Cancellation of capillary and transport energy exchange

Put $E_{\varphi }(t):={\int }_{{\mathbb{T}}^d}(\epsilon \mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2\mathrm{/}2+{\epsilon }^{-1}W(\varphi ))dx$. At fixed $t\in (0,T)$, set $q_h(x):=[\varphi (t+h,x)-\varphi (t,x)]\mathrm{/}h$. By [A1], $q_h\to {\mathrm{\partial }}_t\varphi (t,\cdot )$ and $[\mathrm{\nabla }\varphi (t+h,\cdot )-\mathrm{\nabla }\varphi (t,\cdot )]\mathrm{/}h\to \mathrm{\nabla }{\mathrm{\partial }}_t\varphi (t,\cdot )$ uniformly in $x$. The identity $\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }b{\mathrm{\mid }}^2=(a+b)\cdot (a-b)$ therefore makes the difference quotients of $\epsilon \mathrm{\mid }\mathrm{\nabla }\varphi {\mathrm{\mid }}^2\mathrm{/}2$ converge uniformly to $\epsilon \mathrm{\nabla }\varphi \cdot \mathrm{\nabla }{\mathrm{\partial }}_t\varphi$. Set $R:=\mathrm{\parallel }\varphi (t,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}+1$ and $J:=[-R,R]$. For sufficiently small $h$, both $\varphi (t,\cdot )$ and $\varphi (t+h,\cdot )$ take values in $J$. This closed bounded interval is compact and has the finite-subcover property by [F5]. The function $W^{\mathrm{\prime }}$ is continuous on $J$. For any $\eta >0$, form the family of intervals $(z-r\mathrm{/}2,z+r\mathrm{/}2)$ over all pairs $z\in J$, $r>0$ such that $\mathrm{\mid }{W}^{\mathrm{\prime }}(y)-W^{\mathrm{\prime }}(z)\mathrm{\mid }<\eta \mathrm{/}2$ for every $y\in J$ with $\mathrm{\mid }y-z\mathrm{\mid }<r$. Continuity of $W^{\mathrm{\prime }}$ makes this family a cover of $J$; compactness [F5] gives a finite subcover with pairs $(z_i,r_i)$, and $\delta :={\min }_i{r}_i\mathrm{/}2>0$. If $y,y^{\mathrm{\prime }}\in J$ and $\mathrm{\mid }y-y^{\mathrm{\prime }}\mathrm{\mid }<\delta$, choose $i$ with $\mathrm{\mid }y-z_i\mathrm{\mid }<r_i\mathrm{/}2$; then $\mathrm{\mid }{y}^{\mathrm{\prime }}-z_i\mathrm{\mid }<r_i$, so both values of $W^{\mathrm{\prime }}$ differ from $W^{\mathrm{\prime }}(z_i)$ by less than $\eta \mathrm{/}2$ and hence $\mathrm{\mid }{W}^{\mathrm{\prime }}(y)-W^{\mathrm{\prime }}(y^{\mathrm{\prime }})\mathrm{\mid }<\eta$. Thus $W^{\mathrm{\prime }}$ is uniformly continuous on $J$. Taking $\eta =1$ in the finite-cover argument also bounds $W^{\mathrm{\prime }}$ there: each $y\in J$ lies within $r_i\mathrm{/}2$ of one of finitely many centers $z_i$, giving $\mathrm{\mid }{W}^{\mathrm{\prime }}(y)\mathrm{\mid }<\mathrm{\mid }{W}^{\mathrm{\prime }}(z_i)\mathrm{\mid }+1\mathrm{/}2$. For each $x$ with unequal values, [F4] gives some point $\xi$ between $\varphi (t,x)$ and $\varphi (t+h,x)$ such that the potential difference quotient is $W^{\mathrm{\prime }}(\xi )q_h(x)$; when the values agree the quotient and $q_h(x)$ are both zero. Every such $\xi$ satisfies $\mathrm{\mid }\xi -\varphi (t,x)\mathrm{\mid }\le \mathrm{\parallel }\varphi (t+h,\cdot )-\varphi (t,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }}$. Define $\omega (\rho ):=\sup \{\mathrm{\mid }{W}^{\mathrm{\prime }}(a)-W^{\mathrm{\prime }}(b)\mathrm{\mid }:a,b\in J,\mathrm{\mid }a-b\mathrm{\mid }\le \rho \}$; uniform continuity gives $\omega (\rho )\to 0$ as $\rho \downarrow 0$. Thus the supremum of the difference between the potential quotient and $W^{\mathrm{\prime }}(\varphi )q_h$ is at most $\mathrm{\parallel }{q}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\omega (\mathrm{\parallel }\varphi (t+h,\cdot )-\varphi (t,\cdot ){\mathrm{\parallel }}_{\mathrm{\infty }})\to 0$, and boundedness of $W^{\mathrm{\prime }}$ on $J$ then gives $W^{\mathrm{\prime }}(\varphi )q_h\to W^{\mathrm{\prime }}(\varphi ){\mathrm{\partial }}_t\varphi$ uniformly. These estimates apply to every intermediate point supplied by [F4], so no pointwise selection is needed. Thus the full energy-density difference quotient converges uniformly to $\epsilon \mathrm{\nabla }\varphi \cdot \mathrm{\nabla }{\mathrm{\partial }}_t\varphi +{\epsilon }^{-1}{W}^{\mathrm{\prime }}(\varphi ){\mathrm{\partial }}_t\varphi$. The Riemann integral over $Q_d$ is bounded in absolute value by $(2\pi {)}^d$ times the supremum norm, so differentiation passes through the integral and gives $E_{\varphi }^{\mathrm{\prime }}(t)={\int }_{{\mathbb{T}}^d}(\epsilon \mathrm{\nabla }\varphi \cdot \mathrm{\nabla }{\mathrm{\partial }}_t\varphi +{\epsilon }^{-1}{W}^{\mathrm{\prime }}(\varphi ){\mathrm{\partial }}_t\varphi )dx$.

