One-dimensional mesh and time convergence of convex-splitting phase flows

For $v\in H^1(0,L)$, choose a point $y$ with $\mathrm{\mid }v(y)\mathrm{\mid }\le L^{-1\mathrm{/}2}\mathrm{\parallel }v{\mathrm{\parallel }}_2$; such a point exists since $v$ has a continuous representative. For every $x$, [F8] and Cauchy–Schwarz give $\mathrm{\mid }v(x)\mathrm{\mid }\le \mathrm{\mid }v(y)\mathrm{\mid }+{\int }_0^L\mathrm{\mid }{v}^{\mathrm{\prime }}\mathrm{\mid }\le L^{-1\mathrm{/}2}\mathrm{\parallel }v{\mathrm{\parallel }}_2+\sqrt{L}\mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Thus $H^1(0,L)$ embeds continuously in $L^{\mathrm{\infty }}(0,L)$.

By [F2], $u_h^0=P_h{u}_0$ is uniformly bounded in $H^1$; step 1.1 bounds it in $L^{\mathrm{\infty }}$, so its gradient and quartic terms in [F1] give a uniform initial-energy bound. The energy inequalities in [F3] then give ${\sup }_{h,\tau ,n}\mathrm{\parallel }{u}_h^n{\mathrm{\parallel }}_{H^1}<\mathrm{\infty }$: the gradient term controls $\mathrm{\parallel }{u}_h^{\mathrm{\prime }}{\mathrm{\parallel }}_2$, and the algebraic identity $F_{\epsilon }(s)-(\frac{s^4}{8{\epsilon }^2}-\frac{1}{4{\epsilon }^2})=\frac{(s^2-2{)}^2}{8{\epsilon }^2}\ge 0$ controls $\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_4$ and, by [F8], $\mathrm{\parallel }{u}_h{\mathrm{\parallel }}_2$. Step 1.1 gives ${\sup }_{h,\tau ,n}\mathrm{\parallel }{u}_h^n{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. For Allen–Cahn, the update in [F9] gives $D_{\tau }{u}_h^n=-M{\mu }_h^{n+1}$; hence the summed dissipation in [F3] and $\mathrm{\parallel }w{\mathrm{\parallel }}_{V^{\ast }}\le \mathrm{\parallel }w{\mathrm{\parallel }}_2$ give $\mathrm{\parallel }{D}_{\tau }{u}_{h,\tau }{\mathrm{\parallel }}_{L^2(0,T;V^{\ast })}\le C$. For Cahn–Hilliard, summing [F3] bounds ${\mathrm{\partial }}_x{\mu }_{h,\tau }$ in $L^2$. Its mean is ${\bar{\mu }}_h^{n+1}=L^{-1}{\int }_0^L(F_c^{\mathrm{\prime }}(u_h^{n+1})-F_e^{\mathrm{\prime }}(u_h^n))dx,$ because ${\int }_0^L{A}_h{u}_h^{n+1}dx=0$ and the Riesz gradients in [F9] pair with $1$ as derivatives of the integral potentials in [F1]; this mean is uniformly bounded by the $L^{\mathrm{\infty }}$ bound. For $g={\mu }_h^{n+1}-{\bar{\mu }}_h^{n+1}$, [F8] and Cauchy--Schwarz give $\mathrm{\mid }g(x)\mathrm{\mid }=\mathrm{\mid }{L}^{-1}{\int }_0^L(g(x)-g(y))dy\mathrm{\mid }\le {\int }_0^L\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }\le \sqrt{L}\mathrm{\parallel }{g}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$, so $\mathrm{\parallel }g{\mathrm{\parallel }}_2\le L\mathrm{\parallel }{\mathrm{\partial }}_x{\mu }_h^{n+1}{\mathrm{\parallel }}_2$. Thus ${\mu }_{h,\tau }$ is bounded in $L^2(0,T;H^1)$. For each $v\in V$, the CH update in [F9] gives $D_{\tau }{u}_h^n=-M{A}_h{\mu }_h^{n+1}\in V_h$ and orthogonality gives $(D_{\tau }{u}_h^n,v{)}_{L^2}=(D_{\tau }{u}_h^n,P_{h}v{)}_{L^2}=-M{\int }_0^L({\mu }_h^{n+1}{)}_x(P_{h}v{)}_{x}dx$. The uniform H1 stability of $P_h$ in [F2] therefore bounds the discrete derivative in $L^2(0,T;V^{\ast })$.

