\lambda (x,t)=\lambda ^2 p(\lambda ^2 t, \lambda x),\nonumber \\ {a_0}_\lambda= & {} a_0(\lambda x),\,\, {u_0}_\lambda =\lambda u_0(\lambda x),\quad \lambda >0, \end{aligned}$$
(1.3)

and hence it is important to show the existence of solutions to (1.2) in critical spaces, i.e., the function spaces with norms invariant under the scaling (1.3).

In [7], Danchin proved that if \((a_0, u_0)\) belongs to the critical spaces \(({\dot{B}}^{n/2}_{2,r}({{\mathbb {R}}}^n)\cap L^\infty ({{\mathbb {R}}}^n))\times {\dot{B}}^{n/2-1}_{2,1}({{\mathbb {R}}}^n)\) (\(r\in [1,\infty ]\) for \(n=3\) and \(r=1\) for \(n=2\)) and \(\Vert a_0\Vert _{{\dot{B}}^{n/2}_{2,\infty }}+\Vert u_0\Vert _{{\dot{B}}^{n/2-1}_{2,1}}\) is small enough, then (1.2) has a unique solution \((\rho ,u)\) such that

$$\begin{aligned}&a\in BC([0,T),{\dot{B}}^{n/2}_{2,r}({{\mathbb {R}}}^n))\cap L^\infty (0,T; L^\infty ({{\mathbb {R}}}^n)),\\&u\in BC([0,T),{\dot{B}}^{n/2-1}_{2,1}({{\mathbb {R}}}^n))\cap L^1(0,T; {\dot{B}}^{n/2+1}_{2,1}({{\mathbb {R}}}^n)). \end{aligned}$$

This result was generalized by Abidi [1] to the case where the spaces for (au) are extended to \({\dot{B}}^{n/q}_{q,1}({{\mathbb {R}}}^n)\times {\dot{B}}^{-1+n/q}_{q,1}({{\mathbb {R}}}^n)\), \(1<q<2n\), showing existence for \(1<q<2n\) and uniqueness for \(1<q\le n\); the gap in the uniqueness for \(n<q<2n\) was filled by Danchin and Mucha [10] via Lagrangian approach. Abidi, Gui and Zhang in [2] proved global well-posedness of (1.1) without smallness of initial density variation \(a_0\in B^{3/2}_{2,1}({{\mathbb {R}}}^3)\) for \(u_0\in {\dot{B}}^{1/2}_{2,1}({{\mathbb {R}}}^3)\) whose norm is small depending on \(\Vert a_0\Vert _{B^{3/2}_{2,1}({{\mathbb {R}}}^3)}\); the result was extended in [3] to the case \((a_0,u_0)\in B^{3/q}_{q,1}({{\mathbb {R}}}^3)\times {\dot{B}}^{-1+3/p}_{p,1}({{\mathbb {R}}}^3)\), \(q\in [1,2]\), \(p\in [3,4]\), \(\frac{1}{p}+\frac{1}{q}>\frac{5}{6}\), \(\frac{1}{q}-\frac{1}{p}\le \frac{1}{3}\). Recently, in [23], global well-posedness of (1.1) with \((a_0,u_0)\in B^{3/q}_{q,1}({{\mathbb {R}}}^3)\times {\dot{B}}^{-1+3/p}_{p,1}({{\mathbb {R}}}^3)\), \(1<q\le p<6\), \(\frac{1}{p}+\frac{1}{q}>\frac{1}{2}\), \(\frac{1}{q}-\frac{1}{p}\le \frac{1}{3}\), was proved under the condition that \(\Vert a_0\Vert _{BMO}\) and \(\Vert u_0\Vert _{{\dot{B}}^{-1+3/p}_{p,1}({{\mathbb {R}}}^3)}\) are small enough. Here, we note that consideration of the system (1.1) or (1.2) with initial velocity in a type of critical Besov spaces \({\dot{B}}^s_{q,1}\) is mathematically ideal for showing both existence and uniqueness of a solution in the same class of functions, since in this case the velocity can be found in \(L^1(0,T; C_{Lip}({{\mathbb {R}}}^n))\), where \(C_{Lip}({{\mathbb {R}}}^n)\) denotes the set of all Lipschitz continuous functions on \({{\mathbb {R}}}^n\), thanks to the maximal \(L^1\)-regularity property of the Stokes operator in \({\dot{B}}^s_{q,1}\)-type Besov spaces.

On the other hand, it is a common interest to study (1.2) (equivalently (1.1)) when the velocity does not belong to \(L^1(0,T; C_{Lip}({\mathbb {R}}^n))\) and initial density is merely bounded, cf. [12, 14]. Huang et al. [14] proved existence of a global solution to (1.2) under a smallness condition of \(a_0\in L^\infty ({{\mathbb {R}}}^n)\) and \(u_0\in {\dot{B}}^{-1+n/q}_{q,r}({{\mathbb {R}}}^n)\), \(q\in (1,n), r\in (1,\infty )\), and uniqueness of such solution under a slightly higher regularity assumption on initial velocity \(u_0\); this result is extended to the half-space setting by Danchin and Zhang [12]. The main ideas of [14] and [12] are to employ, for existence, maximal \(L^p\)-regularity for the Stokes operator in Lebesgue spaces and, for uniqueness, a Lagrangian approach which was exploited in [10].

Global well-posedness for initial boundary value problem corresponding to (1.1) on bounded domains \(\Omega \subset {{\mathbb {R}}}^n\) with \(\rho _0\in L^\infty (\Omega )\) was considered in [15] and [11] for some regular and small initial velocity in Sobolev and Besov spaces, while in [19] for small piecewise constant \(a_0\) and \(u_0\in B_{q,\infty }^0(\Omega )\), \(q\ge n\ge 2\).

In this paper, we prove global well-posedness of (1.1) for \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\), \(n\le q<2n\), and \(\rho _0\in L^\infty ({{\mathbb {R}}}^n)\) positive away from 0 provided the corresponding norms of initial velocity and initial density variation is suitably small. For uniqueness of a solution we require a slightly higher regularity of the initial velocity.

Before introducing the main result of the paper, we need to give some notations. For a linear normed space X the notation \(X'\) stands for the dual space of X. We always denote the conjugate number of \(q\in (1,\infty )\) by \(q'\), i.e. \(q'=q/(q-1)\). Let \([\cdot ,\cdot ]_\theta \), \((\cdot ,\cdot )_{\theta ,r}\) and \((\cdot ,\cdot )^0_{\theta ,\infty }\) for \(\theta \in (0,1)\), \(1\le r\le \infty \) be complex, real and continuous interpolation functors, respectively, see [5, 21] for real and complex interpolation functors, and see e.g. [4], §§2.4.4, §2.5, [21] §§1.11.2, page 69 for continuous interpolation functors. We use standard notation \(L^q, H^s_q, B^s_{q,r}\) for Lebesgue spaces, Bessel potential spaces and Besov spaces, respectively, without distinguishing whether or not it is the space of scalar-valued functions or vector-valued functions. For \(q\in (1,\infty )\) and \(s\in {{\mathbb {R}}}\) let \(b^{s}_{q,\infty }({{\mathbb {R}}}^n)\) be the little Nicolskii space defined by the completion of \(H^s_q({{\mathbb {R}}}^n)\) in \(B^s_{q,\infty }({{\mathbb {R}}}^n)\).

Given \(\gamma \in (0,1]\), \(0<T\le \infty \) and Banach space X, let

$$\begin{aligned} L^\infty _\gamma (0,T; X):=\{f: t^{1-\gamma } f\in L^\infty (0,T;X)\}, \quad \Vert f\Vert _{L^\infty _\gamma (0,T; X)}:=\Vert t^{1-\gamma } f(t)\Vert _{L^\infty (0,T; X)}. \end{aligned}$$

The space of all pointwise multipliers in Y is denoted by \({{\mathcal {M}}}(Y)\), i.e.,

$$\begin{aligned} {{\mathcal {M}}}(Y):= \{f: f\varphi \in Y,\varphi \in Y\},\; \Vert f\Vert _{{\mathcal {M}}(Y)} :=\displaystyle \sup _{\Vert \varphi \Vert _{Y}\le 1}\Vert f\varphi \Vert _{Y}. \end{aligned}$$

The characteristic function for a subset G of \({{\mathbb {R}}}^n\) is denoted by \(\chi _G\). We denote the tensor product of two tensors a, b by \(a\otimes b\) and by \(A:B=\sum _{i,j}a_{ij}b_{ij}\) for two matrices \(A=(a_{ij})_{1\le i,j\le n}\) and \(B=(b_{ij})_{1\le i,j\le n}\).

Definition 1.1

Let \(2\le n\le q<2n\), \(0<T\le \infty \) and let \(s\in (0,1)\). We say that a pair of functions \((\rho , u)\) is a solution to (1.1) if it satisfies the followings:

  1. (i)
    $$\begin{aligned} \rho \in L^\infty (0,T; L^\infty ({{\mathbb {R}}}^n)),\; u\in L^\infty _{loc}([0,T), B^{1+n/q-s}_{q,\infty }({{\mathbb {R}}}^n)),\quad \text {div}\,u=0. \end{aligned}$$
    (1.4)
  2. (ii)

    Two identities

    $$\begin{aligned} \int _0^T\int _{{{\mathbb {R}}}^n}(\rho \psi _t+\rho u\cdot \nabla \psi )\,dxdt +\int _{{{\mathbb {R}}}^n} \rho _0\psi (0,\cdot )\,dx=0,\; \forall \psi \in C_0^1([0,T)\times {{\mathbb {R}}}^n), \end{aligned}$$
    (1.5)

    and

    $$\begin{aligned}&\displaystyle \int _0^T\int _{{{\mathbb {R}}}^n}[\rho u\cdot \varphi _t+\mu u\cdot \Delta \varphi +\rho u\otimes u:\nabla \varphi ]\,dxdt +\int _{{{\mathbb {R}}}^n} \rho _0u_0\cdot \varphi (0,\cdot )\,dx=0,\nonumber \\&\quad \forall \varphi \in C_0^\infty ([0,T)\times {{\mathbb {R}}}^n)^n\; (\text {div}\,\varphi =0), \end{aligned}$$
    (1.6)

    hold true.

Note that if \((\rho , u)\) satisfies (1.4), the integrals of (1.5) and (1.6) make a sense due to \(u\in L^\infty _{loc}([0,T)\times {{\mathbb {R}}}^n)\) by Sobolev embedding. If \((\rho , u)\) is a solution to (1.1) in the sense of Definition 1.1, it follows by a standard argument using De-Rham’s lemma that there is a distribution P, associated pressure, such that \(\rho , u\) and P satisfy the equations of (1.1) in the sense of distribution and the initial conditions in (1.1) are satisfied in a weak sense. Hence, if necessary in the below, the triple \((\rho , u, \nabla P)\) will also be called a solution to (1.1).

The main result of the paper is stated as follows:

Theorem 1.1

Let \(0<T\le \infty \) and let \( 2\le n\le q<2 n\). Let

$$\begin{aligned} \rho _0\in L^\infty ({{\mathbb {R}}}^n), \quad \rho _{01}\le \rho _0(x)\le \rho _{02} \end{aligned}$$

with positive constants \(\rho _{0i},i=1,2\), \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\cap B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\), \(r\ge q\), and \(\text {div}\,u_0=0\). Then, for any \(s\in (0,\min \{\frac{n-1}{q},\frac{2n}{q}-1\})\) there are some constants \(\delta _i=\delta _i(q,n,s)>0, i=1,2,\) independent of T, such that if

$$\begin{aligned} \frac{\rho _{02}-\rho _{01}}{\rho _{01}}<\delta _1,\quad \Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }}<\frac{\mu \delta _2}{\rho _{02}}, \end{aligned}$$
(1.7)

then (1.1) has a solution \((\rho ,u,\nabla P)\) satisfying (1.4)–(1.6) and, in addition,

$$\begin{aligned} u_t,\nabla ^2 u, \nabla P \in L^\infty _{s/2}(0,T; B^{-1+n/q-s}_{q,1}({{\mathbb {R}}}^n)\cap B^{-1+n/q-s}_{r,1}({{\mathbb {R}}}^n)). \end{aligned}$$
(1.8)

The solution \((\rho ,u,\nabla P)\) is unique in the class of functions satisfying (1.4) and (1.8) with \(r>q\) and \(s<\frac{n}{q}-\frac{n}{r}\).

Remark 1.2

By Theorem 1.1, in particular, we have an existence result when \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\), while uniqueness result when \(u_0\in B^{-1+n/q+\varepsilon }_{q,\infty }({{\mathbb {R}}}^n)\) for any small \(\varepsilon >0\).

The proof of Theorem 1.1 has two aspects, i.e., existence part and uniqueness part for a solution to the system (1.1).

The existence of a solution is proved via an iterative scheme for the momentum equations with fixed density and the transfer equation with fixed velocity. First, we need to consider solvability for momentum equations, that is,

$$\begin{aligned} \begin{array}{rcl} u_t-\nu \Delta u +(u \cdot \nabla ) u+\nabla p =a(\nu \Delta u -\nabla p),\;\text {div}\,u = 0 \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ u(0,x)=u_0 &{}\text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(1.9)

with \(a\equiv \frac{{\bar{\rho }}}{\rho }-1, {\bar{\rho }}\in [\rho _0,\rho _1],\) fixed; to this end, we rely on maximal \(L^\infty _\gamma \)-regularity of the Stokes operator. It is known that the unsteady Stokes system

$$\begin{aligned} \begin{array}{rcl} v_t-\nu \Delta v +\nabla p = f,\;\text {div}\,v = 0 &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ v(0,x)=v_0 &{}\text {in}&{}{{\mathbb {R}}}^n \end{array} \end{aligned}$$

with \(v_0\in B^{\alpha +\gamma }_{q,\infty }({{\mathbb {R}}}^n)\) and \(f\in L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\) for \(\gamma \in (0,1)\), \(\alpha \in (-\frac{1}{2},1)\), \(2\alpha -1/q\notin {{\mathbb {Z}}}\), \(\alpha +\gamma <1\), has a unique solution such that \(\nabla ^2 v,\nabla p\in L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\), see [18]. In order to apply this property to solvability for (1.9) using a fixed point argument, the term \(a(\nu \Delta u -\nabla p)\) should belong to \(L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\). For this reason, we prove a multiplication property of \(L^\infty \)-functions in little Nicolskii spaces \(b^{\tau }_{q,\infty }, \tau \in (-1+1/q,1/q),\) (also in corresponding Bessel potential and Besov spaces), which is of independent significance. Then, applying the above maximal \(L^\infty _\gamma \)-regularity result for the Stokes operator with \(\gamma =s/2,\alpha =-1/2+n/(2q)-s/2\) for sufficiently small \(s>0\), we obtain solvability for (1.9). Finally, the existence of a solution to (1.1) is proved via an iterative scheme where the density transport equation involves the velocity as a regularization of the solution to (1.9) since it does not belong to \(L^1(0,T; C_{Lip}({{\mathbb {R}}}^n))\), generally.

The uniqueness part of Theorem 1.1 is proved by Lagrangian approach similarly as in [10, 12] and [14], but Theorem 3.3 on the unsteady Stokes system with nonzero divergence and Lemma 5.1 on pointwise multiplication in Besov spaces are essentially used.

The remaining part of the paper is organized as follows. In Sect. 2 we show a multiplication property of \(L^\infty \)-functions in Bessel potential, Besov and little Nicolskii spaces. Section 3 is devoted to solvability for momentum equations involved in (1.1) with fixed variable density. The proof for the existence part of Theorem 1.1 is given in Sect. 4, while the uniqueness part in Sect. 5.

Throughout the paper, we denote the estimate constants appearing in inequalities by the same symbol c or C as long as no confusion arises.

2 A Multiplication Property of \(L^\infty \)-Functions

Lemma 2.1

Let \(\Omega \) be a domain of \({{\mathbb {R}}}^n\), \(n\in {{\mathbb {N}}}\), and let \(1<q<\infty \) and \(s\in (-1+1/q,1/q)\). Then, for any open subset \(\Omega '\) of \(\Omega \)

$$\begin{aligned} \Vert \chi _{\Omega '}f\Vert _{H^s_q(\Omega )}\le \Vert f\Vert _{H^s_q(\Omega )},\quad \forall f\in H^s_q(\Omega ), \end{aligned}$$

where the space \(H^{s}_{q}(\Omega )\) is endowed with the norm of the complex interpolation space \([H^{-1}_{q}(\Omega ), H^1_{q,0}(\Omega )]_{\frac{s+1}{2}}\).

Proof

First, suppose that \(s\in (0,1/q)\). Define the operator \(E_0: L^q(\Omega ')\mapsto L^q(\Omega )\) by \(E_0f:= {\tilde{f}}\), where \({\tilde{f}}\) is the extension of f by zero on \(\Omega \setminus \Omega '\). Then, obviously,

$$\begin{aligned} \Vert E_0 f\Vert _{L^q(\Omega )}\le & {} \Vert f\Vert _{{{\mathcal {L}}}(L^q(\Omega '))},\quad \forall f\in L^q(\Omega '),\\ \Vert E_0 f\Vert _{H^1_{q,0}(\Omega )}\le & {} \Vert f\Vert _{H^1_{q,0}(\Omega ')},\quad \forall f\in H^1_{q,0}(\Omega '), \end{aligned}$$

where and in what follows, \(H^1_{q,0}(G)\) for \(1<q<\infty \), denotes the closures of \(C^\infty _0(G)\) in \(H^1_q(G)\). Hence, by complex interpolation \([L^q,H^1_{q,0}]_s=H_q^s\) for \(s\in (0,1/q)\), we have

$$\begin{aligned} \Vert E_0 f\Vert _{H^s_{q}(\Omega )}\le \Vert f\Vert _{H^s_{q}(\Omega ')},\quad \forall f\in H^s_{q}(\Omega '), \end{aligned}$$

where note that the complex interpolation functor is an exact interpolation functor, cf. [5].

