On the compressible micropolar fluids in a time-dependent domain

Abstract

We investigate compressible micropolar fluids on a time-dependent domain with slip boundary conditions. Our contribution in this paper is threefold. Firstly, we establish the local existence of the strong solution. Secondly, the global existence of weak solutions is shown. The third one is the weak-strong uniqueness principle for slip boundary conditions. There are several new ideas developed by us to overcome the difficulties caused by the coupled terms and slip boundary conditions.

Appendix

Proof of Lemma 4.2and Lemma 5.2. We observe that the stress tension for the velocity is coupled with micro-rotational velocity; however, the stress tension of angular-rotational velocity is decoupled. Hence, we should first construct the micro-rotational velocity; then, we use it to construct the velocity of fluids. We look for the extension of the boundary data satisfying the following conditions:

For $\varvec{w}^b$, it should hold that

$$\begin{aligned} \begin{aligned} \varvec{w}^b(t,y)\cdot {\varvec{n}}(y) & = \left( {\tilde{\varvec{w}}}(t,y)-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) \cdot ({\varvec{n}}(y)-{\varvec{n}}(\varvec{X}(t,y)))\\& \quad +\frac{1}{2}[((\nabla _x\varvec{Y}-{\mathbb {I}})\nabla _y)\wedge {\tilde{\varvec{V}}}]\cdot {\varvec{n}}(\varvec{X}(t,y)),\\ \varvec{w}^b(t,y)\cdot \tau ^k(y) & = \left( {\tilde{\varvec{w}}}(t,y)-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) \cdot (\tau ^k(y)-\tau ^k(\varvec{X}(t,y)))\\& \quad +\frac{1}{2}[((\nabla _x\varvec{Y}-{\mathbb {I}})\nabla _y)\wedge {\tilde{\varvec{V}}}]\cdot \tau ^k(y), \end{aligned} \end{aligned}$$

(10.1)

and

$$\begin{aligned} \begin{aligned}&\left[ (c_a+c_d)\nabla _{y}\varvec{w}^b(t,y)+(c_d-c_a)\nabla _{y}^{{\mathsf {T}}}\varvec{w}^b\right] {\varvec{n}}(y)\cdot \tau ^{k}(y)\\&\quad =\left[ (c_a+c_d)\nabla _{y}\tilde{\varvec{w}}(t,y)({\mathbb {I}}-\nabla _{x}\varvec{Y})+(c_d-c_a)({\mathbb {I}}-\nabla ^{{\mathsf {T}}}_{x}\varvec{Y})\nabla _{y}^{{\mathsf {T}}}\tilde{\varvec{w}}\right] {\varvec{n}}(\varvec{X}(t,y))\cdot \tau ^{k}(\varvec{X}(t,y))\\&\qquad +\left[ (c_a+c_d)\nabla _{y}\tilde{\varvec{w}}(t,y)+(c_d-c_a)\nabla _{y}^{{\mathsf {T}}}\tilde{\varvec{w}}\right] (t,y)\left[ ({\varvec{n}}(y)-{\varvec{n}}(\varvec{X}(t,y)))\cdot \tau ^{k}(\varvec{X}(t,y))\right. \\&\qquad \left.\, +\,{\varvec{n}}(y)\cdot (\tau ^{k}(y)-\tau ^{k}(\varvec{X}(t,y)))\right] . \end{aligned} \end{aligned}$$

(10.2)

For $\varvec{u}^b$, it should hold that

$$\begin{aligned} \begin{aligned} \varvec{u}^{b}(t,y)\cdot {\varvec{n}}(y) & = ({\tilde{\varvec{u}}}-{\tilde{\varvec{V}}})(t,y)\cdot ({\varvec{n}}(y)-{\varvec{n}}(\varvec{X}(t,y)))\\& \quad +(\varvec{V}(t,(\varvec{X}(t,y)))-{\tilde{\varvec{V}}}(t,y))\cdot {\varvec{n}}(\varvec{X}(t,y)),\\ \varvec{u}^{b}(t,y)\cdot \tau ^{k}(y) & = ({\tilde{\varvec{u}}}-{\tilde{\varvec{V}}})(t,y)\cdot (\tau ^{k}(y)-\tau ^{k}(\varvec{X}(t,y)))\\& \quad +(\varvec{V}(t,(\varvec{X}(t,y)))-{\tilde{\varvec{V}}}(t,y))\cdot \tau ^{k}(\varvec{X}(t,y)), \end{aligned} \end{aligned}$$

