# Supplemental Material for “Thermodynamic constraints on the nonequilibrium
response of one-dimensional diffusions”

In this Supplemental Material, we first derive the transformation of
perturbation formulas (3)-(5) [main text]. Then we derive our main predictions
(6)-(8) [main text] quantifying the thermodynamic limits to steady-state
response. Lastly, we demonstrate how our predictions could be obtained from
the results in Ref. Owen et al. (2020) as the limit of a discrete Markov jump
process.

##  I Setup

We have in mind a single degree of freedom with position $x$ diffusing on a
circle of length $L$. Its dynamics as captured by the Fokker-Planck equation
[(1), main text] are determined by two periodic functions $A(x)$ and $B(x)$.
Our central observation is that we can reparameterize these dynamics as

$\displaystyle A(x)$ $\displaystyle=\mu(x)(-U^{\prime}(x)+f),$ (1)
$\displaystyle B(x)$ $\displaystyle=\mu(x).$ (2)

in terms of three new periodic functions, which we call the “energy” $U$,
“mobility” $\mu$, and “force” $f$ in analogy with the equation of motion for a
mesoscopic Brownian particle in a viscous fluid. In terms of the new
parameters, $\mu$, $U$, and $f$, the system’s steady-state distribution
$\pi(x)$ is given by the solution of the equation

$-\partial_{x}[\mu(x)(-U^{\prime}(x)+f)\pi(x)]+\partial_{x}\left[\mu(x)\partial_{x}\pi(x)\right]=0.$
(3)

Even for general choices of the dynamics, this one-dimensional equation with
periodic boundary conditions can always be determined as Gardiner (2004)

$\pi(x)=\frac{e^{-U(x)+fx}}{\mathcal{N}}\left(\int_{0}^{x}e^{U(x^{\prime})-fx^{\prime}-fL-\ln\mu(x^{\prime})}dx^{\prime}+\int_{x}^{L}e^{U(x^{\prime})-fx^{\prime}-\ln\mu(x^{\prime})}dx^{\prime}\right),$
(4)

where ${\mathcal{N}}$ is a normalization constant determined by the
requirement $\int_{0}^{L}\pi(y)dy=1$.

In the following, we will be differentiating the steady-state distribution
with respect to various system parameters. We have found that for organizing
these derivatives it is convenient to introduce an auxiliary function via

$\displaystyle\pi(x)=\frac{1}{\mathcal{N}}\int_{0}^{L}dx^{\prime}{\mathcal{S}}(x^{\prime},x),$
(5)

where

$\displaystyle{\mathcal{S}}(x^{\prime},x)=e^{U(x^{\prime})-U(x)-f(x^{\prime}-x)-\ln\mu(x^{\prime})}\left[e^{-fL}\Theta(x-x^{\prime})+\Theta(x^{\prime}-x)\right],$
(6)
$\displaystyle\mathcal{N}=\int_{0}^{L}\int_{0}^{L}{\mathcal{S}}(x^{\prime},x)\
dx^{\prime}dx,$ (7)

and $\Theta(z)$ is the Heaviside step function that is one for $z>0$ and zero
otherwise. The quantity ${\mathcal{S}}$ acts like a continuous version of the
spanning trees that appear in the Matrix Tree Theorem Aleandri et al. , which
were also key ingredients to the proofs of the thermodynamic bounds in Ref.
Owen et al. (2020).

##  II Transformation of perturbations

We are interested in the response of observables. Our assertion is that any
response can be reformulated in terms of derivatives with respect to our
parameters $\mu$, $U$, and $f$. To verify this, we begin by first deriving the
mapping from $A$ and $B$ to $\mu$, $U$, and $f$. We then use this mapping to
decompose any perturbation into perturbations of $\mu$, $U$, and $f$, thereby
recovering (3)-(5) [main text].

Let us first invert the parameterization in (1) and (2) to find $\mu$, $U$ and
$f$ in terms of $A$ and $B$. First, by definition $\mu(x)=B(x)$. Next, we
eliminate $\mu$ from the transformation equations, by taking the ratio of $A$
and $B$

$\displaystyle\frac{A(x)}{B(x)}=-U^{\prime}(x)+f.$ (8)

Exploiting the periodicity of $U$, we can integrate to single out

$\displaystyle f=\frac{1}{L}\int_{0}^{L}\frac{A(z)}{B(z)}d{z}.$ (9)

Substituting (9) back into (8), leads to a closed equation for $U$ whose
solution is

$\displaystyle
U(x)=-\int_{0}^{x}\frac{A(z)}{B(z)}d{z}+\frac{x}{L}\int_{0}^{L}\frac{A(z)}{B(z)}d{z}+U(0),$
(10)

with $U(0)$ an undetermined constant that has no affect on the dynamics.