On the other hand, for \(r_{\Omega '}\) being the restriction operator onto \(\Omega '\subset \Omega \), we have

$$\begin{aligned} \Vert r_{\Omega '}f\Vert _{H^s_{q}(\Omega ')}\le \Vert f\Vert _{H^s_{q}(\Omega )}, \quad \forall f\in H^s_{q}(\Omega ). \end{aligned}$$

Therefore, since \(\chi _{\Omega '}f=E_0r_{\Omega '}f\), we get

$$\begin{aligned} \Vert \chi _{\Omega '}f\Vert _{H^s_{q}(\Omega )}\le \Vert f\Vert _{H^s_{q}(\Omega )}, \quad \forall f\in H^s_{q}(\Omega ). \end{aligned}$$
(2.1)

The assertion of the lemma for the case \(s\in (-1+1/q,0)\) follows by duality argument using the assertion for \(s\in (0,1/q)\).

Finally, the assertion for the case \(s=0\) directly follows by interpolation. \(\square \)

Lemma 2.2

Let \(\Omega \) be a domain of \({{\mathbb {R}}}^n\), \(n\in {{\mathbb {N}}}\), and let \(M\subset L^\infty (\Omega )\) be the set of all finite or countable summations \(g(x)=\sum _{k}a_k\chi _{\Omega _k}(x)\) such that \(a_k\in {{\mathbb {R}}}\) and \(\Omega _k\) ’s are disjoint Lipschitz subdomains of \(\Omega \) .

Proof

Denote the Lebesgue measure by m. It is known that for every \(f\in (L^\infty (\Omega ))'\) there is a unique \(m-\)absolutely continuous finitely additive set function \(\mu _f\) defined on m-measurable subsets of \(\Omega \) such that

$$\begin{aligned} \langle f, g\rangle _{(L^\infty (\Omega ))', L^\infty (\Omega )}=\int _\Omega g(x)\mu _f(dx),\;\forall g\in L^\infty (\Omega ). \end{aligned}$$
(2.2)

see [22], p.118.

Now, let \(f\in (L^\infty (\Omega ))'\) be such that \(\langle f, g\rangle _{(L^\infty (\Omega ))', L^\infty (\Omega )}=0\) for all \(g\in M\). Note that for the union \(\Pi =\bigcup _j Q_j\) of any finite or countable number of disjoint (open) cubes \(Q_j\subset \Omega \) which are parallel to the coordinate axis of \({{\mathbb {R}}}^n\) one has \(\chi _{\Pi }\in M\) and hence

$$\begin{aligned} \mu _f(\Pi )=\int _\Omega \chi _{\Pi }(x)\mu _f(dx)=0. \end{aligned}$$
(2.3)

Moreover, since \(\mu _f\) is \(m-\)absolutely continuous, for every \(\varepsilon >0\) there is \(\delta =\delta (\varepsilon )>0\) such that

$$\begin{aligned} m(A)<\delta \Rightarrow |\mu _f(A)|<\varepsilon . \end{aligned}$$
(2.4)

Let G be any \(m-\)measurable subset of \(\Omega \). Then, there is a finite or countable number of disjoint cubes \(Q_j\subset \Omega \) such that

$$\begin{aligned} m\big (G\, \ominus \,\Pi \big )<\delta \end{aligned}$$
(2.5)

holds for \(\Pi =\cup _j Q_j\), where “\(\ominus \)” denotes symmetric difference of two sets. In fact, if \(m(G)<\infty \), then, by the definition of the Lebesgue measure, (2.5) obviously holds with some finite disjoint cubes \(Q_j\). Let \(m(G)=\infty \). Then, by \(\sigma \)-finiteness of Lebesgue measure, G can be expressed as a union \(G=\cup _{k=1}^\infty G_k\) of m-measurable sets \(G_k\) with \(m(G_k)<\infty \), where without loss of generality \(G_k\)’s can be regarded as disjoint with each other. For each \(G_k\) one can find \(\Pi _k\subset \Omega \) which is the union of a finite number of disjoint cubes and satisfies \(m(G_k\,\ominus \,\Pi _k)<2^{-k}\delta \). Then, \(\Pi \equiv \cup _{k=1}^\infty \Pi _k\) satisfies (2.5), where \(\Pi _k\)’s also can be regarded as disjoint with each other.

By (2.3)–(2.5), we have

$$\begin{aligned} |\mu _f(G)|\le \big |\mu _f(\Pi )\big |+|\mu _f(G\,\Delta \, \Pi )|<\varepsilon . \end{aligned}$$

Thus we have \(\mu _f(G)=0\), since \(\varepsilon >0\) is arbitrarily taken.

Therefore, by (2.2) we have \( \langle f, g\rangle _{(L^\infty (\Omega ))', L^\infty (\Omega )}=0\) for all \(g\in L^\infty (\Omega )\), and hence \(f=0\). Thus M is dense in \(L^\infty (\Omega )\). \(\square \)

Proposition 2.3

Let \(\Omega \) be a domain of \({{\mathbb {R}}}^n,n\in {{\mathbb {N}}}\), and let \(q\in (1,\infty ), r\in [1,\infty ]\) and \(s\in (-1+1/q,1/q)\). Then, every \(f\in L^\infty (\Omega )\) is a pointwise multiplier of \(H^s_{q}(\Omega ), B^s_{q,r}(\Omega )\) and \(b^s_{q,\infty }(\Omega )\), respectively, and

$$\begin{aligned} \Vert fu\Vert _{Y}\le C\Vert f\Vert _{L^\infty (\Omega )}\Vert u\Vert _{Y},\;\forall u\in Y, \end{aligned}$$
(2.6)

where \(Y=H^s_{q}(\Omega ), B^s_{q,r}(\Omega )\) and \(b^s_{q,\infty }(\Omega )\) and the constant \(C>0\) depends only on qrs and \(\Omega \).

Proof

Let \(f\in L^\infty (\Omega )\) and \(f_k=\sum _i a_{k_i}\chi _{\Omega _{k_i}}, k=1,2,\ldots ,\) be a sequence converging to f in \(L^\infty (\Omega )\) as \(k\rightarrow \infty \), where summation is finite or countable, \(a_{k_i}\in {{\mathbb {R}}}\), and \(\Omega _{k_i}\subset \Omega \) are disjoint Lipschitz subdomains. Note that

$$\begin{aligned} \Vert f_k\Vert _{L^\infty (\Omega )}=\sup _{i}|a_{k_i}|\quad \text {and}\quad \lim _{k\rightarrow \infty }\sup _{i}|a_{k_i}|=\Vert f\Vert _{L^\infty (\Omega )}. \end{aligned}$$

Let the operator \(E_0: L^q(\cup _i\Omega _{k_i})\mapsto L^q(\Omega )\) by \(E_0f:= {\tilde{f}}\), where \({\tilde{f}}\) is the extension of f by zero on \(\Omega \setminus \cup _i\Omega _{k_i}\). We have

$$\begin{aligned} \Vert E_0(f_kv)\Vert _{L^q(\Omega )}= \Vert f_kE_0v\Vert _{L^q(\Omega )}\le \Vert f_k\Vert _{L^\infty (\Omega )}\Vert v\Vert _{L^q(\cup _i\Omega _{k_i})},\,\forall v\in L^q(\cup _i\Omega _{k_i}). \end{aligned}$$
(2.7)

On the other hand, it follows that if \(\varphi \in H^1_{q,0}(\cup _{i}\Omega _{k_i})\), then

$$\begin{aligned} E_0\varphi \in H^1_{q,0}(\Omega ),\quad \Vert E_0\varphi \Vert _{H^1_{q,0}(\Omega )}=\Vert \varphi \Vert _{H^1_{q,0}(\cup _{i}\Omega _{k_i})}. \end{aligned}$$
(2.8)

In fact, since \(\Omega _{k_i}\)’s are disjoint domains, we have

$$\begin{aligned} \Vert \varphi \Vert ^q_{H^1_{q,0}(\cup _{i}\Omega _{k_i})}=\sum _{i}\Vert \varphi |_{\Omega _{k_i}} \Vert ^q_{H_{q,0}^1(\Omega _{k_i})}<\infty . \end{aligned}$$
(2.9)

Moreover,

$$\begin{aligned} E_0\varphi =\sum _i E_{i0}(\varphi |_{\Omega _{k_i}}), \end{aligned}$$
(2.10)

where \(E_{i0}\) is an operator of extension by 0 from \(\Omega _{ki}\) onto \(\Omega \); since \(\Omega _{k_i}\) is a Lipschitz domain, we have

$$\begin{aligned} E_{i0}(\varphi |_{\Omega _{k_i}})\in H^1_{q,0}(\Omega ),\quad \Vert E_{i0}(\varphi |_{\Omega _{k_i}})\Vert _{H^1_{q,0}(\Omega )}=\Vert \varphi |_{\Omega _{k_i}}\Vert _{H^1_{q,0} (\Omega _{k_i})}. \end{aligned}$$
(2.11)

Hence, if \(\cup _i\Omega _{k_i}\) is finite union, then, obviously, \(E_0\varphi \in H^1_{q,0}(\Omega )\) and, by (2.9) and (2.11), we have

$$\begin{aligned} \Vert E_0\varphi \Vert ^q_{H^1_{q,0}(\Omega )}= \sum _i \Vert E_{i0}(\varphi |_{\Omega _{k_i}})\Vert ^q_{H^1_{q,0}(\Omega )} =\sum _i \Vert \varphi |_{\Omega _{k_i}}\Vert ^q_{H^1_{q,0}(\Omega _{k_i})}=\Vert \varphi \Vert ^q_{H^1_{q,0}(\Omega )}. \end{aligned}$$

If \(\cup _i\Omega _{k_i}\) is infinite countable union, then, in view of (2.9),

$$\begin{aligned} \big \Vert \sum _{N_1<i<N_2} E_{i0}(\varphi |_{\Omega _{k_i}})\big \Vert ^q_{H^1_{q,0}(\Omega )} =\sum _{N_1<i<N_2}\Vert \varphi |_{\Omega _{k_i}}\Vert ^q_{H_{q,0}^1(\Omega _{k_i})}\rightarrow 0\quad \text {(as }N_1,N_2\rightarrow \infty ). \end{aligned}$$

Hence, it follows by (2.10) that \(E_0\varphi \in H^1_{q,0}(\Omega )\) and

$$\begin{aligned} \Vert E_0\varphi \Vert ^q_{H^1_{q,0}(\Omega )}= \lim _{N\rightarrow \infty }\big \Vert \sum _{i=1}^N E_{i0}(\varphi |_{\Omega _{k_i}})\big \Vert ^q_{H^1_{q,0}(\Omega )} =\lim _{N\rightarrow \infty }\sum _{i=1}^N \Vert \varphi |_{\Omega _{k_i}}\Vert ^q_{H^1_{q,0}(\Omega _{k_i})}=\Vert \varphi \Vert ^q_{H^1_{q,0}(\Omega )}. \end{aligned}$$

Thus, (2.8) is proved.

For all \( v\in H^1_{q,0}(\cup _i\Omega _{k_i})\), it follows that \(f_k v=\sum _i a_{k_i}\chi _{\Omega _{k_i}}v\in H^1_{q,0}(\cup _{i}\Omega _{k_i})\) and

$$\begin{aligned} \Vert f_k v\Vert ^q_{H^1_{q,0}(\cup _{i}\Omega _{k_i})}= & {} \sum _i |a_{k_i}|^q\Vert v\Vert ^q_{H^1_{q,0}(\Omega _{k_i})}\\\le & {} \sup _i |a_{k_i}|^q\sum _i\Vert v\Vert ^q_{H^1_{q,0}(\Omega _{k_i})}\\= & {} \Vert f_k\Vert ^q_\infty \Vert v\Vert ^q_{H^1_{q,0}(\cup _i\Omega _{k_i})}. \end{aligned}$$

Hence, by (2.8),

$$\begin{aligned} E_0(f_kv)\in H^1_{q,0}(\Omega ),\quad \Vert E_0(f_kv)\Vert _{H^1_{q,0}(\Omega )}\le \Vert f_k\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^1_{q,0}(\cup _i\Omega _{k_i})},\, \forall v\in H^1_{q,0}(\cup _i\Omega _{k_i}). \end{aligned}$$
(2.12)

Now, for the moment let us assume that the space \(H^s_q\) for \(s\in (0,1/q)\) is endowed with norm of the complex interpolation space \([H_q^{-1}, H^1_{q,0}]_{\frac{s+1}{2}}\). Then, since the complex interpolation functor is an exact interpolation functor, cf. [5], it follows by (2.7) and (2.12) that

$$\begin{aligned} \Vert E_0(f_kv)\Vert _{H^s_q(\Omega )}\le \Vert f_k\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^s_{q}(\cup _i\Omega _{k_i})},\, \forall v\in H^s_{q}(\cup _i\Omega _{k_i}). \end{aligned}$$
(2.13)

On the other hand, denoting the restriction operator onto \(\cup _i\Omega _{k_i}\) by R, we have

$$\begin{aligned} E_0R(f_kv)=f_kv, \quad \Vert Rv\Vert _{H^s_q(\cup _i\Omega _{k_i})}\le \Vert v\Vert _{H^s_q(\Omega )},\;\forall v\in H^s_q(\Omega ). \end{aligned}$$
(2.14)

Hence, we get from (2.13), (2.14) that

$$\begin{aligned} \Vert f_kv\Vert _{H^s_q(\Omega )}= & {} \Vert E_0R(f_kv)\Vert _{H^s_q(\Omega )}\\= & {} \Vert E_0(f_kRv)\Vert _{H^s_q(\Omega )}\\\le & {} \Vert f_k\Vert _{L^\infty (\Omega )}\Vert Rv\Vert _{H^s_{q}(\cup _i\Omega _{k_i})}\\\le & {} \Vert f_k\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^s_{q}(\Omega )},\, \forall v\in H^s_{q}(\Omega ), \end{aligned}$$

and, in particular,

$$\begin{aligned} f_k\in {{\mathcal {M}}}(H^s_q(\Omega )),\; \Vert f_k\Vert _{{{\mathcal {M}}}(H^s_q(\Omega ))}\le \Vert f_k\Vert _{L^\infty (\Omega )},\, \forall f_k\in M. \end{aligned}$$
(2.15)

Moreover, by (2.15), we have

$$\begin{aligned} \Vert f_k-f_m\Vert _{{{\mathcal {M}}}(H^s_{q}(\Omega ))}\le \Vert f_k-f_m\Vert _{L^\infty (\Omega )}\rightarrow 0\quad \text {as }k,m\rightarrow \infty . \end{aligned}$$

Thus \(\{f_k\}\) becomes a Cauchy sequence in \({{\mathcal {M}}}(H^s_{q}(\Omega ))\) as well, and \(f_k\) converges to some \({\tilde{f}}\in {\mathcal M}(H^s_{q}(\Omega ))\) as \(k\rightarrow \infty \). Then we have \(f={\tilde{f}}\) since \(f_k\) converges to both f and \({\tilde{f}}\) in the Banach space \({{\mathcal {M}}}(H^s_{q}(\Omega ))+L^\infty (\Omega )\) as \(k\rightarrow \infty \), where note that both the space \({{\mathcal {M}}}(H^s_{q}(\Omega ))\) and \(L^\infty (\Omega )\) are contained in the distribution spaces \({\mathcal D}'(\Omega )\) and hence the sum of the two spaces may be defined. Thus, we have \(L^\infty (\Omega )\hookrightarrow {{\mathcal {M}}}(H^s_{q}(\Omega ))\) and, by (2.15),

$$\begin{aligned} \Vert f\Vert _{{{\mathcal {M}}}(H^s_q(\Omega ))}\le \Vert f\Vert _{L^\infty (\Omega )}, \,\forall f\in L^\infty (\Omega ), \end{aligned}$$

yielding

$$\begin{aligned} \Vert fv\Vert _{H^s_q(\Omega )}\le \Vert f\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^s_q(\Omega )}, \,\forall f\in L^\infty (\Omega ),v\in H^s_q(\Omega ), s\in [0,1/q). \end{aligned}$$

Thus, without the assumption that \(H^s_q(\Omega )\) is endowed with the norm of the complex interpolation space \([H_q^{-1}(\Omega ), H^1_{q,0}(\Omega )]_{\frac{s+1}{2}}\), one generally has

$$\begin{aligned} \Vert fv\Vert _{H^s_q(\Omega )}\le c(q,s,\Omega )\Vert f\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^s_q(\Omega )}, \,\forall f\in L^\infty (\Omega ),v\in H^s_q(\Omega ), s\in [0,1/q). \end{aligned}$$
(2.16)

It then directly follows from (2.16) by standard duality argument that

$$\begin{aligned} \Vert fv\Vert _{H^s_q(\Omega )}\le c(q,s,\Omega )\Vert f\Vert _{L^\infty (\Omega )}\Vert v\Vert _{H^s_q(\Omega )}, \,\forall f\in L^\infty (\Omega ),v\in H^s_q(\Omega ), s\in (-1+1/q,0). \end{aligned}$$
(2.17)

Thus, the assertion of the lemma is proved for \(Y=H^s_q(\Omega )\).

The remaining assertions for \(Y=B^s_{q,r}(\Omega ), 1\le r\le \infty ,\) and \(b^s_{q,\infty }(\Omega )\) follows directly from (2.16) and (2.17) by standard argument using real and continuous interpolation, respectively.

The proof of the lemma is complete. \(\square \)

Remark 2.4

From the proof of Proposition 2.3, it is easy to see that if the Banach spaces \(Y=H^s_{q}(\Omega ), B^s_{q,r}(\Omega )\) and \(b^s_{q,\infty }(\Omega )\) for \(s\in (-1+1/q,1/q)\) are endowed with the norm of the interpolation spaces between \(H^{-1}_q(\Omega )\) and \(H^{1}_{q,0}(\Omega )\) by the complex interpolation functor, real and continuous interpolation functors by K-method, which are exact, respectively, (cf. [5]), then the constant C in (2.6) may be regarded as 1.

3 Solvability for Momentum Equations with Fixed Density

Let \(\rho \in L^\infty (0,T;L^\infty (\Omega ))\) be fixed such that \(\rho _{01}\le \rho (t,x)\le \rho _{02}\) with some positive constants \(\rho _{01},\rho _{02}\) and let \(\frac{1}{\rho }=\frac{1}{{\bar{\rho }}}(1+a)\) with a constant \({\bar{\rho }}\in [\rho _{01}, \rho _{02}]\) and \(a=a(t,x)\in L^\infty ((0,T)\times {{\mathbb {R}}}^n)\). Then the momentum equations in (1.1) may be formally written as

$$\begin{aligned} \begin{array}{rcccl} u_t-\nu \Delta u + (u \cdot \nabla ) u+\nabla p &{}=&{} a(\nu \Delta u-\nabla p) \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,u &{}=&{} 0 \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ u(0,x) &{}=&{} u_0 &{}\text {in}&{}{{\mathbb {R}}}^n. \end{array} \end{aligned}$$
(3.1)

Beforehand, we consider the unsteady Stokes equations with nonzero divergence.

