Finite-rate fluctuation response and Green–Kubo inverse

By [F2], choose an orthonormal eigenbasis of $Q$ with ${\varphi }_0=1$ and $Q{\varphi }_j=-{\lambda }_j{\varphi }_j$ for $j\ge 1$, where ${\lambda }_j>0$. For centered $A,B$, expand $A_c={\sum }_{j\ge 1}{a}_j{\varphi }_j$ and $B_c={\sum }_{j\ge 1}{b}_j{\varphi }_j$. Then $C_{BA}(t)={\sum }_{j=1}^{r-1}{b}_j{a}_j{e}^{-{\lambda }_{j}t}$ and ${\int }_0^{\mathrm{\infty }}{C}_{BA}(t)dt={\sum }_{j=1}^{r-1}{a}_j{b}_j\mathrm{/}{\lambda }_j=⟨B_c,(-Q{)}^{-1}{A}_c{⟩}_{\pi }$. The sum is finite, so the correlation is absolutely integrable. Applying the same expansion to each component of $G_c$ gives $D_{ij}={\sum }_{j^{\mathrm{\prime }}=1}^{r-1}{g}_{i,j^{\mathrm{\prime }}}{g}_{j,j^{\mathrm{\prime }}}\mathrm{/}{\lambda }_{j^{\mathrm{\prime }}}$, a symmetric positive-semidefinite matrix. The two-sided stationary chain in [F2] has bounded, jointly measurable $G_c(X_t)$; its covariance is absolutely integrable by this expansion, so [F3] gives the stated covariance limit and agrees with this matrix.

For a finite time interval, the controlled off-diagonal rate matrix has the uniform expansion $Q^{ϵ}(t)=Q+ϵh(t)K+O({ϵ}^2)$, where $K_{ab}=\frac{\beta }{2}{w}_{ab}(B_b-B_a)$ for $a\mathrm{\ne }b$ and each diagonal entry makes its row sum zero. The remainder is uniform because $E$ is finite and $h$ is bounded. The finite forward equation $p^{\mathrm{\prime }}=p{Q}^{ϵ}(t)$ has a unique solution: its Picard series converges uniformly, since the $n$th iterated integral is bounded by $M^n{t}_f^n\mathrm{/}n!$ when $\mathrm{\parallel }{Q}^{ϵ}(t)\mathrm{\parallel }\le M$. Since $E$ is finite, $h$ is bounded, and the exponential factors are bounded uniformly for $t\in [0,t_f]$ and $\mathrm{\mid }ϵ\mathrm{\mid }\le 1$, choose one $\mathrm{\Lambda }$ larger than every controlled exit rate on this set. Then $Q^{ϵ}(t)+\mathrm{\Lambda }I$ has nonnegative entries; after $r(t)=e^{\mathrm{\Lambda }t}{p}^{ϵ}(t)$, its Picard series has nonnegative terms, so $p^{ϵ}(t)\ge 0$ for all such $ϵ$. The row sums of $Q^{ϵ}$ are zero, hence total mass remains one. Subtracting the unperturbed equation and using $\pi Q=0$ gives $p^{ϵ}-\pi =O(ϵ)$ on $[0,t_f]$ by successive substitution in the bounded linear integral equation. Put $y^{ϵ}=(p^{ϵ}-\pi )\mathrm{/}ϵ$; then $y^{ϵ}(t)={\int }_0^t{y}^{ϵ}(s)Qds+{\int }_0^t{p}^{ϵ}(s)(Q^{ϵ}(s)-Q)\mathrm{/}ϵds$. The second integrand converges uniformly to $\pi h(s)K$, since $p^{ϵ}\to \pi$ uniformly and $(Q^{ϵ}-Q)\mathrm{/}ϵ\to hK$ uniformly. Subtracting the limiting integral equation bounds $\mathrm{\parallel }{y}^{ϵ}-q\mathrm{\parallel }$ by a term tending uniformly to zero plus $M{\int }_0^t\mathrm{\parallel }{y}^{ϵ}-q\mathrm{\parallel },ds$; successive substitution bounds this difference by the forcing error times $e^{M{t}_f}$. Thus $q={\mathrm{\partial }}_{ϵ}{p}^{ϵ}{\mathrm{\mid }}_0$ exists and satisfies $q^{\mathrm{\prime }}=qQ+h\pi K$, $q(0)=0$. Variation of constants, verified by differentiation, gives $q(t)={\int }_0^{t}h(s)\pi K{P}_{t-s}ds$.

For each state $b$, detailed balance in [F1] and the definition of $K$ give $(\pi K{)}_b=\frac{\beta }{2}{\sum }_{a\mathrm{\ne }b}{\pi }_a{w}_{ab}(B_b-B_a)-\frac{\beta }{2}{\pi }_b{\sum }_{c\mathrm{\ne }b}{w}_{bc}(B_c-B_b)=-\beta {\pi }_b(QB{)}_b$. Indeed, replacing ${\pi }_a{w}_{ab}$ by ${\pi }_b{w}_{ba}$ makes the first sum the negative of the second before its prefactor is applied. Thus $q(t)A={\int }_0^{t}h(s)(\pi K)P_{t-s}Ads=-\beta {\int }_0^{t}h(s)⟨Q{B}_c,P_{t-s}{A}_c{⟩}_{\pi }ds$. Self-adjointness of $Q$ and $P_u^{\mathrm{\prime }}=Q{P}_u$ imply $C_{BA}^{\mathrm{\prime }}(u)=⟨Q{B}_c,P_u{A}_c{⟩}_{\pi }$, proving the response formula.

For constant $h$, put $Z_{ϵ}:={\sum }_{c\in E}{\pi }_c{e}^{\beta ϵh{B}_c}>0$ and ${\pi }_a^{ϵ}=Z_{ϵ}^{-1}{\pi }_a{e}^{\beta ϵh{B}_a}$. Multiplying this by $w_{ab}^{ϵ}$ gives a common edge factor proportional to ${\pi }_a{w}_{ab}{e}^{\beta ϵh(B_a+B_b)\mathrm{/}2}$, symmetric in $a,b$ by detailed balance. Hence ${\pi }^{ϵ}$ is stationary. Differentiating its finite normalization sum at $ϵ=0$ gives ${{\mathrm{\partial }}_{ϵ}{\sum }_a{\pi }_a^{ϵ}{A}_a\mid }_0=\beta h(\pi (AB)-\pi (A)\pi (B))$, the stated static susceptibility.