Now, imagine that we apply a small perturbation by changing an external
control parameter $\lambda$. The Fokker-Planck equation is then modified by a
small amount as $A_{\lambda}(x)=A(x)+\lambda\delta A(x)$ and
$B_{\lambda}(x)=B(x)+\lambda\delta B(x)$, where $\delta A(x)$ and $\delta
B(x)$ are periodic functions conjugate to the perturbation. The response of
the steady state to the perturbation in $\lambda$ can then be expressed via
the chain rule as a combination of $A$ and $B$ functional derivatives as

$\displaystyle\frac{\partial}{\partial\lambda}$ $\displaystyle=\int_{0}^{L}dz\
\frac{\partial A_{\lambda}(z)}{\partial\lambda}\frac{\delta}{\delta
A(z)}+\int_{0}^{L}dz\ \frac{\partial
B_{\lambda}(z)}{\partial\lambda}\frac{\delta}{\delta B(z)}$ (11)
$\displaystyle=\int_{0}^{L}dz\ \delta A(z)\frac{\delta}{\delta
A(z)}+\int_{0}^{L}dz\ \delta B(z)\frac{\delta}{\delta B(z)}.$ (12)

Using the mapping in (2), (9), and (10), we can convert the functional
derivatives with respect to $A$ and $B$ via the chain rule as

$\displaystyle\frac{\delta}{\delta A(z)}$ $\displaystyle=\int_{0}^{L}ds\
\frac{\delta U(s)}{\delta A(z)}\frac{\delta}{\delta U(s)}+\int_{0}^{L}ds\
\frac{\delta\mu(s)}{\delta A(z)}\frac{\delta}{\delta\mu(s)}+\frac{\delta
f}{\delta A(z)}\frac{\partial}{\partial f}$ (13)
$\displaystyle=\int_{0}^{L}ds\left(-\frac{\Theta(s-z)}{B(z)}+\frac{s}{L}\frac{1}{B(z)}\right)\frac{\delta}{\delta
U(s)}+\frac{1}{L}\frac{1}{B(z)}\frac{\partial}{\partial f},$ (14)

and

$\displaystyle\frac{\delta}{\delta B(z)}$ $\displaystyle=\int_{0}^{L}ds\
\frac{\delta U(s)}{\delta B(z)}\frac{\delta}{\delta U(s)}+\int_{0}^{L}ds\
\frac{\delta\mu(s)}{\delta B(z)}\frac{\delta}{\delta\mu(s)}+\frac{\delta
f}{\delta B(z)}\frac{\partial}{\partial f}$ (15)
$\displaystyle=\int_{0}^{L}ds\left(\Theta(s-z)\frac{A(z)}{B(z)^{2}}-\frac{s}{L}\frac{A(z)}{B(z)^{2}}\right)\frac{\delta}{\delta
U(s)}+\frac{\delta}{\delta\mu(z)}-\frac{1}{L}\frac{A(z)}{B(z)^{2}}\frac{\partial}{\partial
f}.$ (16)

Substituting these expressions into the $\lambda$-derivative (12), we find
after simplifying the integrals

$\frac{\partial}{\partial\lambda}=\int_{0}^{L}ds\ \delta
U(s)\frac{\delta}{\delta U(s)}+\int_{0}^{L}ds\
\delta\mu(s)\frac{\delta}{\delta\mu(s)}+\delta f\frac{\partial}{\partial f},$
(17)

where

$\displaystyle\delta\mu(s)=\delta B(s)$ (18) $\displaystyle\delta
U(s)=-\int_{0}^{s}dz\ \left(\frac{\delta A(z)B(z)-A(z)\delta
B(z)}{B(z)^{2}}\right)+\frac{s}{L}\int_{0}^{L}dz\ \left(\frac{\delta
A(z)B(z)-A(z)\delta B(z)}{B(z)^{2}}\right)$ (19) $\displaystyle\delta
f=\frac{1}{L}\int_{0}^{L}dz\ \left(\frac{\delta A(z)B(z)-A(z)\delta
B(z)}{B(z)^{2}}\right),$ (20)

recapitulating the functional differentials found in (3)-(5) [main text].
Furthermore, we do not study $\mu$, $U$, and $f$ perturbations explicitly, but
instead constrain linear combinations of these perturbations. Nevertheless, we
can express the response in terms of these linear combinations as

$\frac{\partial}{\partial\lambda}=\int_{0}^{L}ds\ \delta
U(s)\left(\frac{\delta}{\delta
U(s)}+\frac{\delta}{\delta\ln\mu(s)}\right)+\int_{0}^{L}ds\
\big{(}\mu(s)\delta\mu(s)-\delta
U(s)\big{)}\frac{\delta}{\delta\ln\mu(s)}+\delta f\frac{\partial}{\partial f}$
(21)

##  III Energy-mobility perturbations

Our first result in (6) [main text] is an equality for the response of the
steady-state average of an observable $Q(x)$ to a coordinated perturbation in
the energy and mobility: $(\delta/\delta U(z)+\delta/\delta\ln\mu(z))\langle
Q\rangle$.

