Optimal Error Estimates of Semi-implicit Galerkin Method for Time-Dependent Nematic Liquid Crystal Flows

Abstract

This paper focuses on the optimal error estimates of a linearized semi-implicit scheme for the nematic liquid crystal flows, which is used to describe the time evolution of the materials under the influence of both the flow velocity and the microscopic orientation configurations of rod-like liquid crystal flows. Optimal error estimates of the scheme are proved without any restriction of time step by using an error splitting technique proposed by Li and Sun. Numerical results are provided to confirm the theoretical analysis and the stability of the semi-implicit scheme.

Appendix

To get the regularities (2.16)–(2.17) of $\mathbf{u }_{tt}, p_t$ and $\mathbf{b }_{tt}$, where $(\mathbf{u }, p, \mathbf{b })$ is the solution to the problem (1.1)–(1.5) and satisfies (2.10)–(2.14), we need to show that $||\nabla \mathbf{u }_t(x,t)||_{L^2}$ and $||\mathbf{b }(x,t)||_{H^1}$ remain bounded as $t\longrightarrow 0$. In this case, a non-local compatibility condition is needed and can be derived as follows. First, we begin to show that the problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta p_0=\text{ div } (\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)), \quad &{}\text{ in } \ \Omega ,\\ \partial _\mathbf{n }p_0=(\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))\cdot \mathbf{n },\quad &{}\text{ on } \ \partial \Omega , \end{array} \right. \end{aligned}$$

(9.1)

exists a unique $p_0\in H^1(\Omega )\cap M$. In fact, from $\mathbf{f }_0\in \mathbf{H }, \mathbf{u }_0\in \mathbf{D }(A)$ and $\mathbf{b }_0\in \mathbf{H }^3(\Omega )$, it can be shown that $\text{ div } (\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))\in L^2(\Omega )$ and $\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)\in \mathbf{H }(\text{ div } ,\Omega )$. Thus, one has $(\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))\cdot \mathbf{n }|_{\partial \Omega }\in H^{-1/2}(\Omega )$. Finally, we note that the following compatibility condition is satisfied:

$$\begin{aligned}&\displaystyle \int _\Omega \text{ div } (\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)) dx \\&\qquad =\displaystyle \int _{\partial \Omega }(\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))\cdot \mathbf{n }ds \end{aligned}$$

due to the fact $\text{ div } (\Delta \mathbf{u }_0)=0$. From these observations, it follows that the problem (9.1) exists a unique weak solution $p_0\in H^1(\Omega )\cap M$. Moreover, the solution $p_0$ is the limit of the pressure p(x, t) in $H^1(\Omega )\cap M$ as $t\longrightarrow 0$. To make our point precise, we give the following lemma:

Lemma 9.1

Let the initial values $\mathbf{u }_0\in \mathbf{D }(A)$ and $\mathbf{b }_0\in \mathbf{H }^3(\Omega )$ with $|\mathbf{b }_0|=1$ in $\Omega$. Suppose that the solution $(\mathbf{u }, p, \mathbf{b })$ to the problem (1.1)–(1.5) satisfies $||A\mathbf{u }(x,t)-A\mathbf{u }_0(x)||_{L^2}\longrightarrow 0$ and $||\mathbf{b }(x,t)-\mathbf{b }_0(x)||_{H^3}\longrightarrow 0$ as $t\longrightarrow 0$. Then the pressure p(x, t) tends to the solution $p_0$ to the problem (9.1) in the sense that

$$\begin{aligned} ||\nabla p(x,t)-\nabla p_0(x)||_{L^2}\longrightarrow 0 \quad \text{ as } \ t\longrightarrow 0. \end{aligned}$$

Proof

For $t>0$, it follows from (1.1) and $\text{ div } \mathbf{u }=0$ that the pressure $p\in H^1(\Omega )\cap M$ is the weak solution to the problem

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta p=\text{ div } (\mathbf{f }-(\mathbf{u }\cdot \nabla )\mathbf{u }-\lambda \text{ div } (\nabla \mathbf{b }\odot \nabla \mathbf{b })), \quad &{}\text{ in } \ \Omega ,\\ \partial _\mathbf{n }p=(\mu \Delta \mathbf{u }+\mathbf{f }-(\mathbf{u }\cdot \nabla )\mathbf{u }-\lambda \text{ div } (\nabla \mathbf{b }\odot \nabla \mathbf{b }))\cdot \mathbf{n },\quad &{}\text{ on } \ \partial \Omega . \end{array} \right. \end{aligned}$$

