Optimal Error Estimates of Linearized Crank–Nicolson Galerkin Method for Landau–Lifshitz Equation

Abstract

This paper focuses on the optimal error estimates of a linearized Crank–Nicolson scheme for the Landau–Lifshitz (LL) equation describing the evolution of spin fields in continuum ferromagnets. We present a rigorous analysis for the regularity of the local strong solution to LL equation with Neumann boundary conditions. The proof of the optimal error estimates are based upon an error splitting technique proposed by Li and Sun. Numerical results are provided to confirm our theoretical analysis.

Appendix: Proof of Theorem 2.3

Taking $t=0$ at (1.4), we get

$$\begin{aligned} \mathbf{m }_t(0)=\lambda \Delta \mathbf{m }_0 + \mathbf{m }_0\times \Delta \mathbf{m }_0+\lambda |\nabla \mathbf{m }_0|^2\mathbf{m }_0. \end{aligned}$$

By a classical argument, we can prove

$$\begin{aligned} ||\mathbf{m }_t(0)||_{H^3}\le C. \end{aligned}$$

(8.1)

Differentiating (1.4) with respect to t leads to

$$\begin{aligned} \mathbf{m }_{tt}-\lambda \Delta \mathbf{m }_t= \mathbf{m }_t\times \Delta \mathbf{m }+\mathbf{m }\times \Delta \mathbf{m }_t+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_t+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }. \end{aligned}$$

(8.2)

The following theorem improves the regularity (2.11) in Theorem 2.2.

Theorem 8.1

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^1} +\displaystyle \int _0^T ||\mathbf{m }_{tt}||_{L^2}^2+|| \mathbf{m }_t||_{H^2}^2 dt\le C. \end{aligned}$$

(8.3)

Proof

Multiplying (8.2) by $-\Delta \mathbf{m }_t$ and integrating over $\Omega$ yield

$$\begin{aligned}&\ \displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}||\nabla \mathbf{m }_t||_{L^2}^2+\lambda ||\Delta \mathbf{m }_t||_{L^2}^2\nonumber \\&\quad =-(\mathbf{m }_t\times \Delta \mathbf{m },\Delta \mathbf{m }_t)-\lambda (|\nabla \mathbf{m }|^2\mathbf{m }_t,\Delta \mathbf{m }_t)- 2\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m },\Delta \mathbf{m }_t)\nonumber \\&\quad =I_1+I_2+I_3, \end{aligned}$$

(8.4)

where we use $(\mathbf{m }\times \Delta \mathbf{m }_t,\Delta \mathbf{m }_t)=0$. By Hölder inequality, Young inequality, (2.7) and (2.9), $I_1$ is bounded as

$$\begin{aligned} |I_1|&\le ||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }||_{L^2}||\Delta \mathbf{m }_t||_{L^2}\\&\le C||\mathbf{m }_t||_{L^2}^{\frac{4-d}{4}}(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{8}}||\Delta \mathbf{m }_t||_{L^2}\\&\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C ||\mathbf{m }_t||_{L^2}^{\frac{4-d}{2}} (||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{4}}\\&\le 2\varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C ||\mathbf{m }_t||_{L^2}^2, \end{aligned}$$

where $\varepsilon _1>0$ is a small constant determined later. It follows from (2.1) and (2.9) that

$$\begin{aligned} |I_2|\le ||\mathbf{m }_t||_{L^6}||\nabla \mathbf{m }||_{L^6}^2||\Delta \mathbf{m }_t||_{L^2}\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C||\nabla \mathbf{m }_t||_{L^2}^2. \end{aligned}$$

By a similar way, we have

$$\begin{aligned} |I_3|&\le C ||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}||\Delta \mathbf{m }_t||_{L^2}^{1+\frac{d}{6}}+C||\nabla \mathbf{m }_t||_{L^2}||\Delta \mathbf{m }_t||_{L^2}\\&\le \varepsilon _1||\Delta \mathbf{m }_t||_{L^2}^2+C||\nabla \mathbf{m }_t||_{L^2}^2. \end{aligned}$$

Let $\varepsilon _1<\lambda /4$. Then combining these estimates into (8.4), and using Gronwall inequality, we conclude that

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^1} +\displaystyle \int _0^T || \mathbf{m }_t||_{H^2}^2 dt\le C, \end{aligned}$$