Let us proceed by first analyzing the derivatives on
$\mathcal{S}(x^{\prime},x)$ and $\mathcal{N}$:

$\displaystyle\frac{\delta{\mathcal{S}}(x^{\prime},x)}{\delta
U(z)}+\frac{\delta{\mathcal{S}}(x^{\prime},x)}{\delta\ln\mu(z)}=-\delta(x-z){\mathcal{S}}(x^{\prime},x),\qquad\frac{\delta{\mathcal{N}}}{\delta
U(z)}+\frac{\delta{\mathcal{N}}}{\delta\ln\mu(z)}=-{\mathcal{N}}\pi(z).$ (22)

Using these expressions, we then find for the derivative of $\pi(x)$,

$\displaystyle\frac{\delta\pi(x)}{\delta
U(z)}+\frac{\delta\pi(x)}{\delta\ln\mu(z)}$
$\displaystyle=-\frac{1}{\mathcal{N}}\int_{0}^{L}dx^{\prime}\delta(x-z){\mathcal{S}}(x^{\prime},x)-\frac{1}{{\mathcal{N}}^{2}}\left[-{\mathcal{N}}\pi(z)\right]\int_{0}^{L}dx^{\prime}{\mathcal{S}}(x^{\prime},x)$
(23) $\displaystyle=-\pi(x)(\delta(x-z)-\pi(z)).$ (24)

From this expression, we readily obtain equation (6) [main text] for the
response of an observable $\langle Q\rangle=\int_{0}^{L}Q(x)\pi(x)dx$ as

$\frac{\delta\langle Q\rangle}{\delta U(z)}+\frac{\delta\langle
Q\rangle}{\delta\ln\mu(z)}=\int_{0}^{L}Q(x)\left[\frac{\delta\pi(x)}{\delta
U(z)}+\frac{\delta\pi(x)}{\delta\ln\mu(z)}\right]dx=-\pi(z)(Q(z)-\langle
Q\rangle).$ (25)

##  IV Mobility perturbations

Our next prediction, (7) [main text], is a thermodynamic bound on
perturbations with respect to the mobility $\mu(x)$:

$\left|\int_{a}^{b}\frac{\delta\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\delta\ln\mu(z)}dz\right|\leq\tanh(\left|f\right|L/4),$ (26)

where $Q_{1}(x)$, $Q_{2}(x)$ are bounded observables, and their values depend
only on position $x$. Without loss of generality, we can assume that they are
non-negative ($Q(x)\geq 0$), because we can always redefine them by
subtracting off the minimum ($Q^{\prime}(x)=Q(x)-Q_{\rm min}$).

To evaluate the derivative in (26), we proceed by first differentiating
${\mathcal{S}}$ and ${\mathcal{N}}$:

$\frac{\delta{\mathcal{S}}(x^{\prime},x)}{\delta\ln\mu(z)}=-\delta(x^{\prime}-z){\mathcal{S}}(x^{\prime},x),\qquad\frac{\delta{\mathcal{N}}}{\delta\ln\mu(z)}=-\int_{0}^{L}dy\
{\mathcal{S}}(z,y).$ (27)

Consequently,

$\frac{\delta\pi(x)}{\delta\ln\mu(z)}=-\frac{1}{{\mathcal{N}}}{\mathcal{S}}(z,x)+\frac{\pi(x)}{\mathcal{N}}\int_{0}^{L}dy\
{\mathcal{S}}(z,y).$ (28)

With this expression, we can readily obtain an expression for the response of
an observable as

$\displaystyle\frac{\delta\langle Q\rangle}{\delta\ln\mu(z)}$
$\displaystyle=\int_{0}^{L}Q(x)\left[-\frac{1}{{\mathcal{N}}}{\mathcal{S}}(z,x)+\frac{\pi(x)}{\mathcal{N}}\int_{0}^{L}dy\
{\mathcal{S}}(z,y)\right]dx$ (29)
$\displaystyle=-\frac{1}{{\mathcal{N}}}\int_{0}^{L}Q(x){\mathcal{S}}(z,x)dx+\frac{\langle
Q\rangle}{\mathcal{N}}\int_{0}^{L}\ {\mathcal{S}}(z,y)dy.$ (30)

