Unconditional Energy Stability Analysis of a Second Order Implicit–Explicit Local Discontinuous Galerkin Method for the Cahn–Hilliard Equation

Abstract

In this article, we present a second-order in time implicit–explicit (IMEX) local discontinuous Galerkin (LDG) method for computing the Cahn–Hilliard equation, which describes the phase separation phenomenon. It is well-known that the Cahn–Hilliard equation has a nonlinear stability property, i.e., the free-energy functional decreases with respect to time. The discretized Cahn–Hilliard system modeled by the IMEX LDG method can inherit the nonlinear stability of the continuous model. We apply a stabilization technique and prove unconditional energy stability of our scheme. Numerical experiments are performed to validate the analysis. Computational efficiency can be significantly enhanced by using this IMEX LDG method with a large time step.

Appendices

Appendix A: The Proof of Proposition 1

Due to the linearity and finite dimensionality of the problem, it is enough to show that the only solution to (19) and (20) with $u=0$ is $v_h=0$.

$$\begin{aligned}&\big (\sigma _h, \tau \big )_K+\big (v_h, \nabla \cdot \tau \big )_K-<\hat{v}_h, \tau \cdot \mathbf {n}>_{\partial K}=0,\\&\quad \big (\sigma _h, \nabla \eta \big )_K-<\hat{\sigma }_h\cdot \mathbf {n},\eta > _{\partial K}=0. \end{aligned}$$

Taking $\tau =\sigma _h$ and $\eta =v_h$,

$$\begin{aligned}&\big (\sigma _h, \sigma _h\big )_K+\big (v_h, \nabla \cdot \sigma _h\big )_K-<\hat{v}_h, \sigma _h\cdot \mathbf {n}>_{\partial K}=0,\\&\quad \big (\sigma _h, \nabla v_h\big )_K-<\hat{\sigma }_h\cdot \mathbf {n},v_h> _{\partial K}=0. \end{aligned}$$

Applying integration by parts, and adding the two equations, we get

$$\begin{aligned} H_{\partial K}(v_h, \sigma _h)_K=(\sigma _h, \sigma _h)_K. \end{aligned}$$

Summing over K, we can obtain

$$\begin{aligned} \sum \limits _K(\sigma _h, \sigma _h)_K=0 \end{aligned}$$

which implies $\sigma _h=0$. Using the relationship (21), we have $\nabla v_h=0$ on every K and $[v_h]=0$, since $\mu >0$. Then $v_h|_K=C$. Because of $[v_h]=0$, $v_h=C$. However

$$\begin{aligned} \int _{\varOmega } v_hdx=C\cdot |{\varOmega }|=0, \end{aligned}$$

which implies $v_h=0$. Thus we have completed the proof.

Appendix B: The Proof of Lemma 3

Using the LDG discrete “inverse Laplacian” and taking $\tau =\sigma _h^{n+1}$ and $\eta =v_h^{n+1}$ in (22),

$$\begin{aligned}&\big (u^{n+1}-u^{n}, v_h^{n+1}\big )_K-\big (\sigma _h^{n+1}, \nabla v_h^{n+1}\big )_K+<\hat{\sigma }_h^{n+1}\cdot \mathbf {n},v_h^{n+1}> _{\partial K}=0,\\&\quad \big (\sigma _h^{n+1}, \sigma _h^{n+1}\big )_K+\big (v_h^{n+1}, \nabla \cdot \sigma _h^{n+1}\big )_K-<\hat{v}_h^{n+1}, \sigma _h^{n+1}\cdot \mathbf {n}>_{\partial K}=0. \end{aligned}$$

Then

$$\begin{aligned} \big (u^{n+1}-u^{n}, v_h^{n+1})_K+H_{\partial K}(v_h^{n+1}, \sigma _h^{n+1})_K=(\sigma _h^{n+1}, \sigma _h^{n+1})_K. \end{aligned}$$

Summing over K and using the Lemma 1, we can obtain the result (24).

Note that $\delta ^2 u^{n+1}=u^{n+1}-2u^n+u^{n-1}$, clearly

$$\begin{aligned} \left\{ \begin{array}{l} \big (\delta ^2 u^{n+1}, \eta \big )_K=\big (\sigma _h^{n+1}-\sigma _h^{n}, \nabla \eta \big )_K-<(\hat{\sigma }_h^{n+1}-\hat{\sigma }_h^{n})\cdot \mathbf {n},\eta> _{\partial K},\\ \big (\sigma _h^{n+1}, \tau \big )_K=-\big (v_h^{n+1}, \nabla \cdot \tau \big )_K+<\hat{v}_h^{n+1}, \tau \cdot \mathbf {n}>_{\partial K}. \end{array} \right. \end{aligned}$$