(8.5)

where we use (2.10) and (8.1). From (8.2), one has

$$\begin{aligned} ||\mathbf{m }_{tt}||_{L^2}\le&||\Delta \mathbf{m }_t||_{L^2}+||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }||_{L^2}+||\mathbf{m }||_{L^\infty }||\Delta \mathbf{m }_t||_{L^2}\\&+||\nabla \mathbf{m }||_{L^6}^2||\mathbf{m }_t||_{L^6}+||\nabla \mathbf{m }||_{L^6}||\nabla \mathbf{m }_t||_{L^3}\\ \le&C ||\Delta \mathbf{m }_t||_{L^2}+C||\mathbf{m }_t||_{L^2}^{\frac{4-d}{4}}(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{\frac{d}{8}}\\&+C|| \mathbf{m }_t||_{H^1}+C||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}||\Delta \mathbf{m }_t||_{L^2}^{\frac{d}{6}}+C||\nabla \mathbf{m }_t||_{L^2}\\ \le&C||\mathbf{m }_t||_{H^2}+C(||\mathbf{m }_t||_{L^2}^2+||\Delta \mathbf{m }_t||_{L^2}^2)^{1/2}+C||\Delta \mathbf{m }_t||_{L^2}, \end{aligned}$$

which implies that $\mathbf{m }_{tt}\in \mathbf{L }^2(0,T;\mathbf{L }^2(\Omega ))$ after integrating above inequality from 0 to T and using (8.5). $\square$

Taking $t=0$ in (8.2) and using (8.1), we have

$$\begin{aligned} ||\mathbf{m }_{tt}(0)||_{H^1}\le C. \end{aligned}$$

(8.6)

Differentiating (8.2) with respect to t, again, we get

$$\begin{aligned} \mathbf{m }_{ttt}-\lambda \Delta \mathbf{m }_{tt}=&\mathbf{m }_{tt}\times \Delta \mathbf{m }+ 2\mathbf{m }_t\times \Delta \mathbf{m }_t+\mathbf{m }\times \Delta \mathbf{m }_{tt}+2\lambda |\nabla \mathbf{m }_t|^2\mathbf{m }\nonumber \\&+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_{tt}+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{tt})\mathbf{m }+4\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{t})\mathbf{m }_t. \end{aligned}$$

(8.7)

Then we can show the following estimate for $\mathbf{m }_{tt}$.

Theorem 8.2

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_{tt}(t)||_{L^2} +\displaystyle \int _0^T || \mathbf{m }_{tt}||_{H^1}^2 dt\le C. \end{aligned}$$

(8.8)

Proof

Testing (8.7) by $\mathbf{m }_{tt}$ and using $(\mathbf{m }_{tt}\times \Delta \mathbf{m },\mathbf{m }_{tt})=0$ yield

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}|| \mathbf{m }_{tt}||_{L^2}^2+\lambda ||\nabla \mathbf{m }_{tt}||_{L^2}^2\nonumber \\&\quad =(\mathbf{m }\times \Delta \mathbf{m }_{tt},\mathbf{m }_{tt})+2(\mathbf{m }_{t}\times \Delta \mathbf{m }_t, \mathbf{m }_{tt})+\lambda (|\nabla \mathbf{m }|^2\mathbf{m }_{tt}, \mathbf{m }_{tt})\nonumber \\&\qquad +2\lambda (|\nabla \mathbf{m }_t|^2\mathbf{m }, \mathbf{m }_{tt}) +2\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{tt})\mathbf{m },\mathbf{m }_{tt})+4\lambda ((\nabla \mathbf{m }\cdot \nabla \mathbf{m }_{t})\mathbf{m }_t,\mathbf{m }_{tt})\nonumber \\&\quad =I_4+\cdots +I_9. \end{aligned}$$

(8.9)

