On the maximum principle and high-order, delay-free integrators for the viscous Cahn–Hilliard equation

Abstract

The stabilization approach has been known to permit large time-step sizes while maintaining stability. However, it may “slow down the convergence rate” or cause “delayed convergence” if the time-step rescaling is not well resolved. By considering a fourth-order-in-space viscous Cahn–Hilliard (VCH) equation, we propose a class of up to the fourth-order single-step methods that are able to capture the correct physical behaviors with high-order accuracy and without time delay. By reformulating the VCH as a system consisting of a second-order diffusion term and a nonlinear term involving the operator $({I} - \nu \Delta )^{-1}$, we first develop a general approach to estimate the maximum bound for the VCH equation equipped with either the Ginzburg–Landau or Flory–Huggins potential. Then, by taking advantage of new recursive approximations and adopting a time-step-dependent stabilization, we propose a class of stabilization Runge–Kutta methods that preserve the maximum principle for any time-step size without harming the convergence. Finally, we transform the stabilization method into a parametric Runge–Kutta formulation, estimate the rescaled time-step, and remove the time delay by means of a relaxation technique. When the stabilization parameter is chosen suitably, the proposed parametric relaxation integrators are rigorously proven to be mass-conserving, maximum-principle-preserving, and the convergence in the $l^\infty$-norm is estimated with pth-order accuracy under mild regularity assumption. Numerical experiments on multi-dimensional benchmark problems are carried out to demonstrate the stability, accuracy, and structure-preserving properties of the proposed schemes.

Appendices

Appendix 1. Proof of Theorem 2.6

Proof

Consider the Ginzburg–Landau function (10) as an example. For a given $u^0 \in X$ and $t_0 > 0$, we denote $X_{\beta } = \{ v \in X | \Vert v\Vert _{L^\infty } \le \beta \}$ and $C([0, t_0]; X_{\beta }) = \{v: [0, t_0] \rightarrow X_{\beta } | v \text {~is~continuous}\}$. Note that the forward Euler condition (17) is equivalent to the circle condition,

$$\begin{aligned} \Vert \kappa u + {N}(u)\Vert _{L^\infty } \le \kappa \beta , \quad \forall \kappa \ge \frac{1}{\tilde{\tau }_{FE}}, \forall \Vert u\Vert _{L^\infty } \le \beta . \end{aligned}$$

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Letting $\kappa \ge \frac{1}{\tilde{\tau }_{FE}}$, for a given $v \in C([0, t_0]; X_{\beta })$, we define $w: [0, t_0] \rightarrow X$ as the solution to the system

$$\begin{aligned} \left\{ \begin{aligned} w_t&= {L} w - \kappa w + {N}(v) + \kappa v =: {L}_\kappa w + {N}_\kappa (v), \quad t \in [0, t_0], \\ w(0)&= u^0. \end{aligned}\right. \end{aligned}$$

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Then, w is uniquely defined because of the linearity of (70). By Duhamel’s formula, we have

$$\begin{aligned} w(t) = \text {e}^{{L}_\kappa t} u^0 + \int _0^t \text {e}^{{L}_\kappa (t - s)}{N}_\kappa ( v(s)) \text {d} s, \quad t \in [0, t_0]. \end{aligned}$$

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Taking $\Vert \cdot \Vert _{L^\infty }$-norm on both sides of (71) and applying the circle condition (69) yields

$$\begin{aligned} \Vert w(t)\Vert _{L^\infty } \le \text {e}^{-\kappa t} \Vert u^0\Vert _{L^\infty } + \!\int _0^t \text {e}^{-\kappa (t - s)} \Vert {N}_\kappa (v(s))\Vert _{L^\infty } \text {d} s \!\le \! \text {e}^{-\kappa t} \beta + \!\int _0^t \text {e}^{-\kappa (t - s)} \kappa \beta \text {d} s \!=\! \beta , \quad \!t \!\in \! [0, t_0]. \end{aligned}$$

