Equilibrium action time-reversal identity

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.