(9.2)

From $||A\mathbf{u }(x,t)-A\mathbf{u }_0(x)||_{L^2}\longrightarrow 0$ and $||\mathbf{b }(x,t)-\mathbf{b }_0(x)||_{H^3}\longrightarrow 0$ as $t\longrightarrow 0$, we can see that

$$\begin{aligned}&\text{ div } (\mathbf{f }-(\mathbf{u }\cdot \nabla )\mathbf{u }-\lambda \text{ div } (\nabla \mathbf{b }\odot \nabla \mathbf{b }))\\&\longrightarrow \text{ div } (\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)) \end{aligned}$$

in $L^2(\Omega )$, and

$$\begin{aligned}&(\mu \Delta \mathbf{u }+\mathbf{f }-(\mathbf{u }\cdot \nabla )\mathbf{u }-\lambda \text{ div } (\nabla \mathbf{b }\odot \nabla \mathbf{b }))\cdot \mathbf{n }\\&\longrightarrow (\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))\cdot \mathbf{n }\end{aligned}$$

in $H^{-1/2}(\partial \Omega )$ as $t\longrightarrow 0$. These facts imply the desired result. $\square$

Let $p_0\in H^1(\Omega )\cap M$ be defined by the problem (9.1). Then the non-local compatibility conditions are concluded in the following lemma.

Lemma 9.2

Under the assumptions of Lemma 9.1, if $||\nabla \mathbf{u }_t(x,t)||_{L^2}$ and $||\mathbf{b }(x,t)||_{H^1}$ remain bounded as $t\longrightarrow 0$, then there must hold

$$\begin{aligned} \nabla p_0=(\mu \Delta \mathbf{u }_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)),\quad&\text{ on } \ \partial \Omega , \end{aligned}$$

(9.3)

$$\begin{aligned} \nabla (\gamma \Delta \mathbf{b }_0+(\mathbf{u }_0\cdot \nabla )\mathbf{b }_0+\gamma |\nabla \mathbf{b }_0|^2\mathbf{b }_0)\cdot \mathbf{n }=0,\quad&\text{ on } \ \partial \Omega . \end{aligned}$$

(9.4)

Proof

As $t\longrightarrow 0$, it follows from $||A\mathbf{u }(x,t)-A\mathbf{u }_0(x)||_{L^2}\longrightarrow 0$ and $||\mathbf{b }(x,t)-\mathbf{b }_0(x)||_{H^3}\longrightarrow 0$ that

$$\begin{aligned}&\mathbf{u }_t(x,t)\longrightarrow \mu \Delta \mathbf{u }_0-\nabla p_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0)\quad \text{ in } \ \mathbf{L }^2(\Omega ), \end{aligned}$$

(9.5)

$$\begin{aligned}&\mathbf{b }_t(x,t)\longrightarrow \gamma \Delta \mathbf{b }_0+(\mathbf{u }_0\cdot \nabla )\mathbf{b }_0+\gamma |\nabla \mathbf{b }_0|^2\mathbf{b }_0\quad \text{ in } \ \mathbf{H }^1(\Omega ). \end{aligned}$$

(9.6)

If $||\nabla \mathbf{u }_t(x,t)||_{L^2}$ and $||\mathbf{b }(x,t)||_{H^1}$ remain bounded as $t\longrightarrow 0$, then the convergences (9.5) and (9.6) hold weakly in $\mathbf{H }^1(\Omega )$ and $\mathbf{H }^2(\Omega )$, respectively. Thus,

$$\begin{aligned}&\mathbf{u }_t(x,t)|_{\partial \Omega }\longrightarrow (\mu \Delta \mathbf{u }_0-\nabla p_0+\mathbf{f }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0-\lambda \text{ div } (\nabla \mathbf{b }_0\odot \nabla \mathbf{b }_0))|_{\partial \Omega },\\&\nabla \mathbf{b }_t\cdot \mathbf{n }|_{\partial \Omega }\longrightarrow \nabla (\gamma \Delta \mathbf{b }_0+(\mathbf{u }_0\cdot \nabla )\mathbf{b }_0+\gamma |\nabla \mathbf{b }_0|^2\mathbf{b }_0)\cdot \mathbf{n }|_{\partial \Omega }, \end{aligned}$$

hold weakly in $\mathbf{H }^{1/2}(\partial \Omega )$ as $t\longrightarrow 0$. The facts $\mathbf{u }_t|_{\partial \Omega }=0$ and $\nabla \mathbf{b }_t\cdot \mathbf{n }|_{\partial \Omega }=0$ for any $t>0$ imply the desired non-local compatibility conditions (9.3) and (9.4).$\square$

Under these non-local compatibility conditions, we can estimate $\mathbf{u }_t(0)$ and $\mathbf{b }_t(0)$ in $\mathbf{H }^1(\Omega )$ as follows.