(10.3)

and

$$\begin{aligned} \begin{aligned}&\left[ (\mu +\xi )\nabla _{y}\varvec{u}^{b}(t,y)+(\mu -\xi )\nabla _{y}^{{\mathsf {T}}}\varvec{u}^{b}-2\xi A(\varvec{w}^{b})\right] {\varvec{n}}(y)\cdot \tau ^{k}(y)\\&\quad =\left[ (\mu +\xi )\nabla _{y}\tilde{\varvec{u}}(t,y)({\mathbb {I}}-\nabla _{x}\varvec{Y})+(\mu -\xi )({\mathbb {I}}-\nabla ^{{\mathsf {T}}}_{x}\varvec{Y})\nabla _{y}^{{\mathsf {T}}}\tilde{\varvec{u}}\right] {\varvec{n}}(\varvec{X}(t,y))\cdot \tau ^{k}(\varvec{X}(t,y))\\&\qquad +\left[ (\mu +\xi )\nabla _{y}\tilde{\varvec{u}}(t,y)+(\mu -\xi )\nabla _{y}^{{\mathsf {T}}}\tilde{\varvec{u}}-2\xi A(\tilde{\varvec{w}})\right] (t,y)\left[ ({\varvec{n}}(y)-{\varvec{n}}(\varvec{X}(t,y)))\cdot \tau ^{k}(\varvec{X}(t,y))\right. \\&\qquad \left.\, +\,{\varvec{n}}(y)\cdot (\tau ^{k}(y)-\tau ^{k}(\varvec{X}(t,y)))\right] . \end{aligned} \end{aligned}$$

(10.4)

Firstly, we flatten the boundary and choose a proper smooth cutoff function which enjoy the regularity of the boundary. The detail can be found in [25]. Therefore, we can choose

$$\begin{aligned} {\varvec{n}}(y_1,y_2,0)=(0,0,1),\quad \tau ^1(y_1,y_2,0)=(1,0,0), \quad \tau ^2(y_1,y_2,0)=(0,1,0). \end{aligned}$$

Then we construct ${\tilde{\varvec{u}}}^b$ and ${\tilde{\varvec{w}}}^b$ satisfying (10.3) and (10.1), respectively. Moreover, in the following step we use it to define $\varvec{w}^b$. Lastly, with $\varvec{w}^b$ in hand, we define $\varvec{u}^b$.

The construction of ${\tilde{\varvec{u}}}^b$ is directly borrowed from Appendix A in [25]. We list the result as follows

$$\begin{aligned} \Vert {\tilde{\varvec{u}}}^b_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}\le E(T)[1+\Vert ({\tilde{\varvec{u}}}-{\tilde{\varvec{V}}})_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}+\Vert {\tilde{\varvec{u}}}-{\tilde{\varvec{V}}}\Vert _{W^{1,\infty }(0,T;L^2(\Omega _0))}]. \end{aligned}$$

(10.5)

Now, we consider ${\tilde{\varvec{w}}}^b$. Precisely, for any $(y_1,y_2,y_3)\in \Omega _0$, we define

$$\begin{aligned} \begin{aligned} \delta {\varvec{n}}(t,y)={\varvec{n}}(t,(y_1,y_2,0))-{\varvec{n}}(\varvec{X}(t,(y_1,y_2,0))). \end{aligned} \end{aligned}$$

Moreover, the normal component of ${\tilde{\varvec{w}}}^{b}$ can be defined as

$$\begin{aligned} {\tilde{\varvec{w}}}^b\cdot {\varvec{n}}=\left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) (t,y)\cdot \delta {\varvec{n}}(t,y)+\frac{1}{2}[((\nabla _x{\tilde{\varvec{Y}}}-{\mathbb {I}})\nabla _y)\wedge {\tilde{\varvec{V}}}]\cdot {\varvec{n}}(\varvec{X}(t,y)). \end{aligned}$$

(10.6)