We are now in a position to evaluate the derivative in (26):

$\displaystyle\int_{a}^{b}\frac{\delta\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\delta\ln\mu(z)}dz$
$\displaystyle=\int_{a}^{b}\frac{\frac{\delta\langle
Q_{1}\rangle}{\delta\ln\mu(z)}\langle Q_{2}\rangle-\frac{\delta\langle
Q_{2}\rangle}{\delta\ln\mu(z)}\langle Q_{1}\rangle}{\langle
Q_{1}\rangle\langle Q_{2}\rangle}dz.$ (31)

Upon substitution of (30), we see that the terms linear in average of the
observable in (30) cancel, leaving

$\displaystyle\int_{a}^{b}\frac{\delta\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\delta\ln\mu(z)}dz$
$\displaystyle=-\frac{1}{\mathcal{N}}\frac{\left(\int_{a}^{b}\int_{0}^{L}Q_{1}(x)S(z,x)dzdx\right)\langle
Q_{2}\rangle-\left(\int_{a}^{b}\int_{0}^{L}Q_{2}(x)S(z,x)dzdx\right)\langle
Q_{1}\rangle}{\langle Q_{1}\rangle\langle Q_{2}\rangle}$ (32)

To simplify this expression, we note that average of any observable can also
be expressed in terms of ${\mathcal{S}}$ as

$\langle
Q\rangle=\frac{1}{\mathcal{N}}\int_{0}^{L}\int_{0}^{L}Q(x)S(x^{\prime},x)dx^{\prime}dx.$
(33)

Upon substitution of this formula into (32), we find that the result can be
conveniently expressed in terms of the integrals

$\displaystyle q_{1}$
$\displaystyle=\int_{z\in[a,b]}\int_{0}^{L}Q_{1}(x){\mathcal{S}}(z,x)\ dxdz$
(34) $\displaystyle{\bar{q}}_{1}$
$\displaystyle=\int_{z\notin[a,b]}\int_{0}^{L}Q_{1}(x){\mathcal{S}}(z,x)\
dxdz$ (35) $\displaystyle q_{2}$
$\displaystyle=\int_{z\in[a,b]}\int_{0}^{L}Q_{2}(x){\mathcal{S}}(z,x)\ dxdz$
(36) $\displaystyle{\bar{q}}_{2}$
$\displaystyle=\int_{z\notin[a,b]}\int_{0}^{L}Q_{2}(x){\mathcal{S}}(z,x)\
dxdz$ (37)

as

$\int_{a}^{b}\frac{\delta\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\delta\ln\mu(z)}dz=\frac{q_{1}{\bar{q}}_{2}-{\bar{q}}_{2}q_{1}}{({\bar{q}}_{1}+q_{1})({\bar{q}}_{2}+q_{2})}.$
(38)

The notation here is reminiscent of the derivation Ref. Owen et al. (2020),
which will allow us to import those methods directly.

Noting that ${\bar{q}}_{1}$, $q_{1}$, ${\bar{q}}_{2}$, and $q_{2}$ are all
non-negative, the denominator of (38) is bounded by the inequality of
arithmetic and geometric means:

$({\bar{q}}_{1}+q_{1})({\bar{q}}_{2}+q_{2})={\bar{q}}_{1}{\bar{q}}_{2}+{\bar{q}}_{1}q_{2}+q_{1}{\bar{q}}_{2}+q_{1}q_{2}\geq
q_{1}{\bar{q}}_{2}+{\bar{q}}_{1}q_{2}+2\sqrt{{\bar{q}}_{1}{\bar{q}}_{2}q_{1}q_{2}}=(\sqrt{{\bar{q}}_{1}q_{2}}+\sqrt{q_{1}{\bar{q}}_{2}})^{2},$
(39)

where the equality is saturated when ${\bar{q}}_{1}{\bar{q}}_{2}=q_{1}q_{2}$.
The numerator can also be factored

$q_{2}{\bar{q}}_{1}-q_{1}{\bar{q}}_{2}=(\sqrt{{\bar{q}}_{1}q_{2}}-\sqrt{q_{1}{\bar{q}}_{2}})(\sqrt{{\bar{q}}_{1}q_{2}}+\sqrt{q_{1}{\bar{q}}_{2}}).$
(40)