Lemma 9.3

Let $\mathbf{f }_0\in \mathbf{H }$ and $(\mathbf{u }_0, \pi )\in \mathbf{D }(A)\times H^1(\Omega )\cap M$ be determined by the Stokes problem (2.15). Under the assumptions of Lemma 9.1, there holds that $\mathbf{u }_t(0)$ and $\mathbf{b }_t(0)$ belong to $\mathbf{H }^1(\Omega )$.

Proof

Taking $t=0$ at (1.1) deduces to

$$\begin{aligned} \mathbf{u }_t(0)-\mu \Delta \mathbf{u }_0+(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\nabla p_0+ \lambda \text{ div } (\nabla \mathbf{b }_0 \odot \nabla \mathbf{b }_0)=\mathbf{f }_0. \end{aligned}$$

(9.7)

Since $(\mathbf{u }_0, \pi )\in \mathbf{D }(A)\times H^1(\Omega )\cap M$ is determined by the Stokes problem (2.15), then (9.7) can be rewritten as

$$\begin{aligned} \mathbf{u }_t(0) +(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\nabla ( p_0-\pi )+ \lambda \text{ div } (\nabla \mathbf{b }_0 \odot \nabla \mathbf{b }_0)=0. \end{aligned}$$

By using the following formula:

$$\begin{aligned} \text{ div } (\nabla \mathbf{b }\odot \nabla \mathbf{b })=\Delta \mathbf{b }\cdot \nabla \mathbf{b }+\displaystyle \frac{1}{2}\nabla (|\nabla \mathbf{b }|^2), \end{aligned}$$

and applying ${\mathbb {P}}_{\mathbf{H }}$ to the resulting equation, we obtain

$$\begin{aligned} \mathbf{u }_t(0)=-{\mathbb {P}}_\mathbf{H }((\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\lambda \Delta \mathbf{b }_0\cdot \nabla \mathbf{b }_0). \end{aligned}$$

It follows from (2.2) that

$$\begin{aligned} ||\mathbf{u }_t(0)||_{H^1}&=||{\mathbb {P}}_\mathbf{H }((\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\lambda \Delta \mathbf{b }_0\cdot \nabla \mathbf{b }_0)||_{H^1}\\&\le C||(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\lambda \Delta \mathbf{b }_0\cdot \nabla \mathbf{b }_0||_{H^1}. \end{aligned}$$

Note $\mathbf{u }_0\in \mathbf{D }(A)$ and $\mathbf{b }_0\in \mathbf{H }^3(\Omega )$. Then we have

$$\begin{aligned}&||(\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\lambda \Delta \mathbf{b }_0\cdot \nabla \mathbf{b }_0||_{L^2}\\&\quad \le C||\mathbf{u }_0||_{L^\infty }||\nabla \mathbf{u }_0||_{L^2}+C||\nabla \mathbf{b }_0||_{L^\infty }||\Delta \mathbf{b }_0||_{L^2}\\&\quad \le C||A\mathbf{u }_0||_{L^2}||\nabla \mathbf{u }_0||_{L^2}+C||\mathbf{b }_0||_{H^3}||\Delta \mathbf{b }_0||_{L^2}\le C, \end{aligned}$$

and

$$\begin{aligned}&\ ||\nabla ((\mathbf{u }_0\cdot \nabla )\mathbf{u }_0+\lambda \Delta \mathbf{b }_0\cdot \nabla \mathbf{b }_0)||_{L^2}\\&\quad \le ||\nabla \mathbf{u }_0||_{L^4}^2+||\mathbf{u }_0||_{L^\infty }||\nabla ^2\mathbf{u }_0||_{L^2}+||\nabla \mathbf{b }_0||_{L^\infty }||\nabla \Delta \mathbf{b }_0||_{L^2}+||\nabla ^2\mathbf{b }_0||_{L^2}||\Delta \mathbf{b }_0||_{L^2}\\&\quad \le C(||A\mathbf{u }_0||_{L^2}^2+||\mathbf{b }_0||_{H^3}^2)\le C. \end{aligned}$$

