Let
$$\begin{aligned}& \rho^{\varepsilon}_{0}(\mathbf{x})=\textbf{M}^{(2)}_{\varepsilon }( \rho_{0})= \frac{1}{\varepsilon^{2}} \int_{Q_{f}}J\biggl(\frac{|\mathbf{x}-\mathbf {y}|}{\varepsilon}\biggr)\rho_{0}( \mathbf{y})\,dy,\quad \mathbf{x}\in Q_{f}, \\& \rho_{0}^{\varepsilon}\in C^{\infty}(Q_{f}),\qquad \rho_{0}^{\varepsilon }(\mathbf{x})\rightarrow\rho_{0}( \mathbf{x})\quad \mbox{a.e. in }Q^{\pm}_{f}(0), \end{aligned}$$
where
$$\begin{aligned}& J(s)\geq0, \qquad J(s)=0\quad\mbox{for } |s|>1,\\& J(s)=J(-s),\quad J \in C^{\infty }(-\infty, +\infty),\qquad \int_{\mathbb{R}^{2}}J\bigl(|x|\bigr)\,dx=1. \end{aligned}$$
We divide the proof of Theorem 2.1 into several steps.
-
3.1
We show that for each \(\varepsilon>0\) the modified problem
$$\begin{aligned}& \begin{aligned} &\nabla\cdot\mathbb{P}^{\varepsilon}_{f}+ \rho^{\varepsilon}\mathbf {e}+\mathbf{f}=0,\qquad \nabla\cdot \mathbf{u}^{f,\varepsilon}=0, \quad\mathbf{x}\in Q_{f}, 0< t< T, \\ &\nabla\cdot\mathbb{P}^{\varepsilon}_{s}+\mathbf{f}=0,\qquad \nabla \cdot\mathbf{u}^{s,\varepsilon}=0,\quad \mathbf{x}\in Q_{s}, 0< t< T,\\ & \mathbf{u}^{s,\varepsilon} (\mathbf{x},0)=0, \quad\mathbf{x}\in S, \\ &\mathbf{u}^{f,\varepsilon}=\frac{\partial\mathbf {u}^{s,\varepsilon}}{\partial t},\qquad \mathbb{P}^{\varepsilon}_{f} \cdot\mathbf{n}= \mathbb{P}^{\varepsilon}_{s}\cdot\mathbf{n},\quad \mathbf{x}\in S, 0< t< T, \\ &\mathbb{P}^{\varepsilon}_{i}\cdot\mathbf{e}_{1}=0,\quad \mathbf {x}\in S_{i}^{\pm}, i=f,s,\qquad \mathbf{u}^{s,\varepsilon}( \mathbf{x},t)=0, \quad\mathbf {x}\in S^{0}, 0< t< T, \\ &\mathbb{P}_{f}\bigl(\mathbf{u}^{f,\varepsilon},p^{\varepsilon}_{f} \bigr)= 2 \mu\mathbb{D}\bigl(\mathbf{u}^{f,\varepsilon}\bigr)-p^{\varepsilon}_{f} \mathbb{I},\qquad \mathbb{P}_{s} \bigl(\mathbf{u}^{s,\varepsilon},p^{\varepsilon}_{s} \bigr)= 2 \lambda\mathbb{D}\bigl(\mathbf{u}^{s,\varepsilon}\bigr)-p^{\varepsilon }_{s} \mathbb{I}, \end{aligned} \end{aligned}$$
(29)
$$\begin{aligned}& \begin{aligned} &\frac{\partial\rho^{\varepsilon}}{\partial t}+\mathbf {v}^{\varepsilon} \cdot\nabla\rho^{\varepsilon} =0,\quad \mathbf{x}\in Q, 0< t< T, \\ &\rho^{\varepsilon}(\mathbf{x},t)=\rho_{0}^{\pm},\quad \mathbf{x}\in S_{f}^{\pm}, 0< t< T, \\ &\rho^{\varepsilon} (\mathbf{x},0)=\rho_{0}^{\varepsilon }( \mathbf{x}), \quad\mathbf{x}\in Q, \qquad\mathbf{v}^{\varepsilon}=\textbf{M}^{(1)}_{\varepsilon} \bigl(\textbf {M}^{(2)}_{\varepsilon}\bigl(\mathbf{u}^{f,\varepsilon} \bigr) \bigr), \\ & \textbf{M}^{(1)}_{\varepsilon}(\mathbf{v})=\frac{1}{\varepsilon} \int_{0}^{\infty}J \biggl(\frac{|t-\tau|}{\varepsilon} \biggr) \mathbf {v}(\mathbf{x},\tau)\,d\tau, \end{aligned} \end{aligned}$$
(30)
of finding \(\mathbf{u}^{f,\varepsilon}\), \(p^{\varepsilon}_{f}\), \(\rho^{\varepsilon}\), \(\mathbf{u}^{s,\varepsilon}\), \(p^{\varepsilon}_{s}\) has at least one classical solution.
To solve the problem (29), (30) we use the Schauder fixed point theorem [14].
Namely, let \(\mathcal{M}\) be the set of all continuous functions
$$\widetilde{\rho}\in C(\overline{G}), \quad G=Q_{f}\times(0,T), $$
such that
$$ \rho^{-}\leq\widetilde{\rho}(\mathbf{x},t) \leq \rho^{+}. $$
(31)
For fixed \(\varepsilon>0\) we define the following nonlinear operator:
$$ \Phi: \mathcal{M} \rightarrow\mathcal{M},\quad \rho=\Phi( \widetilde{\rho}), $$
(32)
and
-
3.1.1
prove that the linear problem
$$\begin{aligned}& \begin{aligned} &\nabla\cdot\mathbb{P}_{f}+ \widetilde{\rho} ^{\varepsilon} \mathbf {e}+\mathbf{f}=0,\qquad \nabla\cdot \mathbf{u}^{f}=0, \quad\mathbf{x}\in Q_{f}, 0< t< T, \\ &\nabla\cdot\mathbb{P}_{s}+\mathbf{f}=0,\qquad \nabla\cdot \mathbf{u}^{s}=0,\quad \mathbf{x}\in Q_{s}, 0< t< T,\\ & \mathbf{u}^{s} (\mathbf{x},0)=0, \quad\mathbf{x}\in S, \\ &\mathbf{u}^{f}=\frac{\partial\mathbf{u}^{s}}{\partial t},\qquad \mathbb{P}_{f}\cdot \mathbf{n}= \mathbb{P}_{s}\cdot\mathbf{n},\quad \mathbf{x}\in S, 0< t< T, \\ &\mathbb{P}_{i}\cdot\mathbf{e}_{1}=0,\quad \mathbf{x}\in S_{i}^{\pm}, i=f,s,\qquad \mathbf{u}^{s}(\mathbf{x},t)=0,\quad \mathbf{x}\in S^{0}, 0< t< T, \end{aligned} \end{aligned}$$
(33)
$$\begin{aligned}& \begin{aligned} &\frac{\partial\rho}{\partial t}+\mathbf{v}^{\varepsilon} \cdot\nabla \rho =0,\quad \mathbf{x}\in Q, 0< t< T, \\ &\rho(\mathbf{x},t)=\rho_{0}^{\pm}, \quad\mathbf{x}\in S_{f}^{\pm}, 0< t< T,\qquad \rho(\mathbf{x},0)= \rho_{0}^{\varepsilon}(\mathbf{x}),\quad \mathbf{x}\in Q, \end{aligned} \\& \widetilde{\rho} ^{\varepsilon}(\mathbf{x},t)=\textbf {M}^{(1)}_{\varepsilon}(\widetilde{\rho}),\qquad \mathbf{v}^{\varepsilon}= \textbf{M}^{(1)}_{\varepsilon} \bigl(\textbf {M}^{(2)}_{\varepsilon} \bigl(u^{f}\bigr) \bigr), \\ & \mathbb{P}_{f}=2 \mu\mathbb{D}\bigl(\mathbf{u}^{f} \bigr)-p_{f} \mathbb{I},\qquad \mathbb{P}_{s}=2 \lambda \mathbb{D}\bigl(\mathbf{u}^{s}\bigr)-p_{s} \mathbb{I}, \end{aligned}$$
(34)
for given \(\widetilde{\rho}\in\mathcal{M}\) has a unique weak solution \(\{\mathbf{u}^{f}, p_{f}, \mathbf{u}^{s}, p_{s}, \rho=\Phi (\widetilde{\rho})\}\).
The properties of the mollifier \(\textbf{M}^{(2)}_{\varepsilon}\) and continuity equation for u imply the continuity equation for \(\mathbf{v}^{\varepsilon}\):
$$\nabla\cdot\mathbf{v}^{\varepsilon}=0, \quad\mathbf{x}\in Q_{f}, 0< t< T. $$
-
3.1.2
To prove the solvability of (29), (30) we show that the operator Φ is completely continuous and according to Schauder’s fixed point theorem it has at least one fixed point in \(\mathcal{M}\).
