A Third Order Exponential Time Differencing Numerical Scheme for No-Slope-Selection Epitaxial Thin Film Model with Energy Stability

Abstract

In this paper we propose and analyze a (temporally) third order accurate exponential time differencing (ETD) numerical scheme for the no-slope-selection (NSS) equation of the epitaxial thin film growth model, with Fourier pseudo-spectral discretization in space. A linear splitting is applied to the physical model, and an ETD-based multistep approximation is used for time integration of the corresponding equation. In addition, a third order accurate Douglas-Dupont regularization term, in the form of $-A {\Delta t}^2 \phi _0 (L_N) \Delta _N^2 ( u^{n+1} - u^n)$, is added in the numerical scheme. A careful Fourier eigenvalue analysis results in the energy stability in a modified version, and a theoretical justification of the coefficient A becomes available. As a result of this energy stability analysis, a uniform in time bound of the numerical energy is obtained. And also, the optimal rate convergence analysis and error estimate are derived in details, in the $\ell ^\infty (0,T; H_h^1) \cap \ell ^2 (0,T; H_h^3)$ norm, with the help of a careful eigenvalue bound estimate, combined with the nonlinear analysis for the NSS model. This convergence estimate is the first such result for a third order accurate scheme for a gradient flow. Some numerical simulation results are presented to demonstrate the efficiency of the numerical scheme and the third order convergence. The long time simulation results for $\varepsilon =0.02$ (up to $T=3 \times 10^5$) have indicated a logarithm law for the energy decay, as well as the power laws for growth of the surface roughness and the mound width. In particular, the power index for the surface roughness and the mound width growth, created by the third order numerical scheme, is more accurate than those produced by certain second order energy stable schemes in the existing literature.

Appendices

Proof of Lemma 2.4

First, we review the following estimate in Calculus.

Lemma A.1

Suppose that f(x) and g(x) are continuous functions, $f (x) >0$, $g (x) >0$, and $\frac{f (x)}{g (x)}$ is decreasing over $(0, + \infty )$. Define $H(x) = \frac{\int _0^x f (t) dt}{\int _0^x g(t) dt}$. Then H(x) is decreasing over $(0, + \infty )$.

Proof

Denote $F(x) = \int _0^x f (t) dt$, $G(x) = \int _0^x g (t) dt$, so that $H(x) = \frac{F(x)}{G(x)}$. For any $0< x_1 < x_2$, we make a comparison between $H(x_1)$ and $H (x_2)$.

We denote $C_1 = \frac{f(x_1)}{g(x_1)}$. By the decreasing property of $\frac{f (x)}{g (x)}$, we see that $\frac{f (x)}{g (x)} \ge C_1$ for $0\le x \le x_1$, and $\frac{f (x)}{g (x)} \le C_1$ for $x_1 \le x \le x_2$. This in turn implies that

$$\begin{aligned} f (x) \ge C_1 g(x) , \, \, \, \text{ for } 0\le x \le x_1 , \quad f (x) \le C_1 g(x) , \, \, \, \text{ for } x_1\le x \le x_2 . \end{aligned}$$

(A.1)

As a result, we get

$$\begin{aligned} \int _0^{x_1} f (t) dt \ge C_1 \int _0^{x_1} g (t) dt , \quad \int _{x_1}^{x_2} f (t) dt \le C_1 \int _{x_1}^{x_2} g (t) dt . \end{aligned}$$

(A.2)

In turn, if we denote $A_1 = \int _0^{x_1} f (t) dt$, $B_1 = \int _0^{x_1} g (t) dt$, $A_2 = \int _{x_1}^{x_2} f (t) dt$, $B_2 = \int _{x_1}^{x_2} g (t) dt$, we have

$$\begin{aligned} \frac{A_1}{B_1} \ge C_1 \ge \frac{A_2}{B_2} . \end{aligned}$$

(A.3)

Then we arrive at

$$\begin{aligned} H(x_2) = \frac{\int _0^{x_2} f (t) dt}{\int _0^{x_2} g(t) dt} = \frac{A_1 + A_2}{B_1 + B_2} \le \frac{A_1}{B_1} = H(x_1). \end{aligned}$$

(A.4)

This completes the proof for the decreasing property of H(x). $\square$

Next, we proceed into the proof of Lemma 2.4.