The above two estimates imply $\mathbf{u }_t(0)\in \mathbf{H }^1(\Omega )$. Taking $t=0$ in (1.2) leads to

$$\begin{aligned} \mathbf{b }_t(0)=\gamma \Delta \mathbf{b }_0-(\mathbf{u }_0\cdot \nabla )\mathbf{b }_0+\gamma |\nabla \mathbf{b }_0|^2\mathbf{b }_0. \end{aligned}$$

By using a similar method, we can prove

$$\begin{aligned} ||\mathbf{b }_t(0)||_{L^2}&\le \gamma ||\Delta \mathbf{b }_0||_{L^2}+||(\mathbf{u }_0\cdot \nabla )\mathbf{b }_0||_{L^2}+\gamma |||\nabla \mathbf{b }_0|^2\mathbf{b }_0||_{L^2}\\&\le \gamma ||\Delta \mathbf{b }_0||_{L^2}+||\mathbf{u }_0||_{L^\infty }||\nabla \mathbf{b }_0||_{L^2}+\gamma ||\nabla \mathbf{b }_0||_{L^4}^2\le C, \end{aligned}$$

and

$$\begin{aligned} ||\nabla \mathbf{b }_t(0)||_{L^2}&\le \gamma ||\nabla \Delta \mathbf{b }_0||_{L^2}+||\nabla \mathbf{u }_0||_{L^4}||\nabla \mathbf{b }_0||_{L^4}+|| \mathbf{u }_0||_{L^\infty }||\nabla ^2\mathbf{b }_0||_{L^2}\\&\quad +\gamma ||\nabla \mathbf{b }_0||_{L^\infty }|| \nabla ^2\mathbf{b }_0||_{L^2}||\mathbf{b }_0||_{L^\infty }+\gamma || \nabla \mathbf{b }_0||_{L^6}^3\le C, \end{aligned}$$

which imply $\mathbf{b }_t(0)\in \mathbf{H }^1(\Omega )$. $\square$

Based on the results derived in Lemma 9.3, we can show some regularities of $\mathbf{u }_{tt}, p_t$ and $\mathbf{b }_{tt}$ in next theorem.

Theorem 9.1

Under the assumptions of Theorem 2.1 and Lemma 9.3, suppose $\mathbf{f }_t\in \mathbf{L }^\infty (0,T; \mathbf{V }_0^\prime )\cap \mathbf{L }^2(0,T;\mathbf{H })$, then we have

$$\begin{aligned} \mathbf{u }_t\in \mathbf{L }^\infty (0,T^\star ;\mathbf{V })\cap \mathbf{L }^2(0,T^\star ;\mathbf{D }(A)), \ \nabla p_t\in \mathbf{L }^2(0,T^\star ; \mathbf{L }^2(\Omega )),\\ \mathbf{u }_{tt}\in \mathbf{L }^\infty (0,T^\star ;\mathbf{V }_0^\prime )\cap \mathbf{L }^2(0,T^\star ;\mathbf{H }), \ \mathbf{b }_{tt}\in \mathbf{L }^2(0,T^\star ;\mathbf{L }^2(\Omega )), \end{aligned}$$

where $T^\star$ is defined in Theorem 2.1.

Proof

Differentiating (1.1) with respect to t, we get

$$\begin{aligned} \mathbf{u }_{tt}-\mu \Delta \mathbf{u }_t+(\mathbf{u }_t\cdot \nabla )\mathbf{u }+(\mathbf{u }\cdot \nabla )\mathbf{u }_t+\nabla p_t+ 2\lambda \text{ div } (\nabla \mathbf{b }_t \odot \nabla \mathbf{b }) =\mathbf{f }_t. \end{aligned}$$

(9.8)