-
3.1.3
Finally, we prove that the modified problem (
29
), (
30
) has a unique solution.
-
3.2
In this part of the proof we derive uniform bounds for the solutions of the modified problem (
29
), (
30
).
More precisely, we first derive
-
3.2.1
\(L_{2}\)-estimates for the solutions of the modified problem (
29
), (
30
),
and using the Fourier transform’s techniques we find
-
3.2.2
uniform estimates for the solutions of the problem (
29
), (
30
) in Hölder’s spaces.
This part of the proof contains a lot of technical details and is very difficult to follow. Unfortunately, the linear problem that arises is completely novel and requires special consideration. The standard method for classical differential equations consists of:
(1) a Fourier transform of the original problem,
(2) exact representation of the solution of the corresponding linear problem in new variables, and
(3) inverse transformation and derivation of the exact representation of the solution in the original variables.
In our case the linear problem in new variables for the Fourier transform of the solution is still complicated and has no exact representation. This is why we estimate only a Fourier transform of the solution and after that use Parseval’s equality to get estimates for the solution in the original variables.
-
3.3
Finally we derive uniform estimates for the densities and prove the existence of a smooth surface separating the parts of the domain occupied by the two different fluids.
In what follows, by C we denote constants depending only on \(C^{0}\), \(\rho^{\pm}\), and \(\rho_{s}\).
3.1 Solvability of the modified problem (29), (30)
Definition 1
We say that a set of functions \(\{\mathbf{u}^{f,\varepsilon}, p^{\varepsilon}_{f}, \mathbf{u}^{s,\varepsilon}, p^{\varepsilon}_{s}, \rho^{\varepsilon }\}\)
$$\begin{aligned}& \mathbf{u}^{i,\varepsilon}\in L_{\infty} \bigl((0,T);W^{1}_{2}(Q_{i}) \bigr),\qquad p^{\varepsilon}_{i}\in L_{\infty} \bigl((0,T);L_{2}(Q_{i}) \bigr),\quad i=f,s, \\ & \rho^{\varepsilon}\in C^{1}(\overline{Q_{T}}), \quad Q_{T}=Q\times(0,T), \end{aligned}$$
is a weak solution of the problem (29), (30), if it satisfies the integral identity
$$\begin{aligned}& \int_{Q_{f}}\mathbb{P}_{f}\bigl(\mathbf{u}^{f,\varepsilon},p^{\varepsilon}_{f} \bigr): \mathbb{D}(\boldsymbol {\varphi})\,dx+ \int_{Q_{s}}\mathbb{P}_{s} \bigl(\mathbf{u}^{s,\varepsilon },p^{\varepsilon}_{s} \bigr): \mathbb{D}(\boldsymbol {\varphi})\,dx \\& \quad= \int_{Q_{f}}{\rho} ^{\varepsilon} (\mathbf{e}\cdot\boldsymbol {\varphi}) \,dx+ \int_{Q}\mathbf{f}\cdot\boldsymbol {\varphi} \,dx, \end{aligned}$$
(35)
for almost all \(t\in(0,T)\), and for arbitrary smooth functions \(\boldsymbol {\varphi}(\mathbf{x})\), vanishing at \(S^{0}\), and the problem (30) in the usual sense.
3.1.1 The solvability of the problem (33) for given ρ̃
Lemma 1
Under the same conditions of Theorem
2.1
the problem (33) for given
ρ̃
has a unique weak solution,
$$\mathbf{u}^{i}\in L_{\infty} \bigl((0,T);W^{1}_{2}(Q_{i}) \bigr),\quad i=f,s,\qquad \frac{\partial\mathbf{u}^{s}}{\partial t}\in L_{2} \bigl((0,T);W^{1}_{2}(Q_{s}) \bigr), $$
satisfying the following estimates:
$$ \begin{aligned} & \int_{0}^{T} \int_{Q_{f}}\mathbb{D}\bigl(\mathbf{u}^{f}\bigr): \mathbb {D}\bigl(\mathbf{u}^{f}\bigr)\,dx\,dt+ \max_{0< t< T} \int_{Q_{s}}\mathbb{D} \bigl(\mathbf{u}^{s}(\mathbf {x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{s}(\mathbf{x},t) \bigr)\,dx \leq C, \\ &\max_{0< t< T} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f}(\mathbf {x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f}(\mathbf{x},t) \bigr) \,dx\\ &\quad{}+ \int_{0}^{T} \int_{Q_{s}}\mathbb{D}\biggl(\frac{\partial\mathbf {u}^{s}}{\partial t}\biggr): \mathbb{D}\biggl(\frac{\partial\mathbf{u}^{s}}{\partial t}\biggr)\,dx\leq C_{1}(\varepsilon), \\ &\max_{0< t< T} \int_{Q_{f}} \bigl\vert \mathbf{u}^{f}(\mathbf{x},t) \bigr\vert ^{2}\,dx+ \max_{0< t< T} \int_{Q_{s}} \biggl\vert \frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf{x},t) \biggr\vert ^{2}\,dx \leq C_{2}(\varepsilon). \end{aligned} $$
(36)
Proof
First of all, note that due to linearity of the problem it suffices to find corresponding a priori estimates.
To prove the first estimate in (36) we multiply the Stokes equation for \(\mathbf{u}^{f}\) by \(\mathbf{u}^{f}\) and integrate by parts over domain \(Q_{f}\), multiply the Lamé equation for \(\mathbf{u}^{s}\) by \(\frac{\partial\mathbf {u}^{s}}{\partial t}\), integrate by parts over domain \(Q_{s}\), and sum results.
To get the second estimate in (36) we differentiate the Stokes equation for \(\mathbf{u}^{f}\) and the Lamé equation for \(\mathbf{u}^{s}\) with respect to time, multiply the first expression by \(\mathbf{u}^{f}\) and integrate by parts over domain \(Q_{f}\times(0,t_{0})\), multiply the second expression by \(\frac{\partial\mathbf{u}^{s}}{\partial t}\) and integrate by parts over domain \(Q_{s}\times(0,t_{0})\), and sum results:
$$\begin{aligned}& \frac{\mu}{2} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f} (\mathbf {x},t_{0}) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},t_{0}) \bigr)\,dx \\& \qquad{}+\lambda \int_{0}^{t_{0}} \int_{Q_{s}}\mathbb{D} \biggl(\frac{\partial \mathbf{u}^{s}}{\partial t}(\mathbf{x},t) \biggr): \mathbb{D} \biggl(\frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf {x},t) \biggr)\,dx\,dt \\& \quad= \int_{0}^{t_{0}} \int_{Q_{f}}\frac{\partial\widetilde{\rho} ^{\varepsilon }}{\partial t}(\mathbf{x},t) \bigl( \mathbf{u}^{f}(\mathbf{x},t)\cdot\mathbf{e} \bigr) \,dx\,dt\equiv I. \end{aligned}$$
(37)
Thus,
$$\begin{aligned}& \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},t_{0}) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},t_{0}) \bigr)\,dx \\ & \qquad{}+ \int_{0}^{t_{0}} \int_{Q_{s}}\mathbb{D} \biggl(\frac{\partial\mathbf {u}^{s}}{\partial t}(\mathbf{x},t) \biggr): \mathbb{D} \biggl(\frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf {x},t) \biggr)\,dx\,dt \\ & \quad\leq \vert I \vert \leq \int_{0}^{t_{0}} \int_{Q_{f}} \bigl\vert \mathbf {u}^{f}(\mathbf{x},t) \bigr\vert ^{2} \,dx\,dt+C_{0}(\varepsilon). \end{aligned}$$
(38)
To estimate the right-hand side in (38) we introduce a new function \({u}(\mathbf{x},t)\):
$$\mathbf{u}(\mathbf{x},t)=\mathbf{u}^{f}(\mathbf{x},t)\quad \mbox{for } \mathbf{x}\in Q_{f},\quad \mbox{and} \quad\mathbf {u}(\mathbf{x},t)= \frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf{x},t)\quad \mbox{for } \mathbf{x}\in Q_{s}. $$
It is easy to see that \(\mathbf{u}\in W^{1}_{2}(Q)\), \(\mathbf{u}(\mathbf{x},t)=0\) for \(\mathbf{x}\in S^{0}\) (see the boundary condition in (31)), and
$$\begin{aligned}& \int_{Q} \bigl\vert \mathbf{u}(\mathbf{x},t) \bigr\vert ^{2}\,dx= \int _{Q_{f}} \bigl\vert \mathbf{u}^{f}(\mathbf{x},t) \bigr\vert ^{2}\,dx+ \int_{Q_{s}} \biggl\vert \frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf {x},t) \biggr\vert ^{2}\,dx, \\ & \int_{Q}\mathbb{D} \bigl(\mathbf{u}(\mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}(\mathbf{x},t) \bigr)\,dx \\ & \quad= \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f}( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f}(\mathbf{x},t) \bigr)\,dx+ \int_{0}^{t_{0}} \int_{Q_{s}}\mathbb{D} \biggl(\frac{\partial\mathbf {u}^{s}}{\partial t}(\mathbf{x},t) \biggr): \mathbb{D} \biggl(\frac{\partial\mathbf{u}^{s}}{\partial t}(\mathbf {x},t) \biggr)\,dx\,dt. \end{aligned}$$
(39)
Therefore, we may apply the Friedrichs-Poincaré inequality [15]
$$ \int_{Q} \bigl\vert \mathbf{u}(\mathbf{x},t) \bigr\vert ^{2}\,dx \leq C \int_{Q}\mathbb{D} \bigl(\mathbf{u}(\mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}(\mathbf{x},t) \bigr)\,dx, $$
(40)
which together with (38) and (39) imply
$$\begin{aligned}& \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},t_{0}) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},t_{0}) \bigr)\,dx \\ & \quad\leq C \int_{0}^{t_{0}} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf {u}^{f}( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f}(\mathbf{x},t) \bigr)\,dx\,dt+C_{0}(\varepsilon). \end{aligned}$$
In turn, if we put
$$y(t)= \int_{0}^{t} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f} ( \mathbf{x},\tau) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f} (\mathbf{x}, \tau) \bigr) \,dx\,d\tau, $$
we arrive at the differential inequality
$$\frac{dy}{dt}(t) \leq C y(t)+C_{0}(\varepsilon), \quad y(0)=0. $$
This last equation results in
$$\max_{0< t< T}y(t) \leq C_{1}(\varepsilon), $$
and, consequently, the second estimate in (36) follows.
