Existence, uniqueness, mass conservation, and energy decay of a convex-splitting step

In either realization, the quartic leading part $Q_{4,h}$ of $P_{c,h}$ is continuous and strictly positive on the unit sphere of any fixed norm on $X_h$: it is a positive weighted sum of $u_K^4$ in the finite-volume branch and ${\int }_0^L{v}^4$ in the conforming branch, which is positive for every nonzero state. The sphere is compact, so [F3] gives $c_h:={\min }_{\mathrm{\parallel }u\mathrm{\parallel }=1}{Q}_{4,h}(u)>0$. Homogeneity gives $Q_{4,h}(u)\ge c_h\mathrm{\parallel }u{\mathrm{\parallel }}^4$. The formulas in [F4] show that the remaining quadratic term in $P_{c,h}$ is nonnegative, so $P_{c,h}$ is coercive. The second directional derivatives of the two potential functionals are the corresponding weighted sum or integral of $F_c^{\mathrm{\prime }\mathrm{\prime }}(u)w^2$ and $F_e^{\mathrm{\prime }\mathrm{\prime }}(u)w^2$; [F4] makes them nonnegative, so both are convex. Thus both quartic realizations satisfy the coercivity and convexity assumptions.

The Allen–Cahn objective is continuous and coercive: its positive quadratic term is nonnegative, $P_{c,h}$ is coercive, and the remaining term is linear. For any fixed trial value, a nonempty sublevel set is closed and bounded, hence compact in the finite-dimensional space; [F3] gives a minimizer. The quadratic term $\frac{1}{2M\tau }\mathrm{\parallel }v-u^n{\mathrm{\parallel }}_h^2$ is strictly convex, while the remaining nonlinear and quadratic terms are convex by [F2] and the hypotheses, so the minimizer is unique.

Let $a=u^{n+1}$, $b=u^n$, and $\delta =a-b$. Convexity gives $P_{c,h}(a)-P_{c,h}(b)\le ⟨{\mathrm{\nabla }}_h{P}_{c,h}(a),\delta {⟩}_h$ and $-P_{e,h}(a)+P_{e,h}(b)\le -⟨{\mathrm{\nabla }}_h{P}_{e,h}(b),\delta {⟩}_h$. Also $\frac{\kappa }{2}(⟨A_{h}a,a{⟩}_h-⟨A_{h}b,b{⟩}_h)=\kappa ⟨A_{h}a,\delta {⟩}_h-\frac{\kappa }{2}⟨A_h\delta ,\delta {⟩}_h\le \kappa ⟨A_{h}a,\delta {⟩}_h$ by nonnegativity in [F2]. Adding these inequalities gives ${\mathcal{E}}_h(a)-{\mathcal{E}}_h(b)\le ⟨{\mu }^{n+1},\delta {⟩}_h$.

Differentiating $J_{AC}$ in an arbitrary direction $w$ gives $⟨(u^{n+1}-u^n)\mathrm{/}(M\tau )+\kappa A_h{u}^{n+1}+{\mathrm{\nabla }}_h{P}_{c,h}(u^{n+1})-{\mathrm{\nabla }}_h{P}_{e,h}(u^n),w{⟩}_h=0$. Since this holds for every $w$, the vector inside the pairing is zero; multiplying by $-M$ gives the stated Allen–Cahn recurrence.

The Cahn–Hilliard affine set is nonempty because it contains $u^n$. Its objective is continuous and coercive on that affine set by coercivity of $P_{c,h}$ and nonnegativity of the $A_h$ quadratic term. The inverse quadratic term is strictly positive on every nonzero mean-zero direction by [F2], so $J_{CH}$ is strictly convex on the affine set. The compact-sublevel argument of step 1.2 and [F3] therefore gives a unique minimizer.

At the Cahn–Hilliard minimizer, the directional derivative vanishes for every mean-zero direction $w$. Thus ${\mu }^{n+1}+(M\tau {)}^{-1}{A}_h^{†}(u^{n+1}-u^n)$ is orthogonal to the mean-zero subspace and is therefore constant. Applying $A_h$ and using [F2] gives $(u^{n+1}-u^n)\mathrm{/}\tau =-M{A}_h{\mu }^{n+1}$. The affine constraint itself gives $⟨u^{n+1},\mathbf{1}{⟩}_h=⟨u^n,\mathbf{1}{⟩}_h$.

In the Allen–Cahn branch, step 1.2 gives $\delta =-M\tau {\mu }^{n+1}$, so step 1.3 yields ${\mathcal{E}}_h(a)-{\mathcal{E}}_h(b)\le -M\tau \mathrm{\parallel }{\mu }^{n+1}{\mathrm{\parallel }}_h^2$. In the Cahn–Hilliard branch, step 3.1 gives $\delta =-M\tau A_h{\mu }^{n+1}$, so $⟨{\mu }^{n+1},\delta {⟩}_h=-M\tau ⟨A_h{\mu }^{n+1},{\mu }^{n+1}{⟩}_h\le 0$. These are the two asserted energy inequalities.