Multiplying (9.8) by $\mathbf{u }_{tt}$ and integrating over $\Omega$, we have

$$\begin{aligned}&||\mathbf{u }_{tt}||_{L^2}^2+\displaystyle \frac{\mu }{2}\displaystyle \frac{d}{dt} ||\nabla \mathbf{u }_t||_{L^2}^2\\&\quad =(\mathbf{f }_t,\mathbf{u }_{tt})-b(\mathbf{u }_t,\mathbf{u },\mathbf{u }_{tt})-b(\mathbf{u },\mathbf{u }_t,\mathbf{u }_{tt})-2\lambda (\text{ div } (\nabla \mathbf{b }_t \odot \nabla \mathbf{b }),\mathbf{u }_{tt}) \\&\quad \le (||\mathbf{f }_t||_{L^2}+||\nabla \mathbf{u }||_{L^\infty }||\mathbf{u }_t||_{L^2}+||\mathbf{u }||_{L^\infty }||\nabla \mathbf{u }_t||_{L^2}\\&\qquad +2\lambda ||\nabla \mathbf{b }||_{L^\infty }||\mathbf{b }_t||_{H^2}+2\lambda ||\nabla ^2\mathbf{b }||_{L^\infty }||\nabla \mathbf{b }_t||_{L^2})||\mathbf{u }_{tt}||_{L^2}\\&\quad \le \displaystyle \frac{1}{2}||\mathbf{u }_{tt}||_{L^2}^2+C(||\mathbf{f }_t||_{L^2}^2+|| \mathbf{u }||_{W^{2,4}}^2||\mathbf{u }_t||_{L^2}^2+||A\mathbf{u }||_{L^2}^2||\nabla \mathbf{u }_t||^2_{L^2}\\&\qquad +|| \mathbf{b }||_{H^3}^2|| \mathbf{b }_t||_{H^2}^2+|| \mathbf{b }||_{H^4}^2|| \mathbf{b }_t||_{H^1}^2), \end{aligned}$$

where we use $\text{ div } \mathbf{u }_{tt}=0$. It follows from (2.10)–(2.13) that $\mathbf{u }_{tt}\in \mathbf{L }^2(0,T^\star ;\mathbf{H })$ and $\mathbf{u }_t\in \mathbf{L }^\infty (0,T^\star ;\mathbf{V })$ if we integrate the above inequality with respect to t from 0 to $t\le T^\star$ and note that $\mathbf{u }_t(0)\in \mathbf{H }^1(\Omega )$. By using a similar method, multiplying (9.8) by $A\mathbf{u }_{t}$ and integrating over $\Omega$, we have

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt} ||\nabla \mathbf{u }_t||_{L^2}^2+\mu ||A\mathbf{u }_{t}||_{L^2}^2\\&\quad =(\mathbf{f }_t,A\mathbf{u }_{t})-b(\mathbf{u }_t,\mathbf{u },A\mathbf{u }_{t})-b(\mathbf{u },\mathbf{u }_t,A\mathbf{u }_{t})-2\lambda (\text{ div } (\nabla \mathbf{b }_t \odot \nabla \mathbf{b }), A\mathbf{u }_{t})\\&\quad \le \displaystyle \frac{\mu }{2}||A\mathbf{u }_{t}||_{L^2}^2+\displaystyle \frac{C}{\mu }(||\mathbf{f }_t||_{L^2}^2+|| \mathbf{u }||_{W^{2,4}}^2||\mathbf{u }_t||_{L^2}^2+||A\mathbf{u }||_{L^2}^2||\nabla \mathbf{u }_t||^2_{L^2}\\&\qquad +|| \mathbf{b }||_{H^3}^2|| \mathbf{b }_t||_{H^2}^2+|| \mathbf{b }||_{H^4}^2|| \mathbf{b }_t||_{H^1}^2), \end{aligned}$$

which implies $\mathbf{u }_t\in L^2(0,T^\star ;\mathbf{D }(A))$. Multiplying (9.8) by $\mathbf{v }\in \mathbf{V }_{0}$ leads to

$$\begin{aligned} (\mathbf{u }_{tt},\mathbf{v })+\mu (\nabla \mathbf{u }_t,\nabla \mathbf{v })+b(\mathbf{u }_t,\mathbf{u },\mathbf{v })+b(\mathbf{u },\mathbf{u }_t,\mathbf{v })- 2\lambda (\nabla \mathbf{b }_t \odot \nabla \mathbf{b },\nabla \mathbf{v }) =(\mathbf{f }_t,\mathbf{v }). \end{aligned}$$