For the first term of right-hand side in (10.6), we have

$$\begin{aligned} \begin{aligned}&\left[ \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) (t,y)\cdot \delta {\varvec{n}}(t,y)\right] _{tt}\\&\quad =\left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _{tt}\cdot \delta {\varvec{n}}+2\left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _t\cdot (\delta {\varvec{n}})_{t}+\left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) (t,y)\cdot (\delta {\varvec{n}})_{t}. \end{aligned} \end{aligned}$$

(10.7)

Thanks to (1.6), we get

$$\begin{aligned} \delta {\varvec{n}}\thicksim \int _0^T{\tilde{\varvec{V}}},\quad (\delta {\varvec{n}})_t\thicksim {\tilde{\varvec{V}}},\quad (\delta {\varvec{n}})_{tt}\thicksim {\tilde{\varvec{V}}}_{t}. \end{aligned}$$

(10.8)

We only estimate the first term of right-hand side in (10.7); the rest terms can be done in the same way,

$$\begin{aligned} \begin{aligned} \left\| \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _{tt}\delta {\varvec{n}}\right\| _{L^2(0,T;L^2(\Omega _0))} & \le \Vert \delta {\varvec{n}}\Vert _{L^{\infty }(0,T;L^\infty (\Omega _0))}\left\| \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _{tt}\right\| _{L^2(0,T;L^2(\Omega _0))}\\ & \le T\left\| {\tilde{\varvec{V}}}\Vert _{L^{\infty }(0,T;L^(\Omega _0))}\Vert \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _{tt}\right\| _{L^2(0,T;L^2(\Omega _0))}. \end{aligned} \end{aligned}$$

Finally, we get

$$\begin{aligned} \begin{aligned}&\left\| \left[ \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) \cdot \delta {\varvec{n}}(t,y)\right] _{tt}\right\| _{L^2(L^2)}\\&\quad \le C(T+\sqrt{T})\Vert {\tilde{\varvec{V}}}\Vert _{W^{1,\infty }(0,T;L^{\infty }(\Omega _0))}\left[ \left\| \left( {\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right) _{tt}\right\| _{L^2(0,T;L^2(\Omega _0))}\right. \\&\qquad \left.\, +\,\left\Vert \left({\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}}\right)\right\Vert _{W^{1,\infty }(0,T;L^2(\Omega _0))}\right] . \end{aligned} \end{aligned}$$

(10.9)

For the second term of right-hand side in (10.6), we denote $E\varvec{Y}=\nabla _x\varvec{Y}-{\mathbb {I}}$; then, it holds that

$$\begin{aligned} \begin{aligned}&[((E\varvec{Y}\nabla _y)\wedge {\tilde{\varvec{V}}})\cdot {\varvec{n}}(\varvec{X}(t,y))]_{tt}\\&\quad =((E\varvec{Y}_{tt}\nabla _y)\wedge {\tilde{\varvec{V}}})\cdot {\varvec{n}}+2((E\varvec{Y}_{t}\nabla _y)\wedge {\tilde{\varvec{V}}}_t)\cdot {\varvec{n}}+((E\varvec{Y}_t\nabla _y)\wedge {\tilde{\varvec{V}}})\cdot {\varvec{n}}_t\\&\qquad +((E\varvec{Y}\nabla _y)\wedge {\tilde{\varvec{V}}}_{tt})\cdot {\varvec{n}}+2((E\varvec{Y}\nabla _y)\wedge {\tilde{\varvec{V}}}_t)\cdot {\varvec{n}}_t+((E\varvec{Y}\nabla _y)\wedge {\tilde{\varvec{V}}})\cdot {\varvec{n}}_{tt}. \end{aligned} \end{aligned}$$

It is easy to observe that

$$\begin{aligned} \Vert E\varvec{Y}\Vert _{L^{\infty }(0,T;L^\infty (\Omega _0))}\le E(T), \quad \Vert (E\varvec{Y}_t,E\varvec{Y}_{tt})\Vert _{L^{\infty }(0,T;L^\infty (\Omega _0))}\le C, \end{aligned}$$

(10.10)

where E is continuous, $E(0)=0$. Hence, we have

$$\begin{aligned} \begin{aligned}&\Vert [((E\varvec{Y}\nabla _y)\wedge {\tilde{\varvec{V}}})\cdot {\varvec{n}}(\varvec{X}(t,y))]_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}\\&\quad \le E(T)(\Vert \nabla _y{\tilde{\varvec{V}}}\Vert _{W^{1,\infty }(0,T;L^2(\Omega _0))}+\Vert \nabla _y{\tilde{\varvec{V}}}_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}). \end{aligned} \end{aligned}$$