By integrating by parts, we estimate $I_4$ by

$$\begin{aligned} |I_4|\le ||\nabla \mathbf{m }||_{L^6}||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^3} \le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+C||\mathbf{m }_{tt}||_{L^2}^2 \end{aligned}$$

for some small $\varepsilon _2>0$ determined later, where we use Hölder inequality, Young inequality, (2.1), (2.3), (2.9). From (2.1–2.9) and (8.5), other terms in the right-hand side of (8.9) are bounded, respectively, by

$$\begin{aligned} |I_5|&\le ||\mathbf{m }_t||_{L^3}||\Delta \mathbf{m }_{t}||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C ||\Delta \mathbf{m }_{t}||_{L^2}^2,\\ |I_6|&\le ||\nabla \mathbf{m }||_{L^6}^2||\mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{tt}||_{L^2}^2,\\ |I_7|&\le ||\mathbf{m }||_{L^\infty }||\nabla \mathbf{m }_t||_{L^3}||\nabla \mathbf{m }_t||_{L^2}||\mathbf{m }_{tt}||_{L^6}\le C||\nabla \mathbf{m }_t||_{L^2}^{\frac{6-d}{6}}|| \mathbf{m }_t||_{H^2}^{\frac{d}{6}}||\nabla \mathbf{m }_{tt}||_{L^2}\\&\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_t||_{H^2}^2+ C ,\\ |I_8|&\le ||\nabla \mathbf{m }||_{L^6}||\mathbf{m }||_{L^\infty }||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^3}\\&\le C ||\nabla \mathbf{m }_{tt}||_{L^2}||\mathbf{m }_{tt}||_{L^2}+ C ||\mathbf{m }_{tt}||_{L^2}^{\frac{6-d}{6}}||\nabla \mathbf{m }_{tt}||_{L^2}^{1+\frac{d}{6}}\\&\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{tt}||_{L^2}^2,\\ |I_9|&\le ||\nabla \mathbf{m }||_{L^2}||\nabla \mathbf{m }_t||_{L^6}||\mathbf{m }_t||_{L^6}||\mathbf{m }_{tt}||_{L^6}\\&\le C ||\mathbf{m }_t||_{H^2}||\nabla \mathbf{m }_{tt}||_{L^2}\le \varepsilon _2 ||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C || \mathbf{m }_{t}||_{H^2}^2. \end{aligned}$$

Combining these estimates into (8.9) and taking $\varepsilon _2<\lambda /6$, the estimate (8.8) follows from Gronwall inequality. $\square$

To improve the regularity of the solution $\mathbf{m }$ to the LL equation (1.4) and (1.2), the following estimates (8.10) and (8.12) for $\mathbf{m }\times \Delta \mathbf{m }$ are needed.

Lemma 8.1

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || (\mathbf{m }\times \Delta \mathbf{m })(t)||_{H^1} +\displaystyle \int _0^T ||\mathbf{m }\times \Delta \mathbf{m }||_{H^2}^2 dt\le C . \end{aligned}$$

(8.10)

Proof

It follows from (1.5) and (8.3) that

$$\begin{aligned} ||\mathbf{m }\times \Delta \mathbf{m }||_{H^1}&\le ||\mathbf{m }_t||_{H^1}+||\mathbf{m }\times \mathbf{m }_t||_{H^1}\nonumber \\&\le C||\mathbf{m }_t||_{H^1}+||\nabla \mathbf{m }\times \mathbf{m }_t||_{L^2}+C||\mathbf{m }\times \nabla \mathbf{m }_t||_{L^2}\nonumber \\&\le C||\mathbf{m }_t||_{H^1}+C||\mathbf{m }||_{H^2}||\mathbf{m }_t||_{H^1}\le C. \end{aligned}$$

(8.11)

Differentiating (1.5) with respect to $\mathbf{x }$ twice, we obtain

$$\begin{aligned} (1+\lambda ^2)\nabla ^2(\mathbf{m }\times \Delta \mathbf{m })=\nabla ^2\mathbf{m }_t+\lambda (\nabla ^2\mathbf{m }\times \mathbf{m }_t+\nabla \mathbf{m }\times \nabla \mathbf{m }_t+\mathbf{m }\times \nabla ^2 \mathbf{m }_t). \end{aligned}$$

Then one has

$$\begin{aligned}&\displaystyle \int _0^T||\nabla ^2(\mathbf{m }\times \Delta \mathbf{m })||_{L^2}^2dt\\&\quad \le \displaystyle \int _0^T\left( ||\nabla ^2\mathbf{m }_t||_{L^2}^2+||\nabla ^2\mathbf{m }\times \mathbf{m }_t||_{L^2}^2 +||\nabla \mathbf{m }\times \nabla \mathbf{m }_t||_{L^2}^2+||\mathbf{m }\times \nabla ^2 \mathbf{m }_t||_{L^2}^2\right) dt\\&\quad \le C \displaystyle \int _0^T\left( ||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{H^2}^2||\mathbf{m }_t||_{L^\infty }^2 +||\mathbf{m }||_{H^2}^2||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{L^\infty }^2||\mathbf{m }_t||_{H^2}^2\right) dt\le C, \end{aligned}$$

which together with (8.11) completes the proof of (8.10). $\square$

Lemma 8.2

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2} +\displaystyle \int _0^T || (\mathbf{m }\times \Delta \mathbf{m })_t||_{H^1}^2 dt\le C. \end{aligned}$$