The result is

$\left|\int_{a}^{b}\frac{\delta\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\delta\ln\mu(z)}dz\right|\leq\left|\frac{\sqrt{{\bar{q}}_{1}q_{2}}-\sqrt{q_{1}{\bar{q}}_{2}}}{\sqrt{{\bar{q}}_{1}q_{2}}+\sqrt{q_{1}{\bar{q}}_{2}}}\right|=\tanh\left(\frac{1}{4}\left|\ln\frac{q_{1}{\bar{q}}_{2}}{{\bar{q}}_{1}q_{2}}\right|\right).$
(41)

Our last step is to bound the ratio $q_{1}{\bar{q}}_{2}/{\bar{q}}_{1}q_{2}$:

$\displaystyle\frac{q_{1}{\bar{q}}_{2}}{{\bar{q}}_{1}q_{2}}$
$\displaystyle=\frac{\int_{0}^{L}\int_{0}^{L}dx_{1}dx_{0}\
\int_{z_{1}\in[a,b]}dz_{1}\int_{z_{0}\notin[a,b]}dz_{0}\
Q_{1}(x_{0})Q_{2}(x_{1}){\mathcal{S}}(z_{1},x_{0})\mathcal{S}(z_{0},x_{1})}{\int_{0}^{L}\int_{0}^{L}dx_{1}dx_{0}\
\int_{z_{1}\in[a,b]}dz_{1}\int_{z_{0}\notin[a,b]}dz_{0}\
Q_{1}(x_{0})Q_{2}(x_{1})\mathcal{S}(z_{0},x_{0})\mathcal{S}(z_{1},x_{1})}$
(42)
$\displaystyle=\frac{\int_{0}^{L}\int_{0}^{L}dx_{1}dx_{0}\int_{z_{1}\in[a,b]}dz_{1}\int_{z_{0}\notin[a,b]}dz_{0}\
W(x_{0},x_{1},z_{0},z_{1})\frac{\mathcal{S}(z_{1},x_{0})\mathcal{S}(z_{0},x_{1})}{\mathcal{S}(z_{0},x_{0})\mathcal{S}(z_{1},x_{1})}}{\int_{0}^{L}\int_{0}^{L}dx_{1}dx_{0}\int_{z_{1}\in[a,b]}dz_{1}\int_{z_{0}\notin[a,b]}dz_{0}\
W(x_{0},x_{1},z_{0},z_{1})},$ (43)

where we introduced the four-dimensional non-negative weight function

$W(x_{0},x_{1},z_{0},z_{1})=Q_{1}(x_{0})Q_{2}(x_{1})\mathcal{S}(z_{0},x_{0})\mathcal{S}(z_{1},x_{1})\geq
0.$ (44)

Therefore the ratio $q_{1}{\bar{q}}_{2}/{\bar{q}}_{1}q_{2}$ can be viewed as
the weighted average of an observable ($\langle\cdot\rangle_{W}$), which we
can bound by its maximum as

$\displaystyle\frac{q_{1}{\bar{q}}_{2}}{{\bar{q}}_{1}q_{2}}$
$\displaystyle=\left\langle\frac{\mathcal{S}(z_{1},x_{0})\mathcal{S}(z_{0},x_{1})}{\mathcal{S}(z_{0},x_{0})\mathcal{S}(z_{1},x_{1})}\right\rangle_{W}$
(45)
$\displaystyle\leq\max_{\\{z_{0},z_{1},x_{0},x_{1}\\}}\frac{\mathcal{S}(z_{1},x_{0})\mathcal{S}(z_{0},x_{1})}{\mathcal{S}(z_{0},x_{0})\mathcal{S}(z_{1},x_{1})}$
(46)
$\displaystyle=\max_{\\{z_{0},z_{1},x_{0},x_{1}\\}}\frac{\left[e^{-fL}\Theta(x_{0}-z_{1})+\Theta(z_{1}-x_{0})\right]\left[e^{-fL}\Theta(x_{1}-z_{0})+\Theta(z_{0}-x_{1})\right]}{\left[e^{-fL}\Theta(x_{0}-z_{0})+\Theta(z_{0}-x_{0})\right]\left[e^{-fL}\Theta(x_{1}-z_{1})+\Theta(z_{1}-x_{1}))\right]}$
(47) $\displaystyle=e^{\left|f\right|L},$ (48)

where the last equality holds when, for example, $f>0$ and
$z_{0}>x_{1}>z_{1}>x_{0}$. Equation (26) follows immediately.

