First-order scalar-auxiliary-variable Cahn–Hilliard step

Use either Cahn–Hilliard realization of No-flux finite-volume and conforming realizations for phase fields. Write $X_h={\mathbb{R}}^{\mathcal{K}}$ for the finite-volume state space and $X_h=V_h$ for the conforming interval state space; use the specified inner product $⟨\cdot ,\cdot {⟩}_h$ in each case. Set $E_{1,h}=E_{c,h}-E_{e,h}$. Since $F_{\epsilon }(s)=(s^2-1{)}^2\mathrm{/}(4{\epsilon }^2)\ge 0$, this is nonnegative. Explicitly, $E_{1,h}(u)={\sum }_{K\in \mathcal{K}}\mathrm{\mid }K\mathrm{\mid }{F}_{\epsilon }(u_K)\text{in the finite-volume realization},$ and $E_{1,h}(u_h)={\int }_0^L{F}_{\epsilon }(u_h(x))dx\text{in the conforming interval realization}\mathrm{.}$ Its Riesz gradient $g_{1,h}(u)={\mathrm{\nabla }}_h{E}_{1,h}(u)$ is specified in the finite-volume realization by $(g_{1,h}(u){)}_K=f_{\epsilon }(u_K)$, where $f_{\epsilon }=F_{\epsilon }^{\mathrm{\prime }}$. In the conforming interval realization it is the unique $g_{1,h}(u_h)\in V_h$ satisfying $⟨g_{1,h}(u),v_h{⟩}_h={\int }_0^L{f}_{\epsilon }(u_h(x))v_h(x)dx(v_h\in V_h),$ with the exact $L^2$ inner product from that realization.

Inputs: $u^n\in X_h$, $r^n\in \mathbb{R}$, and parameters $\epsilon ,\kappa ,M,\tau ,C_0>0$. Set ${\alpha }_n=(E_{1,h}(u^n)+C_0{)}^{-1\mathrm{/}2}$. Output: the tuple $(u^{n+1},r^{n+1},{\mu }^{n+1})$ obtained by solving the linear equations $\frac{u^{n+1}-u^n}{\tau }=-M{A}_h{\mu }^{n+1},$ ${\mu }^{n+1}=\kappa A_h{u}^{n+1}+{\alpha }_n{r}^{n+1}{g}_{1,h}(u^n),$ $r^{n+1}-r^n=\frac{{\alpha }_n}{2}⟨g_{1,h}(u^n),u^{n+1}-u^n{⟩}_h\mathrm{.}$ The initial scalar may be set to $r^0=\sqrt{E_{1,h}(u^0)+C_0}$. After a step, $r^{n+1}$ is an independent auxiliary variable determined by the third equation; the scheme does not impose $r^{n+1}=\sqrt{E_{1,h}(u^{n+1})+C_0}$.

The separate result The SAV step is uniquely solvable and dissipates a modified energy ↗ proves unique solvability of this linear system and its modified-energy identity.