stochastic lattice gases.

Observe that $I\geq 0$ and $I(\rho, j)=0$ if and only if $j=j^\rho$. Hence the second equation in is equivalent to $I(\rho, j)\leq 0$. By expanding the square we deduce $$\label{ineq-calore}
\int_0^T\!{\mathop{}\!\mathrm{d}}t \int\!{\mathop{}\!\mathrm{d}}x\,
  \Big[ \frac 12 \frac 1{\rho(t)} j(t) \cdot D^{-1} j(t)  
  + \frac 12 \frac 1{\rho(t)}   \nabla\rho(t) \cdot D \nabla \rho(t)
  + \frac  1{\rho(t)} \nabla \rho(t) \cdot j(t) \Big]\leq 0.$$ Since $(\nabla \rho)/\rho =\nabla \log\rho$, integrating by parts and using the continuity equation, the last term is the total derivative of $H(\rho(t))$.

We now introduce the *Fisher information* $E$ as the Dirichlet form of square root, namely $$E(\rho)=\frac 12\int\! {\mathop{}\!\mathrm{d}}x  \frac 1{\rho}   \nabla\rho \cdot D \nabla
  \rho = 2\int \,{\mathop{}\!\mathrm{d}}x\, \nabla\sqrt\rho\cdot D\nabla\sqrt\rho.$$ Let also the *kinematic term* $R$ be the functional on the set the path $(\rho(t),j(t))$ defined by $$R (\rho, j)=\frac 12\int_0^T\! {\mathop{}\!\mathrm{d}}t\,\int\!{\mathop{}\!\mathrm{d}}x\,
    \frac 1{\rho(t)} j(t) \cdot D^{-1} j(t),$$ then reads $$\label{gfhe}
 H(\rho(T))+\int_0^T \!{\mathop{}\!\mathrm{d}}t\,  E(\rho(t))+ R (\rho, j) \leq H(\rho(0))$$ which is the gradient flow formulation of the heat equation that we will use here.

We now specify the precise formulation in which we consider a family of probabilities $\mu_t({\mathop{}\!\mathrm{d}}x)=\rho(t,x){\mathop{}\!\mathrm{d}}x$, $t\in [0, T]$, while the currents are the vector valued measures $J({\mathop{}\!\mathrm{d}}t, {\mathop{}\!\mathrm{d}}x)=j(t,x){\mathop{}\!\mathrm{d}}t {\mathop{}\!\mathrm{d}}x$. Given $T>0$ 