(10.11)

Combining (10.9) and (10.11), we obtain

$$\begin{aligned} \begin{aligned}&\Vert ({\tilde{\varvec{w}}}^b\cdot {\varvec{n}})_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}\\&\quad \le E(t)[1+\Vert ({\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}})_{tt}\Vert _{L^2(0,T;L^2(\Omega _0))}+\Vert ({\tilde{\varvec{w}}}-\frac{1}{2}\mathrm {curl}_y{\tilde{\varvec{V}}})\Vert _{W^{1,\infty }(0,T;L^2(\Omega _0))}]. \end{aligned} \end{aligned}$$

(10.12)

The tangential parts have similar bounds.

Now we construct $\varvec{w}^b$ and rewrite (10.1) in a compact form

$$\begin{aligned} {\tilde{\varvec{w}}}^{b}=E^{1}{\tilde{\varvec{w}}}+E^2, \end{aligned}$$

(10.13)

where $E^1$ and $E^2$ are sufficiently regular matrix and vector functions. Meanwhile, we rewrite (10.2) as

$$\begin{aligned} \begin{aligned} (c_a+c_d)w_{1,y_3}^b+(c_d-c_a)w_{3,y_1}^b=\sum _{i,j=1}^{3}B_{ij}^1(t,x){{\tilde{w}}}_{i,y_j},\\ (c_a+c_d)w_{2,y_3}^b+(c_d-c_a)w_{3,y_2}^b=\sum _{i,j=1}^{3}C_{ij}^1(t,x){{\tilde{w}}}_{i,y_j}. \end{aligned} \end{aligned}$$

(10.14)

First, we take $w_3^b={\tilde{w}}_3^b$. Next we construct $w_1^b$, since $w_2^b$ can be constructed in the same way. In particular,

$$\begin{aligned} w_{3,y_1}^b={\tilde{w}}_{3,y_1}^b=\sum _{i=1}^3E_{3i}^1w_{i,y_1}+\sum _{i=1}^3E_{3i,y_1}^1w_{i}+E_{3,y_1}^2. \end{aligned}$$

(10.15)

Substituting the above identity into (10.14), we have

$$\begin{aligned} \begin{aligned} (c_a+c_d)w_{1,y_3}^b & = \sum _{i,j=1}^3B^1_{ij}(t,x)w_{i,y_j}-(c_d-c_a)\sum _{i=1}^3E_{3i}^1w_{i,y_1}\\&\quad -(c_d-c_a)\sum _{i=1}^3E_{3i,y_1}^1w_{i}-(c_d-c_a)E_{3,y_1}^2\\ & = \sum _{i,j=1}^3\bar{B}^1_{ij}(t,x)w_{i,y_j}-(c_d-c_a)\sum _{i=1}^3E_{3i,y_1}^1w_{i}-(c_d-c_a)E_{3,y_1}^2. \end{aligned} \end{aligned}$$

(10.16)

Then we divide $w_1^b$ into two parts such that $w_1^b=w_1^{b1}+w_1^{b2}$ where

$$\begin{aligned} \begin{aligned} (c_a+c_d)w_{1}^{b1}(y) & = \left( \sum _{i=1}^3E_{1i}^1w_i+E_{1}^2\right) (y)+2\sum _{i,j}{\tilde{B}}_{ij}^1w_{i}((y_1,y_2,0)+y_3\varvec{e}_{j})\\&\quad -2\sum _{i,j}{\tilde{B}}^1_{ij}w_{i}\left( (y_1,y_2,0)+\frac{y_3}{2}\varvec{e}_{j}\right) , \end{aligned} \end{aligned}$$

(10.17)

here

$$\begin{aligned} \begin{aligned} {\tilde{B}}_{ij}^1 & = \bar{B}_{ij}^1,\quad j\ne 3,\\ {\tilde{B}}_{i3}^1 & = \bar{B}_{i3}^1-E_{1i}^1,\quad i=1,2,3. \end{aligned} \end{aligned}$$

(10.18)