The third estimate in (36) follows from (40). □
3.1.2 Solvability of the modified problem (29), (30)
Lemma 2
Under the same conditions of Theorem
2.1
the problem (29), (30) has a unique weak solution.
Proof
For given ρ̃ we may find the solution \(\mathbf {u}^{f}=\Phi_{1}(\widetilde{\rho})\) of the problem (33), then solve the initial boundary value problem (34) and find \(\rho=\Phi_{2}(\mathbf{u}^{f})=\Phi(\widetilde{\rho})\) such that
$$ \rho^{-}\leq\rho(\mathbf{x},t) \leq\rho^{+}, \quad \rho\in C^{2,1}(\overline{G}). $$
(41)
The first estimate follows from the maximum principle and shows that Φ transforms \(\mathcal{M}\) into itself, and the smoothness of ρ follows from existence theorems for parabolic equations with smooth coefficients ([13], p.320).
So, if we prove the continuity of the operator Φ, then Φ would be completely continuous due to (41). Finally, the Schauder fixed point theorem [14] permits us to find a fixed point of the operator Φ and solve the problem (29), (30).
The continuity of the linear operator \(\Phi_{1}\)
$$\Phi_{1} : \mathcal{M} \rightarrow\mathcal{B}= L_{\infty} \bigl((0,T); W^{1}_{2}(Q_{f}) \bigr), $$
follows from estimates (34).
The nonlinear operator \(\Phi_{2}\) is also continuous.
In fact, let \(\mathbf{u}^{f}_{1}, \mathbf{u}^{f}_{2}\in L_{\infty} ((0,T); W^{1}_{2}(Q_{f}) )\). Then
$$\mathbf{v}^{\varepsilon}_{i}=\textbf{M}^{(1)}_{\varepsilon} \bigl(\textbf{M}^{(2)}_{\varepsilon} \bigl(\mathbf{u}^{f}_{i} \bigr) \bigr) \in C^{\infty}(\overline{G}), $$
and for the differences
$$\rho=\rho_{1}-\rho_{2},\qquad \rho_{i}= \Phi_{2}\bigl(\mathbf{u}^{f}_{i}\bigr),\qquad \mathbf{v}=\mathbf{v}^{\varepsilon}_{1}-\mathbf {v}^{\varepsilon}_{2}, $$
one has
$$ \begin{aligned} &\frac{\partial\rho}{\partial t}+ \mathbf{v}_{1}^{\varepsilon}\cdot \nabla\rho =\mathbf{v}\cdot\nabla \rho_{2},\quad \mathbf{x}\in Q, 0< t< T, \\ &\rho(\mathbf{x},t)=0,\quad \mathbf{x}\in S_{f}^{\pm}, 0< t< T, \\ &\rho(\mathbf{x},0)=0,\quad \mathbf{x}\in Q. \end{aligned} $$
(42)
Therefore
$$ \vert \rho_{1}-\rho_{2} \vert ^{(0)}_{G}=|\rho|^{(0)}_{G}\leq C_{1}(\varepsilon) |\mathbf{v}|^{(1)}_{G}\leq C_{2}(\varepsilon) \bigl\Vert \mathbf {u}^{f}_{1}- \mathbf{u}^{f}_{2} \bigr\Vert _{\mathcal{B}}, $$
(43)
which proves the complete continuity of the operator Φ and the solvability of the problem (29).
To prove the uniqueness of the problem (29), (30) we suppose that there exist two different solutions \((\mathbf{u}^{f,\varepsilon}_{i}, \mathbf{u}^{s,\varepsilon }_{i}, \rho^{\varepsilon}_{i})\), \(i=1,2\).
Then the difference \(\{\mathbf{u}^{f}, \mathbf{u}^{s}, \rho\}\), \(\mathbf{u}^{f}=\mathbf{u}^{f,\varepsilon}_{1}-\mathbf {u}^{f,\varepsilon}_{2}\), \(\mathbf{u}^{s}=\mathbf{u}^{s,\varepsilon}_{1}-\mathbf {u}^{s,\varepsilon}_{2}\), \(\rho=\rho^{\varepsilon}_{1}-\rho^{\varepsilon}_{2}\), satisfies the following initial boundary value problem:
$$\begin{aligned}& \nabla\cdot\mathbb{P}_{f}+\rho\mathbf{e}=0,\qquad \nabla\cdot \mathbf{u}^{f}=0,\quad \mathbf{x}\in Q_{f}, 0< t< T, \\& \nabla\cdot\mathbb{P}_{s}=0,\qquad \nabla\cdot\mathbf{u}^{s}=0, \quad \mathbf{x}\in Q_{s}, 0< t< T,\qquad \mathbf{u}^{s}( \mathbf{x},0)=0, \quad\mathbf{x}\in S, \\& \mathbf{u}^{f}=\frac{\partial\mathbf{u}^{s}}{\partial t},\qquad \mathbb{P}_{f}\cdot \mathbf{n}= \mathbb{P}_{s}\cdot\mathbf{n},\quad \mathbf{x}\in S, 0< t< T, \\& \mathbb{P}_{i}\cdot\mathbf{e}_{1}=0, \quad\mathbf{x}\in S_{i}^{\pm}, i=f,s, \qquad\mathbf{u}^{s}(\mathbf{x},t)=0,\quad \mathbf{x}\in S^{0}, 0< t< T, \\& \frac{\partial\rho}{\partial t}+\mathbf{v}_{1}^{\varepsilon}\cdot \nabla\rho = \mathbf{v}^{\varepsilon}\cdot\nabla\rho_{2}^{\varepsilon},\quad \mathbf{x}\in Q, 0< t< T, \\& \rho(\mathbf{x},t)=0, \quad\mathbf{x}\in S_{f}^{\pm}, 0< t< T, \\& \rho(\mathbf{x},0)=0,\quad \mathbf{x}\in Q, \\& \mathbf{v}^{\varepsilon}=\textbf{M}^{(1)}_{\varepsilon} \bigl(\textbf {M}^{(2)}_{\varepsilon}(\mathbf{u}) \bigr),\qquad \mathbb{P}_{f}=2 \mu\mathbb{D}\bigl(\mathbf{u}^{f}\bigr)-p_{f} \mathbb{I}, \qquad \mathbb{P}_{s}=2 \lambda\mathbb{D}\bigl(\mathbf{u}^{s} \bigr)-p_{s} \mathbb{I}. \end{aligned}$$
Now we multiply the dynamic equation for \(\mathbf{u}^{f}\) by \(\mathbf{u}^{f}\) and integrate by parts over domain \(Q_{f}\times(0,t_{0})\), the dynamic equation for \(\mathbf{u}^{s}\) by \(\frac{\partial\mathbf{u}^{s}}{\partial t}\) and integrate by parts over domain \(Q_{s}\times(0,t_{0})\) the equation for ρ by ρ and integrate by parts over domain \(Q_{f}\times(0,t_{0})\), and sum all results:
$$\begin{aligned}& \mu \int_{0}^{t_{0}} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf {u}^{f}( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f}(\mathbf{x},t) \bigr)\,dx\,dt \\& \qquad{}+\frac{1}{2} \int_{Q_{f}} \bigl\vert \rho(\mathbf{x},t_{0}) \bigr\vert ^{2}\,dx+ \frac{\lambda}{2} \int_{Q_{s}}\mathbb{D} \bigl(\mathbf{u}^{s} ( \mathbf{x},t_{0}) \bigr): \mathbb{D} \bigl(\mathbf{u}^{s}( \mathbf{x},t_{0}) \bigr)\,dx \\& \quad= \int_{0}^{t_{0}} \int_{Q_{f}}\rho(\mathbf{x},t) \bigl(\mathbf{u}^{f} ( \mathbf{x},t)\mathbf{e} \bigr) \,dx\,dt\equiv I_{0}. \end{aligned}$$