3.1 Unsteady Stokes Equations

Consider the Cauchy problem for the unsteady Stokes equations

$$\begin{aligned} \begin{array}{rcl} u_t-\nu \Delta u +\nabla p = f \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,u = g \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ u(0,x)=u_0 &{}\text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(3.2)

where \(n\ge 2\) and \(0<T\le \infty \).

Lemma 3.1

There is a unique solution operator \({{\mathcal {R}}}\) for the problem

$$\begin{aligned} -\nu \Delta u+\nabla p =0,\;\text {div}\,u = g\quad \text {in }{{\mathbb {R}}}^n, \end{aligned}$$
(3.3)

such that \(u={{\mathcal {R}}}g\) for \(g\in {{\mathcal {S}}'}\) becomes a solution to (3.3) in the sense of distribution and for \(q\in (1, \infty )\) and \(\alpha \in {{\mathbb {R}}}\)

$$\begin{aligned} \Vert {{\mathcal {R}}}\text {div}\,h\Vert _{F^{\alpha }_{q}({{\mathbb {R}}}^n)} \le c\Vert h\Vert _{F^{\alpha }_{q}({{\mathbb {R}}}^n)},\forall h\in F^{\alpha }_{q}({{\mathbb {R}}}^n), \end{aligned}$$
(3.4)

where the constant \(c>0\) is independent of \(q,\nu \) and \(\alpha \), and \(F_q^\alpha \in \{H^\alpha _q, B^\alpha _{q,r} (1\le r\le \infty ), b^\alpha _{q,\infty }\}\).

Proof

It is formally checked that \(u=\Delta ^{-1}\nabla g\) and \(\nabla p=\nu \nabla g\) uniquely solve (3.3). Defining \({{\mathcal {R}}}\) in the space of tempered distributions \({{\mathcal {S}}'}({{\mathbb {R}}}^n)\) by

$$\begin{aligned} {{\mathcal {R}}}g:=\Delta ^{-1}\nabla g, \end{aligned}$$

then we have

$$\begin{aligned} {{\mathcal {R}}}\in {{\mathcal {L}}}({\dot{H}}^\alpha _q({{\mathbb {R}}}^n), {\dot{H}}^{\alpha +1}_q({{\mathbb {R}}}^n)),\;\Vert {{\mathcal {R}}}\Vert _{{\mathcal L}({\dot{H}}^\alpha _q({{\mathbb {R}}}^n), {\dot{H}}^{\alpha +1}_q({{\mathbb {R}}}^n))}\le c,\;\forall \alpha \in {{\mathbb {R}}},\forall q\in (1,\infty ), \end{aligned}$$
(3.5)

with constant \(c>0\) independent of q, \(\nu \) and \(\alpha \), which follows directly by the classical Fourier multiplier theorem. In particular, from (3.5) we have

$$\begin{aligned} {{\mathcal {R}}}\text {div}\,\in {\mathcal L}(H^\alpha _q({{\mathbb {R}}}^n)),\;\forall \alpha \in {{\mathbb {R}}},\forall q\in (1,\infty ). \end{aligned}$$
(3.6)

Thus, (3.4) follows from (3.6) by real and continuous interpolation in view of

$$\begin{aligned} (H^{\alpha _1}_{q}, H^{\alpha _2}_{q})_{\theta ,r} =B^{\alpha }_{q,r},\;(H^{\alpha _1}_{q}, H^{\alpha _2}_{q})^0_{\theta ,\infty } =b^{\alpha }_{q,\infty }, \end{aligned}$$

for \(\alpha =(1-\theta )\alpha _1+\theta \alpha _2, \alpha _1, \alpha _2\in {{\mathbb {R}}},\theta \in (0,1)\) and \(r\in [1,\infty ]\).

The proof is complete. \(\square \)

Lemma 3.2

Let \(0<T\le \infty \), \(\gamma \in (0,1)\) and \(\alpha \le 0\). If \(w_t,\nabla ^2 w\in L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)\), \(1<q<\infty \), then \(w\in L^\infty _{1-\theta +\gamma }(0,T; X)\) for all \(\theta \in (\gamma ,1)\) and the estimate

$$\begin{aligned} \Vert w\Vert _{L^\infty _{1-\theta +\gamma }(0,T; X)}\le c(\gamma ,\theta )\Vert w_t,\nabla ^2 w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty })},\;\forall \theta \in (\gamma ,1), \end{aligned}$$

holds true, where \(X:={\dot{B}}^{2\theta }_{q,1}({{\mathbb {R}}}^n)+{\dot{B}}^{2(\alpha +\theta )}_{q,1}({{\mathbb {R}}}^n)\). More precisely, it holds

$$\begin{aligned} \Vert w_1\Vert _{L^\infty _{1-\theta +\gamma }(0,T;{\dot{B}}^{2\theta }_{q,1})}+ \Vert w_2\Vert _{L^\infty _{1-\theta +\gamma }(0,T;{\dot{B}}^{2(\alpha +\theta )}_{q,1})} \le c(\gamma ,\theta )\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} \end{aligned}$$

where \(w_1:=({\hat{w}}\chi _{|\xi |\le 1})^{\vee },\,\,w_2:=({\hat{w}}\chi _{|\xi |> 1})^{\vee }\), and \({\hat{\varphi }}\) and \({\varphi }^\vee \), respectively, denote the Fourier and inverse Fourier transform of \(\varphi \) and \(\chi _{|\xi |\le 1}\), \(\chi _{|\xi |> 1}\) denote the characteristic functions of the sets \(\{\xi : |\xi |\le 1\}\), \(\{\xi : |\xi |> 1\}\), respectively.

Proof

First, consider the case \(\alpha <0\). Then, by the assumption of w, we have

$$\begin{aligned} w_{1t},\nabla ^2 w_1\in L^\infty _\gamma (0,T; L^q({{\mathbb {R}}}^n)), \quad w_{2t},\nabla ^2 w_2\in L^\infty _\gamma (0,T; {\dot{B}}^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)), \end{aligned}$$

and

$$\begin{aligned} \Vert w_{1t},\nabla ^2 w_1\Vert _{L^\infty _\gamma (0,T; L^q({{\mathbb {R}}}^n))}\le & {} \Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}, \nonumber \\ \Vert w_{2t},\nabla ^2 w_2\Vert _{L^\infty _\gamma (0,T; {\dot{B}}^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}\le & {} \Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$
(3.7)

Using these inequalities, we get by [21], Theorem 1.8.2 that

$$\begin{aligned}&w_1\in L^\infty (0,T; (L^q({{\mathbb {R}}}^n), {\dot{H}}^2_q({{\mathbb {R}}}^n))_{\gamma ,\infty }) =L^\infty (0,T; {\dot{B}}^{2\gamma }_{q,\infty }({{\mathbb {R}}}^n)),\nonumber \\&\quad \Vert w_1\Vert _{L^\infty (0,T; {\dot{B}}^{2\gamma }_{q,\infty }({{\mathbb {R}}}^n))} \le c\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} \end{aligned}$$
(3.8)

and that

$$\begin{aligned}&w_2\in L^\infty (0,T; ({\dot{B}}^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n), {\dot{B}}^{2\alpha +2}_{q,\infty }({{\mathbb {R}}}^n))_{\gamma ,\infty }) =L^\infty (0,T; {\dot{B}}^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n)),\nonumber \\&\quad \Vert w_2\Vert _{L^\infty (0,T; {\dot{B}}^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n))} \le c\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$
(3.9)

Then, by interpolation of (3.8) and the first relation of (3.7), we get for almost all \(t\in (0,T)\) that

$$\begin{aligned}&w_1(t)\in ({\dot{B}}^{2\gamma }_{q,\infty }({{\mathbb {R}}}^n), {\dot{H}}^{2}_{q}({{\mathbb {R}}}^n))_{\frac{\theta -\gamma }{1-\gamma },1} \subset ({\dot{B}}^{2\gamma }_{q,\infty }({{\mathbb {R}}}^n), {\dot{B}}^{2}_{q,\infty }({{\mathbb {R}}}^n))_{\frac{\theta -\gamma }{1-\gamma },1}= {\dot{B}}^{2\theta }_{q,1}({{\mathbb {R}}}^n),\\&\quad \Vert w_1(t)\Vert _{{\dot{B}}^{2\theta }_{q,1}}\le \Vert w_1(t)\Vert _{{\dot{B}}^{2\gamma }_{q,\infty }}^{\frac{1-\theta }{1-\gamma }} \Vert w_1(t)\Vert _{{\dot{H}}^{2}_{q}}^{\frac{\theta -\gamma }{1-\gamma }}\le c(\gamma ,\theta )t^{\gamma -\theta }\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}, \end{aligned}$$

yielding

$$\begin{aligned} \Vert w_1\Vert _{L^\infty _{1-\theta +\gamma }(0,T;{\dot{B}}^{2\theta }_{q,1})}\le c(\gamma ,\theta )\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$

On the other hand, by interpolation of (3.9) and the second relation of (3.7), we get for almost all \(t\in (0,T)\) that

$$\begin{aligned}&w_2(t)\in ({\dot{B}}^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n), {\dot{B}}^{2+2\alpha }_{q,\infty }({{\mathbb {R}}}^n))_{\frac{\theta -\gamma }{1-\gamma },1}= {\dot{B}}^{2(\alpha +\theta )}_{q,1}({{\mathbb {R}}}^n),\\&\quad \Vert w_2(t)\Vert _{{\dot{B}}^{2(\alpha +\theta )}_{q,1}}\le \Vert w_2(t)\Vert _{{\dot{B}}^{2(\alpha +\gamma )}_{q,\infty }}^{\frac{1-\theta }{1-\gamma }} \Vert w_2(t)\Vert _{{\dot{B}}^{2+2\alpha }_{q,\infty }}^{\frac{\theta -\gamma }{1-\gamma }}\le c(\gamma ,\theta )t^{\gamma -\theta }\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}, \end{aligned}$$

yielding

$$\begin{aligned} \Vert w_2\Vert _{L^\infty _{1-\theta +\gamma }(0,T;{\dot{B}}^{2(\alpha +\theta )}_{q,1})}\le c(\gamma ,\theta )\Vert w_t,\nabla ^2w\Vert _{L^\infty _\gamma (0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$

Since \(w:=w_1+w_2\), the proof of the theorem is complete. \(\square \)

Theorem 3.3

Let \(q\in (1,\infty )\), \(\gamma \in (0,1)\), \(0<T\le \infty \), \(\alpha \in (-\frac{1}{2},0]\), \(2\alpha -1/q\notin {{\mathbb {Z}}}\), and \(\alpha +\gamma <1\). Let \(u_0\in B^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n), \text {div}\,u_0=0\) and \(f\in L^\infty _\gamma (0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\) and \(\nabla g\in L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\), \(g=\text {div}\,R\) with some distribution \(R=R(t,\cdot )\) such that \(R_t\in L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\) and \(R(0)=0\). Then the problem (3.2) has a unique solution \((u,\nabla p)\) such that

$$\begin{aligned}&u_t, \nabla ^2u, \nabla p\in L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)), \nonumber \\&\quad \Vert u_t, \nu \nabla ^2 u, \nabla p\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty })} \le c(\Vert f,\nu \nabla g,R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })} +\nu ^{\gamma }\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }}) \end{aligned}$$
(3.10)

with constant \(c>0\) depending only on qns and independent of T.

In addition, the solution u to (3.2) satisfies \(u\in L^\infty _{1-\theta +\gamma }(0,T; X)\), where

$$\begin{aligned} X:={\dot{B}}^{2\theta }_{q,1}({{\mathbb {R}}}^n)+{\dot{B}}^{2(\alpha +\theta )}_{q,1}({{\mathbb {R}}}^n), \end{aligned}$$

and the estimate

$$\begin{aligned} \nu ^\theta \Vert u\Vert _{L^\infty _{1-\theta +\gamma }(0,T; X)} \le c(\Vert f,\nu \nabla g,R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })} +\nu ^{\gamma }\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }}), \; \forall \theta \in (\gamma ,1), \end{aligned}$$
(3.11)

with constant \(c=c(q,n,s)>0\) independent of T and \(\nu \).

Remark 3.4

If \(T<\infty \), then it easily follows that

$$\begin{aligned} \Vert u\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha +2}_{q,\infty }({{\mathbb {R}}}^n))} \le c(\gamma ,\nu ) \max \{1,T\} \Vert u_t,\nabla ^2u\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$

Hence, if \(T<\infty \) is assumed in Theorem 3.3, then \(u\in L^\infty _{1-\theta +\gamma }(0,T; B^{2(\alpha +\theta )}_{q,1}({{\mathbb {R}}}^n))\) and

$$\begin{aligned} \Vert u\Vert _{L^\infty _{1-\theta +\gamma }(0,T; B^{2(\alpha +\theta )}_{q,1})}\le & {} c(\gamma ,\nu )\max \{1,T\}(\Vert f,\nabla g,R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })}\nonumber \\&\quad +\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }}), \; \forall \theta \in (\gamma ,1), \end{aligned}$$
(3.12)

holds.

Proof of Theorem 3.3

First, assume that \(\nu =1\). Let \(w(t)={{\mathcal {R}}}g(t)=\Delta ^{-1}\nabla g(t), t\in (0,T)\), where \({{\mathcal {R}}}\) is the solution operator for (3.3) given by Lemma 3.1. Then, \(\text {div}\,w(t)=g(t)\) for all \(t\in (0,T)\). It is easily checked that

$$\begin{aligned} \nabla ^2 w \in L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n),\; \Vert \nabla ^2 w\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty })} \le c\Vert \nabla g\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })}. \end{aligned}$$
(3.13)

Moreover, since \(w_t={{\mathcal {R}}}g_t={{\mathcal {R}}}\text {div}\,R_t\) and \(R_t\in L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\), we get by Lemma 3.1 that

$$\begin{aligned} w_t\in L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)), \Vert w_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })}\le c\Vert R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })}. \end{aligned}$$
(3.14)

Now, with the introduction of the new unknown \(U=u-w\), the problem (3.2) is reduced to a divergence-free problem, that is,

$$\begin{aligned} \begin{array}{rlccl} U_t-\Delta U +\nabla p &{}= &{}F \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,U &{}=&{} 0 \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ U(0,x)&{}=&{}u_0&{}\text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(3.15)

where \(F:=f-w_t+\Delta w\in L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))\). It follows by [18], Corollary 4.13 (i) that the problem (3.15) has a unique solution U satisfying

$$\begin{aligned}&U_t, \nabla ^2 U, \nabla p\in L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)),\\&\quad \Vert U_t,\nabla ^2 U, \nabla p\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} \le c(\Vert F\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} +\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n)}), \end{aligned}$$

which yields by (3.13) and (3.14) that for \(u:=U+w\)

$$\begin{aligned}&u_t, \nabla ^2 u, \nabla p\in L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n)),\\&\Vert u_t,\nabla ^2 u, \nabla p\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))}\\&\quad \le c(\Vert F,w_t,\nabla ^2w\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} +\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n)})\\&\quad \le c(\Vert f, \nabla g, R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty }({{\mathbb {R}}}^n))} +\Vert u_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }({{\mathbb {R}}}^n)}). \end{aligned}$$

Then, the estimate (3.11) with \(\nu =1\) follows directly by Lemma 3.2 and by the above proved fact.

Finally, assuming \(\nu \ne 1\), let us prove (3.10) and (3.11). Notice that the rescaling transform

$$\begin{aligned} ({\tilde{u}},{\tilde{g}})(t,x):=\frac{1}{\nu } (u,g)\left( \frac{t}{\nu },x\right) ,\,\, ({\tilde{p}}, {\tilde{f}})(t,x):=\frac{1}{\nu ^{2}} (p,f)\left( \frac{t}{\nu },x\right) ,\,\, {\tilde{u}}_0(x):=\frac{u_0(x)}{\nu } \end{aligned}$$
(3.16)

reduces the system (3.2) with \(\nu \ne 1\) to the case with \(\nu =1\) and that \({\tilde{g}}_t(t,x)=\text {div}\,{\tilde{R}}(t,x)\), where \({\tilde{R}}(t,x):=\frac{1}{\nu ^2} R(\frac{t}{\nu },x)\). Then we have

$$\begin{aligned}&\Vert {\tilde{u}}_t, \nabla ^2 {\tilde{u}}, \nabla {\tilde{p}}\Vert _{L^\infty _{\gamma }(0,\nu T; b^{2\alpha }_{q,\infty })} +\Vert {\tilde{u}}\Vert _{L^\infty _{1-\theta +\gamma }(0,\nu T; X)}\\&\quad \le c(\Vert {\tilde{f}},\nabla {\tilde{g}}, {\tilde{R}}_t\Vert _{L^\infty _{\gamma }(0,\nu T;b^{2\alpha }_{q,\infty })} +\Vert {\tilde{u}}_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }}) \end{aligned}$$

with \(c=c(q,n,\alpha ,\gamma ,\theta )\) independent of \(\nu \) and T. Here we have

$$\begin{aligned}&\Vert {\tilde{u}}_t, \nabla ^2 {\tilde{u}},\nabla {\tilde{p}}\Vert _{L^\infty _{\gamma }(0,\nu T; b^{2\alpha }_{q,\infty })}\\&\quad = \frac{1}{\nu ^2}\nu ^{1-\gamma } \Vert \left( \frac{t}{\nu }\right) ^{1-\gamma }u_t\left( \frac{t}{\nu },x\right) , \left( \frac{t}{\nu }\right) ^{1-\gamma }\nu \nabla ^2u\left( \frac{t}{\nu },x\right) , \left( \frac{t}{\nu }\right) ^{1-\gamma }\nabla p\left( \frac{t}{\nu },x\right) \Vert _{L^\infty (0,\nu T; b^{2\alpha }_{q,\infty })}\\&\quad =\nu ^{-1-\gamma } \Vert u_t, \nu \nabla ^2u,\nabla p\Vert _{L^\infty _{\gamma }(0,T; b^{2\alpha }_{q,\infty })},\\&\Vert {\tilde{u}}\Vert _{L^\infty _{1-\theta +\gamma }(0,\nu T; X)}= \nu ^{-1+\theta -\gamma }\Vert u\Vert _{L^\infty _{1-\theta +\gamma }(0,T; X))},\\&\Vert {\tilde{u}}_0\Vert _{B^{2(\alpha +\gamma )}_{q,\infty }}=\nu ^{-1}\Vert u_0 \Vert _{B^{2(\alpha +\gamma )}_{q,\infty }} \end{aligned}$$

and

$$\begin{aligned} \Vert {\tilde{f}},\nabla {\tilde{g}}, {\tilde{R}}_t\Vert _{L^\infty _{\gamma }(0,\nu T;b^{2\alpha }_{q,\infty })} =\nu ^{-1-\gamma }\Vert f,\nu \nabla g, R_t\Vert _{L^\infty _{\gamma }(0,T;b^{2\alpha }_{q,\infty })}. \end{aligned}$$

Hence, we get (3.10) and (3.11) for the general case \(\nu \ne 1\).