##  V Force perturbation

The final prediction is a bound on the force response, (8) [main text],
reproduced here in a slightly modified form

$\displaystyle\left|\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial f}\right|\leq L.$ (49)

To organize the derivatives with respect to the force $f$, we will find it
convenient to use the function

${\mathcal{O}}(x^{\prime},x)=(x^{\prime}-x+L)\Theta(x-x^{\prime})+(x^{\prime}-x)\Theta(x^{\prime}-x),$
(50)

which we note for later use is bounded $0\leq{\mathcal{O}}\leq L$. Then, we
have

$\displaystyle\frac{{\partial\mathcal{S}}(x^{\prime},x)}{\partial
f}=-{\mathcal{O}}(x^{\prime},x){\mathcal{S}}(x^{\prime},x),\qquad\frac{{\partial\mathcal{N}}}{\partial
f}=-\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
{\mathcal{O}}(x^{\prime},x){\mathcal{S}}(x^{\prime},x),$ (51)

so that

$\frac{\partial\pi(x)}{\partial
f}=-\frac{1}{\mathcal{N}}\int_{0}^{L}dx^{\prime}{\mathcal{O}}(x^{\prime},x){\mathcal{S}}(x^{\prime},x)+\frac{\pi(x)}{\mathcal{N}}\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx^{\prime\prime}\
{\mathcal{O}}(x^{\prime},x^{\prime\prime}){\mathcal{S}}(x^{\prime},x^{\prime\prime}).$
(52)

As a result the response of an observable can be expressed as

$\frac{\partial\langle Q\rangle}{\partial
f}=-\frac{1}{\mathcal{N}}\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
Q(x){\mathcal{O}}(x^{\prime},x){\mathcal{S}}(x^{\prime},x)+\frac{\langle
Q\rangle}{\mathcal{N}}\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
{\mathcal{O}}(x^{\prime},x){\mathcal{S}}(x^{\prime},x).$ (53)

With these formulas in hand, we can now address the derivative in (49). Upon
substitution of (53) into (49), we find that the second terms in (53) linear
in the average of the observables cancel, resulting in the expression

$\displaystyle\left|\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial
f}\right|=\left|\frac{\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
{\mathcal{O}}(x^{\prime},x)Q_{2}(x){\mathcal{S}}(x^{\prime},x)}{\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
Q_{2}(x){\mathcal{S}}(x^{\prime},x)}-\frac{\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
{\mathcal{O}}(x^{\prime},x)Q_{1}(x){\mathcal{S}}(x^{\prime},x)}{\int_{0}^{L}\int_{0}^{L}dx^{\prime}dx\
Q_{1}(x){\mathcal{S}}(x^{\prime},x)}\right|,$ (54)

after simplification using the definition of ${\mathcal{N}}$. A particularly
useful interpretation presents itself after we note that
$Q(x){\mathcal{S}}(x^{\prime},x)\geq 0$. Therefore each ratio can be
interpreted as a normalized average of ${\mathcal{O}}$ with observable-
dependent weight $Q(x){\mathcal{S}}(x^{\prime},x)\geq 0$, which we denote as
$\langle\cdot\rangle_{Q}$. The result is that we can express (54) as

$\displaystyle\left|\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial
f}\right|=\big{|}\langle{\mathcal{O}}\rangle_{Q_{2}}-\langle{\mathcal{O}}\rangle_{Q_{1}}\big{|}\leq
L,$ (55)

where the bound follows from $0\leq{\mathcal{O}}\leq L$, completing the
derivation.

##  VI Connection to Markov Jump Processes

The aim of this section is to discuss the relationship between the present
study and previous work on thermodynamic limitations to steady-state response
in discrete Markov jump processes Owen et al. (2020).

###  VI.1 Review of thermodynamic limits to response for discrete Markov jump
processes

As we are only interested in stochastic processes on a ring, we will introduce
the ideas and results from Ref. Owen et al. (2020) specialized to this
context.

We have in mind a system of $N$ discrete states at positions $x_{i}=i\Delta x$
around a ring of length $L=N\Delta x$. We label these states as of
$i=0,\dots,N$, where we identify the redundant state $i=N$ with $i=0$ to
enforce the periodic boundary conditions. The probability to find the system
in state $i$ at time $t$ is then governed by the Master equation Gardiner
(2004)

$\dot{p}_{i}(t)=\sum_{j=0}^{N-1}W_{ij}p_{j}(t),$ (56)

where the off-diagonal entries of the transition rate matrix $W_{ij}$ specify
the probability per unit time to jump from state $j$ to state $i$, and
$W_{ii}=-\sum_{j\neq i}W_{ji}$. As only nearest-neighbor hops are allowed, the
only nonzero transition rates are those for which $i$ and $j$ differ by one;
thus, all rates are of the form $W_{i+1,i}$ or $W_{i-1,i}$, corresponding to
‘right’ and ‘left’ hops. As the state space is irreducible, a unique
stationary distribution $\pi_{i}$ exists and can be obtained as the solution
of