Introducing the new function
$$\begin{aligned}& \mathbf{w}(\mathbf{x},t)= \int_{0}^{t}\mathbf{u}^{f} (\mathbf{x}, \tau)\,d\tau, \quad\mathbf{x}\in Q_{f}, \qquad\mathbf{w}(\mathbf{x},t)= \mathbf{u}^{s}(\mathbf{x},t),\quad \mathbf{x}\in Q_{s}, \\& \mathbf{u}(\mathbf{x},t)=0, \quad\mathbf{x}\in S^{0}, \end{aligned}$$
in the same way as before (see estimates (36)) we get
$$|I_{0}|\leq\delta \int_{0}^{t_{0}} \int_{Q_{f}} \bigl\vert \mathbf {u}^{f}(\mathbf{x},t) \bigr\vert ^{2}\,dx\,dt+ C(\delta) \int_{0}^{t_{0}} \int_{Q_{f}} \bigl\vert \rho(\mathbf{x},t) \bigr\vert ^{2}\,dx\,dt $$
for arbitrary small \(\delta>0\), and
$$\int_{Q_{f}} \bigl\vert \rho(\mathbf{x},t_{0}) \bigr\vert ^{2}\,dx \leq C(\delta) \int_{0}^{t_{0}} \int_{Q_{f}} \bigl\vert \rho(\mathbf{x},t) \bigr\vert ^{2}\,dx\,dt,\qquad \int_{Q_{f}} \bigl\vert \rho(\mathbf{x},0) \bigr\vert ^{2}\,dx=0. $$
The Gronwall inequality results in \(\rho(\mathbf{x},t)=0\) almost everywhere in G. □
3.2 Uniform bounds for the solutions of the problem (29), (30)
3.2.1
\(L_{2}\)-Estimates for the solutions of the problem (29), (30)
Lemma 3
Under the conditions of Theorem
2.1
for the solution
\(\mathbf{u}^{\varepsilon}\)
of the problem (29) one gets the following estimates:
$$\begin{aligned}& \int_{0}^{T} \int_{Q_{f}}\mathbb{D}\bigl(\mathbf{u}^{f,\varepsilon }\bigr): \mathbb{D}\bigl(\mathbf{u}^{f,\varepsilon}\bigr)\,dx\,dt+ \max_{0< t< T} \int_{Q_{s}}\mathbb{D} \bigl(\mathbf{u}^{s,\varepsilon }( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{s,\varepsilon}(\mathbf{x},t) \bigr)\,dx\leq C, \\& \max_{0< t< T} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon }( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon}(\mathbf{x},t) \bigr)\,dx+ \int_{0}^{T} \int_{Q_{s}}\mathbb{D}\biggl(\frac{\partial\mathbf {u}^{s,\varepsilon}}{\partial t}\biggr): \mathbb{D}\biggl(\frac{\partial\mathbf{u}^{s,\varepsilon}}{\partial t}\biggr)\,dx\leq C, \\& \max_{0< t< T} \int_{Q_{f}} \bigl\vert \mathbf{u}^{f,\varepsilon}(\mathbf {x},t) \bigr\vert ^{2}\,dx+ \max_{0< t< T} \int_{Q_{s}} \biggl\vert \frac{\partial\mathbf{u}^{s,\varepsilon }}{\partial t}(\mathbf{x},t) \biggr\vert ^{2}\,dx\leq C. \end{aligned}$$
(44)
Proof
The proof of these estimates almost exactly repeats the proof of estimates (36). The single difference is in the estimation of the term I in (37):
$$\begin{aligned}& \frac{\mu}{2} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon }( \mathbf{x},t_{0}) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon}( \mathbf{x},t_{0}) \bigr)\,dx \\& \qquad{}+\lambda \int_{0}^{t_{0}} \int_{Q_{s}}\mathbb{D} \biggl(\frac{\partial \mathbf{u}^{s,\varepsilon}}{ \partial t}(\mathbf{x},t) \biggr):\mathbb{D} \biggl(\frac{\partial \mathbf{u}^{s,\varepsilon}}{ \partial t}(\mathbf{x},t) \biggr)\,dx\,dt \\& \quad= \int_{0}^{t_{0}} \int_{Q_{f}}\frac{\partial{\rho} ^{\varepsilon }}{\partial t}(\mathbf{x},t) \bigl( \mathbf{u}^{f,\varepsilon}(\mathbf{x},t)\cdot\mathbf {e} \bigr) \,dx\,dt\equiv I^{\varepsilon}. \end{aligned}$$
To estimate \(I^{\varepsilon}\) we use the differential equation for \(\rho^{\varepsilon}\) in (30):
$$\begin{aligned} \bigl\vert I^{\varepsilon} \bigr\vert =& \biggl\vert \int_{0}^{t_{0}} \int_{Q_{f}}\bigl(\mathbf {u}^{f,\varepsilon}\cdot\mathbf{e}\bigr) \nabla\cdot\bigl(\rho^{\varepsilon}\mathbf{v}^{\varepsilon}\bigr) \,dx\,dt \biggr\vert \\ =& \biggl\vert \int_{0}^{t_{0}} \int_{Q_{f}} \bigl(\rho^{\varepsilon}\mathbf {v}^{\varepsilon} \bigr)\cdot \nabla\mathbf{u}^{f,\varepsilon}\cdot\mathbf{e} \,dx\,dt \biggr\vert \\ \leq & \int_{0}^{t_{0}} \int_{Q_{f}}\mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon }( \mathbf{x},t) \bigr): \mathbb{D} \bigl(\mathbf{u}^{f,\varepsilon}(\mathbf{x},t) \bigr)\,dx\,dt+ C \int_{0}^{t_{0}} \int_{Q_{f}} \bigl(\mathbf{u}^{f,\varepsilon }(\mathbf{x},t) \bigr)^{2}\,dx\,dt+C, \end{aligned}$$
where we have used the evident estimates for the mollifiers \(\textbf {M}^{(1)}_{\varepsilon}\) (\(\textbf{M}^{(2)}_{\varepsilon}\))
$$\int_{0}^{t_{0}} \int_{Q_{f}} \bigl(\mathbf{v}^{\varepsilon }(\mathbf{x},t) \bigr)^{2}\,dx\,dt\leq C \int_{0}^{t_{0}} \int_{Q_{f}} \bigl(\mathbf{u}^{f,\varepsilon }(\mathbf{x},t) \bigr)^{2}\,dx\,dt. $$
The rest of the proof is the same as for (36). □
Lemma 4
Under the conditions of Theorem
2.1
let
\(\mathbf {u}^{f,\varepsilon}\)
be the solution of the problem (29). Then
\(\mathbb{P}_{f}(\mathbf{u}^{f,\varepsilon},p^{\varepsilon}_{f})\in L_{\infty} ((0,T);L_{2}(Q_{f}) )\),
$$ \max_{0< t< T} \int_{Q_{f}} \bigl\vert p_{f}^{\varepsilon}( \mathbf{x},t) \bigr\vert ^{2}\,dx \leq C, $$
(45)
\(\mathbb{P}_{s}(\mathbf{u}^{s,\varepsilon},p^{\varepsilon}_{s})\in L_{\infty} ((0,T);L_{2}(Q_{s}) )\),
$$ \max_{0< t< T} \int_{Q_{s}} \bigl\vert p_{s}^{\varepsilon}( \mathbf{x},t) \bigr\vert ^{2}\,dx \leq C, $$
(46)
and for any
\(\Omega_{f} \subset Q_{f}\)
and
\(\Omega_{s} \subset Q_{s}\)
$$\mathbf{u}^{f,\varepsilon}\in L_{\infty} \bigl((0,T);W^{2}_{m}( \Omega _{f}) \bigr),\qquad \mathbf{u}^{s,\varepsilon}\in L_{\infty} \bigl((0,T);W^{2}_{m}(\Omega _{s}) \bigr) $$
for all
\(m>2\).