Therefore, $w \in C([0, t_0]; X_{\beta })$. Next by defining a mapping ${M}: C([0, t_0]; X_{\beta }) \rightarrow C([0, t_0]; X_{\beta })$ as ${M}(v) = w$ through (70), we show that ${M}$ is a contraction for sufficiently small $t_0$. Assuming that $v_1, v_2 \in C([0, t_0]; X_{\beta })$, $w_1 = {M}(v_1)$ and $w_2 = {M} (v_2)$, we then obtain

$$\begin{aligned} w_1(t) - w_2(t) = \int _0^t \text {e}^{{L}_\kappa (t - s)}[ {N}_\kappa (v_1(s)) - {N}_\kappa (v_2(s))] \text {d} s. \end{aligned}$$

Since ${N}(u)$ satisfies the Lipschitz condition (20), we derive

$$\begin{aligned} \begin{aligned} \Vert {N}_\kappa (v_1(s)) \!-\! {N}_\kappa (v_2(s))\Vert _{L^\infty }&\!\le \! \Vert {N}( v_1(s)) \!-\! {N}(v_2(s))\Vert _{L^\infty } +\! \Vert \kappa v_1(s) - \kappa v_2(s)\Vert _{L^\infty } \\&\le (l_N + \kappa ) \Vert v_1(s) - v_2(s)\Vert _{L^\infty }. \end{aligned} \end{aligned}$$

Noting that ${L} = \frac{\epsilon ^2}{\nu } \Delta$ is the generator of a contraction semigroup with respect to the supremum norm on X [20], then it holds that

$$\begin{aligned} \begin{aligned} \Vert w_1(t) - w_2(t)\Vert _{L^\infty }&\le \int _0^{t} \text {e}^{-\kappa (t - s)} (l_N +\kappa ) \Vert v_1(s) - v_2(s)\Vert _{L^\infty } \text {d} s \\&\le \frac{l_N + \kappa }{\kappa } (1 - \text {e}^{-\kappa t_0})\Vert v_1 - v_2\Vert _{C([0, t_0]; X)}, \quad \forall t \in [0, t_0]. \end{aligned} \end{aligned}$$

If $t_0 < \frac{1}{\kappa }\ln \frac{l_N + \kappa }{l_N}$, we have $\frac{l_N + \kappa }{\kappa }(1 - \text {e}^{-\kappa t_0}) < 1$, and

$$\begin{aligned} \Vert {M}(v_1) - {M}(v_2)\Vert _{C([0, t_0]; X)} < \Vert v_1 - v_2\Vert _{C([0, t_0]; X)}, \end{aligned}$$

and then ${M}$ is a contraction. Since $X_{\beta }$ is closed in X, $C([0,t_0]; X_{\beta })$ is thus complete with respect to the metric induced by the norm $\Vert \cdot \Vert _{C([0, t_0]; X)}$, and then Banach’s fixed point theorem gives a unique fixed point $u \in C([0, t_0]; X_{\beta })$ of ${M}(u) = u$, which is the unique solution to Eq. (8). Continuing the process gives global existence of the unique solution $u \in C([0, T]; X_{\beta })$. $\square$

Appendix 2. Some non-negative RK Butcher tableaux

$$\begin{aligned}{} & {} \mathrm {RK(1, 1):} \begin{array}{c|c} 0 &{} 0 \\ \hline 1 &{} 1 \end{array}, \quad \quad{} & {} \mathrm {RK(2, 2):} \begin{array}{c|cc} 0 &{} 0 &{} 0\\ 1 &{} 1 &{} 0\\ \hline 1 &{} \frac{1}{2} &{} \frac{1}{2} \end{array}, \quad \quad \quad \\ \nonumber{} & {} \text {Forward Euler scheme, {C} = 1}{} & {} \text {Heun's second-order scheme [63], {C} = 1} \end{aligned}$$

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$$\begin{aligned} \begin{aligned}&\quad \begin{array} {c|ccc} 0 &{} 0 &{} 0 &{} 0 \\ 1 &{} 1 &{} 0 &{} 0 \\ \frac{1}{2} &{} \frac{1}{4} &{} \frac{1}{4} &{} 0\\ \hline 1 &{} \frac{1}{6} &{} \frac{1}{6} &{} \frac{2}{3} \end{array},&\begin{array}{c|cccc} 0 &{} 0 &{} 0 &{} 0 &{} 0\\ \frac{1}{2} &{} \frac{1}{2} &{} 0 &{} 0 &{} 0\\ \frac{1}{2} &{} 0 &{} \frac{1}{2} &{} 0 &{} 0 \\ 1 &{} 0 &{} 0&{} 1 &{} 0 \\ \hline 1 &{} \frac{1}{6} &{} \frac{1}{3}&{} \frac{1}{3} &{} \frac{1}{6} \end{array}, \quad \quad \quad \quad \\&\text {RK(3,3) [63], {C} = 1} \quad \quad&\text {RK(4, 4)}, {C} = 0 \quad \quad \quad \quad \end{aligned} \end{aligned}$$