The proof of the theorem is complete. \(\square \)

3.2 Momentum Equations with Fixed Variable Density

Let us consider the existence for the momentum equations with fixed density. We need the following lemma for the proof of Theorem 3.6 which is the main result of this section.

Lemma 3.5

Let \(\Omega \) be a domain of \({{\mathbb {R}}}^n,n\in {{\mathbb {N}}}\), \( n\le q<2n\), \(s\in (0,\frac{2n}{q}-1)\) and \(q\le r\). Then,

  1. (i)

    \(\forall \xi \in (1-\frac{n}{q},\min \{-s+\frac{n}{q},1-\frac{n}{q}+\frac{n}{r}\})\);

    $$\begin{aligned}&{\dot{H}}^{\xi -1+n/q}_{q}(\Omega )\cdot {\dot{H}}^{-\xi -s+n/q}_{r}(\Omega ) \hookrightarrow H^{-s-1+n/q}_{r}(\Omega ),\nonumber \\&{\dot{H}}^{\xi -1+n/q}_{r}(\Omega )\cdot {\dot{H}}^{-\xi -s+n/q}_{q}(\Omega ) \hookrightarrow H^{-s-1+n/q}_{r}(\Omega ). \end{aligned}$$
    (3.17)
  2. (ii)

    \(\forall \xi \in (1-\frac{n}{q},\min \{-s+\frac{n}{q},1-\frac{n}{q}+\frac{n}{r}\})\), \(\forall \eta \in (0,1-\xi )\), \(\forall \zeta \in (0,\xi +s)\);

    $$\begin{aligned}&\big ({\dot{H}}^{\xi -1+n/q+\eta }_{q}+{\dot{H}}^{\xi -1+n/q}_{q}\big )\cdot \big ({\dot{H}}^{-\xi -s+n/q+\zeta }_{r}+{\dot{H}}^{-\xi -s+n/q}_{r}\big ) \hookrightarrow H^{-s-1+n/q}_{r},\nonumber \\&\big ({\dot{H}}^{\xi -1+n/q+\eta }_{r}+{\dot{H}}^{\xi -1+n/q}_{r}\big )\cdot \big ({\dot{H}}^{-\xi -s+n/q+\zeta }_{q}+{\dot{H}}^{-\xi -s+n/q}_{q}\big ) \hookrightarrow H^{-s-1+n/q}_{r}. \end{aligned}$$
    (3.18)

Proof

In the proof, we omit the symbol \(\Omega \) for the domain. Under the assumption on qrs and \(\xi \), we have \(0<\xi -1+n/q<n/q\) and \(0<-\xi -s+n/q<n/r\). Hence, we have

$$\begin{aligned} {\dot{H}}^{\xi -1+n/q}_{q}\hookrightarrow L^{p_1},\; {\dot{H}}^{-\xi -s+n/q}_{r}\hookrightarrow L^{p_2},\; H^{s+1-n/q}_{r',0}\hookrightarrow L^{p_3}, \end{aligned}$$

where

$$\begin{aligned} \xi -1=-\frac{n}{p_1},\; -\xi -s+\frac{n}{q}-\frac{n}{r}=-\frac{n}{p_2},\; s+1-\frac{n}{q}-\frac{n}{r'}=-\frac{n}{p_3}. \end{aligned}$$
(3.19)

Note that \(\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=1\). Hence, the first relation of (3.17) follows by Hölder’s inequality. In the same way, we get the second relation of (3.17).

Then, (3.18) is an easy consequence of (3.17). Indeed, we have

$$\begin{aligned}&\big ({\dot{H}}^{\xi -1+n/q+\eta }_{q}+{\dot{H}}^{\xi -1+n/q}_{q}\big )\cdot \big ({\dot{H}}^{-\xi -s+n/q+\zeta }_{r}+{\dot{H}}^{-\xi -s+n/q}_{r}\big )\\&\quad \hookrightarrow {\dot{H}}^{\xi -1+n/q+\eta }_{q}\cdot {\dot{H}}^{-\xi -s+n/q+\zeta }_{r}+ {\dot{H}}^{\xi -1+n/q}_{q}\cdot {\dot{H}}^{-\xi -s+n/q}_{r}, \end{aligned}$$

where, by (3.17),

$$\begin{aligned} {\dot{H}}^{\xi -1+n/q}_{q}\cdot {\dot{H}}^{-\xi -s+n/q}_{r}\hookrightarrow H^{-s-1+n/q}_{r} \end{aligned}$$

and, with \(p_1,p_2, p_3\) in (3.19) and some positive numbers \(\varepsilon _i,i=1\sim 4,\)

$$\begin{aligned}&{\dot{H}}^{\xi -1+n/q+\eta }_{q}\cdot {\dot{H}}^{-\xi -s+n/q+\zeta }_{q} \hookrightarrow L^{p_1+\varepsilon _1}\cdot L^{p_2+\varepsilon _2}\\&\quad \hookrightarrow L^{p'_3+\varepsilon _3}\hookrightarrow H^{-s-1+n/q-\varepsilon _4}_{r}\hookrightarrow H^{-s-1+n/q}_{r}, \end{aligned}$$

proving the first relation of (3.18). The second relation of (3.18) can be proved in the same way.

Thus the lemma is proved. \(\square \)

Theorem 3.6

Let \(n\ge 2\), \(0<T\le \infty \). Let \(n\le q<2 n\), \(s\in (0,\min \{\frac{n-1}{q},\frac{2n}{q}-1\})\) and \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\cap B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\), \(q\le r\), \(\text {div}\,u_0=0\) and \(a\in L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))\). Then, there are some constants \(\delta _i=\delta _i(q,r,n,s)>0, i=1,2,\) and \(M=M(q,r,n,s)>0\) independent of T such that if

$$\begin{aligned} \Vert a\Vert _{L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))}<\delta _1,\quad \Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)}<\delta _2\nu , \end{aligned}$$

then the nonlinear problem (3.1) has a solution \((u,\nabla p)\) satisfying

$$\begin{aligned} u_t, \nabla ^2u, (u\cdot \nabla )u, \nabla p\in L^\infty _{s/2}(0,T; B^{-1+n/q-s}_{q,1}({{\mathbb {R}}}^n))\cap B^{-1+n/q-s}_{r,1}({{\mathbb {R}}}^n)). \end{aligned}$$

In addition, \( u\in L^\infty _{1-\theta +s/2}(0,T;Y),\;\forall \theta \in (\frac{s}{2},1),\) where

$$\begin{aligned} Y=Y_\theta :={\dot{B}}^{2\theta }_{q,1}({{\mathbb {R}}}^n)\cap {\dot{B}}^{2\theta }_{r,1}({{\mathbb {R}}}^n)+ {\dot{B}}^{2\theta -s-1+n/q}_{q,1}({{\mathbb {R}}}^n)\cap {\dot{B}}^{2\theta -s-1+n/q}_{r,1}({{\mathbb {R}}}^n), \end{aligned}$$
(3.20)

and the estimate

$$\begin{aligned} \nu ^{\theta }\Vert u\Vert _{L^\infty _{1-\theta +s/2}(0,T; Y)} +\Vert u_t, \nabla p\Vert _{L^\infty _{s/2}(0,T; B^{-1+n/q-s}_{q,1}\cap B^{-1+n/q-s}_{r,1})}\le M\nu ^{1+s/2} \end{aligned}$$
(3.21)

holds true. The solution \((u,\nabla p)\) is unique in the class of functions satisfying (3.21) with sufficiently small \(M>0\) on the right-hand side.

Proof

The proof is based on a linearization and a fixed point argument.

Consider the linear system (3.2) with arbitrarily fixed \(f\in L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty }({{\mathbb {R}}}^n)\cap b^{-1+n/q-s}_{r,\infty }({{\mathbb {R}}}^n))\) and \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\cap B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\) with \(q\le r\). Then, by Theorem 3.3 with \(\alpha =\frac{1}{2}(-1+\frac{n}{q}-s)\) the system (3.2) has a unique solution \((u,\nabla p)\) such that

$$\begin{aligned} u_t, \nabla ^2u, \nabla p\in L^\infty _{s/2}(0,T; b^{-1+n/q-s}_{q,\infty }({{\mathbb {R}}}^n)\cap b^{-1+n/q-s}_{r,\infty }({{\mathbb {R}}}^n)). \end{aligned}$$

Moreover, \(u\in L^\infty _{1-\theta +s/2}(0,T; Y_\theta )\) holds for all \(\theta \in (\frac{s}{2},1)\) and \(Y_\theta \) given by (3.20), and the estimate

$$\begin{aligned}&\nu ^{\theta }\Vert u\Vert _{L^\infty _{1-\theta +s/2}(0,T; Y_\theta )} + \Vert u_t, \nabla p\Vert _{L^\infty _{s/2}(0,T; b^{-1+n/q-s}_{r,\infty })} \nonumber \\&\quad \le {\tilde{C}}(\Vert f\Vert _{L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{r,\infty })} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{r,\infty }}) \end{aligned}$$
(3.22)

holds with constant \({\tilde{C}}>0\) depending on \(q,r,n,\theta \) and s. Note that \(r=q\) is allowed in (3.22). Now, fix \(\theta _1\in \big (\frac{1}{2}(s+1-\frac{n}{q}),\min \{\frac{n}{2q},\frac{1}{2} (s+1-\frac{n}{q}+\frac{n}{r})\}\big )\). Then

$$\begin{aligned} \xi :=2\theta _1-s,\;\eta :=s+1-\frac{n}{q},\; \zeta :=\eta . \end{aligned}$$
(3.23)

satisfy the assumption of Lemma 3.5. Let \( \theta _2:=1-\frac{\xi }{2}\in (\frac{s}{2},1)\). Then,

$$\begin{aligned} \xi -1+\frac{n}{q}= & {} 2\theta _1-s-1+\frac{n}{q},\;\xi -1+\frac{n}{q}+\eta =2\theta _1,\\ -\xi -s+\frac{n}{q}= & {} 2\theta _2-s-2+\frac{n}{q},\;-\xi -s+\frac{n}{q}+\zeta =2\theta _2-1. \end{aligned}$$

Let

$$\begin{aligned} {{\mathcal {X}}}:=L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty }({{\mathbb {R}}}^n)\cap b^{-1+n/q-s}_{r,\infty }({{\mathbb {R}}}^n)). \end{aligned}$$

Then, for \(f_1,f_2\in {{\mathcal {X}}}\) and almost all \(t\in (0,T)\) we get, using (2.7) and the first relation of (3.17) of Lemma 3.5 with \(\xi ,\eta \) and \(\zeta \) of (3.23) and Theorem 3.3 that

$$\begin{aligned}&\Vert (u_{f_1}\cdot \nabla )u_{f_2}(t)\Vert _{b^{-1+n/q-s}_{r,\infty }}\nonumber \\&\quad \le c\Vert u_{f_1}(t)\Vert _{({\dot{H}}^{\xi -1+n/q+\eta }_{q}+{\dot{H}}^{\xi -1+n/q}_{q})} \Vert \nabla u_{f_2}(t)\Vert _{({\dot{H}}^{-\xi -s+n/q+\zeta }_{r}+{\dot{H}}^{-\xi -s+n/q}_{r})}\nonumber \\&\quad \le c\Vert u_{f_1}(t)\Vert _{({\dot{H}}^{2\theta _1}_{q}+{\dot{H}}^{2\theta _1-s-1+n/q}_{q})} \Vert u_{f_2}(t)\Vert _{({\dot{H}}^{2\theta _2}_{r}+{\dot{H}}^{2\theta _2-s-1+n/q}_{r})}\nonumber \\&\quad \le c \nu ^{-\theta _1}t^{-\theta _1+s/2}(\Vert {f_1}\Vert _{L^\infty _{s/2} (0,T;b^{-1+n/q-s}_{q,\infty })} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})\cdot \nonumber \\&\qquad \cdot \nu ^{-\theta _2}t^{-\theta _2+s/2}(\Vert {f_2}\Vert _{{\mathcal {X}}} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{r,\infty }}), \nonumber \\&\quad \le c \nu ^{-1-\frac{s}{2}}t^{-1+\frac{s}{2}} (\Vert {f_1}\Vert _{L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty })} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})\cdot (\Vert {f_2}\Vert _{{\mathcal {X}}} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{r,\infty }}),\qquad \end{aligned}$$
(3.24)

where we used that \(\theta _1+\theta _2=1+s/2\). Interchanging the role of \(f_1\) and \(f_2\) and using the second relation of (3.17), we can get the counterpart inequality of (3.24) where the role of \(f_1\) and \(f_2\) are interchanged. Thus,

$$\begin{aligned}&\Vert (u_{f_1}\cdot \nabla )u_{f_2}\Vert _{{\mathcal {X}}} +\Vert (u_{f_2}\cdot \nabla )u_{f_1}\Vert _{{\mathcal {X}}} \nonumber \\&\quad \le c\nu ^{-1-s/2}(\Vert {f_1}\Vert _{L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty })} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})\cdot (\Vert {f_2}\Vert _{{\mathcal {X}}} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{r,\infty }}) \end{aligned}$$
(3.25)

holds with \(c=c(q,r,n,s)\).

Moreover, using Proposition 2.3 in view of

$$\begin{aligned} -1+\frac{1}{r}\le -1+\frac{1}{q}<-s-1+\frac{n}{q}<\frac{1}{r}\le \frac{1}{q} \end{aligned}$$
(3.26)

due to \(0<s<\frac{n-1}{q}\) and \(q\ge n\), we have

$$\begin{aligned} \Vert a(\nu \Delta u_f-\nabla p_f)\Vert _{{\mathcal {X}}}\le & {} c(q,r,n,s)\Vert a\Vert _{L^\infty (0,T;L^\infty )} \Vert \nu \Delta u_f-\nabla p_f\Vert _{{\mathcal {X}}}\nonumber \\\le & {} c(q,r,n,s)\Vert a\Vert _{L^\infty (0,T;L^\infty )}(\Vert f\Vert _{{\mathcal {X}}} +\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }}). \end{aligned}$$
(3.27)

Now, define the mapping \(\Phi \) from \({{\mathcal {X}}}\) to itself by

$$\begin{aligned} \Phi f:= -(u_f\cdot \nabla )u_f + a (\nu \Delta u_f-\nabla p_f), \end{aligned}$$

where \((u_f,\nabla p_f)\) is the unique solution to the system (3.2) corresponding to \(u_0\) and f. If we show that \(\Phi \) has a fixed point \({\tilde{f}}\in {{\mathcal {X}}}\), then \((u_{{\tilde{f}}},\nabla p_{{\tilde{f}}})\) is obviously a solution to the system (3.1). Let

$$\begin{aligned} G_K:=\{f\in {{\mathcal {X}}}: \Vert f\Vert _{L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty })}\le K\},\;K>0. \end{aligned}$$

Obviously, \(G_K\) is a closed subset of \({\mathcal {X}}\). If \(f\in G_{K}\), then, by (3.25) with \(r=q\) and (3.27),

$$\begin{aligned}&\Vert \Phi (f)\Vert _{L^\infty _{s/2}(0,T;b^{-1+n/q-s}_{q,\infty })} \nonumber \\&\quad \le C_1(K+\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{r,\infty }}) (\Vert a\Vert _{L^\infty (0,T;L^\infty )} +\nu ^{-1-s/2}K+\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }}) \end{aligned}$$
(3.28)

holds, where \(C_1=C_1(q,r,n,s)\).