Proof

The function $g_0 (x)$ could be represented as

$$\begin{aligned} g_0 (x) = \frac{1 - \mathrm{e}^{-x}}{x} = \frac{\int _0^x \mathrm{e}^{-t} \, dt}{\int _0^x 1 \, dt} . \end{aligned}$$

(A.5)

On the other hand, $\frac{\mathrm{e}^{-x}}{1} = \mathrm{e}^{-x}$ is a decreasing function over $(0, +\infty )$. By Lemma A.1, we conclude that $g_0 (x)$ is decreasing.

Similarly, $g_1 (x)$ could be rewritten as

$$\begin{aligned} g_1 (x) = \frac{x - (1 - \mathrm{e}^{-x} )}{x^2} = \frac{\int _0^x ( 1 - \mathrm{e}^{-t} ) \, dt}{\int _0^x 2t \, dt} . \end{aligned}$$

(A.6)

Since $\frac{1 - \mathrm{e}^{-x}}{2 x} = \frac{1}{2} g_0 (x)$ is a decreasing function over $(0, +\infty )$, an application of Lemma A.1 reveals that $g_1 (x)$ is also decreasing.

For $g_2 (x)$, we look at its rewritten form

$$\begin{aligned} g_2 (x) = \frac{x^2 - 2 (x - (1 - \mathrm{e}^{-x} ) )}{x^3} = \frac{\int _0^x 2 ( t - (1 - \mathrm{e}^{-t} ) ) \, dt}{\int _0^x 3 t^2 \, dt} . \end{aligned}$$

(A.7)

Since $\frac{2 ( x - (1 - \mathrm{e}^{-x} ) )}{3 x^2} = \frac{2}{3} g_1 (x)$ is a decreasing function over $(0, +\infty )$, an application of Lemma A.1 reveals that $g_2 (x)$ is also decreasing. This finishes the proof of the first part of Lemma 2.4.

As a direct consequence of their decreasing property, we see that $g_0 (x) \le g_0 (0) = 1$, $g_1 (x) \le g_1 (0) = \frac{1}{2}$ and $g_2 (x) \le g_2 (0) = \frac{1}{3}$,     $\forall x > 0$.

In turn, we observe that

$$\begin{aligned}&\frac{g_1 (x)}{g_0 (x)} \le \frac{g_1 (0)}{g_0 (2)} = \frac{\frac{1}{2}}{\frac{1 - \mathrm{e}^{-2}}{2} } = \frac{1}{1 - \mathrm{e}^{-2}} , \quad \text{ for } \, x \le 2 , \end{aligned}$$

(A.8)

$$\begin{aligned}&\frac{g_1 (x)}{g_0 (x)} = \frac{1 - \frac{1 - \mathrm{e}^{-x}}{x} }{1 - \mathrm{e}^{-x} } \le \frac{1}{1 - \mathrm{e}^{-2} } , \quad \text{ for } \, x \ge 2. \end{aligned}$$

(A.9)

For the function $\frac{g_2 (x)}{g_0 (x)}$, we have

$$\begin{aligned}&\frac{g_2 (x)}{g_0 (x)} \le \frac{g_2 (0)}{g_0 (2)} = \frac{\frac{1}{3}}{\frac{1 - \mathrm{e}^{-2}}{2} } = \frac{2}{3(1 - \mathrm{e}^{-2})} , \quad \text{ for } \, x \le 2, \end{aligned}$$

(A.10)

$$\begin{aligned}&\frac{g_2 (x)}{g_0 (x)} = \frac{1 - 2 \frac{1 - \frac{1 - \mathrm{e}^{-x}}{x} }{x} }{1 - \mathrm{e}^{-x} } \le \frac{1}{1 - \mathrm{e}^{-2} } , \quad \text{ for } \, x \ge 2. \end{aligned}$$