(8.12)

Proof

Differentiating (1.5) with respect to t yields

$$\begin{aligned} (1+\lambda ^2)(\mathbf{m }\times \Delta \mathbf{m })_t=\mathbf{m }_{tt}+\lambda \mathbf{m }\times \mathbf{m }_{tt}. \end{aligned}$$

From (2.9) and (8.8), it is easy to show

$$\begin{aligned} \max \limits _{t\in [0,T]}|| (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}\le C. \end{aligned}$$

Moreover, from (2.9) and (8.8) one has

$$\begin{aligned}&\displaystyle \int _0^T ||\nabla (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}^2 dt\\&\quad \le \displaystyle \int _0^T ||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\nabla \mathbf{m }\times \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }\times \nabla \mathbf{m }_{tt}||_{L^2}^2 dt\\&\quad \le \displaystyle \int _0^T C||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }||_{H^2}^2||\nabla \mathbf{m }_{tt}||_{L^2}^2+||\mathbf{m }||_{L^\infty }^2|| \nabla \mathbf{m }_{tt}||_{L^2}^2 dt \le C, \end{aligned}$$

which completes the proof of (8.12). $\square$

Based upon the above results, The regularities of the solution $\mathbf{m }$ can be improved.

Theorem 8.3

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }(t)||_{H^3} \le C. \end{aligned}$$

(8.13)

Proof

Differentiating (1.4) with respect to $\mathbf{x }$ yields:

$$\begin{aligned} -\lambda \nabla \Delta \mathbf{m }=\nabla (\mathbf{m }\times \Delta \mathbf{m })-\nabla \mathbf{m }_t+2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }+\lambda |\nabla \mathbf{m }|^2\nabla \mathbf{m }. \end{aligned}$$

From (2.8), we have

$$\begin{aligned} ||\nabla ^3\mathbf{m }||_{L^2}&\le ||\nabla \mathbf{m }_t||_{L^2}+||\nabla (\mathbf{m }\times \Delta \mathbf{m })||_{L^2}+||\Delta \mathbf{m }||_{L^2} +||(\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }||_{L^2}+|||\nabla \mathbf{m }|^2\nabla \mathbf{m }||_{L^2}\\&\le C\left( ||\nabla \mathbf{m }_t||_{L^2}+||\nabla (\mathbf{m }\times \Delta \mathbf{m })||_{L^2}+||\Delta \mathbf{m }||_{L^2} +||\nabla ^2\mathbf{m }||_{L^3}||\nabla \mathbf{m }||_{L^6}+||\nabla \mathbf{m }||_{L^6}^2 \right) \\&\le C ||\nabla ^2\mathbf{m }||_{L^2}^{\frac{6-d}{6}}||\nabla ^3\mathbf{m }||_{L^2}^{\frac{d}{6}}+ C \le \displaystyle \frac{1}{2}||\nabla ^3\mathbf{m }||_{L^2}+ C||\nabla ^2\mathbf{m }||_{L^2}+ C, \end{aligned}$$

which leads to $||\nabla ^3\mathbf{m }||_{L^2}\le C$ and $\mathbf{m }\in \mathbf{L }^\infty (0,T;\mathbf{H }^3(\Omega ))$. $\square$

An alternative to (8.2 ) is

$$\begin{aligned} \mathbf{m }_{tt}-\lambda \Delta \mathbf{m }_t= (\mathbf{m }\times \Delta \mathbf{m })_t+\lambda |\nabla \mathbf{m }|^2\mathbf{m }_t+ 2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }. \end{aligned}$$

(8.14)

We can derive the following estimates for $\mathbf{m }_t$ and $\mathbf{m }_{tt}$.

