Sharp Error Estimate of an Implicit BDF2 Scheme with Variable Time Steps for the Phase Field Crystal Model

Abstract

A fully implicit two-step backward differentiation formula (BDF2) scheme with variable time steps is considered for solving the phase field crystal (PFC) model by combining with pseudo-spectral method in space. We show that the BDF2 scheme inherits a modified energy dissipation law under a mild ratio restriction A1, i.e., $0<r_k:=\tau _k/\tau _{k-1}< r_{\max }\approx 4.8645$, which justifies the thermodynamic consistency of the PFC model numerically. In addition, the optimal second-order convergence of the proposed scheme is also established under the ratio restriction A1. The proof involves the tools of DOC and DCC kernels, and some generalized properties of the DOC kernels. As far as we know, the ratio restriction A1 required in our results is mildest so far for the variable time-step BDF2 scheme for calculating PFC model. Numerical examples are provided to demonstrate our theoretical results.

Appendices

Appendix

The proof of Lemma 4.6

Proof

Before proving Lemma 4.6, we give some definitions of matrices as follows

$$\begin{aligned} B_2:=\left( \begin{array}{cccc} b^{(1)}_0&{}&{}&{}\\ b^{(2)}_1&{}b^{(2)}_0&{}&{}\\ &{}\ddots &{}\ddots &{}\\ &{}&{}b^{(n)}_1&{}b^{(n)}_0 \end{array} \right) _{n\times n}\ \text {and}\quad \Theta _2:=\left( \begin{array}{cccc} \theta ^{(1)}_0&{}&{}&{}\\ \theta ^{(2)}_1&{}\theta ^{(2)}_0&{}&{}\\ \vdots &{}\ddots &{}\ddots &{}\\ \theta ^{(n)}_{n-1}&{}\cdots &{}\theta ^{(n)}_1&{}\theta ^{(n)}_0 \end{array} \right) _{n\times n,} \end{aligned}$$

which are the matrix forms of BDF2 kernels $b^{(n)}_{n-k}$ and DOC kernels $\theta _{n-k}^{(n)}$. From the definition (4.3), the matrices $B_2$ and $\Theta _2$ satisfy the following relation.

$$\begin{aligned} \Theta _2 = B_2^{-1}. \end{aligned}$$

(A.1)

Define the diagonal matrix $\Lambda _\tau :=\text {diag}(\sqrt{\tau _1},\cdots ,\sqrt{\tau _n})$ and $\tilde{B}_2:=\Lambda _\tau B_2\Lambda _\tau$ and the symmetric matrices

$$\begin{aligned} B:=B_2+B_2^T \quad \text {and} \quad \tilde{B}:=\tilde{B}_2+\tilde{B}^T_2 \quad \text {and} \quad \Theta :=\Theta _2+\Theta _2^T. \end{aligned}$$

It follows from Lemmas 3.1 and 4.1 and [10, Lemmas 4.3] that matrices $B_2$ and $\theta _2$ and $\tilde{B}$ are positive definite. Since B is a real symmetric positive definite matrix, from the standard Cholesky decomposition, there exists a non-singular matrix L such that $B=L^TL$. Hence, for any $\varvec{v}:=\{v^k\}_{k=1}^n,\varvec{w}:=\{w^k\}_{k=1}^n$, one has

$$\begin{aligned} \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}w^kv^j&=\varvec{w}^T\Theta _2\varvec{v}=\varvec{w}^T\Theta _2(LB_2^{-1})^{-1}LB_2^{-1}\varvec{v}\nonumber \\&\le \frac{\varepsilon }{2}\varvec{w}^T\Theta _2B_2L^{-1}(L^{-1})^TB_2^T\Theta _2^T\varvec{w}+\frac{1}{2\varepsilon }\varvec{v}^T(B_2^{-1})^TL^TLB_2^{-1}\varvec{v}\nonumber \\&=\frac{\varepsilon }{2}\varvec{w}^T(\Theta _2B_2)(L^TL)^{-1}(\Theta _2B_2)^T\varvec{w}+\frac{1}{2\varepsilon }\varvec{v}^T(B_2^{-1})^TBB_2^{-1}\varvec{v}\nonumber \\&=\frac{\varepsilon }{2}\varvec{w}^TB^{-1}\varvec{w}+\frac{1}{2\varepsilon }\varvec{v}^T\Theta \varvec{v}, \end{aligned}$$