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RK(5, 4)[65], ${C} \approx 1.508$, $c \approx [0, 0.3918, 0.5861, 0.4745, 0.9350]^T$:

$$\begin{aligned} \left\{ \begin{aligned} u_{n, 1}&= u_{n, 0} + 0.391752226571890 \tau g(u_{n, 0}), \\ u_{n, 2}&= 0.444370493651235 u_{n, 0} + 0.555629506348765 u_{n, 1} + 0.368410593050371 \tau g(u_{n, 1}), \\ u_{n, 3}&= 0.620101851488403 u_{n, 0} + 0.379898148511597u_{n, 2} + 0.251891774271694 \tau g(u_{n, 2}), \\ u_{n, 4}&= 0.178079954393132 u_{n, 0} + 0.821920045606868 u_{n, 3} + 0.544974750228521 \tau g(u_{n, 3}), \\ u^{n+1}&= 0.517231671970585 u_{n, 2} + 0.096059710526147 u_{n, 3} + 0.063692468666290 \tau g(u_{n, 3}) \\&\quad + 0.386708617503268 u_{n, 4} + 0.226007483236906 \tau g(u_{n, 4}). \end{aligned}\right. \end{aligned}$$

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Appendix 3. Third- and fourth-order RK Butcher tableaux with non-decreasing parametric abscissas

RK$^+(3, 3)$ [36], ${C} = \frac{3}{4}$, $c = [0, \frac{2}{3}, \frac{2}{3}]^T$:

$$\begin{aligned} \left\{ \begin{aligned} u_{n, 1}&= u_{n, 0} + \frac{2}{3}\tau g(u_{n, 0}), \\ u_{n, 2}&= \frac{2}{3} u_{n, 0} + \frac{1}{3}[u_{n, 1} + \frac{4}{3} \tau g(u_{n, 1})], \\ u_{n, 3}&= \frac{59}{128} u_{n, 0} + \frac{15}{128}[u_{n, 0} + \frac{4}{3} \tau g(u_{n, 0})] + \frac{27}{64}[u_{n, 2} + \frac{4}{3} \tau g(u_{n, 2})]. \end{aligned} \right. \end{aligned}$$

RK$^+(5, 4)$[36], ${C} \!=\! 1.346586417284006$, $c \approx \! [0, 0.4549, 0.5165, 0.5165, 0.9903]^T$:

$$\begin{aligned} \left\{ \begin{aligned} u_{n, 1}&= 0.387392167970373 u_{n, 0} + 0.612607832029627[ u_{n, 0} + \frac{\tau }{{C}} g(u_{n, 0})], \\ u_{n, 2}&= 0.568702484115635 u_{n, 0} + 0.431297515884365[ u_{n, 1} + \frac{\tau }{{C}} g(u_{n, 1})], \\ u_{n, 3}&= 0.589791736452092 u_{n, 0} + 0.410208263547908[u_{n, 2} + \frac{\tau }{{C}} g(u_{n, 2})], \\ u_{n, 4}&= 0.213474206786188 u_{n, 0} + 0.786525793213812[u_{n, 3} + \frac{\tau }{{C}} g(u_{n, 3})], \\ u^{n+1}&= 0.270147144537063 u_{n, 0} + 0.029337521506634[u_{n, 0} + \frac{\tau }{{C}} g(u_{n, 0})] \\&\quad + 0.239419175840559[u_{n, 1} + \frac{\tau }{{C}} g(u_{n, 1})] + 0.227000995504038[ u_{n, 3} \\&\quad + \frac{\tau }{{C}} g(u_{n, 3})] + 0.234095162611706[u_{n, 4} + \frac{\tau }{{C}} g(u_{n, 4})]. \end{aligned} \right. \end{aligned}$$

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