Theorem 8.4

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_t(t)||_{H^2} +\displaystyle \int _0^T || \mathbf{m }_t||_{H^3}^2 dt\le C. \end{aligned}$$

(8.15)

Proof

In view of (2.5), (8.8) and (8.12), it is easy to show

$$\begin{aligned} ||\nabla ^2\mathbf{m }_t||_{L^2}&\le C\left( ||\mathbf{m }_{tt}||_{L^2}+||(\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+|||\nabla \mathbf{m }|^2\mathbf{m }_t||_{L^2}+||(\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }||_{L^2}\right) \\&\le C \left( ||\mathbf{m }_{tt}||_{L^2}+||(\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+||\mathbf{m }||_{H^3}^2||\mathbf{m }_t||_{L^2}+ ||\mathbf{m }||_{H^3}||\nabla \mathbf{m }_t||_{L^2}\right) \le C, \end{aligned}$$

which leads to $\mathbf{m }_t\in \mathbf{L }^\infty (0,T;\mathbf{H }^2(\Omega ))$. Differentiating (8.14) with respect to $\mathbf{x }$ yields:

$$\begin{aligned} \nabla \mathbf{m }_{tt}-\lambda \nabla \Delta \mathbf{m }_t&= \nabla (\mathbf{m }\times \Delta \mathbf{m })_t+ 2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m })\mathbf{m }_t+\lambda |\nabla \mathbf{m }|^2\nabla \mathbf{m }_t\\&\quad +2\lambda (\nabla ^2\mathbf{m }\cdot \nabla \mathbf{m }_t)\mathbf{m }+2\lambda (\nabla \mathbf{m }\cdot \nabla ^2\mathbf{m }_t)\mathbf{m }+2\lambda (\nabla \mathbf{m }\cdot \nabla \mathbf{m }_t)\nabla \mathbf{m }. \end{aligned}$$

By using (2.8), we get

$$\begin{aligned} ||\nabla ^3\mathbf{m }_t||_{L^2}&\le C\left( ||\nabla \mathbf{m }_{tt}||_{L^2}+|| \nabla (\mathbf{m }\times \Delta \mathbf{m })_t||_{L^2}+||\mathbf{m }||_{H^3}^2||\nabla \mathbf{m }_t||_{L^2}\right. \\&\quad \left. +||\Delta \mathbf{m }_t||_{L^2}+||\mathbf{m }||_{H^3}|| \mathbf{m }_t||_{H^2}\right) , \end{aligned}$$

which together with (8.8) and (8.12) yields $\mathbf{m }_t\in \mathbf{L }^2(0,T;\mathbf{H }^3(\Omega ))$. $\square$

Theorem 8.5

Suppose $\mathbf{m }_0\in \mathbf{H }^5(\Omega )$. Then we have

$$\begin{aligned} \max \limits _{t\in [0,T]} || \mathbf{m }_{tt}(t)||_{H^1} +\displaystyle \int _0^T || \mathbf{m }_{tt}||_{H^2}^2 dt\le C. \end{aligned}$$

(8.16)

Proof

Testing (8.7) by $-\Delta \mathbf{m }_{tt}$ and using Young inequality yield

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}||\nabla \mathbf{m }_{tt}||_{L^2}^2+\lambda ||\Delta \mathbf{m }_{tt}||_{L^2}^2\\&\quad \le \left( ||\mathbf{m }_{tt}||_{L^6}||\mathbf{m }||_{H^3}+||\mathbf{m }_t||_{L^\infty }||\Delta \mathbf{m }_t||_{L^2}+||\nabla \mathbf{m }||_{L^\infty }^2||\mathbf{m }_{tt}||_{L^2}\right. \\&\qquad \left. +||\mathbf{m }_t||_{H^2}^2+||\mathbf{m }||_{H^3}^2||\nabla \mathbf{m }_{tt}||_{L^2}+||\mathbf{m }||_{L^\infty }||\mathbf{m }_t||_{L^\infty }||\nabla \mathbf{m }_t||_{L^2}\right) ||\Delta \mathbf{m }_{tt}||_{L^2}\\&\quad \le \displaystyle \frac{\lambda }{2}||\Delta \mathbf{m }_{tt}||_{L^2}^2+C||\nabla \mathbf{m }_{tt}||_{L^2}^2+ C. \end{aligned}$$

Integrating the above inequality from 0 to T and using (8.6), we complete the proof of (8.16). $\square$