Then it is easy to show

$$\begin{aligned} ||\mathbf{u }_{tt}||_{V_0^\prime }\le C( ||\nabla \mathbf{u }_t||_{L^2}+||\nabla \mathbf{u }||_{L^2}||\nabla \mathbf{u }_t||_{L^2}+||\nabla \mathbf{b }||_{L^\infty }||\nabla \mathbf{b }_t||_{L^2} +||\mathbf{f }_t||_{V_0^\prime }), \end{aligned}$$

which implies $\mathbf{u }_{tt}\in \mathbf{L }^\infty (0,T^\star ;\mathbf{V }_0^\prime )$. To estimate $\nabla p_t$, we use (9.8) to deduce that

$$\begin{aligned} ||\nabla p_t||_{L^2}&\le ||\mathbf{u }_{tt}||_{L^2}+\mu ||\Delta \mathbf{u }_t||_{L^2}+||(\mathbf{u }_t\cdot \nabla )\mathbf{u }||_{L^2}+||(\mathbf{u }\cdot \nabla )\mathbf{u }_t||_{L^2}\\&\quad +2\lambda ||\text{ div } (\nabla \mathbf{b }_t \odot \nabla \mathbf{b })||_{L^2}+||\mathbf{f }_t||_{L^2}\\&\le C(||\mathbf{u }_{tt}||_{L^2}+||A\mathbf{u }_t||_{L^2}+|\mathbf{f }_t||_{L^2}+||\nabla \mathbf{u }||_{L^\infty }||\mathbf{u }_t||_{L^2}+||\mathbf{u }||_{L^\infty }||\nabla \mathbf{u }_t||_{L^2}\\&\quad +||\nabla \mathbf{b }||_{L^\infty }|| \mathbf{b }_t||_{H^2}+||\nabla ^2\mathbf{b }||_{L^\infty }||\nabla \mathbf{b }_t||_{L^2}). \end{aligned}$$

From $\mathbf{u }_{tt}\in \mathbf{L }^2(0,T^\star ;\mathbf{H })$ and $\mathbf{u }_t\in L^2(0,T^\star ;\mathbf{D }(A))$ shown in the previous paragraph, it is easily seen that $\nabla p_t\in \mathbf{L }^2(0,T^\star ; \mathbf{L }^2(\Omega ))$ after integrating the above inequality from 0 to $t\le T^\star$ and using (2.10)–(2.13).

Differentiating (1.2) with respect to t yields

$$\begin{aligned} \mathbf{b }_{tt}-\gamma \Delta \mathbf{b }_t+(\mathbf{u }_t\cdot \nabla )\mathbf{b }+(\mathbf{u }\cdot \nabla )\mathbf{b }_t=\gamma |\nabla \mathbf{b }|^2\mathbf{b }_t+2\gamma (\nabla \mathbf{b }\cdot \nabla \mathbf{b }_t)\mathbf{b }. \end{aligned}$$

It follows from (2.4)–(2.9) that

$$\begin{aligned} ||\mathbf{b }_{tt}||_{L^2}&\le \gamma ||\Delta \mathbf{b }_t||_{L^2}+||(\mathbf{u }_t\cdot \nabla )\mathbf{b }||_{L^2}+|| (\mathbf{u }\cdot \nabla )\mathbf{b }_t||_{L^2}\\&\quad +\gamma || |\nabla \mathbf{b }|^2\mathbf{b }_t||_{L^2}+2\gamma || (\nabla \mathbf{b }\cdot \nabla \mathbf{b }_t)\mathbf{b }||_{L^2}\\&\le C(||\mathbf{b }_t||_{H^2}+||\mathbf{u }_t||_{L^2}||\nabla \mathbf{b }||_{L^\infty }+||\mathbf{u }||_{L^\infty }||\nabla \mathbf{b }_t||_{L^2}\\&\quad +||\nabla \mathbf{b }||_{L^\infty }^2||\mathbf{b }_t||_{L^2}+||\nabla \mathbf{b }||_{L^\infty }||\nabla \mathbf{b }_t||_{L^2})\\&\le C(||\mathbf{b }_t||_{H^2}+||\mathbf{u }_t||_{L^2}||\mathbf{b }||_{H^3}+||A\mathbf{u }||_{L^2}||\nabla \mathbf{b }_t||_{L^2}\\&\quad +||\mathbf{b }||_{H^3}^2||\mathbf{b }_t||_{L^2}+||\mathbf{b }||_{H^3}||\nabla \mathbf{b }_t||_{L^2}), \end{aligned}$$

which completes the proof of $\mathbf{b }_{tt}\in \mathbf{L }^2(0,T^\star ;\mathbf{L }^2(\Omega ))$ by using (2.10)–(2.13). $\square$