(A.11)

Therefore, the inequalities $\frac{g_1 (x)}{g_0 (x)} \le \frac{1}{1 - \mathrm{e}^{-2} }$, $\frac{g_2 (x)}{g_0 (x)} \le \frac{1}{1 - \mathrm{e}^{-2} }$ are valid. This finishes the proof of Lemma 2.4. $\square$

Proof of Proposition 2.5

An application of Parseval equality to the discrete Fourier expansions for f and ${{\mathcal {G}}}^{(0)}_N f$, given by (2.33) and (2.37), respectively, leads to

$$\begin{aligned} \Vert f \Vert _2^2 = L^2 \sum _{k,\ell =-K}^K | {\hat{f}}_{k,\ell } |^2 , \quad \left\| {{\mathcal {G}}}^{(0)}_N f \right\| _2^2 = L^2 \sum _{k,\ell =-K}^K \frac{{\Delta t}\Lambda _{k,\ell }}{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} } | {\hat{f}}_{k,\ell } |^2 . \end{aligned}$$

(B.1)

Meanwhile, the following observation is made:

$$\begin{aligned} \frac{{\Delta t}\Lambda _{k,\ell }}{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} } = \frac{1}{g_0 ({\Delta t}\Lambda _{k,\ell })} . \end{aligned}$$

(B.2)

With an application of Lemma 2.4, we obtain

$$\begin{aligned} 1 = \frac{1}{g_0 (0)} \le \frac{1}{g_0 (x) } = \frac{x}{ 1 - \mathrm{e}^{-x} } \le 1+x , \quad \forall x > 0 . \end{aligned}$$

(B.3)

This in turn implies that

$$\begin{aligned} 1 \le \frac{{\Delta t}\Lambda _{k,\ell }}{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} } \le 1+ {\Delta t}\Lambda _{k,\ell } , \quad \text{ for } \text{ any } \, k, \, \ell . \end{aligned}$$

(B.4)

Its combination with (B.1) reveals that

$$\begin{aligned} \Vert f \Vert _2^2 \le \left\| {{\mathcal {G}}}^{(0)}_N f \right\| _2^2\le & {} L^2 \sum _{k,\ell =-N}^N ( 1+ {\Delta t}\Lambda _{k,\ell }) | {\hat{f}}_{k,\ell } |^2 \nonumber \\= & {} \Vert f \Vert _2^2 + {\Delta t}( \varepsilon ^2 \Vert \Delta _N f \Vert _2^2 + \kappa \Vert \nabla _N f \Vert _2^2 ) , \end{aligned}$$

(B.5)

which in turn results in (2.40), by taking $C_1 = 1$.

The proof of the first inequality of (2.41) follows a similar form of Fourier analysis, combined with the following identity:

$$\begin{aligned} \frac{{\Delta t}\Lambda _{k,\ell }}{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} } \ge {\Delta t}\Lambda _{k,\ell } , \quad \forall k, \ell . \end{aligned}$$

(B.6)

For the second inequality of (2.41), we begin with the following identities

$$\begin{aligned} (L_N f)_{i,j}= & {} \sum _{k,\ell =-K}^K \Lambda _{k, \ell } {\hat{f}}_{k,\ell } \mathrm{e}^{2 \pi i ( k x_i + \ell y_j)/L}, \\ (- \Delta _N \mathrm{e}^{- L_N {\Delta t}} f )_{i,j}= & {} \sum _{k,\ell =-K}^K ( - \lambda _{k, \ell }) \mathrm{e}^{- {\Delta t}\Lambda _{k, \ell } } {\hat{f}}_{k,\ell } \mathrm{e}^{2 \pi i ( k x_i + \ell y_j)/L}, \end{aligned}$$

so that

$$\begin{aligned} \left\langle L_N f , - \Delta _N \mathrm{e}^{- L_N {\Delta t}} f \right\rangle = L^2 \sum _{k, \ell =-K}^K ( - \lambda _{k, \ell }) \Lambda _{k, \ell } \mathrm{e}^{- {\Delta t}\Lambda _{k, \ell } } | {\hat{f}}_{k,\ell } |^2 \ge 0 , \end{aligned}$$

(B.7)

since $- \lambda _{k, \ell } \ge 0$, $\Lambda _{k, \ell } \ge 0$, and $\mathrm{e}^{- {\Delta t}\Lambda _{k, \ell } } \ge 0$, for any $k, \ell$.