Applying $\partial _{y_3}$ to (10.17), we get

$$\begin{aligned} \begin{aligned} (c_a+c_d)w_{1,y_3}^{b1}(y)&=\sum _{i=1}^3E_{1i}^1w_{i,y_3}(y)+\sum _{i=1}^3E_{1i,y_3}^1w_{i}(y)+E^2_{1,y_3}+\sum _{i,j}{\tilde{B}}_{ij}^1w_{i,y_j}(y)\\&=\sum _{i,j}\bar{B}_{ij}^1w_{i,y_j}+\sum _{i}E_{1i,y_3}^1w_i+E^2_{1,y_3}. \end{aligned} \end{aligned}$$

(10.19)

So, subtracting (10.19) from (10.16) gives that

$$\begin{aligned} w_{1,y_3}^{b2}=-\frac{1}{c_a+c_d}\left( \sum _{i}E_{1i}^1w_{i,y_3}+(c_d-c_a)\sum _{i}E_{3i,y_1}^1w_{i}+(c_d-c_a)E_{3,y_1}^2+E^2_{1,y_3}\right) :=P_{w}(y). \end{aligned}$$

(10.20)

Hence, we can define $w_{1}^{b2}$ as follows

$$\begin{aligned} w_{1}^{b2}=\int _0^{y_3}P_{w}(y_1,y_2,s)\text{d}s. \end{aligned}$$

(10.21)

Finally, $w_{1}^b=w_{1}^{b1}+w_{1}^{b2}$ is constructed as required relations. With $\varvec{w}^b$ in hand, we can construct $\varvec{u}^b$ in the same way which depends on $\varvec{w}^b$. We also rewrite (10.3) in a compact way

$$\begin{aligned} {\tilde{\varvec{u}}}^{b}=E^{3}{\tilde{\varvec{u}}}+E^4, \end{aligned}$$

(10.22)

where $E^3$ and $E^4$ are regular matrix and vector functions. And, (10.4) gives that

$$\begin{aligned} \begin{aligned} (\mu +\xi )u_{1,y_3}^b+(\mu -\xi )u_{3,y_1}^b+w_2^b=\sum _{i,j=1}^{3}B_{ij}^2(t,x){{\tilde{u}}}_{i,y_j}+D^1({\tilde{\varvec{w}}}),\\ (\mu +\xi )u_{2,y_3}^b+(\mu -\xi )u_{3,y_2}^b-w_1^b=\sum _{i,j=1}^{3}C_{ij}^2(t,x){{\tilde{u}}}_{i,y_j}+D^2({\tilde{\varvec{w}}}). \end{aligned} \end{aligned}$$

(10.23)

Also, we take $u_3^b={{\tilde{u}}}_3^b$, then

$$\begin{aligned} \begin{aligned} (\mu +\xi )u_{1,y_3}^b & = \sum _{i,j=1}^{3}B_{ij}^2(t,x){{\tilde{u}}}_{i,y_j}-(\mu -\xi ){{\tilde{u}}}_{3,y_1}^b-w_2^b+D^1({\tilde{\varvec{w}}})\\ & = \sum _{i,j=1}^3B^2_{ij}(t,x){{\tilde{u}}}_{i,y_j}-(\mu -\xi )\left( \sum _{i=1}^3E_{3i}^3{{\tilde{u}}}_{i,y_1}+\sum _{i=1}^3E_{3i,y_1}^3{{\tilde{u}}}_{i}+E_{3,y_1}^4\right) \\& \quad -w_2^b+D^1({\tilde{\varvec{w}}})\\ & = \sum _{i,j=1}^3\bar{B}^2_{ij}(t,x){{\tilde{u}}}_{i,y_j}-(\mu -\xi )\sum _{i=1}^3E_{3i,y_1}^3{{\tilde{u}}}_{i}-(\mu -\xi )E_{3,y_1}^4-w_2^b+D^1({\tilde{\varvec{w}}}). \end{aligned} \end{aligned}$$

(10.24)