On the other hand, if \(f_1,f_2\in G_{K}\), then

$$\begin{aligned} \Phi (f_1)-\Phi (f_2)= & {} [-((u_{f_1}-u_{f_2})\cdot \nabla )u_{f_1}-(u_{f_2}\cdot \nabla )(u_{f_1}-u_{f_2})] \nonumber \\&+[a(\nu \Delta (u_{f_1}-u_{f_2})-\nabla (p_{f_1}-p_{f_2}))] =:\text {(I)+(II)}. \end{aligned}$$
(3.29)

Note that \((u_{f_1}-u_{f_2}, \nabla (p_{f_1}- p_{f_2}))\) is a solution to (3.2) with zero initial value and \(g\equiv 0\) and right-hand side \({f_1}-{f_2}\). Hence, by (3.25), the first part (I) on the right-hand side of (3.29) is estimated as

$$\begin{aligned} \Vert \text {(I)}\Vert _{{\mathcal {X}}}\le 2C_2\nu ^{-1-s/2}\Vert f_1-f_2\Vert _{{\mathcal {X}}} (K+\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }}) \end{aligned}$$

and, by Proposition 2.3 and (3.10) of Theorem 3.3, the part (II) is estimated as

$$\begin{aligned} \Vert \text {(II)}\Vert _{{\mathcal {X}}}\le c\Vert a\Vert _{L^\infty (0,T;L^\infty )} \Vert \Delta (u_{f_1}-u_{f_2}),\nabla (p_{f_1}-p_{f_2})\Vert _{{\mathcal {X}}}\le C_2\Vert a\Vert _{L^\infty (0,T;L^\infty )}\Vert f_1-f_2\Vert _{{\mathcal {X}}}, \end{aligned}$$

in view of (3.26), where \(C_2=C_2(q,r,n,s)>0\). Finally, we have

$$\begin{aligned} \Vert \Phi (f_1)-\Phi (f_2)\Vert _{{\mathcal {X}}}\le C_2 (2\nu ^{-1-s/2}K+\Vert a\Vert _{L^\infty (0,T;L^\infty )} +2\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }} )\Vert f_1-f_2\Vert _{{\mathcal {X}}}. \end{aligned}$$
(3.30)

Now, in view of (3.28) and (3.30), consider the inequality

$$\begin{aligned} \left\{ \begin{array}{l} C_0(K+\nu ^{s/2}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})(\nu ^{-1-s/2}K +\Vert a\Vert _{L^\infty (0,T;L^\infty )} +\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})<K,\\ C_0 (2\nu ^{-1-s/2}K+\Vert a\Vert _{L^\infty (0,T;L^\infty )} +2\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }})<1, \end{array} \right. \end{aligned}$$
(3.31)

where \(C_0:=\max \{C_1,C_2\}\). By elementary calculations, it follows that, if

$$\begin{aligned} \epsilon :=\Vert a\Vert _{L^\infty (0,T;L^\infty )}+\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }} <\frac{1}{4C_0}, \end{aligned}$$
(3.32)

then for any

$$\begin{aligned} K\in (k_1,k_2) \end{aligned}$$
(3.33)

with

$$\begin{aligned} k_1=\frac{1-2\epsilon C_0-\sqrt{1-4\epsilon C_0}}{2C_0\nu ^{-1-s/2}},\;k_2=\frac{1-2\epsilon C_0}{2C_0\nu ^{-1-s/2}} \end{aligned}$$
(3.34)

the inequality (3.31) holds true. In other words, if (3.32) and (3.33) are satisfied, then \(\Phi (G_{K})\subset G_{K}\) and \(\Phi : G_{K}\mapsto G_{K}\) is a contraction mapping. Thus, by the Banach fixed point theorem \(\Phi \) has a unique fixed point \({\tilde{f}}\) in \(G_K\), and \(u=u_{{\tilde{f}}}\) is a solution to (3.1).

Tracking the above arguments, we can infer that the constant \(C_0=C_0(q,r,n,s)\) in (3.31) and hence the constants \(k_{1,2}\) in (3.34) are continuously dependent on s. Denote \(k_{1,2}\) in (3.34) by \(k_{1,2}(s)\). Then, one can choose sufficiently small \(\alpha =\alpha (q,r,n,s)>0\) so that \(\max \{k_1(s-\alpha ), k_1(s+\alpha )\}<\min \{k_2(s-\alpha ), k_2(s+\alpha )\}\). Let

$$\begin{aligned} {\mathcal {{{\tilde{X}}}}}:={{\mathcal {X}}}_+ \cap {{\mathcal {X}}}_-, \end{aligned}$$

where \({{\mathcal {X}}}_\pm :=L^\infty _{(s-\alpha )/2}(0,T; b^{-s-1+n/q \pm \alpha }_{q,\infty }({{\mathbb {R}}}^n)\cap b^{-s-1+n/q\pm \alpha }_{r,\infty }({{\mathbb {R}}}^n))\) and the norm in \({\mathcal {{{\tilde{X}}}}}\) is given as maximum of the two norms. We construct the mapping \(\Psi : {\mathcal {\tilde{X}}}\mapsto {\mathcal {{{\tilde{X}}}}}\) by

$$\begin{aligned} \Psi f:= -(u_f\cdot \nabla )u_f + a (\nu \Delta u_f-\nabla p_f), \end{aligned}$$

where \((u_f,\nabla p_f)\) is the unique solution to (3.2) corresponding to \(u_0\) and \(f\in {\mathcal {{{\tilde{X}}}}}\). Then, it is obvious from the argument of the proof of (i) (see (3.30)–(3.33)) that \(\Psi \) maps

$$\begin{aligned} B_K:=\{g\in {\mathcal {{{\tilde{X}}}}}: \Vert g\Vert _{L^\infty _{(s-\alpha )/2}(0,T;b^{-s-1+n/q+\alpha }_{q,\infty })}\le K, \Vert g\Vert _{L^\infty _{(s+\alpha )/2}(0,T;b^{-s-1+n/q-\alpha }_{q,\infty })}\le K\} \end{aligned}$$

into \(B_K\), (note that \(B_K\) is closed in \({\mathcal {\tilde{X}}}\)) and becomes a contraction mapping on \(B_K\) provided that

$$\begin{aligned} \Vert a\Vert _{L^\infty (0,T;L^\infty )} +\nu ^{-1}\Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }}<\min \left\{ \frac{1}{4C_0(q,n,s-\alpha )}, \frac{1}{4C_0(q,n,s+\alpha )}\right\} , \end{aligned}$$

see (3.32), and that

$$\begin{aligned} K\in \big (\max \{k_1(s-\alpha ), k_1(s+\alpha )\},\min \{k_2(s-\alpha ), k_2(s+\alpha )\}\big ), \end{aligned}$$

see (3.33). Hence, \(\Psi \) has a unique fixed point \({\bar{f}}\) and \((u_{{\bar{f}}},\nabla p_{{\bar{f}}})\) becomes a solution to (3.1). Note that \(K<\frac{\nu ^{1+s/2}}{2C_0}\). It follows from (3.22), (3.32) and (3.33) that the solution u satisfies

$$\begin{aligned} \nu ^{\theta }\Vert u\Vert _{L^\infty _{1-\theta +(s\pm \alpha )/2}(0,T; Y)} +\Vert u_t, \nabla p\Vert _{L^\infty _{(s\pm \alpha )/2}(0,T;b^{-1+n/q-s\mp \alpha }_{q,\infty }\cap b^{-1+n/q-s\mp \alpha }_{r,\infty })}\le M\nu ^{1+s/2} \end{aligned}$$
(3.35)

with \(M\equiv \frac{5{\tilde{C}}}{8C_0}\).

Thus, (3.21) follows from (3.35) by the real interpolation relation \((\cdot ,\cdot )_{1/2,1}\). Here, we recall the fact that

$$\begin{aligned} \big (L^\infty _{\gamma _1}(0,T;X_1), L^\infty _{\gamma _2}(0,T;X_2)\big )_{\theta ,1} \hookrightarrow L^\infty _{\gamma }\big (0,T;(X_1,X_2)_{\theta ,1}\big ), \gamma =(1-\theta )\gamma _1+\theta \gamma _2,0<\theta <1, \end{aligned}$$

for Banach interpolation couple \((X_1,X_2)\) and that

$$\begin{aligned} (b^{\kappa -\varepsilon }_{r,\infty }, b^{\kappa +\varepsilon }_{r,\infty })_{1/2,1}=(B^{\kappa -\varepsilon }_{r,\infty }, B^{\kappa +\varepsilon }_{r,\infty })_{1/2,1}=B^{\kappa }_{r,1},\;\kappa \in {{\mathbb {R}}}, \end{aligned}$$

which follows by [5], Theorem 3.4.2 (d), Theorem 6.4.5 (1).

Let \(K'\equiv \frac{K}{3}\) and the norm of solution \((u, \nabla p)\) in (3.21) is bounded by \(K'\). Then

$$\begin{aligned} \Vert u_t-\nu \Delta u+\nabla p\Vert _{L^\infty _{s/2}(0,T; b^{-s-1+n/q}_{q,\infty }}\le K. \end{aligned}$$

Hence, in view of uniqueness of the fixed point of \(\Phi \) in \(B_K\), \((u,\nabla p)\) must be the only solution satisfying the inequality (3.21) with \(K'\) on the right-hand side.

Thus, the theorem is proved. \(\square \)

4 Proof of the Main Result: Existence Part

Let \(2\le n\le q<2n\), \(0<T\le \infty \) and \(s\in (0,\min \{\frac{n-1}{q},\frac{2n}{q}-1\})\). Suppose that

$$\begin{aligned} \rho _0(x)\in L^\infty ({{\mathbb {R}}}^n), 0<\rho _{01}\le & {} \rho _0(x)\le \rho _{02}, \;x\in {{\mathbb {R}}}^n,\nonumber \\ u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\cap B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n), q\le & {} r,\;\text {div}\,u_0=0. \end{aligned}$$
(4.1)

In order to prove existence of solutions to the system (1.1), we construct an iterative scheme. Let \(\eta _m\in C^\infty ({{\mathbb {R}}}^n), m\in {{\mathbb {N}}},\) be mollifiers such that

$$\begin{aligned} \eta _m(x)\ge 0,\,\,\eta _m(x)=\eta _m(-x),\,\, \mathrm{supp}\,\eta _m\subset \big \{x: |x| <\frac{1}{m}\big \},\,\, \int _{{{\mathbb {R}}}^n} \eta _m(x)\,dx=1. \end{aligned}$$

For \(m=1,2,\ldots \), let us construct an iterative scheme for (1.1) as

$$\begin{aligned} \left\{ \begin{array}{rll} \rho _{mt}+(u^{(m-1)}\cdot \nabla )\rho _m=0,\,\,\rho _{m}(0,x)&{}=\rho _0(x), &{}\,\text {in }(0,T)\times {{\mathbb {R}}}^n,\\ u_{mt}-\nu \Delta u_m + (u_m \cdot \nabla ) u_m+\nabla p_m &{}= a_m(\nu \Delta u_m-\nabla p_m), &{}\, \text {in }(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,u_m &{}= 0,&{}\; \text {in } (0,T)\times {{\mathbb {R}}}^n,\\ u_m(0,x) &{}= u_0, &{}\,\text {in }{{\mathbb {R}}}^n, \end{array} \right. \end{aligned}$$
(4.2)

where \(\nu :=\frac{\mu }{{\bar{\rho }}}\) with constant \({\bar{\rho }}\in [\rho _{01},\rho _{02}]\) fixed, \(u^{(0)}\equiv 0\) and \(a_m(t,x):=\frac{{\bar{\rho }}}{\rho _m(t,x)}-1\), \(u^{(m)}=\eta _m\star u_m\), \(m\in {{\mathbb {N}}}\), where “\(\star \)” is used to denote the convolution.

Remark 4.1

For \(f\in H^1_q({{\mathbb {R}}}^n)\) (or \(H^{-1}_q({{\mathbb {R}}}^n)\)), \(q\in (1,\infty )\), it holds \(\Vert \eta _m\star f-f\Vert _{H^1_q({{\mathbb {R}}}^n)}\rightarrow 0\) (or \(\Vert \eta _m\star f-f\Vert _{H^{-1}_q({{\mathbb {R}}}^n)}\rightarrow 0\) ) as \(m\rightarrow \infty \). Let \({\mathcal F}^s_{q}\in \{b_{q,\infty }^s, B_{q,r}^s, 1\le r\le \infty \}\). Then it follows by interpolation that

$$\begin{aligned} \eta _m\star w \rightarrow w\,\,\text {in }{\mathcal F}^s_{q}({{\mathbb {R}}}^n),\,\, \forall w\in {\mathcal F}^s_{q}({{\mathbb {R}}}^n), -1< s<1,\;(\text {as }m\rightarrow \infty ). \end{aligned}$$

Moreover, by the Banach-Steinhaus theorem, it follows that

$$\begin{aligned} \Vert \eta _m\star w\Vert _{{{\mathcal {F}}}^s_{q}({{\mathbb {R}}}^n)}\le C\Vert w\Vert _{{\mathcal F}^s_{q}({{\mathbb {R}}}^n)},\,\, \forall w\in {\mathcal F}^s_{q}({{\mathbb {R}}}^n),\,\,\forall s\in (-1,1), \end{aligned}$$

with \(C=C(q,s)>0\) independent of \(m\in {{\mathbb {N}}}\).

Furthermore, if \(u_m\in L^\infty _{s/2}(0,T; b^{2-s}_{q,\infty }({{\mathbb {R}}}^n))\) for some \(q\ge n\) and \(s\in (0,2)\), then \(u^{(m)}(t)\in C^\infty ({{\mathbb {R}}}^n)\) for almost all \(t\in (0,T),\) and, in particular,

$$\begin{aligned} u^{(m)}\in L^1_{loc}([0,T), W^{1,\infty }({{\mathbb {R}}}^n)). \end{aligned}$$

We have the following lemma.

Lemma 4.2

Let \(n\ge 2\), \(n\le q<2n\) and \(0<T\le \infty \). Suppose that (4.1) holds. Then for any \(s\in (0, \min \{\frac{n-1}{q}, \frac{2n}{q}-1\})\) there are some constants \(\delta _i=\delta _i(q,r,n,s)>0, i=1,2,\) and \(M=M(q,r,n,s)>0\) independent of \(m\in {{\mathbb {N}}}\) with the following property: If

$$\begin{aligned} \frac{\rho _{02}-\rho _{01}}{\rho _{01}}<\delta _1,\quad \Vert u_0\Vert _{B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)}<\delta _2\nu , \end{aligned}$$

the iterative system (4.2) has a solution \(\{(\rho _m,u_m, \nabla p_m):m\in {{\mathbb {N}}}\}\) such that

$$\begin{aligned} \rho _m \in L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n)) \end{aligned}$$

and

$$\begin{aligned}&u_{mt},\nabla ^2 u_m, \nabla p_m\in L^\infty _{s/2}(0,T; B^{-s-1+n/q}_{q,1}({{\mathbb {R}}}^n)\cap B^{-s-1+n/q}_{r,1}({{\mathbb {R}}}^n)),\nonumber \\&u_m\in L^\infty _{1-\theta +s/2}(0,T;Y_\theta ),\,\forall \theta \in (s/2,1), \end{aligned}$$
(4.3)

and the estimates

$$\begin{aligned} \Vert \rho _m\Vert _{L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))}&=\Vert \rho _0\Vert _{L^\infty ({{\mathbb {R}}}^n)},\; \Vert a_m\Vert _{L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))} \nonumber \\&\le \frac{\rho _{02}-\rho _{01}}{\rho _{01}},\nu ^\theta \Vert u_m\Vert _{L^\infty _{1-\theta +s/2}(0,T;Y_\theta )} \nonumber \\&\quad +\Vert u_{mt}, \nu \nabla ^2 u_m, \nabla p_m\Vert _{L^\infty _{s/2}(0,T; B^{-s-1+n/q}_{q,1}\cap B^{-s-1+n/q}_{r,1})}\le M\nu ^{1+s/2},\;\forall \theta \in (s/2,1),\nonumber \\ \end{aligned}$$
(4.4)

hold true.

Proof

By [6], Theorem II.3, it follows that when \(v\in L^1(0,T; W^{1,\infty }({{\mathbb {R}}}^n))\), \(\text {div}\,v=0\), and \(\rho _0\in L^\infty ({{\mathbb {R}}}^n)\), the transport equation

$$\begin{aligned} \rho _t+(v\cdot \nabla )\rho =0\quad \text {in }(0,T)\times {{\mathbb {R}}}^n,\quad \rho (0)=\rho _0\quad \text {in }{{\mathbb {R}}}^n, \end{aligned}$$

has a unique solution \(\rho \in L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))\) such that \(\Vert \rho \Vert _{L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))}=\Vert \rho _0\Vert _{L^\infty ({{\mathbb {R}}}^n)}\).

This fact, together with Theorem 3.6, yields the assertions of the lemma. \(\square \)

Remark 4.3

Let \(n\ge 2\), \(0<T \le \infty \), \(n\le q\le r\) and let \(s\in (0,1)\). Let \(v\in L^\infty _{1-\theta +s/2}(0,T; Y_\theta )\) for all \(\theta \in (s/2,1)\), where \(Y_\theta \) is given by (3.20). Then, if we take \(\theta =\theta _1\) as

$$\begin{aligned} \theta _1\in \Big (\frac{s}{2}+\frac{1}{4}, \frac{s}{2}+\frac{1}{3}\Big ), \end{aligned}$$
(4.5)

we get for any finite \(T'\le T\) and bounded domain \(G\subset {{\mathbb {R}}}^n\) that

$$\begin{aligned} v\in L^\infty _{1-\theta _1+s/2}(0,T; Y_{\theta _1})\hookrightarrow L^{3}(0,T';{\dot{H}}^{2\theta _1-s-1+n/q}_q(G)) \hookrightarrow L^{3}(0,T';L^{2n}(G)), \end{aligned}$$
(4.6)

and

$$\begin{aligned} \Vert v\Vert _{L^{3}(0,T';L^{2n}(G))} \le c(q_1,T',G)\Vert v\Vert _{L^\infty _{1-\theta _1+s/2}(0,T; Y_{\theta _1})}. \end{aligned}$$
(4.7)

On the other hand, if we take \(\theta =\theta _2\in \big (\frac{s+1}{2}, \frac{s}{2}+\frac{2}{3}\big )\), then for any finite \(T'\le T\) and bounded domain \(G\subset {{\mathbb {R}}}^n\) we have

$$\begin{aligned} \nabla v\in L^{3/2}(0,T'; L^{2n/(2n-1)}(G)),\; \Vert \nabla v\Vert _{L^{3/2}(0,T'; L^{2n/(2n-1)}(G))}\le c\Vert v\Vert _{L^\infty _{1-\theta _2+s/2}(0,T; Y_{\theta _2})}. \end{aligned}$$
(4.8)