Similarly, taking $\eta =v_h^{n+1}$ and $\tau =\sigma _h^{n+1}-\sigma _h^{n}$,

$$\begin{aligned}&\big (\delta ^2u^{n+1}, v_h^{n+1}\big )_K-\big (\sigma _h^{n+1}-\sigma _h^{n}, \nabla v_h^{n+1}\big )_K+<(\hat{\sigma }_h^{n+1}-\hat{\sigma }_h^{n})\cdot \mathbf {n},v_h^{n+1}> _{\partial K}=0,\\&\quad \big (\sigma _h^{n+1}, \sigma _h^{n+1}-\sigma _h^{n}\big )_K+\big (v_h^{n+1}, \nabla \cdot (\sigma _h^{n+1}-\sigma _h^{n})\big )_K-<\hat{v}_h^{n+1}, (\sigma _h^{n+1}-\sigma _h^{n})\cdot \mathbf {n}>_{\partial K}=0. \end{aligned}$$

So that

$$\begin{aligned}&\big (\delta ^2 u^{n+1}, v_h^{n+1})_K+H_{\partial K}(v_h^{n+1}, \sigma _h^{n+1}-\sigma _h^{n}) \\&\quad =\left( \sigma _h^{n+1}, \sigma _h^{n+1}-\sigma _h^{n}\right) _K. \end{aligned}$$

Summing over K and using the Lemma 1, we can obtain the result (25). This completes the proof.

Appendix C: The Proof of the Lemma 7

Taking $\rho =\varepsilon ^{2}u^1, \mathbf q =-\varepsilon ^{2}\mathbf {w}^1, \phi =p^1-r^0, {\psi }=\varepsilon ^{2}(\mathbf {s}_1^{0}-\mathbf s _2^1)$ in Eqs. (51), and using the same analysis as the Lemma 6, we have

$$\begin{aligned} \frac{\varepsilon ^2}{t_1}\big (u^1-u^0, u^1\big )+(p^1,p^1-r^0)=0. \end{aligned}$$

Summing up over K,

$$\begin{aligned} \frac{\varepsilon ^2}{t_1}\cdot \Big (\Vert u^1\Vert ^2-\Vert u^0\Vert ^2\Big )+\Vert p^1\Vert ^2\le \frac{1}{2}\big (\Vert p^1\Vert ^2+\Vert r^0\Vert ^2\big ), \end{aligned}$$

So we have the result

$$\begin{aligned} \Vert u^1\Vert ^2\le \Vert u^0\Vert ^2+\frac{t_1}{\varepsilon ^2}\Vert f(u^0)\Vert ^2, \end{aligned}$$

and

$$\begin{aligned} \Vert p^1\Vert ^2\le \frac{\varepsilon ^2}{t_1}\Vert u^0\Vert ^2+\Vert f(u^0)\Vert ^2, \end{aligned}$$

By the definition of the discrete Laplacian operator,

$$\begin{aligned} (p^1,\phi )=-\varepsilon ^2(-\Delta _h u^1, \phi ), \end{aligned}$$

Then Taking $\phi =-\Delta _h u^1$, we obtain

$$\begin{aligned} \Vert \Delta _h u^1\Vert ^2\le \frac{1}{\varepsilon ^2 t_1 }\Vert u^0\Vert ^2+\frac{1}{\varepsilon ^4}\Vert f(u^0)\Vert ^2, \end{aligned}$$

Next, taking $\rho =v_h^1=(-\Delta )^{-1}(u^1-u^0), \mathbf q =\sigma _h^1, \phi =u^1-u^0,\xi =u^1-u^0$ in Eqs. (51), and

$$\begin{aligned} (\mathbf w ^{1}-\mathbf w ^{0}, \mathbf w ^{1})_K=\big (\nabla (u^1-u^0), \mathbf w ^{1})_K+H_{\partial K}(u^1-u^0,\mathbf w ^{1})-<\hat{\mathbf{w }}^{1}\cdot \mathbf {n},u^1-u^0>_{\partial K}, \end{aligned}$$

using the same argument as Eqs. (36)–(48), then

$$\begin{aligned} \Big (\frac{u^{1}-u^0}{t_1}, v_h^1\Big )_K+\varepsilon ^2(\mathbf w ^{1}-\mathbf w ^{0}, \mathbf w ^{1})_K+(f(u^0), u^1-u^0)_K-\varepsilon ^2H_{\partial K}(u^1-u^0,\mathbf w ^{1})=0. \end{aligned}$$

Using Lemmas 1, 3, and summing on K,

$$\begin{aligned}&\frac{1}{t_1}\Vert \sigma _h^1\Vert ^2+\frac{\varepsilon ^2}{2}\Vert \mathbf w ^{1}\Vert ^2+\sum \limits _{K}(F(u^1),1)\\&\quad \le E(u^0)+|f^{\prime }(u^0)|\cdot \Vert u^1-u^0\Vert ^2\\&\quad \le C\Big ({E(u^0), \Vert u^0\Vert _{H^4}} \Big ). \end{aligned}$$

So that

$$\begin{aligned} \Vert \mathbf {w}^{1}\Vert ^2 \le C\Big ({E(u^0), \Vert u^0\Vert _{H^4}} \Big ) \frac{1}{\varepsilon ^2}, \end{aligned}$$

Finally, by the Lemma 4,

$$\begin{aligned} \Vert u^1\Vert _{\infty }\le C\sqrt{1+\varepsilon ^{-1}}\cdot \sqrt{\log \left( 1+\left( \frac{1}{\varepsilon ^2 t_1} +\varepsilon ^{-4}\right) ^{\frac{1}{2}}\right) } . \end{aligned}$$