Hydrodynamic fluctuation action and nonequilibrium quasipotential

For $0<u<1$ and sufficiently small nonzero $h$, the first product quotient is $\frac{(u+h)\log (u+h)-u\log u}{h}=u\frac{\log (u+h)-\log u}{h}+\log (u+h)$, which tends to $1+\log u$ by [F3]. With $v=1-u$, the second quotient is $\frac{(v-h)\log (v-h)-v\log v}{h}=v\frac{\log (v-h)-\log v}{h}-\log (v-h)$, which tends to $-1-\log v$ by the same derivative formula. Thus $f^{\mathrm{\prime }}(u)=\log u-\log (1-u)$. By [F4] applied to $\log$ composed with $u\mapsto 1-u$, the derivative of $\log (1-u)$ is $-1\mathrm{/}(1-u)$; [F3] gives the derivative of $\log u$ as $1\mathrm{/}u$. Hence $f^{\mathrm{\prime }\mathrm{\prime }}(u)=1\mathrm{/}u+1\mathrm{/}(1-u)=1\mathrm{/}[u(1-u)]$.

The reversed fields remain smooth, periodic, and in the same density range. Also ${\mathrm{\partial }}_t{\rho }^{\theta }+\mathrm{\nabla }\cdot j^{\theta }=-({\mathrm{\partial }}_t\rho +\mathrm{\nabla }\cdot j)(T-t,x)=0$, so they are admissible for the same action.

By the reversed-path definitions, ${\mathcal{I}}_{[0,T]}({\rho }^{\theta },j^{\theta })=\frac{1}{4}{\int }_0^T{\int }_{{\mathbb{T}}^d}\frac{\mathrm{\mid }-j(T-t,x)+\mathrm{\nabla }\rho (T-t,x){\mathrm{\mid }}^2}{\chi (\rho (T-t,x))}dxdt$. Substitute $s=T-t$: $dt=-ds$ and the limits $t=0,T$ become $s=T,0$, so this equals $\frac{1}{4}{\int }_0^T{\int }_{{\mathbb{T}}^d}\frac{\mathrm{\mid }-j(s,x)+\mathrm{\nabla }\rho (s,x){\mathrm{\mid }}^2}{\chi (\rho (s,x))}dxds$. After renaming $s$ to $t$, subtract this expression from the forward action. At the common point $(t,x)$, the resulting integrand is $\frac{\mathrm{\mid }j+\mathrm{\nabla }\rho {\mathrm{\mid }}^2-\mathrm{\mid }-j+\mathrm{\nabla }\rho {\mathrm{\mid }}^2}{4\chi (\rho )}=\frac{j\cdot \mathrm{\nabla }\rho }{\chi (\rho )}=\mathrm{\nabla }{f}^{\mathrm{\prime }}(\rho )\cdot j$, by quadratic expansion and the coordinatewise chain rule with $f^{\mathrm{\prime }\mathrm{\prime }}(\rho )=1\mathrm{/}\chi (\rho )$. All integrands are continuous on the compact cylinder since $\chi (\rho )\ge \epsilon (1-\epsilon )>0$.

For each coordinate $i$, the integral of ${\mathrm{\partial }}_i(f^{\mathrm{\prime }}(\rho )j_i)$ over $[0,1{]}^d$ is zero because its endpoint values in the $i$th coordinate agree by periodicity. Summing gives ${\int }_{{\mathbb{T}}^d}\mathrm{\nabla }{f}^{\mathrm{\prime }}(\rho )\cdot jdx=-{\int }_{{\mathbb{T}}^d}{f}^{\mathrm{\prime }}(\rho )\mathrm{\nabla }\cdot jdx={\int }_{{\mathbb{T}}^d}{f}^{\mathrm{\prime }}(\rho ){\mathrm{\partial }}_t\rho dx$. By [F4] this last integrand is the time derivative of $f(\rho )$; it is continuous on the compact cylinder, so the fundamental theorem of calculus makes its time difference quotients converge uniformly in space and justifies differentiating the spatial integral. Hence the expression is $\frac{d}{dt}{\int }_{{\mathbb{T}}^d}f(\rho (t,x))dx$.

Integrating the identity from step 2.2 over $t\in [0,T]$ and using the integrand difference from step 2.1 gives the stated endpoint formula.