We define $u_1^b=u_{1}^{b1}+u_{1}^{b2}$ where

$$\begin{aligned} \begin{aligned} (\mu +\xi )u_{1}^{b1}(y) & = \left( \sum _{i=1}^3E_{1i}^3{{\tilde{u}}}_i+E_{1}^4\right) (y)+2\sum _{i,j}{\tilde{B}}_{ij}^2{{\tilde{u}}}_{i}((y_1,y_2,0)+y_3\varvec{e}_{j})\\& \quad -2\sum _{i,j}{\tilde{B}}^2_{ij}{{\tilde{u}}}_{i}\left( (y_1,y_2,0)+\frac{y_3}{2}\varvec{e}_{j}\right) , \end{aligned} \end{aligned}$$

(10.25)

with

$$\begin{aligned} \begin{aligned} {\tilde{B}}_{ij}^2 & = \bar{B}_{ij}^2,\quad j\ne 3,\\ {\tilde{B}}_{i3}^2 & = \bar{B}_{i3}^2-E_{1i}^3,\quad i=1,2,3. \end{aligned} \end{aligned}$$

(10.26)

Hence, we have

$$\begin{aligned} \begin{aligned} u_{1,y_3}^{b2} & = -\frac{1}{\mu +\xi }\left( \sum _{i}E_{1i}^3{{\tilde{u}}}_{i,y_3} +E_{1,y_3}^4+(\mu -\xi )\sum _{i}E_{3i,y_1}^3u_{i}\right. \\&\quad \left.\, +\,(\mu -\xi )E^4_{3,y_1} +w_2^b+D^1({\tilde{\varvec{w}}})\right) \\ & := \,P_{u}(y). \end{aligned} \end{aligned}$$

(10.27)

Finally, we can define $u_{1}^{b2}$ as follows

$$\begin{aligned} u_{1}^{b2}=\int _0^{y_3}P_{u}(y_1,y_2,s)\mathrm {d}s. \end{aligned}$$

(10.28)

We notice that $u_1^{b}=u_1^{b1}+u_1^{b2}$ satisfies (10.3) and (10.4).

Note that $B^1_{ij}$, $B^2_{ij}$, $C^1_{ij}$ and $C^2_{ij}$ are made of $\tau (y)-\tau (\varvec{X}(y))$, ${\varvec{n}}(y)-{\varvec{n}}(\varvec{X}(y))$ and $\nabla _x\varvec{Y}-{\mathbb {I}}$. Thanks to (10.8) and (10.10), we obtain the estimate for $w_1^{b1}$ and $u_1^{b1}$. When estimates come to $w_1^{b2}$ and $u_1^{b2}$, we use the fact

$$\begin{aligned} \nabla _y E^{i}\thicksim \int _0^T\nabla _y{\tilde{\varvec{V}}},\quad i=1,2,3,4. \end{aligned}$$

Therefore, repeating all the argument as before, we can derive the bound for $\Vert \varvec{u}_{tt}^b\Vert _{L^2(0,T;L^2(\Omega _0))}$ and $\Vert \varvec{w}_{tt}^b\Vert _{L^2(0,T;L^2(\Omega _0))}$. Other components in ${\mathcal {Y}}(T)$ are obtained in a similar way; we omit the details.

Lemma 4.1

[11] (Fundamental lemma) Let $\rho \in L^\infty (0,T;L^2(B))$, $\rho \ge 0$, $\varvec{u}\in L^2(0,T;W_{0}^{1,2}(B;{\mathbb {R}}^3))$ be a weak solution of the continuity equation,

satisfying

$$\begin{aligned} \int _{B}(\rho (\tau ,\cdot )\phi (\tau , \cdot )-\rho _0\phi (0,\cdot ))\mathrm {d}x=\int _0^\tau \int _{B}(\rho \phi _t+\rho \varvec{u}\cdot \nabla _{x}\phi )\mathrm {d}x\mathrm {d}t \end{aligned}$$

for any $\tau \in [0,T]$ and any test function $\phi \in C_c^1([0,T]\times {\mathbb {R}}^3)$.

In addition, assume that $(\varvec{u}-\varvec{V})(\tau , \cdot )\cdot \varvec{n}|_{\Gamma _{\tau }}=0$ for $\text{ a.a. } \tau \in (0, T),$ and that $\rho _0\in L^2({\mathbb {R}}^3)$, $\rho _0\ge 0$, $\rho _0|_{B\backslash \Omega _0}=0$. Then

$$\begin{aligned} \rho (\tau ,\cdot )|_{B\backslash \Omega _{\tau }}=0,\quad \forall \tau \in [0, T]. \end{aligned}$$