Remark 4.4

Let \(2\le n\le q< 2n\), \(s\in (0,\min \{\frac{n-1}{q},\frac{2n}{q}-1\})\). Let \(\{(\rho _m,u_m, \nabla p_m):m\in {{\mathbb {N}}}\}\) be solutions to the iterative system (4.2) whose existence is guaranteed by Lemma 4.2. Then,

$$\begin{aligned} u_m\in L^\infty _{1-\theta +s/2}(0,T;Y_\theta )\subset L^\infty _{1-\theta +s/2}(0,T;H^{2\theta -s-1+n/q}_{q,loc}({{\mathbb {R}}}^n)) ,\,\forall \theta \in (s/2,1), \end{aligned}$$
(4.9)

and so is for \(u^{(m)}\). Therefore, if

$$\begin{aligned} 0<2\theta -s-1+n/q<1/q, \end{aligned}$$
(4.10)

then by Proposition 2.3 we get for \(v=u_m\) or \(u^{(m)}\) that

$$\begin{aligned}&\Big |\displaystyle \int _{{{\mathbb {R}}}^n} (v(t,x)\cdot \nabla )\rho _m(t,x)\varphi (x)\,dx\Big |\\&\quad =\Big |\displaystyle \int _G \rho _m(t,x) v(t,x)\cdot \nabla \varphi (x)\,dx\Big |\\&\quad \le \Vert \rho _m(t)\Vert _{L^\infty ({{\mathbb {R}}}^n)}\Vert v(t)\Vert _{H^{2\theta -s-1+n/q}_{q}(G)} \Vert \nabla \varphi \Vert _{H^{-2\theta +s+1-n/q}_{q'}(G)}\\&\quad \le \Vert \rho _0\Vert _{L^\infty ({{\mathbb {R}}}^n)}\Vert v(t)\Vert _{H^{2\theta -s-1+n/q}_{q}(G)} \Vert \varphi \Vert _{H^{-2\theta +s+2-n/q}_{q'}({{\mathbb {R}}}^n)},\; \forall \varphi \in C_0^\infty ({{\mathbb {R}}}^n), \end{aligned}$$

for almost all \(t\in (0,T)\). Therefore, if (4.10) holds, then

$$\begin{aligned} (u_m\cdot \nabla )\rho _m,\; (u^{(m-1)}\cdot \nabla )\rho _m=-\rho _{mt} \in L^\infty _{1-\theta +s/2}(0,T; H^{2\theta -s-2+n/q}_{q}({{\mathbb {R}}}^n)),\;m\in {{\mathbb {N}}}. \end{aligned}$$
(4.11)

Proof of Theorem 1.1: Existence part

Let \(s\in (0, \min \{\frac{n-1}{q},\frac{2n}{q}-1\})\) and let \(\{(\rho _m, u_m, \nabla p_m):m\in {{\mathbb {N}}}\}\) be the solutions to the iterative system (4.2), the existence of which is given by Lemma 4.2. By Lemma 4.2\(\{u_m\}\) is bounded in \(L^\infty _{1-\theta +s/2}(0,T; Y_\theta )\) for all \(\theta \in (s/2,1)\), and \(\{u_{mt},\nabla p_m\}\), \(\{\rho _m\}\) are bounded in \(L^\infty _{s/2}(0,T; b^{-s-1+n/q}_{q,\infty }({{\mathbb {R}}}^n))\) and \(L^\infty ((0,T)\times {{\mathbb {R}}}^n)\), respectively. Note that

$$\begin{aligned} {\dot{B}}^{-\kappa }_{r,\infty }({{\mathbb {R}}}^n)=({\dot{B}}^{\kappa }_{r',1}({{\mathbb {R}}}^n))',\; B^{-\kappa }_{r,1}({{\mathbb {R}}}^n)=\big (b^{\kappa }_{r',\infty }({{\mathbb {R}}}^n)\big )',\,\kappa \in {{\mathbb {R}}}. \end{aligned}$$

Therefore, it follows by standard arguments that \(\{\rho _m, u_m, \nabla p_m\}\) has a subsequence (denote it by the same symbols) such that

$$\begin{aligned} \begin{array}{ll} u_{m} \rightharpoonup u &{}\text {in } L^\infty _{1-\theta +s/2}(0,T; Y_\theta )\,\,(\text {*-weakly as }m\rightarrow \infty ),\\ \nabla ^2 u_{m} \rightharpoonup \nabla ^2 u &{}\text {in } L^\infty _{s/2}(0,T; B^{-s-1+n/q}_{q,1}({{\mathbb {R}}}^n)\cap B^{-s-1+n/q}_{r,1}({{\mathbb {R}}}^n)) \,\,(\text {*-weakly as }m\rightarrow \infty ),\\ u_{{m}t} \rightharpoonup u_{t} &{}\text {in } L^\infty _{s/2}(0,T; B^{-s-1+n/q}_{q,1}({{\mathbb {R}}}^n)\cap B^{-s-1+n/q}_{r,1}({{\mathbb {R}}}^n))\,\,(\text {*-weakly as }m\rightarrow \infty ),\\ \bar{\rho }\nabla p_{{m}} \rightharpoonup \nabla P &{}\text {in } L^\infty _{s/2}(0,T; B^{-s}_{q,1}(\Omega )^n)\,\,(\text {*-weakly as }m\rightarrow \infty ),\\ \rho _{m} \rightharpoonup \rho &{}\text {in } L^\infty (0,T; L^\infty ({{\mathbb {R}}}^n))\,\,(\text {*-weakly as }m\rightarrow \infty ) \end{array} \end{aligned}$$
(4.12)

for some u, \(\rho \) and distribution P. Moreover, it follows from (4.6) and (4.7) that

$$\begin{aligned} u_{m} \rightharpoonup u \quad \text {in } L^{3}(0,T'; {\dot{H}}^{2\theta _1-s-1+n/q}_{q}(G)) \,\,(\text {weakly as }m\rightarrow \infty ) \end{aligned}$$
(4.13)

for all \(\theta _1\in (s/2,1)\) satisfying (4.5), finite \(T'\le T\) and bounded \(G\subset {{\mathbb {R}}}^n\).

We shall show that \((\rho , u,\nabla P)\) is a solution to (1.1) satisfying the assertion of Theorem 1.1.

By (4.12), obviously, \((\rho , u, \nabla P)\) satisfies (1.4) and (1.8).

Next, in order to prove (1.6), note that by (4.3), (4.4) the sequence \(\{u_{mt}\}\) weakly converges in \(L^{\alpha }(0,T; H^{-s-1+n/q}_q(G))\) for some \(\alpha >1\). Therefore, in view of (4.13) and compact embedding \({\dot{H}}^{2\theta _1-s-1+n/q}_{q}(G)\hookrightarrow \hookrightarrow L^{2n}(G)\), we get by a compactness theorem ([20], Ch.3, Theorem 2.1) that

$$\begin{aligned} u_{m}\rightarrow u\quad \text {in } L^{3}(0,T'; L^{2n}(G)) \end{aligned}$$
(4.14)

for any finite \(T'\le T\) and bounded G as \(m\rightarrow \infty \). Moreover, we get from (4.8) that \(\{\nabla u_{m}\}\) is bounded in \(L^{3/2}(0,T'; L^{(2n)/(2n-1)}(G))\).

Rewriting the second equation of the system (4.2), we have

$$\begin{aligned} u_{mt}-\frac{\mu }{\rho _{m}}\Delta u_{m}+(u_{m}\cdot \nabla )u_{m} +\frac{{\bar{\rho }}}{\rho _{m}}\nabla p_{m}=0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \rho _{m}u_{mt}-\mu \Delta u_{m}+\rho _{m}(u_{m}\cdot \nabla )u_{m}+{\bar{\rho }}\nabla p_{m}=0 \end{aligned}$$
(4.15)

in view of

$$\begin{aligned} \rho _{m}\in L^\infty (0,T; L^\infty ({{\mathbb {R}}}^n))),\,u_{mt}, (u_{m}\cdot \nabla )u_{m}, \nabla p_{m}\in L^\infty _{s/2}(0,T; B^{-s}_{q,1}({{\mathbb {R}}}^n)). \end{aligned}$$

Let \(\varphi \in C_0^\infty ([0,T)\times {{\mathbb {R}}}^n)^n\), \(\text {div}\,\varphi =0\) be given arbitrarily such that \(\mathrm{supp}\,\varphi \subset [0,T')\times G\) with finite \(T'>0\) and bounded \(G\subset {{\mathbb {R}}}^n\). Since each term of (4.15) belongs to \(L^1(0,T; B^{-s-1+n/q}_{q,1}({{\mathbb {R}}}^n))\) by (4.3) and \((b^{-s-1+n/q}_{q,\infty }({{\mathbb {R}}}^n))'=B^{s+1-n/q}_{q',1}({{\mathbb {R}}}^n)\), we can test (4.15) with \(\varphi \) to obtain

$$\begin{aligned} \displaystyle \int _0^T\langle \rho _{m} u_{mt}-\mu \Delta u_{m}+\rho _{m} (u_{m}\cdot \nabla ) u_{m},\varphi \rangle _{B^{-s-1+n/q}_{q',1}({{\mathbb {R}}}^n),b^{s+1-n/q}_{q,\infty }({{\mathbb {R}}}^n)}\,dt=0. \end{aligned}$$

By (4.9) we have

$$\begin{aligned} u_m\in L^\infty _{1-\tau +s/2}(0,T; H^{2\tau -s-1+n/q}_{q}(G)),\;\forall m\in {{\mathbb {N}}}, \end{aligned}$$

with \(\tau =s-\theta +3/2-n/q\) being \(s/2<\tau <1\), which holds true if (4.10) is satisfied. Hence, if \((\tau -s/2)p'<1\), i.e., if \(0<1/p<\theta -s/2+n/q-1/2\) and (4.10) is satisfied, then

$$\begin{aligned} u_{m}\in L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q}(G))\subset L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)), \end{aligned}$$

where note that \(-2\theta +s+2-n/q>0\). On the other hand, by (4.11), we have \(\rho _{mt}\varphi \in L^p(0,T;H^{2\theta -s-2+n/q}_{q}({{\mathbb {R}}}^n))\) for \(\theta \) satisfying (4.10) and \(\theta -s/2<1/p<\infty \). Therefore, if

$$\begin{aligned} p\in \big ( \frac{1}{\theta -s/2+n/q-1/2},\frac{1}{\theta -s/2}\big ) \end{aligned}$$
(4.16)

and (4.10) is satisfied, then we have

$$\begin{aligned} \displaystyle \int _0^T\langle \rho _{m} u_{mt}, \varphi \rangle \,dt= & {} -\langle u_{m},\rho _{mt}\varphi \rangle _{L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)),L^p (0,T;H^{2\theta -s-2+n/q}_{q}(G))}\\&-\displaystyle \int _0^T\int _{{{\mathbb {R}}}^n}\rho _{m}u_{m}\cdot \varphi _t\,dxdt -\int _{{{\mathbb {R}}}^n} \rho _0u_0\cdot \varphi (0,\cdot )\,dx. \end{aligned}$$

Moreover, for p and \(\theta \) satisfying (4.16) and (4.10) we get by (4.11) that

$$\begin{aligned}&\displaystyle \int _0^T \langle \rho _{m} (u_{m}\cdot \nabla ) u_{m},\varphi \rangle _{B^{-s}_{q,1}({{\mathbb {R}}}^n), b^s_{q',\infty }({{\mathbb {R}}}^n)}\,dt =-\int _0^T\int _G \rho _{m} u_{m}\otimes u_{m}\cdot \nabla \varphi \,dxdt\\&-\langle u_{m}, (u_{m}\cdot \nabla ) \rho _{m}\varphi \rangle _{L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)),L^p (0,T;H^{2\theta -s-2+n/q}_{q}(G))}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} 0= & {} \int _0^T\int _{{{\mathbb {R}}}^n}[\rho _{m} u_{m}\cdot \varphi _t+\mu u_{m}\cdot \Delta \varphi +\rho _{m} u_{m}\otimes u_{m}\cdot \nabla \varphi ]\,dxdt +\int _{{{\mathbb {R}}}^n} \rho _0u_0\cdot \varphi (0,\cdot )\,dx\nonumber \\&+\langle u_{m},(\rho _{mt}+(u_{m}\cdot \nabla )\rho _{m})\varphi \rangle _{L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)),L^p(0,T;H^{2\theta -s-2+n/q}_{q}(G))}\nonumber \\= & {} \int _0^T\int _{G}[\rho _{m} u_{m}\cdot \varphi _t+\mu u_{m}\cdot \Delta \varphi +\rho _{m} u_{m}\otimes u_{m}\cdot \nabla \varphi ]\,dxdt +\int _{G} \rho _0u_0\cdot \varphi (0,\cdot )\,dx\nonumber \\&+\langle u_{m},((u_{m}-u^{(m-1)})\cdot \nabla )\rho _{m}\varphi \rangle _{L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)),L^p(0,T;H^{2\theta -s-2+n/q}_{q}(G))}. \end{aligned}$$
(4.17)

For the estimate of the last term on the right-hand side of (4.17) we need the following lemma.

Lemma 4.5

Let \(T'\le T\) be finite and \(G\subset {{\mathbb {R}}}^n\) be a bounded domain. Let p and \(\theta \) satisfy (4.16) and put

$$\begin{aligned} R_m:=\langle u_{m},((u_{m}-u^{(m-1)})\cdot \nabla )\rho _{m}\varphi \rangle _{L^{p'}(0,T';H^{-2\theta +s+2-n/q}_{q'}(G)),L^p(0,T;H^{2\theta -s-2+n/q}_{q}(G))}. \end{aligned}$$

Then \(R_m\) tends to 0 as \(m\rightarrow \infty \).

Proof

By Remark 4.3 and Lemma 4.2 we have

$$\begin{aligned} |R_m|= & {} |\langle \rho _{m}(u_{m}-u^{(m-1)}), \nabla (u_{m}\cdot \varphi )\rangle _{Q_{{\tilde{T}}}}|\\\le & {} \Vert \rho _0\Vert _\infty \Vert u_{m}-u^{(m-1)}\Vert _{L^{3}(0,T';L^{2n}(G))} \Vert \nabla (u_{m}\cdot \varphi )\Vert _{L^{3/2}(0,T';L^{2n/(2n-1)}(G))}. \end{aligned}$$

Then, since \(\Vert \nabla (u_{m}\cdot \varphi )\Vert _{L^{3/2}(0,T';L^{2n/(2n-1)}(G))}\) is bounded with respect to \(m\in {{\mathbb {N}}}\), see (4.8), the proof of the lemma will be complete if we show that

$$\begin{aligned} \Vert u_{m}-u^{(m-1)}\Vert _{L^{3}(0,T';L^{2n}(G))}\rightarrow 0\quad (m\rightarrow \infty ). \end{aligned}$$
(4.18)

Note that

$$\begin{aligned} \Vert u_{m}-u^{(m-1)}\Vert _{L^{3}(0,T';L^{2n}(G))}\le \Vert u_{m}-u\Vert _{L^{3}(0,T';L^{2n}(G))} +\Vert u^{(m-1)}-u\Vert _{L^{3}(0,T';L^{2n}(G))}, \end{aligned}$$

where the first term on the right-hand side tends to 0 as \(m\rightarrow \infty \) due to (4.14).

In order to show

$$\begin{aligned} \Vert u^{(m-1)}-u\Vert _{L^{3}(0,T';L^{2n}(G))}\rightarrow 0\quad (m\rightarrow \infty ), \end{aligned}$$
(4.19)

we write

$$\begin{aligned} u^{(m-1)}-u=\eta _{m-1}\star (u_{m-1}-u) +(\eta _{m-1}\star u-u). \end{aligned}$$

Here, \(L^{3}(0,T';L^{2n}(G))\)-norm of \(\eta _{m-1}\star u-u\) obviously tends to 0 as \(m\rightarrow \infty \). The \({L^{3}(0,T';L^{2n}(G))}\)-norm of \(\eta _{m-1}\star (u_{m-1}-u)\) goes to zero as \(m\rightarrow \infty \) thanks to (4.14) and the fact that \(\Vert \eta _{m}\star \,\,\cdot \Vert _{{{\mathcal {L}}}(L^{3}(0,T';L^{2n}(G)))}\) is uniformly bounded with respect to \(m\in {{\mathbb {N}}}\). Thus, (4.19) and, consequently, (4.18) are proved. The proof of the lemma comes to end. \(\square \)

Let us continue the proof of existence part of Theorem 1.1. In (4.17), we get easily that

$$\begin{aligned} \int _0^T\int _{G}(\rho _{m} u_{m}\cdot \varphi _t+\mu u_{m}\cdot \Delta \varphi )\,dxdt \rightarrow \int _0^T\int _{{{\mathbb {R}}}^n}(\rho u\cdot \varphi _t+\mu u\cdot \Delta \varphi )\,dxdt \end{aligned}$$

as \(m\rightarrow \infty \) due to \(*\)-weak convergence \(\rho _{m} \rightharpoonup \rho \) in \(L^\infty (0,T;L^\infty ({{\mathbb {R}}}^n))\) and strong convergence \(u_{m} \rightarrow u\) in \(L^{3}(0,T'; L^{2n}(G))\), see (4.12) and (4.14). By (4.14) we have \(u_{m}\otimes u_{m}\rightarrow u\otimes u\) in \(L^{1}((0,T')\times G)\) as \(m\rightarrow \infty \) and therefore,

$$\begin{aligned} \int _0^T\int _{G} \rho _{m} (u_{m}\otimes u_{m})\cdot \nabla \varphi \,dxdt \rightarrow \int _0^T\int _{{{\mathbb {R}}}^n} \rho (u\otimes u)\cdot \nabla \varphi \,dxdt\quad \text {as }m\rightarrow \infty . \end{aligned}$$

Thus, letting \(m\rightarrow \infty \) in (4.17), it follows that \((\rho ,u)\) satisfies (1.6).

Finally, in order to show (1.5), test the first equation of (4.2) with arbitrary \(\psi \in C_0^1([0,T)\times {{\mathbb {R}}}^n)\) to get

$$\begin{aligned} \int _0^T\int _{{{\mathbb {R}}}^n}(\rho _{m}\psi _t+\rho _{m} u^{(m-1)}\cdot \nabla \psi )\,dxdt +\int _{{{\mathbb {R}}}^n} \rho _0\psi (0,\cdot )\,dx=0. \end{aligned}$$
(4.20)

Obviously,

$$\begin{aligned} \int _0^T\int _{{{\mathbb {R}}}^n} \rho _{m}\psi _t\,dxdt\rightarrow \int _0^T\int _{{{\mathbb {R}}}^n} \rho \psi _t\,dxdt\quad (m\rightarrow \infty ). \end{aligned}$$

Moreover, in view of (4.19) and (4.12), we have

$$\begin{aligned}&\big |\displaystyle \int _0^T\int _{{{\mathbb {R}}}^n}(\rho _{m} u^{(m-1)}-\rho u)\cdot \nabla \psi \,dxdt\big |\\&\quad \le \big |\displaystyle \int _0^T\int _{{{\mathbb {R}}}^n}(u^{(m-1)}- u)\cdot (\rho _{m} \nabla \psi )\,dxdt\big | + \big |\displaystyle \int _0^T\int _{{{\mathbb {R}}}^n}(\rho _{m}-\rho )u\cdot \nabla \psi \,dxdt\big |\rightarrow 0 \end{aligned}$$

as \(m\rightarrow \infty \). Therefore, \((\rho ,u)\) satisfies (1.5) in the limiting case \(m\rightarrow \infty \) in (4.20).