The piecewise-constant reconstructions take values in $V=H^1(0,L)$ and have uniformly bounded $L^2(0,T;V)$ norms by step 2.1. The interval embedding in [F5] supplies $V⋐H$, and [F4] supplies the evolution triple. The time steps tend to zero and step 2.1 supplies the discrete-derivative bound, so [F6] gives a subsequence converging strongly in $L^2(0,T;L^2)$. Applying [F7] to the same bounded $L^2(0,T;H^1)$ sequence and taking a further subsequence gives weak $L^2(H^1)$ convergence to the same $u$, since the strong $L^2$ limit is unique. The Cahn–Hilliard chemical potentials are bounded in $L^2(H^1)$ by step 2.1, so [F7] also gives a weak limit $\mu$.

Let $w_{h,\tau }(t)=u_h^{n+1}-u_h^n$ on $t\in (t_n,t_{n+1})$. Then $w_{h,\tau }=\tau D_{\tau }{u}_{h,\tau }$, so step 2.1 gives $\mathrm{\parallel }{w}_{h,\tau }{\mathrm{\parallel }}_{L^2(0,T;V^{\ast })}\le C\tau$; the uniform H1 bound gives $\mathrm{\parallel }{w}_{h,\tau }{\mathrm{\parallel }}_{L^2(0,T;V)}\le C$. For every $w\in V$, the evolution-triple pairing and Cauchy--Schwarz give $\mathrm{\parallel }w{\mathrm{\parallel }}_H^2=⟨w,w{⟩}_{V^{\ast },V}\le \mathrm{\parallel }w{\mathrm{\parallel }}_{V^{\ast }}\mathrm{\parallel }w{\mathrm{\parallel }}_V$. Integrating and applying Cauchy--Schwarz in time shows $\mathrm{\parallel }{w}_{h,\tau }{\mathrm{\parallel }}_{L^2(0,T;H)}^2\le C\tau$, so adjacent time-level reconstructions differ by a term tending to zero in $L^2((0,T)\times (0,L))$. By step 1.1 both adjacent states are uniformly bounded in $L^{\mathrm{\infty }}$; the polynomial split derivatives $F_c^{\mathrm{\prime }}$ and $F_e^{\mathrm{\prime }}$ are Lipschitz on this common bounded range. Therefore their values at adjacent levels differ by a term tending to zero in $L^2$, and strong convergence from step 3.1 implies convergence of each split term, and hence $f_{\epsilon }$, strongly in $L^2$ to its value at $u$.

Let $\varphi \in C^1([0,T]\times [0,L])$ vanish at $T$. The sets $\{\varphi (t):0\le t\le T\}$ and $\{{\mathrm{\partial }}_t\varphi (t):0\le t\le T\}$ are compact in $H^1$ and $L^2$, respectively, by continuity in those norms. Uniform operator bounds and strong convergence of $P_h$ in [F2] imply uniform projector convergence on each compact set: cover the set by finitely many balls of radius $\delta$, use convergence at their centers, and bound the errors inside each ball by the uniform operator norm plus $1$. Uniform continuity of ${\mathrm{\partial }}_t\varphi$ then shows the projected time difference quotients converge in $L^2$ to ${\mathrm{\partial }}_t\varphi$. Use $P_h\varphi (t_{n+1})$ as the spatial test in the discrete equation [F9] and sum by parts in time. The terminal term vanishes exactly because $\varphi (T)=0$; the initial term is $(P_h{u}_0,P_h\varphi (t_1))=(u_0,P_h\varphi (t_1))\to (u_0,\varphi (0))$. Strong convergence of the tests, weak convergence of $u_{h,\tau }$ in $L^2(H^1)$ from step 3.1, and strong convergence of the nonlinear terms from step 4.1 permit passage in every sum, yielding the Allen–Cahn identity in the Statement.

In the Cahn–Hilliard branch, apply the same test passage to the first mixed equation [F9]. The weak $L^2(H^1)$ convergence of ${\mu }_{h,\tau }$ from step 3.1 and strong convergence of the tests give $-\int u{\varphi }_t+M\int {\mu }_x{\varphi }_x=\int u_0\varphi (0)$. For any $\psi \in H^1(0,L)$ and $\eta \in L^2(0,T)$, test the second mixed equation [F9] with $P_h\psi$, multiply by $\eta (t)$, and integrate in time. By [F2], $P_h\psi \to \psi$ in $H^1$; weak convergence of ${\mu }_{h,\tau }$ and $u_{h,\tau }$ in $L^2(H^1)$ and strong convergence of the split potential in $L^2$ from step 4.1 pass the identity to the integrated form in the Statement. Finally, the discrete mean is constant by [F3]. Strong convergence in $L^2((0,T)\times (0,L))$ makes the mean functions converge in $L^2(0,T)$ by Cauchy–Schwarz in space, so the limit has that mean for almost every time.