Since $\rho$ is smooth and periodic, repeated integration by parts gives $(1+4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2{)}^m\widehat{\rho }(k)=\widehat{(1-\mathrm{\Delta }{)}^m\rho }(k)$, so $\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }\le C_m(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$ for every integer $m\ge 1$. The number of lattice points with $\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}=r$ is at most $(2r+1{)}^d-(2r-1{)}^d\le C_d{r}^{d-1}$. Thus for every multi-index $\alpha$, choosing $m>(d+\mathrm{\mid }\alpha \mathrm{\mid })\mathrm{/}2$ makes ${\sum }_k(2\pi \mathrm{\mid }k\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }$ finite, so the Fourier series and its differentiated series converge uniformly on compact time intervals including $s=0$. For $s\ge 1$, the same shell bound and the Fourier formula in [F2] give $\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }({\rho }_s-\bar{\rho }){\mathrm{\parallel }}_{\mathrm{\infty }}\le e^{-4{\pi }^{2}s}{\sum }_{k\mathrm{\ne }0}(2\pi \mathrm{\mid }k\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }\widehat{\rho }(k)\mathrm{\mid }{e}^{-4{\pi }^2(\mathrm{\mid }k{\mathrm{\mid }}^2-1)}\le C_{\alpha ,\rho }{e}^{-4{\pi }^{2}s}$. The zero Fourier coefficient is $\widehat{\rho }(0)=\bar{\rho }$, proving the asserted smoothness and $C^{\mathrm{\infty }}$ convergence.

The heat profile stays in $[\epsilon ,1-\epsilon ]$: the affine rescaling $v_0=(\rho -\epsilon )\mathrm{/}(1-2\epsilon )$ lies in $[0,1]$, and its Fourier heat profile $v_s$ preserves $[0,1]$ by [F2]. The multiplier is linear and fixes constants, so ${\rho }_s=\epsilon +(1-2\epsilon )v_s$; its mean is $\bar{\rho }$ by the zero-mode calculation in step 1.1.

Define $a(t)=0$ for $t\le 0$ and $a(t)=e^{-1\mathrm{/}t}$ for $t>0$, and set $\zeta (t)=a(t)\mathrm{/}(a(t)+a(1-t))$ on $[0,1]$. For each derivative order, the right derivatives of $a$ at zero are sums of terms $t^{-m}{e}^{-1\mathrm{/}t}$, which tend to zero; hence $a$ is smooth and flat at zero. The denominator is positive on $[0,1]$, $\zeta (0)=0$, $\zeta (1)=1$, and $\zeta$ is flat at both endpoints in the sense that all derivatives of $\zeta$ vanish at zero and all derivatives of $\zeta -1$ vanish at one. Define ${\sigma }_t^S=\bar{\rho }+\zeta (t)({\rho }_{S+1-t}-\bar{\rho })$. This is smooth by [F2], has mean $\bar{\rho }$, has values in $[\epsilon ,1-\epsilon ]$ by step 2.1, starts at $\bar{\rho }$, and ends at ${\rho }_S$.

Put $g_t={\mathrm{\partial }}_t{\sigma }_t^S$, which has zero mean. Define the smooth periodic potential ${\varphi }_t$ by its Fourier coefficients ${\widehat{\varphi }}_t(0)=0$ and ${\widehat{\varphi }}_t(k)={\widehat{g}}_t(k)\mathrm{/}(4{\pi }^2\mathrm{\mid }k{\mathrm{\mid }}^2)$ for $k\mathrm{\ne }0$, and set ${\kappa }_t^S=\mathrm{\nabla }{\varphi }_t$. Repeated integration by parts applied to the smooth $g_t$ gives rapid coefficient decay, so the series for ${\varphi }_t$ and all its space and time derivatives converge uniformly. Termwise differentiation gives $\mathrm{\Delta }{\varphi }_t=-g_t$, hence ${\mathrm{\partial }}_t{\sigma }^S+\mathrm{\nabla }\cdot {\kappa }^S=0$. At $t=1$, $g_1=-\mathrm{\Delta }{\rho }_S$, so ${\varphi }_1={\rho }_S-\bar{\rho }$ and ${\kappa }_1^S=\mathrm{\nabla }{\rho }_S$. Flatness of $\zeta -1$ at $1$ makes every derivative of ${\sigma }^S$ agree there with the reversed heat segment; the fixed linear Fourier inverse gives the same matching for ${\kappa }^S$.

For $k\mathrm{\ne }0$, the Fourier coefficient of $\mathrm{\nabla }{\varphi }_t$ has magnitude $\mathrm{\mid }{\widehat{g}}_t(k)\mathrm{\mid }\mathrm{/}(2\pi \mathrm{\mid }k\mathrm{\mid })$. Repeated integration by parts gives $\mathrm{\mid }{\widehat{g}}_t(k)\mathrm{\mid }\le C_m\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}$; choosing $m$ with $2m>d-1$ makes ${\sum }_{k\mathrm{\ne }0}(1+\mathrm{\mid }k{\mathrm{\mid }}^2{)}^{-m}\mathrm{/}\mathrm{\mid }k\mathrm{\mid }$ finite by the lattice-shell bound in step 1.1, hence $\mathrm{\parallel }{\kappa }_t^S{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_m^{\mathrm{\prime }}\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}$. The formula for $g_t$ and the $C^{\mathrm{\infty }}$ decay in step 1.1, together with ${\mathrm{\partial }}_s{\rho }_s=\mathrm{\Delta }{\rho }_s$ from [F2], imply ${\sup }_{0\le t\le 1}\mathrm{\parallel }{g}_t{\mathrm{\parallel }}_{C^{2m}}\to 0$ and ${\sup }_{0\le t\le 1}\mathrm{\parallel }\mathrm{\nabla }{\sigma }_t^S{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ as $S\to \mathrm{\infty }$. By step 2.1, $\chi ({\sigma }_t^S)\ge \epsilon (1-\epsilon )$; since the space-time domain has measure one, [F1] bounds the connector action by $\frac{1}{4\epsilon (1-\epsilon )}{\sup }_{t,x}\mathrm{\mid }{\kappa }_t^S+\mathrm{\nabla }{\sigma }_t^S{\mathrm{\mid }}^2$, which tends to zero.