Thus, the proof of existence part of Theorem 1.1 is completed. \(\square \)

5 Proof of the Main Result: Uniqueness Part

Lemma 5.1

Let \(\Omega \) be a Lipschitz domain of \({{\mathbb {R}}}^n\), \(n\ge 2\), and let \(n\le q<r<2n\) and \(s\in (0,\frac{n}{q}-\frac{n}{r})\). Then the following statements hold:

  1. (i)

    \(\forall f\in b^{-s+n/q}_{r,\infty }(\Omega ), \varphi \in B^{s+1-n/q}_{r',1}(\Omega )\);

    $$\begin{aligned} \Vert f\varphi \Vert _{B^{s+1-n/q}_{r',1}(\Omega )}\lesssim \Vert f\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )} \Vert \varphi \Vert _{B^{s+1-n/q}_{r',1}(\Omega )}. \end{aligned}$$
  2. (ii)

    \(\forall f\in b^{-s+n/q}_{r,\infty }(\Omega ), g\in b^{-s-1+n/q}_{r,\infty }(\Omega )\);

    $$\begin{aligned} \Vert f g\Vert _{b^{-s-1+n/q}_{r,\infty }(\Omega )}\le l_1\Vert f\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )}\Vert g\Vert _{b^{-s-1+n/q}_{r,\infty }(\Omega )}. \end{aligned}$$
  3. (iii)

    \(\forall f,g\in b^{-s+n/q}_{r,\infty }(\Omega )\);

    $$\begin{aligned} \Vert f g\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )}\le l_2\Vert f\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )}\Vert g\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )}. \end{aligned}$$
  4. (iv)

    \(\forall f\in {\dot{B}}^{-s+n/q}_{r,\infty }(\Omega ), g\in B^{-1+n/q}_{r,\infty }(\Omega )\);

    $$\begin{aligned} \Vert f g\Vert _{B^{-s-1+n/q}_{r,\infty }(\Omega )} \lesssim \Vert f\Vert _{{\dot{B}}^{-s+n/r}_{r,\infty }(\Omega )}\Vert g\Vert _{B^{-1+n/q}_{r,\infty }(\Omega )}. \end{aligned}$$

Proof

Proof of (i): It is clear from \(r>n\) and Sobolev and Hölder inequality that

$$\begin{aligned} {\dot{H}}^1_n(\Omega )\cdot H^1_{r'}(\Omega )\hookrightarrow H^1_{r'}(\Omega ). \end{aligned}$$
(5.1)

Moreover, we have

$$\begin{aligned} {\dot{B}}^{n/\zeta }_{\zeta ,1}(\Omega )\cdot L^{r'}(\Omega )\hookrightarrow L^{r'}(\Omega ), \end{aligned}$$
(5.2)

where \(\zeta :=(\frac{1}{r}-\frac{s+1-n/q}{n})^{-1}(-s+\frac{n}{q})>1\) holds true thanks to \(s<n/q-n/r\) and \(r<2n\).

From (5.1) and (5.2) we get by bilinear real interpolation that

$$\begin{aligned} ({\dot{B}}^{n/\zeta }_{\zeta ,1}(\Omega ), {\dot{H}}^1_n(\Omega ))_{s+1-n/q,1}\cdot (L^{r'}(\Omega ), H^1_{r'}(\Omega ))_{s+1-n/q,1} \hookrightarrow (L^{r'}(\Omega ), H^1_{r'}(\Omega ))_{s+1-n/q,1}, \end{aligned}$$

cf. [5, 21]. Hence, in view of

$$\begin{aligned} \frac{-s+n/q}{\zeta }+\frac{s+1-n/q}{n}=\frac{1}{r},\quad \left( -s+\frac{n}{q}\right) \cdot \frac{n}{\zeta }+\left( s+1-\frac{n}{q}\right) =\frac{n}{r}, \end{aligned}$$

we have

$$\begin{aligned} {\dot{B}}^{n/r}_{r,1}(\Omega )\cdot B^{s+1-n/q}_{r',1}(\Omega )\hookrightarrow B^{s+1-n/q}_{r',1}(\Omega ), \end{aligned}$$
(5.3)

which implies the assertion (i) in view of \(b^{-s+n/q}_{r,\infty }(\Omega )\hookrightarrow B^{n/r}_{r,1}(\Omega )\subset {\dot{B}}^{n/r}_{r,1}(\Omega )\) due to \(-s+n/q>n/r\).

Proof of (ii): Using duality argument in view of \((B^{-s-1+n/q}_{r',1}(\Omega ))'=B^{s+1-n/q}_{r,\infty }(\Omega )\) due to \(0<s+1-n/q<1/r'\), it follows directly from (i) that

$$\begin{aligned} \Vert f g\Vert _{B^{-s-1+n/q}_{r,\infty }(\Omega )}\le k_0(q,r,s,\Omega )\Vert f\Vert _{b^{-s+n/q}_{r,\infty }(\Omega )}\Vert g\Vert _{B^{-s-1+n/q}_{r,\infty }(\Omega )} \end{aligned}$$

for all \(f\in b^{-s+n/q}_{r,\infty }(\Omega )\) and \(g\in B^{-s-1+n/q}_{r,\infty }(\Omega )\). Then, the assertion (ii) follows by a density argument.

Proof of (iii): The assertion (iii) can be proved exactly in the same way as for the assertion (i). In fact, using

$$\begin{aligned} H^1_n(\Omega )\cdot H^1_{r}(\Omega )\hookrightarrow H^1_{r}(\Omega ),\; B^{n/\sigma }_{\sigma ,1}(\Omega )\cdot L^{r}(\Omega )\hookrightarrow L^{r}(\Omega ), \end{aligned}$$

where \(\frac{s+1-n/q}{\sigma }+\frac{-s+n/q}{n}=\frac{1}{r}\), we get by bilinear continuous interpolation \((\cdot ,\cdot )^0_{-s+n/q,\infty }\) that

$$\begin{aligned} b^{n/r}_{r,\infty }(\Omega )\cdot b^{-s+n/q}_{r,\infty }(\Omega )\hookrightarrow b^{-s+n/q}_{r,\infty }(\Omega ), \end{aligned}$$

which yields the assertion (iii).

Proof of (iv): Note that \(s+1-n/q-n/r<0\). It then easily follows by Sobolev embedding theorem that

$$\begin{aligned} L^r(\Omega )\cdot B^{s+1-n/q}_{r',1}(\Omega )\hookrightarrow H^{s+1-n/q-n/r}_{r'}(\Omega ), \end{aligned}$$

which together with (5.3) yields by real interpolation \((\cdot ,\cdot )_{{\tilde{\theta }},1}\) with \({\tilde{\theta }}=\frac{r}{n}(-s+\frac{n}{r})\) that

$$\begin{aligned} {\dot{B}}^{-s+n/r}_{r,1}(\Omega )\cdot B^{s+1-n/q}_{r',1}(\Omega )\hookrightarrow B^{1-n/q}_{r',1}(\Omega ). \end{aligned}$$

Hence, by duality, the assertion (iv) is proved. \(\square \)

Proof of Theorem 1.1: Uniqueness part

The uniqueness proof relies on a Lagrangian coordinates approach following the idea of [12].

First let us recall some facts concerning Lagrangian coordinates. Let u be a vector field such that

$$\begin{aligned} u\in L^1_{loc}([0,\infty ), C_{Lip}({{\mathbb {R}}}^n)) \end{aligned}$$
(5.4)

and let X(ty) be the (unique) solution to the ordinary differential system:

$$\begin{aligned} \frac{dX}{dt}=u(t,X)\quad t\in (0,\infty ), \quad X(0)=y\in {{\mathbb {R}}}^n. \end{aligned}$$
(5.5)

The unique solution X to (5.5), that is,

$$\begin{aligned} X(t,y)=y+\int _0^t u(\tau ,X(\tau ,y))\,d\tau ,\;t\in (0,\infty ), \end{aligned}$$
(5.6)

determines a unique continuous semiflow, i.e., \(t\rightarrow X(t,y)\) for each y is continuous and \(X(0,\cdot )=\text {Id}\), \(X(t+s,y)=X(t, X(s,y))\) for \(t,s>0\). Then, Eulerian coordinates x and Lagrangian coordinates y are related by

$$\begin{aligned} x=X(t,y). \end{aligned}$$

Note that \(W^{1,\infty }({{\mathbb {R}}}^n)\subset C_{Lip}({{\mathbb {R}}}^n)\). Moreover, if \(\text {div}\,u=0\), then \(X(t,\cdot )\) for each \(t>0\) is a \(C^1\)-diffeomorphism and measure preserving due to the Jacobian \(|D_y X(t,y)|=1\). Let \(Y(t,\cdot )\) be the inverse mapping of \(X(t,\cdot )\), then

$$\begin{aligned} D_xY(t,x)=(D_yX(t,y))^{-1}=:A(t,y), \end{aligned}$$

where and in what follows we use the notation \((\nabla u)_{i,j}=(\partial _iu^j)_{1\le i,j\le n}, Du=(\nabla u)^T\). Let \(v(t,y):=u(t, X(t,y))\). Then,

$$\begin{aligned} \nabla _x u(t,x)=A^T \nabla _y v(t,y),\quad \text {div}_x u(x,t)=\text {div}_y(A v(t,y)), \end{aligned}$$
(5.7)

see [10], ”Appendix”. In view of (5.7), we use the notation

$$\begin{aligned} \nabla _u w:= A^T \nabla _y w,\quad \text {div}_u w:= \text {div}_y(Aw ), \quad \Delta _uw:=\text {div}_u(\nabla _u w). \end{aligned}$$

Now, let \((\rho , u,\nabla P)\) be a solution to (1.1) satisfying (1.4) and (1.8) with \(r>q\) and \(s\in (0, \frac{n}{q}-\frac{n}{r})\). Then, it follows from \(u_t\in L^\infty _{s/2}(0,T; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\) and \(u_0\in B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\subset b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\) that \(u\in L^\infty _{s/2}(0,T'; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\) for any finite \(T'\le T\), which together with \(\nabla ^2 u\in L^\infty _{s/2}(0,T; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\) yields that \(u\in L^\infty _{s/2}(0,T'; b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\). Therefore, we have

$$\begin{aligned} u\in L^1_{loc}([0,T),b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\subset L^1_{loc}([0,T), W^{1,\infty }({{\mathbb {R}}}^n)) \end{aligned}$$
(5.8)

by Sobolev embedding \(b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\subset W^{1,\infty }({{\mathbb {R}}}^n)\) in view of \(-s+1+n/q-n/r>1\), and thus u satisfies (5.4).

For \(a(t,x):=\frac{{\bar{\rho }}}{\rho (t,x)}-1\), \((t,x)\in (0,T)\times {{\mathbb {R}}}^n\), and

$$\begin{aligned} (b,v,Q)(t,y):=(a,u,P)(t,X(t,y)),\,\,(t,y)\in (0,T)\times {{\mathbb {R}}}^n, \end{aligned}$$

where X(ty) is given by (5.6), we get from (1.5) that \(b_t=0\) in the sense of distribution. In fact, given any \({\tilde{\psi }}\in C^1_0((0,T)\times {{\mathbb {R}}}^n)\) and \(\psi (t,x):={\tilde{\psi }}(t,Y(t,x))\), we have \({\tilde{\psi }}(t,y)=\psi (t, X(t,y))\) and \(\psi \in C^1_0((0,T)\times {{\mathbb {R}}}^n)\). Hence, we have

$$\begin{aligned} \int _{(0,T)\times {{\mathbb {R}}}^n}b{\tilde{\psi }}_t\,dydt= & {} \int _{(0,T)\times {{\mathbb {R}}}^n}a(t, X(t,y)) \frac{\partial \psi }{\partial t}(t,X(t,y))\,dydt\\= & {} \int _{(0,T)\times {{\mathbb {R}}}^n}a(t, X(t,y)) (\psi _t+u\cdot \nabla \psi )(t,X(t,y))\,dydt\\= & {} \int _{(0,T)\times {{\mathbb {R}}}^n}a(t,x) (\psi _t+u\cdot \nabla \psi )(t,x)|D_y X(t,y)|\,dxdt\\= & {} \int _{(0,T)\times {{\mathbb {R}}}^n}a(t,x) (\psi _t+u\cdot \nabla \psi )(t,x)\,dxdt=0. \end{aligned}$$

Therefore, \(b(t,y)\equiv b(0,y)\equiv a_0(y)\) for each \(y\in {{\mathbb {R}}}^n\).

Assuming \(\mu =1\) without loss of generality, \(\{v,Q\}\) solves the system:

$$\begin{aligned} \begin{array}{rcl} v_t-(1+b)(\Delta _u v -\nabla _u Q) = 0 \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}_u v = 0 \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ v(0,y)=u_0&{}\text {in}&{}{{\mathbb {R}}}^n. \end{array} \end{aligned}$$
(5.9)

Now, in order to prove uniqueness of solutions, let \((\rho _i, u_i, \nabla P_i)\), \(i=1,2\), be two solutions to (1.1) which satisfy (1.4) and (1.8) with \(r>q\) and \(s\in (0,\frac{n}{q}-\frac{n}{r})\). Note that \(r<2n\) may be assumed without loss of generality in view of the initial assumption \(q <2n\) and \(u_0\in B^{-1+n/q}_{q,\infty }({{\mathbb {R}}}^n)\cap B^{-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)\subset B^{-1+n/q}_{{\tilde{r}},\infty }({{\mathbb {R}}}^n)\), \(\forall {\tilde{r}}\in (q,r)\).

For \(i=1,2\), let \(X_i\) be the semiflow corresponding to \(u_i\) (see (5.6)) and let \((b_i, v_i, Q_i)\) be the corresponding density perturbation, velocity and pressure in the Lagrangian coordinates. Then, for \(\delta v=v_1-v_2, \delta Q=Q_1-Q_2\) we get from (5.9) that

$$\begin{aligned} \begin{array}{rclll} (\delta v)_t-\Delta \delta v +\nabla \delta Q &{}=&{} a_0(\Delta \delta v-\nabla \delta Q)+\delta F \,\, &{}\text {in}&{}(0,T)\times {{\mathbb {R}}}^n,\\ \text {div}\,\delta v &{}=&{} \delta g \equiv \text {div}\,\delta R \,\, &{}\text {in} &{}(0,T)\times {{\mathbb {R}}}^n,\\ \delta v(0,y)&{}=&{} 0&{} \text {in}&{}{{\mathbb {R}}}^n, \end{array} \end{aligned}$$
(5.10)

where

$$\begin{aligned} \delta F= & {} \delta f_1+ \delta f_2,\nonumber \\ \delta f_1:= & {} (1+a_0)[(\text {Id}-A_2^T)\nabla \delta Q-\delta A\nabla Q_1]\quad \text {with}\quad \delta A:= A_1-A_2, \nonumber \\ \delta f_2:= & {} (1+a_0)\text {div}\,[(A_2^T A_2-\text {Id})\nabla \delta v-(A_2^T A_2-A_1^T A_1)\nabla v_1], \nonumber \\ \delta g:= & {} (\text {Id}-A_2^T): \nabla \delta v-(\delta A)^T:\nabla v_1, \nonumber \\ \delta R:= & {} (\text {Id}-A_2)\delta v-\delta A v_1=(\text {Id}-A_1)v_1-(\text {Id}-A_2)v_2. \end{aligned}$$
(5.11)

Note that \(\delta R(0)=0\). Therefore, by Theorem 3.3 with \(q\equiv r\), \(\gamma \equiv s/2, \alpha \equiv -s-1+n/q\), we have

$$\begin{aligned}&\Vert \delta v_t,\nabla ^2 \delta v, \nabla \delta Q\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\quad \le K \Vert a_0(\nu \Delta \delta v-\nabla \delta Q),\delta F, \nabla \delta g, (\delta R)_t\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}, \forall t\in (0,T), \end{aligned}$$
(5.12)

with constant \(K>0\) independent of t and \(\nu \). In particular, if \(t\le 1\), it follows from (5.12) that

$$\begin{aligned}&\Vert \delta v_t, \nabla \delta Q\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} +\Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })} +\Vert \delta v\Vert _{L^\infty _{1-\theta +s/2}(0,t;B^{2\theta -s-1+n/q}_{r,1})}\nonumber \\&\quad \le K \Vert a_0(\nu \Delta \delta v-\nabla \delta Q),\delta F, \nabla \delta g, (\delta R)_t\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })},\;\forall \theta \in (s/2,1), \end{aligned}$$
(5.13)

see Remark 3.4.

From now on, let us get estimate of the right-hand side of (5.12). Thanks to (5.8), there is some \(m_0>0\) such that, if \(0\le t_1<t_2<1, t_2-t_1<m_0\),

$$\begin{aligned}&\displaystyle \big \Vert \int _{t_1}^{t_2} \nabla v_i(t,y)\,dt\big \Vert _{L^\infty ({{\mathbb {R}}}^n)}<\frac{1}{2},\nonumber \\&\displaystyle \int _{t_1}^{t_2} \Vert v_i(t,y)\Vert _{b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n)}\,dt \le c(s)t^{s/2} \Vert v_i\Vert _{L^\infty _{s/2}(0,1;b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n)}<\frac{1}{2l}\le \frac{1}{2},i=1,2, \end{aligned}$$
(5.14)

where \(l:=\max \{l_1, l_2\}\) with constants \(l_1\), \(l_2\) appearing in the estimates of Lemma 5.1. Throughout the proof, we assume that \(0<t<m_0\).