Proof
Let be a test function in the integral identity (35). Then this identity takes the form
$$ \int_{Q_{f}} p_{f}^{\varepsilon} \nabla\cdot\boldsymbol { \varphi} \,dx= \int_{Q_{f}} \bigl(2 \mu\mathbb{D}\bigl(\mathbf{u}^{f,\varepsilon } \bigr):\mathbb{D}(\boldsymbol {\varphi}) -\bigl(\mathbf{f}+{\rho} ^{\varepsilon} \mathbf{e}\bigr)\cdot\boldsymbol {\varphi} \bigr) \,dx. $$
(47)
Now we choose φ as a solution of the problem
$$\begin{aligned}& \boldsymbol {\varphi}=\boldsymbol {\varphi}_{0}+\nabla\psi, \\& \triangle\psi=p_{f}^{\varepsilon}, \quad\mathbf{x}\in Q_{f},\qquad \psi (\mathbf{x},t)=0, \quad\mathbf{x}\in S, 0< t< T, \\& \nabla\cdot\boldsymbol {\varphi}_{0}=0,\quad \mathbf{x}\in Q_{f}, \qquad \nabla \psi(\mathbf{x},t)+ \boldsymbol {\varphi}_{0}(\mathbf{x},t)=0, \quad \mathbf{x}\in S, 0< t< T. \end{aligned}$$
The above problem has a unique solution [16] and
$$ \max_{0< t< T} \int_{Q_{f}} \bigl\vert \nabla\boldsymbol {\varphi}(\mathbf {x},t) \bigr\vert ^{2}\,dx \leq C \bigl\Vert p_{f}^{\varepsilon} \bigr\Vert ^{2}_{2,Q_{f}}. $$
(48)
Thus (44), (47), and (48) result in
$$ \max_{0< t< T} \int_{Q_{f}} \bigl\vert p_{f}^{\varepsilon}( \mathbf{x},t) \bigr\vert ^{2}\,dx \leq C, $$
(49)
and \(\mathbb{P}_{f}(\mathbf{u}^{f,\varepsilon },p^{\varepsilon}_{f})\in L_{\infty} ((0,T);L_{2}(Q_{f}) )\).
Coming back to (47) we conclude that \(\mathbb{P}_{f}(\mathbf{u}^{f,\varepsilon },p^{\varepsilon}_{f})\in L_{\infty} ((0,T);W^{1}_{2}(Q_{f}) )\), and (47) is equivalent to the Stokes equation
$$ \mu\triangle\mathbf{u}^{f,\varepsilon}-\nabla p_{f}^{\varepsilon}+ \mathbf{f}+{\rho} ^{\varepsilon} \mathbf{e}=0,\quad \mathbf {x}\in Q_{f}, 0< t< T. $$
(50)
The right-hand side \(\mathbf{F}=\mathbf{f}+{\rho} ^{\varepsilon } \mathbf{e}\) of the differential equation belongs to \(L_{\infty}(G)\). Therefore we may use the same arguments as in [1] and conclude that for any \(\Omega\subset Q_{f}\)
$$\mathbf{u}^{f,\varepsilon}\in L_{\infty} \bigl((0,T);W^{2}_{m}( \Omega ) \bigr)\quad \mbox{for any } m>2. $$
Now we apply the same arguments for the solid component:
$$ \max_{0< t< T} \int_{Q_{s}} \bigl\vert p_{s}^{\varepsilon}( \mathbf{x},t) \bigr\vert ^{2}\,dx \leq C, $$
(51)
and \(\mathbb{P}_{s}(\mathbf{u}^{s,\varepsilon },p^{\varepsilon}_{s})\in L_{\infty} ((0,T);L_{2}(Q_{s}) )\). □
3.2.2 Uniform estimates for the solutions of the problem (29), (30) in Hölder’s spaces
Lemma 5
Under the conditions of Theorem
2.1
let
\(\mathbf {u}^{f,\varepsilon}\)
be the solution of the problem (29). Then
$$ \max_{0< t< T} \bigl\vert \mathbf{u}^{f,\varepsilon}( \cdot,t) \bigr\vert ^{(1+\alpha )}_{Q^{(2\delta)}_{f}}\leq C(\alpha,\delta) $$
(52)
for
\(0<\delta<\delta_{0}\)
with sufficiently small
\(\delta_{0}\), and any
α, \(0<\alpha<1\).
Proof
The domain \(Q^{(\delta)}_{f}\) consists of two disconnected parts. For each part the proof is the same. Note also that
$$\max_{0< t< T} \bigl\vert \mathbf{u}^{f,\varepsilon}(\cdot,t) \bigr\vert ^{(1+\alpha)}_{\Omega ^{(f,\delta)}}\leq C(\alpha,\delta) $$
for any domain \(\Omega^{(f,\delta)}\subset Q^{(\delta)}_{f}\) with the distance to the solid part \(Q_{s}\) greater than δ.
So, we may restrict ourselves only to the part in \(x_{2}<0\) and the following domains \(\Omega^{(\delta)}\).
For \(0<\delta<\frac{1}{2}\) we put
$$\begin{aligned}& \Omega^{(\delta)}=\biggl\{ \mathbf{x}\in Q: -1+\delta< x_{1}< 1- \delta, -\frac{1}{2}+\delta< x_{2}< -\frac{1}{2}-\delta\biggr\} ,\\& \Omega^{(\delta )}_{f}=\Omega^{(\delta)}\cap Q_{f}. \end{aligned}$$
Let \(\zeta^{(\delta)}(\mathbf{x})\) be infinitely smooth functions such that \(\zeta^{(\delta)}(\mathbf{x})=1\) for \(\mathbf{x} \in\Omega^{(2\delta)}\), and \(\zeta^{(\delta )}(\mathbf{x})=0\) for \(\mathbf{x} \in Q\backslash\Omega^{(\delta)}\).
The functions \(\mathbf{u}^{f,\varepsilon,\delta}=\mathbf{u}^{f,\varepsilon }\zeta^{(\delta)}\), \(p_{f}^{\varepsilon,\delta}=p^{\varepsilon}_{f}\zeta^{(\delta)}\), \(\mathbf{u}^{s,\varepsilon,\delta}=\mathbf{u}^{s,\varepsilon }\zeta^{(\delta)}\), and \(p_{s}^{\varepsilon,\delta}=p_{s}^{\varepsilon}\zeta^{(\delta)}\) satisfy in \(\mathbb{R}^{2}\) for \(t>0\) the following linear problem:
$$ \begin{aligned} &\mu\triangle\mathbf{u}^{f,\varepsilon,\delta}- \nabla p_{f}^{\varepsilon,\delta}= \mathbf{F}^{\varepsilon,\delta},\qquad \nabla \cdot\mathbf{u}^{f,\varepsilon,\delta}= {\varphi}^{\varepsilon,\delta},\quad x_{2}>- \frac{1}{2}, 0< t< T; \\ &\lambda\triangle\mathbf{u}^{s,\varepsilon,\delta}-\nabla p_{s}^{\varepsilon,\delta}= \mathbf{F}^{\varepsilon,\delta},\qquad \nabla\cdot\mathbf{u}^{s,\varepsilon,\delta}= { \varphi}^{\varepsilon,\delta},\quad x_{2}< -\frac{1}{2}, 0< t< T; \\ &\mu\biggl(\frac{\partial}{\partial x_{2}}u^{f,\varepsilon,\delta}_{1}+ \frac{\partial}{\partial x_{1}}u^{f,\varepsilon,\delta}_{2}\biggr)= \lambda\biggl( \frac{\partial}{\partial x_{2}}u^{s,\varepsilon,\delta}_{1}+ \frac{\partial}{\partial x_{1}}u^{s,\varepsilon,\delta}_{2} \biggr)+\psi _{1}, \\ &\mu\frac{\partial}{\partial x_{2}}u^{f,\varepsilon,\delta }_{2}+p_{f}^{\varepsilon,\delta}= \lambda\frac{\partial}{\partial x_{2}}u^{s,\varepsilon,\delta }_{2}+p_{s}^{\varepsilon,\delta}+ \psi_{2}, \\ &u^{f,\varepsilon,\delta}_{1}=\frac{\partial}{\partial t} u^{s,\varepsilon,\delta}_{1}, \qquad u^{f,\varepsilon,\delta}_{2}=\frac{\partial}{\partial t}u^{s,\varepsilon,\delta}_{2}, \quad x_{2}=\frac{1}{2}, 0< t< T, \\ &\mathbf{u}^{s,\varepsilon,\delta}(\mathbf{x},0)=0,\quad x_{2}=- \frac{1}{2}. \end{aligned} $$
(53)
Here
$$ \begin{aligned} &\mathbf{F}^{\varepsilon,\delta}= \zeta^{(\delta)}\bigl(\rho^{\varepsilon }\mathbf{e}+\mathbf{f}\bigr)+ \mu \nabla\mathbf{u}^{f,\varepsilon,\delta}\cdot\nabla\zeta^{(\delta )}+p^{\varepsilon,\delta}_{f} \nabla\zeta^{(\delta)}, \\ & {\varphi}^{\varepsilon,\delta}=\mathbf{u}^{f,\varepsilon,\delta }\cdot\nabla \zeta^{(\delta)},\quad x_{2}>-\frac{1}{2}; \\ &\psi_{1}=\frac{\partial\zeta^{(\delta)}}{\partial x_{1}}\bigl(\mu u^{f,\varepsilon,\delta}_{2}- \lambda u^{s,\varepsilon,\delta}_{2}\bigr)+\frac{\partial\zeta^{(\delta )}}{\partial x_{2}}\bigl(\mu u^{f,\varepsilon,\delta}_{1}- \lambda u^{s,\varepsilon,\delta}_{1}\bigr), \\ &\psi_{2}=\frac{\partial\zeta^{(\delta)}}{\partial x_{2}}\bigl(\mu u^{f,\varepsilon,\delta}_{2}- \lambda u^{s,\varepsilon,\delta}_{2}\bigr),\quad x_{2}=- \frac{1}{2}; \\ &\mathbf{F}^{\varepsilon,\delta}=\zeta^{(\delta)}\mathbf{f}+ \lambda\nabla \mathbf{u}^{s,\varepsilon,\delta}\cdot\nabla\zeta ^{(\delta)}+p^{\varepsilon,\delta}_{s} \nabla\zeta^{(\delta)},\qquad {\varphi}^{\varepsilon,\delta}=\mathbf{u}^{s,\varepsilon,\delta } \cdot\nabla\zeta^{(\delta)} x_{2}< -\frac{1}{2}; \\ &\max_{0< t< T} \bigl\Vert \mathbf{F}^{\varepsilon,\delta}(\cdot,t) \bigr\Vert _{2,\mathbb{R}^{2}}+ \max_{0< t< T} \bigl\Vert \boldsymbol { \varphi}^{\varepsilon,\delta}(\cdot,t) \bigr\Vert ^{(1)}_{2,\mathbb{R}^{2}}\leq C(\delta). \end{aligned} $$
(54)
For the sake of simplicity we denote all constants independent of ε as C (or \(C(\delta)\)), and omit for the moment the indices ε and δ.
