Phase-field and dissipative gradient-flow computation — Examples

Put $a=(\sqrt{2\kappa }\epsilon {)}^{-1}$. Differentiation gives $u_{\epsilon }^{\mathrm{\prime }}=a(1-u_{\epsilon }^2)$ and $u_{\epsilon }^{\mathrm{\prime }\mathrm{\prime }}=-2a^2{u}_{\epsilon }(1-u_{\epsilon }^2)=2a^2(u_{\epsilon }^3-u_{\epsilon })$. Since $2\kappa a^2={\epsilon }^{-2}$, $-\kappa u_{\epsilon }^{\mathrm{\prime }\mathrm{\prime }}+f_{\epsilon }(u_{\epsilon })=0$ by [F1].

The limits of $u_{\epsilon }$ at $\pm \mathrm{\infty }$ are $\pm 1$. Also $1-u_{\epsilon }^2={\mathrm{sech}}^2(ax)$, so the energy density is $\frac{1}{2{\epsilon }^2}{\mathrm{sech}}^4(ax)$, which is integrable because it decays exponentially at both ends. Hence the profile has finite energy and joins the two wells.

Since $\tanh$ is strictly increasing and $\tanh (\mathrm{artanh}(1\mathrm{/}2))=1\mathrm{/}2$, the endpoints of $\mathrm{\mid }{u}_{\epsilon }\mathrm{\mid }\le 1\mathrm{/}2$ are $x=\pm \sqrt{2\kappa }\epsilon \mathrm{artanh}(1\mathrm{/}2)$. Subtracting them gives $W_{\epsilon }$ as stated.

Place sample points at $(j+\frac{1}{2})h$ for $j\in \mathbb{Z}$. Their distance from the front center is at least $h\mathrm{/}2>W_{\epsilon }\mathrm{/}2$, so every sample point lies outside the transition interval from step 1.1. The grid therefore records no sample with $\mathrm{\mid }{u}_{\epsilon }\mathrm{\mid }\le 1\mathrm{/}2$.

On one cell, [F2] gives $A_h=0$ and ${\mathcal{E}}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(u)=F_1(u)$. Direct substitution using [F1] gives $F_1(1)=0$ and $F_1(0)=1\mathrm{/}4$.

The modified energies of the displayed pairs are ${\tilde{E}}^0=(r^0{)}^2=1$ and ${\tilde{E}}^1=(r^1{)}^2=1\mathrm{/}4$ by [F3], so the modified energy strictly decreases while the physical energy strictly increases by step 1.1. Hence the proposed implication is false. The pairs are not claimed to satisfy the SAV update, so this refutes only the logical implication from modified-energy decay alone.

The initial mean is zero, so mass conservation [F2] forces $u^1=(a,-a)$. Since $A_h(a,-a)=(2a,-2a)$ and $F_c^{\mathrm{\prime }}(s)=s^3+s$, $F_e^{\mathrm{\prime }}(s)=2s$, the first chemical-potential component is ${\mu }_1=2a+(a^3+a)-2=a^3+3a-2$. The first state equation gives $a-1=-2{\mu }_1$, hence $2a^3+7a-5=0$. The polynomial is strictly increasing because its derivative is $6a^2+7>0$, and its values at $0$ and $1$ are $-5$ and $4$; therefore it has exactly one root $a\in (0,1)$.

The initial gradient energy is $\frac{1}{2}⟨A_h{u}^0,u^0{⟩}_h=2$ and both potential terms vanish, so $E_h(u^0)=2$. For $u^1=(a,-a)$ the gradient energy is $2a^2$ and the potential energy is $\frac{1}{2}(1-a^2{)}^2$. Since $0<a<1$, $(1-a^2{)}^2<1-a^2$, hence $E_h(u^1)<2=E_h(u^0)$. This is the exact finite ledger for the step.