Put $K(t):=\frac{1}{2}{\int }_{{\mathbb{T}}^d}\mathrm{\mid }u{\mathrm{\mid }}^{2}dx$. The $C^1$ time regularity in [A1] makes $(u(t+h)+u(t))\cdot (u(t+h)-u(t))\mathrm{/}(2h)$ converge uniformly in space to $u\cdot {\mathrm{\partial }}_{t}u$, so differentiation passes through the Riemann integral by its supremum-norm bound and $K^{\mathrm{\prime }}(t)={\int }_{{\mathbb{T}}^d}u\cdot {\mathrm{\partial }}_{t}udx$. Substitute the momentum equation from [F1] to obtain $K^{\mathrm{\prime }}=-\int u\cdot ((u\cdot \mathrm{\nabla })u)dx-\int u\cdot \mathrm{\nabla }\pi dx+\nu \int u\cdot \mathrm{\Delta }udx-\int \varphi u\cdot \mathrm{\nabla }{\mu }_{c}dx$.

Apply [F2] with scalar $g={\mathrm{\partial }}_t\varphi$ and vector $F=\epsilon \mathrm{\nabla }\varphi$. Since $\mathrm{div}(\epsilon \mathrm{\nabla }\varphi )=\epsilon \mathrm{\Delta }\varphi$, step 1.1 becomes $E_{\varphi }^{\mathrm{\prime }}(t)={\int }_{{\mathbb{T}}^d}{\mu }_c{\mathrm{\partial }}_t\varphi dx$ by the chemical-potential definition [F3].

For each $i,j$, the factorization $\frac{u_i(x+h{e}_j{)}^2-u_i(x{)}^2}{2h}=\frac{u_i(x+h{e}_j)+u_i(x)}{2}\frac{u_i(x+h{e}_j)-u_i(x)}{h}$ and taking $h\to 0$, differentiability gives ${\mathrm{\partial }}_j(u_i^2\mathrm{/}2)=u_i{\mathrm{\partial }}_j{u}_i$; summing the finite identities yields $u\cdot ((u\cdot \mathrm{\nabla })u)=u\cdot \mathrm{\nabla }(\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2)$. By [F2], this term integrates to $-\int (\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2)\mathrm{div}udx=0$; applying [F2] with $g=\pi$ and $F=u$ gives the pressure contribution $-\int u\cdot \mathrm{\nabla }\pi dx=\int \pi \mathrm{div}udx=0$ by [F1]. Applying [F2] componentwise with $g=u_i$, $F=\mathrm{\nabla }{u}_i$ gives $\int u\cdot \mathrm{\Delta }udx=-\int \mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^{2}dx$. Therefore step 1.2 yields $K^{\mathrm{\prime }}(t)=-\nu {\int }_{{\mathbb{T}}^d}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^{2}dx-{\int }_{{\mathbb{T}}^d}\varphi u\cdot \mathrm{\nabla }{\mu }_{c}dx$.

Substitute the phase equation from [F1] into step 2.1. The transport contribution is $-{\int }_{{\mathbb{T}}^d}{\mu }_{c}u\cdot \mathrm{\nabla }\varphi dx$, and [F2] with $g={\mu }_c$, $F=\mathrm{\nabla }{\mu }_c$ gives ${\int }_{{\mathbb{T}}^d}{\mu }_c\mathrm{\Delta }{\mu }_{c}dx=-{\int }_{{\mathbb{T}}^d}\mathrm{\mid }\mathrm{\nabla }{\mu }_c{\mathrm{\mid }}^{2}dx$. Hence $E_{\varphi }^{\mathrm{\prime }}(t)=-{\int }_{{\mathbb{T}}^d}{\mu }_{c}u\cdot \mathrm{\nabla }\varphi dx-M{\int }_{{\mathbb{T}}^d}\mathrm{\mid }\mathrm{\nabla }{\mu }_c{\mathrm{\mid }}^{2}dx$.

For each component, the difference quotient of ${\mu }_c{u}_j$ can be written as $\frac{{\mu }_c(x+h{e}_j)-{\mu }_c(x)}{h}{u}_j(x+h{e}_j)+{\mu }_c(x)\frac{u_j(x+h{e}_j)-u_j(x)}{h}$; taking $h\to 0$ and summing over $j$ proves $\mathrm{div}({\mu }_{c}u)=u\cdot \mathrm{\nabla }{\mu }_c+{\mu }_c\mathrm{div}u=u\cdot \mathrm{\nabla }{\mu }_c$ by [F1]. The stated regularity makes $\varphi$ and ${\mu }_{c}u$ periodic $C^1$ fields, so [F2] applies with $g=\varphi$ and $F={\mu }_{c}u$ and gives ${\int }_{{\mathbb{T}}^d}{\mu }_{c}u\cdot \mathrm{\nabla }\varphi dx=-{\int }_{{\mathbb{T}}^d}\varphi u\cdot \mathrm{\nabla }{\mu }_{c}dx$. Consequently the transport term in step 3.1 cancels the capillary-work term in step 2.2.

Since $E=K+E_{\varphi }$, adding steps 3.1 and 2.2 and using the cancellation in step 4.1 gives the asserted identity for every $t\in (0,T)$.