Let $(u,j)$ be any admissible connection of duration $T$ from $\bar{\rho }$ to $\rho$. By [F4], ${\mathcal{I}}_{[0,T]}(u,j)={\mathcal{H}}_{\bar{\rho }}(\rho )+{\mathcal{I}}_{[0,T]}(u^{\theta },j^{\theta })$, because both endpoints have mean $\bar{\rho }$ and [F3] identifies the endpoint free-energy difference with ${\mathcal{H}}_{\bar{\rho }}(\rho )$. The reversed action is nonnegative by [F1], so every admissible connection has action at least ${\mathcal{H}}_{\bar{\rho }}(\rho )$.

Let ${\rho }_s$ be the Fourier heat profile with datum $\rho$. Under [A1], [F6] applies; because $\rho$ is smooth, its Fourier multiplier profile is smooth up to $s=0$, solves ${\mathrm{\partial }}_s{\rho }_s=\mathrm{\Delta }{\rho }_s$, and preserves mean $\bar{\rho }$. Apply the range-preservation assertion of [F6] to $v=(\rho -\epsilon )\mathrm{/}(1-2\epsilon )$, which lies in $[0,1]$. Linearity of the Fourier multiplier and its fixing constants give ${\rho }_s=\epsilon +(1-2\epsilon )v_s$, hence $\epsilon \le {\rho }_s\le 1-\epsilon$ for every $s\ge 0$.

Taking the infimum over all finite-duration connections in [F2] yields ${\mathcal{V}}_{\bar{\rho }}(\rho )\ge {\mathcal{H}}_{\bar{\rho }}(\rho )$.

On the forward heat path $({\rho }_s,-\mathrm{\nabla }{\rho }_s)$, $0\le s\le S$, the continuity equation holds because ${\mathrm{\partial }}_s{\rho }_s=\mathrm{\Delta }{\rho }_s$ and $\mathrm{\nabla }\cdot (-\mathrm{\nabla }{\rho }_s)=-\mathrm{\Delta }{\rho }_s$, and the action is zero because $-\mathrm{\nabla }{\rho }_s+\mathrm{\nabla }{\rho }_s=0$. Its time reverse on $[0,S]$ has current $+\mathrm{\nabla }{\rho }_{S-t}$, and [F4] gives its action as $\int f(\rho )dx-\int f({\rho }_S)dx$. Shift this reversed path to $[1,S+1]$ by $t\mapsto t-1$; substitution in the action integral shows that its action is unchanged, and the resulting path has density ${\rho }_{S+1-t}$ and current $+\mathrm{\nabla }{\rho }_{S+1-t}$. The heat equation verifies continuity on the shifted interval.

Prepend the connector from [F5] on $[0,1]$. Its endpoint derivatives match the reversed heat segment, so the concatenation is a smooth admissible fixed-mean connection from $\bar{\rho }$ to $\rho$ of duration $S+1$. By [F2], ${\mathcal{V}}_{\bar{\rho }}(\rho )\le {\mathcal{I}}_{[0,1]}({\sigma }^S,{\kappa }^S)+\int f(\rho )dx-\int f({\rho }_S)dx$.

As $S\to \mathrm{\infty }$, [F5] gives ${\mathcal{I}}_{[0,1]}({\sigma }^S,{\kappa }^S)\to 0$ and ${\rho }_S\to \bar{\rho }$ uniformly; continuity of $f$ on $[\epsilon ,1-\epsilon ]$ then gives $\int f({\rho }_S)dx\to f(\bar{\rho })$. By [F7], the fixed-mean identity in [F3] is ${\mathcal{H}}_{\bar{\rho }}(\rho )={\int }_{{\mathbb{T}}^d}[\rho (\log \rho -\log \bar{\rho })+(1-\rho )(\log (1-\rho )-\log (1-\bar{\rho }))]dx={\int }_{{\mathbb{T}}^d}[\rho \log (\rho \mathrm{/}\bar{\rho })+(1-\rho )\log ((1-\rho )\mathrm{/}(1-\bar{\rho }))]dx$. Thus ${\mathcal{V}}_{\bar{\rho }}(\rho )\le {\mathcal{H}}_{\bar{\rho }}(\rho )$; combining with the lower bound in step 2.1 proves the claimed equality. The constructed horizons tend to infinity; no finite-time minimizer is asserted.