$\sum_{j=0}^{N-1}W_{ij}\pi_{j}=0.$ (57)

Thermodynamics is included in the model by identifying the log-ratio of rates
around cycles as the thermodynamic force driving the system out of
equilibrium. As a ring only has a single cycle, the sole thermodynamic force
is

$F_{C}=\ln\frac{W_{0,N-1}\cdots W_{2,1}W_{1,0}}{W_{N-1,0}\cdots
W_{1,2}W_{0,1}}.$ (58)

Reference Owen et al. (2020) introduced a parameterization of the transition
rate matrix in terms of vertex parameters $E_{i}$, symmetric edge parameters
$B_{i+1,i}=B_{i,i+1}$, and asymmetric edge parameters $F_{i+1,i}=-F_{i,i+1}$:

$\displaystyle W_{i+1,i}=e^{-(B_{i+1,i}-E_{i}-F_{i+1,i}/2)},\qquad
W_{i-1,i}=e^{-(B_{i,i-1}-E_{i}+F_{i,i-1}/2)}.$ (59)

Nonequilibrium effects are included in this parameterization solely through
the asymmetric edge parameters, which can be seen by substituting this
decomposition into the definition of the thermodynamic force (58) to conclude

$F_{C}=\sum_{i=0}^{N-1}F_{i+1,i}.$ (60)

The main predictions of Ref. Owen et al. (2020) are then a series of
equalities and inequalities for the derivative of the steady state
distribution with respect to these three parameter families. Here, we present
forms most relevant for our present discussion.

_Vertex parameters_ : An equality for vertex parameter perturbations can be
obtained from Eq. (13) of Ref. Owen et al. (2020),

$\displaystyle\sum_{j=0}^{N-1}V_{j}\frac{\partial\langle Q\rangle}{\partial
E_{j}}=-\sum_{j=0}^{N-1}V_{j}\pi_{j}(Q_{j}-\langle Q\rangle)=-{\rm Cov}(Q,V),$
(61)

where the state space observable $V_{i}$ is the conjugate coordinate to the
perturbation.

_Symmetric edge parameters_ : If we perturb the $B_{i+1,i}$ of all the edges
between a pair of nodes at positions $x_{a}=a\Delta x$ and $x_{b}=b\Delta x$
then Eq. (20) of Ref. Owen et al. (2020) predicts

$\displaystyle\left|\sum_{i=a}^{b-1}\frac{\partial\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial B_{i+1,i}}\right|$
$\displaystyle\leq\tanh(\left|F_{C}\right|/4).$ (62)

_Asymmetric edge parameters_ : For perturbations of all the $F_{i+1,i}$ all
the way around the ring, one can deduce from Eq. (21) of Ref. Owen et al.
(2020) using techniques in that paper an equality of the form,

$\displaystyle\left|\sum_{i=0}^{N-1}\frac{\partial\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial F_{i+1,i}}\right|$
$\displaystyle\leq N,$ (63)

although this expression does not explicitly appear.

It is the continuous limits of these formulas that are operative for diffusion
processes. To make this connection, we first have to develop the mapping
between this discrete Markov jump process and its limit as a continuous
diffusion process, obtained as the spacing between lattice points tends to
zero, $\Delta x\to 0$.

###  VI.2 Discrete approximation of a continuous diffusion process

Motivated by the structure of the decomposition of the transition rate matrix
in (59), we now look to construct a Markov jump process that has a well
defined continuous limit as a diffusion process, and that maintains that
structure.

To begin, we first introduce a smooth probability density $\rho(x,t)$ such
that the probability the system is between $x_{i}-\Delta x/2$ and
$x_{i}+\Delta x/2$ at time $t$ is given by $\rho(x_{i},t)=p_{i}(t)/\Delta x$.
We also introduce three more smooth functions of space $E(x)$, $B(x)$ and $f$
such that

$\displaystyle E(x_{i})=E_{i}$ (64) $\displaystyle
B(x_{i})=B_{i+1,i}=B_{i,i+1}$ (65) $\displaystyle f\Delta
x=F_{i+1,i}=-F_{i,i+1}.$ (66)

Notice that we have assigned the “location” of $B_{i+1,i}$ to the position
with the smaller index. In addition, to have a well-defined limit the
asymmetric edge parameters need to be linear in $\Delta x$, and a constant
value is sufficient to include all possible nonconservative effects. In terms
of these functions, the transition rates (59) become

$\displaystyle W_{+}(x_{i})\equiv W_{i+1,i}=e^{-(B(x_{i})-E(x_{i})-f\Delta
x/2)}$ (67) $\displaystyle W_{-}(x_{i})\equiv
W_{i-1,i}=e^{-(B(x_{i-1})-E(x_{i})+f\Delta x/2)}=e^{-(B(x_{i}-\Delta
x)-E(x_{i})+f\Delta x/2)},$ (68)

and thermodynamic force (58) simplifies to $F_{C}=\sum_{i=0}^{N-1}f\Delta
x=fL$.