Now we reduce (53) to homogeneous differential equations by introducing new functions \(\{\mathbf{w}^{f}, r_{f}, \mathbf{w}^{s}, r_{s}\}\) as a solution to the following problem:
$$\begin{aligned}& \begin{aligned} & \mu\triangle\mathbf{w}^{f}-\nabla r_{f}=\mathbf{F},\qquad \nabla\cdot\mathbf{w}^{f}={\varphi}, \quad x_{2}+\frac{1}{2}>0, 0< t< T; \\ & \lambda\triangle\mathbf{w}^{s}-\nabla r_{s}=\mathbf{F}, \qquad \nabla\cdot\mathbf{w}^{s}={\varphi},\quad x_{2}+ \frac{1}{2}< 0, 0< t< T; \\ & \mu\biggl(\frac{\partial w^{f}_{1}}{\partial x_{2}}+ \frac{\partial w^{f}_{2}}{\partial x_{1}}\biggr)= \lambda\biggl( \frac{\partial w^{s}_{1}}{\partial x_{2}}+ \frac{\partial w^{s}_{2}}{\partial x_{1}}\biggr)+\psi_{1}, \\ & \mu\frac{\partial w^{f}_{2}}{\partial x_{2}}+r_{f}= \lambda\frac{\partial w^{s}_{2}}{\partial x_{2}}+r_{s}+ \psi_{2}, \\ & w^{f}_{1}=\frac{\partial w^{s}_{1}}{\partial t},\qquad w^{f}_{2}= \frac{\partial w^{s}_{2}}{\partial t},\quad x_{2}=-\frac{1}{2}, 0< t< T,\qquad \mathbf{w}^{s}(\mathbf{x},0)=0,\quad x_{2}=-\frac{1}{2}. \end{aligned} \end{aligned}$$
(55)
Thus, for
$$\mathbf{v}=\mathbf{u}-\mathbf{w},\qquad q=p-r, $$
one has
$$ \begin{aligned} &\mu\triangle\mathbf{v}^{f}- \nabla q_{f}=0,\qquad \nabla\cdot\mathbf{v}^{f}=0,\quad x_{2}+\frac{1}{2}>0, 0< t< T; \\ &\lambda\triangle\mathbf{v}^{s}-\nabla q_{s}=0,\qquad \nabla \cdot\mathbf{v}^{s}=0,\quad x_{2}+\frac{1}{2}< 0, 0< t< T; \\ &\mu\biggl(\frac{\partial{v}^{f}_{1}}{\partial x_{2}}+\frac{\partial {v}^{f}_{2}}{\partial x_{1}}\biggr)= \lambda\biggl( \frac{\partial{v}^{s}_{1}}{\partial x_{2}}+\frac{\partial {v}^{s}_{2}}{\partial x_{1}}\biggr), \\ &\mu\frac{\partial{v}^{f}_{2}}{\partial x_{2}}+q_{f}= \lambda\frac{\partial{v}^{s}_{2}}{\partial x_{2}}+q_{s}, \\ &\frac{\partial{v}^{s}_{1}}{\partial t}={v}^{f}_{1}+\varphi_{1}, \qquad \frac{\partial{v}^{s}_{2}}{\partial t}={v}^{f}_{2}+\varphi_{2}, \quad x_{2}=-\frac{1}{2}, 0< t< T, \\ &\mathbf{v}^{s}(\mathbf{x},0)=0, \quad x_{2}=- \frac{1}{2}. \end{aligned} $$
(56)
Here
$$\boldsymbol {\varphi}(x_{1},t)= \bigl(\varphi_{1}(x_{1},t), \varphi _{2}(x_{1},t) \bigr)= \mathbf{w}^{f}(x_{1},0,t)- \frac{\partial\mathbf {w}^{s}}{\partial t}(x_{1},0,t). $$
Note that due to the homogeneous boundary condition in (55) for \(\mathbf{w}^{s}\) at S
$$\frac{\partial\mathbf{w}^{s}}{\partial t}(x_{1},0,t)\equiv0 \quad\mbox{and}\quad \varphi_{i}(x_{1},t)= w^{f}_{i}(x_{1},0,t) \qquad \mbox{for } i=1,2 \mbox{ and } t>0. $$
To solve (56) we apply the Fourier transform
$$\widehat{v}(\xi,x_{2},t)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty }v(x_{1},x_{2},t)e^{-i\xi x_{1}} \,dx_{1} $$
with respect the variable \(x_{1}\), and we get the following system of ordinary differential equations in the variable \(x_{2}\):
$$\begin{aligned}& \begin{aligned} &\mu\frac{\partial^{2} \widehat{v}^{ f}_{1}}{\partial x^{2}}-\mu\xi ^{2}\widehat{v}^{ f}_{1}+ i \xi \widehat{p}_{f}=0, \\ &\mu\frac{\partial^{2} \widehat{v}^{ f}_{2}}{\partial x^{2}}-\mu\xi ^{2}\widehat{v}^{ f}_{2}- \frac{\partial\widehat{p}_{f}}{\partial x_{2}}=0,\qquad \frac{\partial^{2}\widehat{p}_{f}}{\partial x^{2}}-\xi^{2}\widehat {p}_{f}=0, \\ &\widehat{v}^{ f}_{1}=-\frac{i}{\xi} \frac{\partial\widehat{v}^{ f}_{2}}{\partial x_{2}},\quad x_{2}+\frac{1}{2}>0, \end{aligned} \end{aligned}$$
(57)
$$\begin{aligned}& \begin{aligned} &\lambda\frac{\partial^{2} \widehat{v}^{ s}_{1}}{\partial x^{2}}-\lambda \xi^{2}\widehat{v}^{ s}_{1}+ i \xi \widehat{p}_{f}=0, \\ &\lambda\frac{\partial^{2} \widehat{v}^{ s}_{2}}{\partial x^{2}}-\lambda\xi^{2}\widehat{v}^{ s}_{2}- \frac{\partial\widehat{p}_{s}}{\partial x_{2}}=0,\qquad \frac{\partial^{2}\widehat{p}_{s}}{\partial x^{2}}-\xi^{2}\widehat {p}_{s}=0, \\ &\widehat{v}^{ s}_{1}=-\frac{i}{\xi} \frac{\partial\widehat{v}^{ s}_{2}}{\partial x_{2}},\quad x_{2}+\frac{1}{2}< 0, \end{aligned} \end{aligned}$$
(58)
$$\begin{aligned}& \begin{aligned} &\mu\frac{\partial\widehat{v}^{ f}_{1}}{\partial x_{2}}-\mu i \xi \widehat{v}^{ f}_{2}= \lambda\frac{\partial\widehat{v}^{ s}_{1}}{\partial x_{2}}-\lambda i \xi \widehat{v}^{ s}_{2}, \\ &\mu\frac{\partial\widehat{v}^{ f}_{2}}{\partial x_{2}}+\widehat{p}_{f}= \lambda\frac{\partial\widehat{v}^{ s}_{2}}{\partial x_{2}}+ \widehat {p}_{s}, \\ &\widehat{v}^{ f}_{1}+\widehat{\varphi}_{1}= \frac{\partial\widehat {v}^{ s}_{1}}{\partial t},\qquad \widehat{v}^{ f}_{2}+\widehat{ \varphi}_{2}=\frac{\partial\widehat{v}^{ s}_{2}}{\partial t},\quad x_{2}=- \frac{1}{2}, \\ &\widehat{v}^{ s}_{1}(\xi,x_{2},0)= \widehat{v}^{ s}_{2}(\xi,x_{2},0)=0. \end{aligned} \end{aligned}$$