Note that

$$\begin{aligned} L^\infty ({{\mathbb {R}}}^n)\subset {\mathcal M}(b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n)) \end{aligned}$$
(5.15)

due to \(-1+1/r<-s-1+n/q<1/r\) and by Proposition 2.3. Hence, \(a_0(y)=\frac{{\bar{\rho }}-\rho _0(y)}{\rho _0(y)}\) implies that there is some \(\delta _1>0\) independent of \(\nu \) such that if

$$\begin{aligned} \frac{\rho _{02}-\rho _{01}}{\rho _{01}}<\delta _1, \end{aligned}$$
(5.16)

then

$$\begin{aligned} \Vert a_0(\nu \Delta \delta v-\nabla \delta Q)\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \le \frac{1}{2K}\Vert \nu \nabla ^2 \delta v, \nabla \delta Q\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}, \end{aligned}$$
(5.17)

where K is the constant appearing in (5.12).

On the one hand, we have

$$\begin{aligned} A_i(t,y)=(\text {Id}+C_i(t))^{-1}=\sum _{k\ge 0}(-1)^k C_i(t)^k, i=1,2, \end{aligned}$$
(5.18)

where

$$\begin{aligned} C_i(t):=\int _0^t D v_i(\tau )\,d\tau , i=1,2. \end{aligned}$$

Hence one has

$$\begin{aligned} \delta A(t)=h_1(t)\int _0^t D\delta v (\tau )\,d\tau ,t\in (0,m_0), \end{aligned}$$
(5.19)

with

$$\begin{aligned} h_1(t):=\sum _{k\ge 1}(-1)^k\sum _{j=0}^{k-1}C_1(t)^jC_2(t)^{k-1-j},\,\,t \in (0,m_0). \end{aligned}$$
(5.20)

By elementary calculation using the first relation of (5.14) we have

$$\begin{aligned} \Vert h_1\Vert _{L^\infty (0,t;L^\infty )}<1 \end{aligned}$$
(5.21)

with a generic constant C. Hence, it follows from (5.15) and \(v\in L^\infty _{s/2}(0,T'; b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n))\) for any finite \(T'\le T\) that

$$\begin{aligned} \Vert \delta A\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} \Vert h_1(t)\Vert _{L^\infty (0,t;L^\infty )}\Vert \int _0^t D\delta v(\tau )\,d\tau \Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\nonumber \\\le & {} ct^{s/2}\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \end{aligned}$$
(5.22)

with \(c=c(q,r,s)>0\).

By definition of \(C_i\) and the second relation of (5.14) we have

$$\begin{aligned} {\bar{C}}_i(t):=\Vert C_i\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })}<\frac{1}{2k_2(q,r,s)}, \end{aligned}$$
(5.23)

and, in particular, by Lemma 5.1 (iii)

$$\begin{aligned} \Vert h_1\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le \sum _{k\ge 1}\sum _{j=0}^{k-1}{\bar{C}}_1(t)^j{\bar{C}}_2(t)^{k-1-j}\le c(q,r,s). \end{aligned}$$
(5.24)

Then, using (5.24) and Lemma 5.1 (iii), we have

$$\begin{aligned} \Vert \delta A\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })}\le & {} \Vert h_1\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })} \Vert \int _0^t D\delta v d\tau \Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })}\nonumber \\\le & {} ct^{s/2}\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} \end{aligned}$$
(5.25)

with \(c=c(q,r,s)>0\).

Next, let us get estimate of \(\Vert \nabla \delta g\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\). From (5.11) one has

$$\begin{aligned}&\Vert \nabla \delta g\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\quad \le \Vert \nabla A^T_2\otimes \nabla \delta v, (\text {Id}-A^T_2) \otimes \nabla ^2 \delta v, \nabla \delta A^T\otimes \nabla v_1, \delta A^T\otimes \nabla ^2 v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}, \end{aligned}$$
(5.26)

where the right-hand side can be estimated as below.

Since \(DA_i=\sum _{k\ge 1}(-1)^kkC_i^{k-1}DC_i\) due to (5.18), we get by Lemma 5.1 (ii) and (5.23) that

$$\begin{aligned} \Vert \nabla A^T_i\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}\le & {} \sum _{k\ge 1}k{\bar{C}}_i^{k-1}(t)\Vert DC_i\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })} \nonumber \\\le & {} c\Vert \nabla ^2 v_i\Vert _{L^1(0,t; b^{-s-1+n/q}_{r,\infty })} \nonumber \\\le & {} ct^{s/2}\Vert \nabla ^2 v_i\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}, i=1,2, \end{aligned}$$
(5.27)

and

$$\begin{aligned} \Vert \nabla A^T_2\otimes \nabla \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}\le & {} c\Vert \nabla A^T_2\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty })} \Vert \nabla \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })} \nonumber \\\le & {} ct^{s/2}\Vert \nabla ^2 v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })} \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{q,\infty })} \end{aligned}$$
(5.28)

with \(c=c(q,r,s)\). Furthermore, by (5.18), Lemma 5.1 (iii) and (5.23) we have

$$\begin{aligned} \Vert \text {Id}-A_i\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty }({{\mathbb {R}}}^n))}= & {} \Vert \sum _{k\ge 1}(-1)^kC_i(t)^k\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })}\nonumber \\\le & {} c(q,r,s)\Vert \nabla v_i\Vert _{L^1(0,t, b^{-s+n/q}_{r,\infty })}\nonumber \\\le & {} c(q,r,s)t^{s/2}\Vert v_i\Vert _{L^\infty _{s/2}(0,t, b^{-s+1+n/q}_{r,\infty })}, i=1,2, \end{aligned}$$
(5.29)

and, consequently,

$$\begin{aligned}&\Vert (\text {Id}-A^T_2) \otimes \nabla ^2 \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\quad \le c\Vert \text {Id}-A^T_2\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })} \Vert \nabla ^2 \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{q,\infty })}\nonumber \\&\quad \le ct^{s/2}\Vert v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{q,\infty })} \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })} \end{aligned}$$
(5.30)

with \(c=c(q,r,s)\).

Note that

$$\begin{aligned} D\delta A(t)= h_1(t) \int _0^t D^2\delta v (\tau )\,d\tau + h_2(t)\int _0^t D\delta v (\tau )\,d\tau , \end{aligned}$$

where

$$\begin{aligned} h_2:=Dh_1= \sum _{k\ge 1}(-1)^k\sum _{j=0}^{k-1}\big (jC_1^{j-1}C_2^{k-1-j}DC_1+ (k-1-j)C_1^{j}C_2^{k-2-j}DC_2\big ). \end{aligned}$$

Then, we get by Lemma 5.1 (ii), (5.23) and the second relation of (5.14) that

$$\begin{aligned} \Vert h_2\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty })}\le & {} c\Vert \nabla ^2 v_1,\nabla ^2 v_2\Vert _{L^1(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\\le & {} ct^{s/2}\Vert \nabla ^2v_1,\nabla ^2 v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \end{aligned}$$

with \(c=c(q,r,s)\). Hence, by Lemma 5.1 (ii),

$$\begin{aligned} \Vert D\delta A\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} \Vert h_1\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty }))} \Vert \int _0^t D^2\delta v (\tau )\,d\tau \Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty })} \\&+ c\Vert h_2\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })} \Vert \int _0^t D\delta v (\tau )\,d\tau \Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}\\\le & {} ct^{s/2}(1+\Vert \nabla ^2v_1,\nabla ^2 v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}) \Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} \end{aligned}$$

with \(c=c(q,r,s)\). Then, by Lemma 5.1 (ii) we have

$$\begin{aligned} \Vert \nabla \delta A^T\otimes \nabla v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} c\Vert \nabla \delta A\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty })} \Vert \nabla v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })}\nonumber \\\le & {} ct^{s/2} \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}, \end{aligned}$$
(5.31)

where the constant c depends on qrs and \(\Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}\).

By (5.25) and Lemma 5.1 (ii) we have

$$\begin{aligned} \Vert \delta A^T\otimes \nabla ^2 v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} \Vert \delta A\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })} \Vert \nabla ^2 v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\\le & {} ct^{s/2} \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}, \end{aligned}$$
(5.32)

where the constant c depends on qrs and \(\Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}\).

Thus, from (5.26), (5.28), (5.30), (5.31) and (5.32) we get

$$\begin{aligned} \Vert \nabla \delta g\Vert _{L^\infty _{s/2}(0,t;b^{-s}_{q,\infty }(\Omega ))} \le \eta _1(t)\Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{2-s}_{q,\infty }(\Omega ))} \end{aligned}$$
(5.33)

with some \(\eta _1(t)\) such that \(\eta _1(t)\rightarrow 0\) as \(t\rightarrow +0\).

Following the same procedure as the derivation of (5.33), we can get estimate

$$\begin{aligned} \Vert \delta f_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \le ct^{s/2}\Vert \nabla ^2\delta v, \nabla \delta Q\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \end{aligned}$$
(5.34)

under the condition (5.16), where \(c>0\) depends on qrs, \(\Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\) and \(\Vert \nabla Q_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\); we omit the details here.

On the other hand, we have

$$\begin{aligned}&\Vert \delta f_2\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \nonumber \\&\quad \le \Vert \text {div}\,[(A_2^T A_2-\text {Id})\nabla \delta v-(A_2^T A_2-A_1^T A_1)\nabla v_1]\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}. \end{aligned}$$
(5.35)

Here, by Lemma 5.1 (ii) we have

$$\begin{aligned}&\Vert \text {div}\,[(A_2^T A_2-\text {Id})\nabla \delta v]\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \\&\quad \le (\Vert \nabla (A_2^T A_2)\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })} +\Vert A_2^T A_2-\text {Id}\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })}) \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}. \end{aligned}$$

Note that, by Lemma 5.1 (ii), (iii), (5.27) and (5.14),

$$\begin{aligned} \Vert \nabla (A_i^T A_i)\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} 2 \Vert \nabla A_i\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })} \Vert A_i\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })}\\\le & {} c(q,r,s) t^{s/2} \Vert v_i\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}, i=1,2, \end{aligned}$$

and, by Lemma 5.1 (iii),

$$\begin{aligned}&\Vert A_2^T A_2-\text {Id}\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \\&\quad \le ct^{s/2}\Vert v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })}(1+t^{s/2}\Vert v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })}) \end{aligned}$$

in view of \(A_2^T(t) A_2(t)-\text {Id}= \int _0^t (Dv_2+Dv_2^T)\,d\tau + \int _0^tDv_2^T\,d\tau \cdot \int _0^tDv_2\,d\tau \). Therefore,

$$\begin{aligned}&\Vert \text {div}\,[(A_2^T A_2-\text {Id})\nabla \delta v]\Vert _{L^\infty _{s/2} (0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \nonumber \\&\quad \le c t^{s/2} \Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} (\Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })}+1) \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}. \end{aligned}$$
(5.36)

Furthermore, using \(A_2^T A_2-A_1^T A_1= A_2^T \delta A+ \delta A^T A_1\), Lemma 5.1 (ii), (iii), (5.25) and the second relation of (5.14), we have

$$\begin{aligned}&\Vert \text {div}\,[(A_2^T A_2-A_1^T A_1)\nabla v_1]\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \nonumber \\&\quad \le \Vert \nabla (A_2^T \delta A+ \delta A^T A_1)\otimes \nabla v_1, (A_2^T \delta A+ \delta A^T A_1)\otimes \nabla ^2 v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\quad \le c\big (\Vert \delta A\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} \Vert \nabla A_1\otimes \nabla v_1, \nabla A_2\otimes \nabla v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\qquad +\Vert A_1, A_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })} \Vert \nabla \delta A\otimes \nabla v_1, \delta A\otimes \nabla ^2 v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })}\big )\nonumber \\&\quad \le c\big (\Vert \delta A\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} (\Vert \nabla A_1\otimes \nabla v_1, \nabla A_2\otimes \nabla v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\qquad +\Vert A_1, A_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })}\Vert v_1\Vert _{L^\infty _{s/2} (0,t;b^{-s+1+n/q}_{r,\infty })})\big ) \nonumber \\&\quad \le c\Vert \delta A\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} \Vert A_1,A_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })} \Vert v_1\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}\nonumber \\&\quad \le ct^{s/2}\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} \Vert v_1, v_2\Vert ^2_{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}. \end{aligned}$$
(5.37)

Thus, from (5.35)–(5.37) we have

$$\begin{aligned}&\Vert \delta f_2\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\\&\quad \le ct^{s/2} \Vert v_1, v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })} (\Vert v_1,v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}+1) \Vert \delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}, \end{aligned}$$

which together with (5.34) yields

$$\begin{aligned} \Vert \delta F\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \le \eta _2(t) \Vert \nabla ^2\delta v, \nabla \delta Q\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \end{aligned}$$
(5.38)

with some \(\eta _2(t)\) such that \(\eta _2(t)\rightarrow 0\) as \(t\rightarrow +0\).

Finally, let us get estimate of \(\Vert \delta R_t\Vert \). Recall that \(\delta R=(\text {Id}-A_2)\delta v-\delta A v_1\) and hence \((\delta R)_t=-A_{2t}\delta v+(\text {Id}-A_2)\delta v_t-\delta A_t v_1-\delta A v_{1t}\). Since we have

$$\begin{aligned} A_{it}= -A_i^2(t)Dv_i, \;i=1,2, \end{aligned}$$

see (5.18), it follows by (5.23) and Lemma 5.1 (ii) that

$$\begin{aligned} \Vert A_{2t}\delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}= & {} \Vert A_2^2Dv_2\delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\\le & {} c \Vert A_2^2 Dv_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })} \cdot \Vert \delta v\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\\le & {} c t^{s/2}\Vert v_2\Vert _{L^\infty _{s/2}(0,t;b^{-s+1+n/q}_{r,\infty })}\Vert \delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}, \end{aligned}$$
(5.39)

where we used that

$$\begin{aligned} \Vert \delta v\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}\le ct^{s/2} \Vert \delta v_t\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}. \end{aligned}$$

On the other hand, using (5.29), we have

$$\begin{aligned}&\Vert (\text {Id}-A_2)\delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })} \le c \Vert \text {Id}-A_2\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} \Vert \delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\quad \le c t^{s/2}\Vert v_2\Vert _{L^\infty (0,t; b^{-s+1+n/q}_{r,\infty })} \Vert \delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}. \end{aligned}$$
(5.40)

It is easy to see that \(\delta A_t= -A_1^2 D\delta v-\delta A(A_1+A_2)Dv_2\). By (5.23) and (5.25), we have

$$\begin{aligned}&\Vert \delta A(A_1+A_2)Dv_2 v_1\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\\&\quad \le \Vert \delta A\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} \Vert (A_1+A_2)Dv_2\Vert _{L^\infty _{s/2}(0,t; b^{-s+n/q}_{r,\infty })}\Vert v_1\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}\\&\quad \le ct^{s/2}\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} \Vert \nabla ^2 v_2\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })} \Vert v_{1}, v_2\Vert _{L^\infty (0,t; b^{-s-1+n/q}_{r,\infty })}. \end{aligned}$$

Moreover,

$$\begin{aligned}&\Vert A_1^2 D\delta vv_1\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\nonumber \\&\quad \le c\Vert A_1^2\Vert _{L^\infty (0,t;b^{-s+n/q}_{r,\infty })}\Vert D\delta v\otimes v_1\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\quad \le c\Vert D\delta v\otimes (v_1-v_1(0)), D\delta v\otimes v_1(0)\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\quad \le c\Vert D\delta v\Vert _{L^\infty _{s/2}(0,t;b^{-s+n/q}_{r,\infty })} \Vert v_1-v_1(0)\Vert _{L^\infty (0,t;b^{-s-1+n/q}_{r,\infty })} +c\Vert D\delta v\otimes v_1(0)\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \nonumber \\&\quad \le c\Vert \nabla ^2 \delta v,\delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\cdot t^{s/2}\Vert v_{1t}\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\qquad +c\Vert D\delta v\Vert _{L^\infty _{1-\theta +s/2}(0,t;B^{2\theta -s-2+n/q}_{r,1})} \Vert v_1(0)\Vert _{L^\infty _\theta (0,t;B^{-1+n/q}_{r,\infty })}\nonumber \\&\quad \le ct^{s/2}\Vert v_{1t}\Vert _{L^\infty _{s/2}(0,t;b^{-s-1+n/q}_{r,\infty })} \Vert \nabla ^2 \delta v,\delta v_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\&\qquad +ct^{1-\theta }\Vert v_1(0)\Vert _{B^{-1+n/q}_{r,\infty }} \Vert \delta v\Vert _{L^\infty _{1-\theta +s/2}(0,t;B^{2\theta -s-1+n/q}_{r,1})}, \end{aligned}$$
(5.41)

where we used that \(B^{2\theta -s-2+n/q}_{r,1}\cdot B^{-1+n/q}_{r,\infty } \hookrightarrow B^{-s-1+n/q}_{r,\infty }\) for \(\theta =1-\frac{1}{2}(\frac{n}{q}-\frac{n}{r})\) thanks to Lemma 5.1 (iv).

Finally, in view of the expression \(\delta A(t)=h_1(t)\int _0^t D\delta v\,d\tau \) and (5.25), we get by Lemma 5.1 (ii) that

$$\begin{aligned} \Vert \delta A v_{1t}\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))}\le & {} c\Vert \delta A\Vert _{L^\infty (0,t; b^{-s+n/q}_{r,\infty })} \Vert v_{1t}\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}\nonumber \\\le & {} ct^{s/2}\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })} \Vert v_{1t}\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty })}. \end{aligned}$$
(5.42)

Thus, from (5.39)–(5.42) it follows that

$$\begin{aligned} \Vert \delta R_t\Vert _{L^\infty _{s/2}(0,t; b^{-s-1+n/q}_{r,\infty }({{\mathbb {R}}}^n))} \le \eta _3(t)\big ( \Vert \delta v\Vert _{L^\infty (0,t; B^0_{q,\infty }(\Omega ))} +\Vert \delta v\Vert _{L^\infty _{s/2}(0,t; b^{-s+1+n/q}_{r,\infty })}\big ) \end{aligned}$$
(5.43)

with some \(\eta _3(t)\) converging to 0 as \(t\rightarrow +0\).

Summarizing, we can conclude from (5.13), (5.17), (5.33), (5.38) and (5.43) that \(\delta v(t)=0, \delta Q(t)=0\) for all \(t\in (0,T_1)\) with some \(T_1>0\). Then, by standard continuation argument, it can be shown that \(\delta v(t)=0, \delta Q(t)=0\) for all \(t\in (0,T)\).

Now, the proof of the uniqueness part of Theorem 1.1 comes to end. \(\square \)