(A.2)

where the Young’s inequality and (A.1) are used. For the first term of the right-hand side of (A.2), one finds

$$\begin{aligned} \varvec{w}^TB^{-1}\varvec{w}&=\varvec{w}^TL^{-1}(L^T)^{-1}\varvec{w}=\Vert (L^T)^{-1}\varvec{w}\Vert ^2_2\nonumber \\&=\Vert (L^T)^{-1}(LB_2^{-1})^{-1}LB_2^{-1}\varvec{w}\Vert ^2_2 \; \le \; \Vert (L^T)^{-1} B_2L^{-1}\Vert _2^2\varvec{w}^T\Theta \varvec{w}, \end{aligned}$$

(A.3)

where $\Vert \cdot \Vert _2$ denotes the $l^2$-norm in the space $\mathbb {R}^n$. Set $\tilde{L}:=L\Lambda _\tau$ and it is easy to check $\tilde{B}=\Lambda _\tau B\Lambda _\tau = \tilde{L}^T\tilde{L}$. Then, we have

$$\begin{aligned} \Vert (L^T)^{-1} B_2L^{-1}\Vert _2^2=\Vert (\tilde{L}^T)^{-1}\tilde{B_2}\tilde{L}^{-1}\Vert _2^2 \le \Vert \tilde{L}^{-1}\Vert ^4_2\Vert \tilde{B}_2\Vert _2^2\le \mathcal {Q}_\delta , \end{aligned}$$

(A.4)

where the constant $\mathcal {Q}_\delta$ is defined by

$$\begin{aligned} \mathcal {Q}_{\delta }:=\max _{n\ge 1}\Vert \tilde{B_2}\Vert _2^2\Vert \tilde{L}^{-1}\Vert _2^4=\max _{n\ge 1}\frac{\lambda _{\max }(\tilde{B_2}^T\tilde{B_2})}{\lambda _{\min }^2(\tilde{B})}. \end{aligned}$$

(A.5)

Combining (A.4) and (A.3) with (A.2), we finally arrive at

$$\begin{aligned} \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}v^kw^j\le \epsilon \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}v^kv^j+ \frac{\mathcal {Q}_{\delta }}{\epsilon }\sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}w^kw^j,\quad \forall \epsilon >0, \end{aligned}$$

(A.6)

where we use the fact that $\varvec{v}^T\Theta \varvec{v}=2\varvec{v}^T\Theta _2\varvec{v}=\sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}v^kv^j$. The proof is completed.

The proof of Lemma 4.7

Proof

Choosing $v^k=u^k,w^j=\Delta u^j$, $\epsilon =\frac{4\mathcal {Q}_\delta ^2}{\eta }$ and $v^k=\nabla u^k,w^j=\nabla \Delta u^j, \epsilon =\frac{2\mathcal {Q}_{\delta }}{\eta }$ in (A.6), one has

$$\begin{aligned} \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(u^k,\Delta u^j)&\le \frac{4\mathcal {Q}_\delta ^2}{\eta } \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(u^k,u^j)+ \frac{\eta }{4\mathcal {Q}_{\delta }}\sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\Delta u^k,\Delta u^j),\end{aligned}$$

(B.1)

$$\begin{aligned} 2\sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\nabla u^k,\nabla \Delta u^j)&\le \frac{4\mathcal {Q}_{\delta }}{\eta }\sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\nabla u^k,\nabla u^j)\nonumber \\&\quad +\eta \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\nabla \Delta u^k,\nabla \Delta u^j). \end{aligned}$$

(B.2)

It follows from the summation-by-parts formula (2.5) that

$$(u^k,\Delta u^j)=-(\nabla u^k,\nabla u^j), (\nabla u^k,\nabla \Delta u^j)=-(\Delta u^k,\Delta u^j).$$

Thus, inserting (B.1) into (B.2), we arrive at

$$\begin{aligned} \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\Delta u^k,\Delta u^j)\le \frac{16\mathcal {Q}_{\delta }^3}{\eta ^2} \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(u^k, u^j)+ \eta \sum _{k=1}^n\sum _{j=1}^k\theta ^{(k)}_{k-j}(\nabla \Delta u^k,\nabla \Delta u^j). \end{aligned}$$

(B.3)

The proof is completed.