(59)
Solutions of (57) and (58) have a very simple form:
$$ \begin{aligned} &\widehat{p}_{f}=c ^{f}_{p} e^{-|\xi|z},\qquad \widehat{v}^{ f}_{1}=i \biggl(\frac{|\xi|}{\xi}c ^{f}_{v}-\frac{1}{2 \mu \xi}\bigl(1-z |\xi|\bigr) c ^{f}_{p} \biggr)e^{-|\xi|z}, \\ &\widehat{v}^{ f}_{2}= \biggl(c ^{f}_{v}+ \frac{z}{2 \mu} c ^{f}_{p} \biggr) e^{-|\xi|z}, \end{aligned} $$
(60)
where \(z=|x_{2}+\frac{1}{2}|\),
$$ \begin{aligned} &\widehat{p}_{s}=c ^{s}_{p} e^{|\xi|z},\qquad \widehat{v}^{ s}_{1}=-i \biggl(\frac{|\xi|}{\xi}c ^{s}_{v}+\frac{1}{2 \lambda\xi}\bigl(1+z |\xi|\bigr) c ^{s}_{p} \biggr)e^{|\xi|z}, \\ &\widehat{v}^{ s}_{2}= \biggl(c ^{f}_{v}+ \frac{z}{2 \mu} c ^{f}_{p} \biggr) e^{-|\xi|z}. \end{aligned} $$
(61)
To define the functions \(c ^{f}_{p}\), \(c ^{f}_{v}\), \(c ^{s}_{p}\), and \(c ^{s}_{v}\), we use the boundary conditions (59):
$$\begin{aligned}& \begin{aligned} &c ^{s}_{v}+ \frac{1}{2\lambda|\xi|} c ^{s}_{p}= \frac{\mu}{\lambda} c ^{f}_{v}-\frac{1}{2 \lambda|\xi|} c ^{f}_{p}, \\ &c ^{s}_{v}+\frac{3}{2 \lambda|\xi|} c ^{s}_{p}=- \frac{\mu}{\lambda}c ^{f}_{v}+ \frac{3}{2 \lambda|\xi|} c ^{f}_{p}, \end{aligned} \end{aligned}$$
(62)
$$\begin{aligned}& \begin{aligned} &\frac{\partial c ^{s}_{v}}{\partial t}+ \frac{1}{2 \lambda|\xi|} \frac{\partial c ^{s}_{p}}{\partial t}= - c ^{f}_{v}+\frac{1}{2 \mu|\xi|} c ^{f}_{p}+i \frac{|\xi|}{\xi} \widehat{ \varphi}_{1}, \\ &\frac{\partial c ^{s}_{v}}{\partial t}=c ^{f}_{v}+\widehat{ \varphi}_{2},\quad c ^{s}_{v}(\xi,0)=c ^{s}_{p}(\xi,0)=0. \end{aligned} \end{aligned}$$
(63)
The first system (62) gives us values \(c ^{s}_{v}\) and \(c ^{s}_{p}\) as a combination of \(c ^{f}_{v}\) and \(c ^{f}_{p}\):
$$ c ^{s}_{v}=2 \frac{\mu}{\lambda} c ^{f}_{v}-\frac{3}{2 \lambda|\xi|} c ^{f}_{p}, \qquad c ^{s}_{p}=-2 \mu|\xi| c ^{f}_{v}+2 c ^{f}_{p}. $$
(64)
Taking into account (63) and (64) we define \(c ^{-}_{v}\) and \(c ^{-}_{r}\) from the Cauchy problem for the following system of ordinary differential equations:
$$ \begin{aligned} &\frac{\partial c ^{f}_{v}}{\partial t}=-4 \frac{\lambda}{\mu} c ^{f}_{v}+ \frac{3 \lambda}{4 \mu^{2} |\xi|} c ^{f}_{p}+ 3 i \frac{\lambda|\xi| }{\mu\xi} \widehat{ \varphi}_{1}-\frac{\lambda }{\mu} \widehat{\varphi}_{2}, \\ &\frac{\partial c ^{f}_{p}}{\partial t}=-6 \lambda|\xi| c ^{f}_{v}+2 \frac{\lambda}{\mu} c ^{f}_{p}+ 4 i \xi\widehat{ \varphi}_{1}-2 \lambda|\xi| \widehat{\varphi}_{2}, \\ &c ^{f}_{v}(\xi,0)=c ^{f}_{p}( \xi,0)=0. \end{aligned} $$
(65)
The last equation is equivalent to the Cauchy problem for the second order ordinary differential equation with coefficient \(k^{2}=32 \frac{\lambda}{\mu}\) independent of ξ:
$$ \begin{aligned} &\frac{\partial^{2} c ^{f}_{v}}{\partial t^{2}}-k^{2} c ^{f}_{v}= 3 \frac{\lambda}{\mu} \widehat{ \varphi}_{2}+ \frac{\partial\widehat{\varphi}_{1}}{\partial t}-4 \frac{\lambda}{\mu} \widehat{ \varphi}_{1}, \\ &c ^{f}_{v}(\xi,0)=0,\qquad \frac{\partial c ^{f}_{v}}{\partial t}(\xi,0)= 3 i \frac{\lambda|\xi| }{\mu\xi} \widehat{\varphi}_{1}(\xi,0)- \frac{\lambda}{\mu} \widehat{\varphi}_{2}(\xi,0). \end{aligned} $$
(66)
Thus,
$$ c ^{j}_{l}(\xi,t)=\sum _{m=f,s}\sum_{i=1}^{2} \biggl(Z^{j,m}_{l,i}(t)\widehat{w}^{m}_{i}( \xi,0,0) + \int_{0}^{t}G^{j,m}_{l,i}(t-\tau) \widehat{w}^{m}_{i}(\xi,0,\tau)\,d\tau \biggr) $$
(67)
for \(j=f,s\) and \(l=v,p\).
Note that \(\widehat{w}^{k}_{i}(\xi,0,0)\), \(z^{j,i}_{l}\) and \(g^{j,i}_{l}\) are known functions, and the functions \(z^{j,i}_{l}\) and \(g^{j,i}_{l}\) are infinitely smooth in t.
Gathering all these issues one has for \(j=f,s\) and \(l=1,2\)
$$\begin{aligned} \widehat{v} ^{j}_{l}(\xi,z,t) =& \sum _{m=f,s}\sum_{i=1}^{2} \biggl\{ \biggl( \biggl(Z^{j,m}_{l,i,0}(z,t)+ \frac{|\xi|}{\xi}Z^{j,m}_{l,i,1}(z,t) \biggr) \widehat{w}^{m}_{i}(\xi ,0,0) \biggr) e^{-|\xi|z} \\ &{}+ \biggl( \int_{0}^{t} \biggl(G^{j,m}_{l,i,0}(z,t- \tau)+ \frac{|\xi|}{\xi}G^{j,m}_{l,i,1}(z,t-\tau) \biggr) \widehat{w}^{m}_{i}(\xi ,0,\tau)\,d\tau \biggr)\biggr\} e^{-|\xi|z}, \end{aligned}$$
(68)
where \(Z^{j,m}_{l,i,k}(z,t)\) and \(G^{j,m}_{l,i,k}(z,t)\), \(k=0,1\), are linear in \(z\geq0\) and infinitely smooth in t.