The proof of (2.42) and (2.43) could be carried out in the same manner; the details are left to interested readers.

For the analysis of $G^{(1)}_N f$, with its discrete Fourier expansion given by (2.35), an application of Parseval equality gives

$$\begin{aligned} \Vert f \Vert _2^2 = L^2 \sum _{k,\ell =-K}^K | {\hat{f}}_{k,\ell } |^2 , \quad \left\| G^{(1)}_N f \right\| _2^2 = L^2 \sum _{k,\ell =-K}^K \left( \frac{1 - \frac{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} }{{\Delta t}\Lambda _{k,\ell }} }{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} } \right) ^2 | {\hat{f}}_{k,\ell } |^2 . \nonumber \\ \end{aligned}$$

(B.8)

On the other hand, the following observation is available:

$$\begin{aligned} \frac{1 - \frac{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,\ell }} }{{\Delta t}\Lambda _{k,\ell }} }{1 - \mathrm{e}^{- {\Delta t}\Lambda _{k,l}} } = \frac{ g_1 ({\Delta t}\Lambda _{k,\ell }) }{ g_0 ({\Delta t}\Lambda _{k,\ell }) } \le \frac{1}{1 - \mathrm{e}^{-2}} , \end{aligned}$$

(B.9)

with an application of Lemma 2.4 in the second step. Its substitution into (B.8) results in

$$\begin{aligned} \Vert f \Vert _2^2 = L^2 \sum _{k,\ell =-K}^K | {\hat{f}}_{k,\ell } |^2 , \quad \left\| G^{(1)}_N f \right\| _2^2 \le \left( \frac{1}{1 - \mathrm{e}^{-2}} \right) ^2 \Vert f \Vert _2^2 . \end{aligned}$$

(B.10)

Then we have proved the first inequality of (2.44), with $C_4 = \frac{1}{1 - \mathrm{e}^{-2}}$.

The analysis of $G^{(2)}_N f$ could be carried out in the same manner, and the second inequality of (2.44) is with $C_5 = \frac{1}{1 - \mathrm{e}^{-2}}$. This finishes the proof of Proposition 2.5.

Proof of Proposition 3.2

We begin with the following expansion:

$$\begin{aligned} \nabla _N \left( f_N (U^k) - f_N (u^k) \right) = \nabla _N \nabla _N \cdot \left( g_N (U^k) - g_N (u^k) \right) . \end{aligned}$$

(C.1)

Meanwhile, we denote $U_N^k$, $u_N^k$ and $e_N^k$ as the continuous extension of $U^k$, $u^k$ and $e^k$, as the formula given by (2.15). We notice that $g_N$ is defined in the sense of collocation way, at a point-wise level. Since $g_N (U^k) - g_N (u^k)$ is the grid point interpolation of $g (U_N^k) - g (u_N^k)$, we apply (2.18) (in Lemma 2.2) to control the aliasing error. Subsequently, we arrive at

$$\begin{aligned} \Vert \nabla _N \nabla _N \cdot ( g_N (U^k) - g_N (u^k) ) \Vert _2= & {} \Vert \nabla \nabla \cdot P_c^N ( g (U_N^k) - g (u_N^k) ) \Vert \nonumber \\\le & {} C \Vert P_c^N ( g (U_N^k) - g (u_N^k) ) \Vert _{H^2} \nonumber \\\le & {} C \Vert g (U_N^k) - g (u_N^k) \Vert _{H^2} , \end{aligned}$$