For \(z=0\) we get
$$\begin{aligned} \widehat{v} ^{j}_{l}(\xi,0,t) =& \sum _{m=f,s}\sum_{i=1}^{2} \biggl\{ \biggl( \biggl(Z^{j,m}_{l,i,0}(0,t)+ \frac{|\xi|}{\xi}Z^{j,m}_{l,i,1}(0,t) \biggr) \widehat{w}^{m}_{i}(\xi ,0,0) \biggr) \\ &{}+ \biggl( \int_{0}^{t} \biggl(G^{j,m}_{l,i,0}(0,t- \tau)+ \frac{|\xi|}{\xi}G^{j,m}_{l,i,1}(0,t-\tau) \biggr) \widehat{w}^{m}_{i}(\xi ,0,\tau)\,d\tau \biggr)\biggr\} \end{aligned}$$
and
$$\max_{0< t< T} \bigl\Vert \widehat{\mathbf{v}} ^{j}( \cdot,0,t) \bigr\Vert ^{(l-\frac {1}{2})}_{2,\mathbb{R}}\leq C_{0} \biggl(1+\sum_{m=f,s}\max_{0< t< T} \bigl\Vert \widehat{\mathbf {w}}^{m}(\cdot,0,t) \bigr\Vert ^{(l-\frac{1}{2})}_{2,\mathbb{R}} \biggr). $$
Due to the Parseval equality
$$\begin{aligned} \max_{0< t< T} \bigl\Vert \mathbf{v} ^{j}(\cdot,0,t) \bigr\Vert ^{(l-\frac{1}{2})}_{2,\mathbb {R}} =& \max _{0< t< T} \bigl\Vert \widehat{\mathbf{v}} ^{j}( \cdot,0,t) \bigr\Vert ^{(l-\frac {1}{2})}_{2,\mathbb{R}} \\ \leq & C_{0} \biggl(1+\sum_{m=f,s}\max _{0< t< T} \bigl\Vert \widehat{\mathbf {w}}^{m}( \cdot,0,t) \bigr\Vert ^{(l-\frac{1}{2})}_{2,\mathbb{R}} \biggr) \\ =& C_{0} \biggl(1+\sum_{m=f,s}\max _{0< t< T} \bigl\Vert \mathbf{w}^{m}(\cdot,0,t) \bigr\Vert ^{(l-\frac{1}{2})}_{2,\mathbb{R}} \biggr). \end{aligned}$$
(69)
Therefore (see (24))
$$\begin{aligned} \max_{0< t< T} \bigl\Vert \mathbf{v} ^{j}(\cdot,t) \bigr\Vert ^{(l)}_{2,\mathbb {R}^{2}_{j}} \leq & C_{2} \max_{0< t< T} \bigl\Vert \mathbf{v} ^{j}(\cdot,t) \bigr\Vert ^{(l-\frac {1}{2})}_{2,\mathbb{R}}\leq C_{2} C_{0} \biggl(1+\sum_{m=f,s} \max_{0< t< T} \bigl\Vert \mathbf{w}^{m}(\cdot,0,t) \bigr\Vert ^{(l-\frac{1}{2})}_{2,\mathbb{R}} \biggr) \\ \leq & C_{2} C_{0} \biggl(1+C_{1} \sum _{m=f,s}\max_{0< t< T} \bigl\Vert \mathbf{w} ^{m}(\cdot,t) \bigr\Vert ^{(l)}_{2,\mathbb{R}^{2}_{j}} \biggr), \quad j=f,s. \end{aligned}$$
(70)
Coming back to the previous notations i taking into account (54) and the definitions of \(\mathbf{v} ^{j,\varepsilon,\delta}\) and \(\mathbf{w} ^{j,\varepsilon,\delta}\), we get from (70) for \(l=2\)
$$ \max_{0< t< T} \bigl\Vert \mathbf{u} ^{j,\varepsilon}(\cdot,t) \bigr\Vert ^{(2)}_{2,Q^{(\delta )}_{j}}\leq C_{1}(\delta),\quad j=f,s. $$
(71)
Now we repeat all this with the function \(\zeta^{(2\delta)}\) and domain \(\Omega^{(2\delta)}\): \(\zeta^{(2\delta)}(\mathbf{x})=1\) for \(\mathbf{x} \in\Omega ^{(4\delta)}\), and \(\zeta^{(2\delta)}(\mathbf{x})=0\) for \(\mathbf{x} \in Q\backslash\Omega^{(2\delta)}\).
Namely, (71) implies for (55)
$$ \max_{0< t< T} \bigl\Vert \mathbf{F}^{\varepsilon,2\delta}( \cdot,t) \bigr\Vert ^{(1)}_{2,\mathbb{R}^{2}}+ \max_{0< t< T} \bigl\Vert \boldsymbol {\varphi}^{\varepsilon,\delta}(\cdot,t) \bigr\Vert ^{(2)}_{2,\mathbb{R}^{2}}\leq C_{2}(\delta) $$
(72)
and, consequently,
$$\begin{aligned}& \max_{0< t< T} \bigl\Vert \mathbf{v} ^{j\varepsilon,2\delta}(\cdot,t) \bigr\Vert ^{(3)}_{2,\mathbb{R}^{2}_{j}}\leq C_{3} \max_{0< t< T} \bigl\Vert \mathbf{w} ^{j\varepsilon,2\delta}(\cdot,t) \bigr\Vert ^{(3)}_{2,\mathbb{R}^{2}_{j}} \leq C_{3}(\delta),\quad j=f,s, \end{aligned}$$
(73)
$$\begin{aligned}& \max_{0< t< T} \bigl\Vert \mathbf{u} ^{j,\varepsilon}(\cdot,t) \bigr\Vert ^{(3)}_{2,Q^{(2\delta)}_{j}}\leq C_{4}(\delta),\quad j=f,s. \end{aligned}$$
(74)
The corresponding embedding theorem \(W^{3}_{2}(\Omega )\rightarrow C^{1+\alpha}(\Omega)\) for \(0<\alpha<1\) ([13], Theorem 2.1, p.61, Chapter 2)
$$ \max_{0< t< T} \bigl\vert \mathbf{u} ^{j,\varepsilon}(\cdot,t) \bigr\vert ^{(1+\alpha )}_{Q^{(2\delta)}_{j}}\leq C \max _{0< t< T} \bigl\Vert \mathbf{u} ^{j,\varepsilon}(\cdot,t) \bigr\Vert ^{(3)}_{2,Q^{(2\delta)}_{j}}\leq C_{5}(\delta),\quad j=f,s $$
(75)
proves (52) and the statement of the lemma. □
3.3 Uniform bounds for density
Let \(\Gamma^{\varepsilon}(t)\subset Q_{f}\) be a smooth surface obtained by moving the initial position \(\Gamma(0)\) along the trajectories of the velocity field \(\mathbf {v}^{\varepsilon}\):
$$\frac{d\mathbf{x}}{dt}=\mathbf{v}^{\varepsilon}(\mathbf{x},t),\quad \mathbf{x}(0)= \boldsymbol {\xi}, \boldsymbol {\xi}\in\Gamma(0). $$
As we have mentioned above, the time T is chosen from the condition
$$\operatorname{dist} \bigl(\Gamma^{\varepsilon}(t), S^{\pm} \bigr)>0. $$
Moreover, we suppose that
$$ \operatorname{dist} \bigl(\Gamma^{\varepsilon}(t), S^{\pm} \bigr)>\varepsilon. $$
(76)
Lemma 6
Under the conditions of Theorem
2.1
let
\({\rho} ^{\varepsilon}\)
be the solution of the problem (30). Then
$$ \max_{0< t< T} \Biggl( \int_{Q^{(\delta)}_{f}} \biggl\vert \frac{\partial\rho ^{\varepsilon}}{\partial t}(\mathbf{x},t) \biggr\vert \,dx+ \sum_{i=1}^{2} \int_{Q^{(\delta)}_{f}} \biggl\vert \frac{\partial\rho^{\varepsilon }}{\partial x_{i}}(\mathbf{x},t) \biggr\vert \,dx \Biggr) \leq C(\delta). $$
(77)
Proof
Let \(q_{i}=\frac{\partial\rho^{\varepsilon}}{\partial x_{i}}\).
Then
$$\frac{\partial q_{i}}{\partial t}+\mathbf{v}^{\varepsilon}\cdot \nabla q_{i}= \sum _{j=1}^{n}a_{ij}q_{j}, \quad \mathbf{x}\in Q, 0< t< T, $$
where \(a_{ij}=-\frac{\partial v^{\varepsilon }_{j}}{\partial x_{i}}\), \(i,j=1,2\), and
$$ \max_{0< t< T} \bigl\vert a_{ij}( \cdot,t) \bigr\vert ^{(1)}_{2,Q^{(2\delta)}_{j}}\leq C(\delta). $$
(78)
Note that \(q_{i}\equiv0\) near the boundaries \(S^{\pm}\), and \(S^{0}\). This follows from the supposition on the behavior of the boundary \(\Gamma^{\varepsilon}(t)\), and the choice of the time T.
Multiplying the equation for \(q_{i}\) by \(\frac{q_{i}}{(q_{i}^{2}+\delta^{2})^{\frac{1}{2}}}\) and integrating by parts over Q we arrive at the equality
$$ \frac{d}{dt} \int_{Q}\bigl(q_{i}^{2}+ \delta^{2}\bigr)^{\frac{1}{2}}\,dx= - \int_{Q}\sum_{j=1}^{2} a_{ij}q_{j}\frac{q_{i}}{(q_{i}^{2}+\delta ^{2})^{\frac{1}{2}}}\,dx, $$
(79)
and, consequently, the inequality
$$ \frac{d y}{dt}\leq C(\delta) y,\quad y(0)\leq C(\delta) $$
(80)
for \(y=\sum_{i=1}^{2}\int_{Q}(q_{i}^{2}+\delta^{2})^{\frac {1}{2}}\,dx\).
The Gronwall inequality provides estimates (77) for \(q_{i}, i=1,2\), and the transport equation (30) provides estimate (77) for the time derivative of \(\rho^{\varepsilon}\). □
Passage to non-smooth initial data, existence of a regular free boundary, existence of the maximal time interval and uniqueness of the solution are proved in the same way as in the previous papers [10] and [1].