(C.2)

in which the fact that $2 > \frac{d}{2} =1$ has been used. On the other hand, we have the following expansion for $g (U_N^k) - g (u_N^k)$, in a similar form as (2.77):

$$\begin{aligned} g (U_N^k) - g (u_N^k) = \frac{\nabla e_N^k}{1+|\nabla u_N^k|^2} + \frac{ \nabla (U_N^k + u_N^k) \cdot \nabla e_N^k}{(1+|\nabla U_N^k |^2) (1+|\nabla u_N^k|^2)} \nabla U_N^k + \kappa \nabla e_N^k . \end{aligned}$$

(C.3)

A repeated application of Hölder inequality and Sobobev inequality leads to the following estimates:

$$\begin{aligned}&\left\| \frac{\nabla e_N^k}{1+|\nabla u_N^k|^2} + \kappa \nabla e_N^k \right\| _{H^2} \nonumber \\&\quad \le C ( \Vert U_N^k \Vert _{H^3}^2 + \Vert u_N^k \Vert _{H^3}^2 ) ( \Vert \nabla e_N^k \Vert + \Vert \Delta e_N^k \Vert + \Vert \nabla \Delta e_N^k \Vert ) , \end{aligned}$$

(C.4)

$$\begin{aligned}&\left\| \frac{ \nabla (U_N^k + u_N^k) \cdot \nabla e_N^k}{(1+|\nabla U_N^k |^2) (1+|\nabla u_N^k|^2)} \nabla U_N^k \right\| _{H^2} \nonumber \\&\quad \le C ( \Vert U_N^k \Vert _{H^3}^2 + \Vert u_N^k \Vert _{H^3}^2 ) ( \Vert \nabla e_N^k \Vert + \Vert \Delta e_N^k \Vert + \Vert \nabla \Delta e_N^k \Vert ) . \end{aligned}$$

(C.5)

Meanwhile, with the a-priori assumption (3.12), we have

$$\begin{aligned}&\Vert U_N^k \Vert _{H^3} \le C*, \nonumber \\&\Vert u_N^k \Vert _{H^3} \le \Vert U_N^k \Vert _{H^3} + \Vert e_N^k \Vert _{H^3} \le \Vert U_N^k \Vert _{H^3} + C_6 \Vert \nabla \Delta e_N^k \Vert \le C^* + C_6 := {\tilde{C}}_1 , \end{aligned}$$

(C.6)

in which $C_6$ is a constant associated with elliptic regularity: $\Vert e_N^k \Vert _{H^3} \le C_6 \Vert \nabla \Delta e_N^k \Vert$, since $\int _\Omega e_N^k \, d \mathbf{x} =0$. Then we arrive at

$$\begin{aligned} \Vert \nabla _N ( f_N (U^k) - f_N (u^k) ) \Vert _2= & {} \Vert \nabla _N \nabla _N \cdot ( g_N (U^k) - g_N (u^k) ) \Vert _2 \nonumber \\\le & {} C \Vert g (U_N^k) - g (u_N^k) \Vert _{H^2} \nonumber \\\le & {} C ( \Vert U_N^k \Vert _{H^3}^2 + \Vert u_N^k \Vert _{H^3}^2 ) ( \Vert \nabla e_N^k \Vert + \Vert \Delta e_N^k \Vert + \Vert \nabla \Delta e_N^k \Vert ) \nonumber \\\le & {} C ( (C^*)^2 + {\tilde{C}}_1^2 ) C_6 \Vert \nabla \Delta e_N^k \Vert \nonumber \\\le & {} C ( (C^*)^2 + {\tilde{C}}_1^2 ) C_6 \Vert \nabla _N \Delta _N e^k \Vert . \end{aligned}$$

(C.7)

As a result, (3.13) has been established, by taking $C_0^{(2)} = C ( (C^*)^2 + {\tilde{C}}_1^2 ) C_6$. This finishes the proof of Proposition 3.2.