With this setup the procedure to carry out the limit $\Delta x\to 0$ is as
follows: We substitute these definitions into the Master equation (56), expand
for small $\Delta x$, and then diffusively rescale time $t\to t/\Delta x^{2}$.
The result is the Fokker-Planck equation

$\displaystyle\partial_{t}\rho(x,t)$
$\displaystyle=-\partial_{x}\left[e^{-{\bm{(}}B(x)-E(x){\bm{)}}}(E^{\prime}(x)-f)\rho(x,t)\right]+\partial_{x}\left[e^{-{\bm{(}}B(x)-E(x){\bm{)}}}\partial_{x}\rho(x,t)\right].$
(69)

This is of the form in (1) [main text]. Codifying the observation that
interesting results in the discrete case correspond to separate perturbations
in the $E$, $B$, and $f$ functions, then suggests the decomposition introduced
in (2) [main text] via the identification

$\mu(x)=\exp{\bm{(}}E(x)-B(x){\bm{)}},\quad U(x)=E(x).$ (70)

###  VI.3 Diffusion limits of thermodynamic bounds

Having established a consistent discretization of our diffusion process, we
turn to utilizing the thermodynamic bounds for discrete Markov processes (61)
- (63) to prove the analogous thermodynamic bounds for the continuous limit.

In make this connection, we will repeatedly face the situation where we have
to convert a derivative with respect to a finite collection of variables, like
the $\\{E_{i}\\}$ or $\\{B_{i+1,i}\\}$, into a functional derivative as the
spacing tends to zero ($\Delta x\to 0$). In preparation for these
calculations, we first present this relationship in general and then exploit
it in the following. To this end, let us consider two smooth functions $f(x)$
and $g(x)$, and the functional ${\mathcal{I}}[f]$. In the discrete picture, we
only evaluate these functions at the positions $x_{i}$, with values $f(x_{i})$
and $g(x_{i})$. The functional is then a function
${\mathcal{I}}{\bm{(}}\\{f(x_{i})\\}{\bm{)}}$ of the finite set of values
$\\{f(x_{i})\\}$, but is assumed to tend smoothly to
${\mathcal{I}}{\bm{(}}\\{f(x_{i})\\}{\bm{)}}\to{\mathcal{I}}[f]$ as $\Delta
x\to 0$. With this setup, as $\Delta x\to 0$ the definitions of the derivative
and functional derivative are connected by

$\displaystyle\lim_{\Delta x\to
0}\sum_{i=0}^{N-1}g(x_{i})\frac{\partial{\mathcal{I}}{\bm{(}}\\{f(x_{i})\\}{\bm{)}}}{\partial
f(x_{i})}=\int_{0}^{L}g(x)\frac{\delta{\mathcal{I}}[f]}{\delta f(x)}dx.$ (71)

Let us now address each type of perturbation in turn.

_Vertex parameters_ : For the vertex derivatives, we first replace
$E_{i}=E(x_{i})$ and $V_{i}=V(x_{i})$, and then take the continuous limit

$\lim_{\Delta x\to 0}\sum_{j=0}^{N-1}V_{j}\frac{\partial\langle
Q\rangle}{\partial E_{j}}=\lim_{\Delta x\to
0}\sum_{j=0}^{N-1}V(x_{j})\frac{\partial\langle Q\rangle}{\partial
E(x_{j})}=\int_{0}^{L}V(x)\frac{\delta\langle Q\rangle}{\delta E(x)}dx,$ (72)

where we used (71) with ${\mathcal{I}}=\langle Q\rangle$, $f(x)=E(x)$, and
$g(x)=V(x)$. Inserting this expression into (61) and applying the
identification $U(x)=E(x)$ (70), we arrive at expression equivalent to Eq.
(14) [main text].

_Symmetric edge parameters_ : When we perturb all the symmetric edge
parameters between positions $x_{a}=a\Delta x$ and $x_{b}=b\Delta x$, we
obtain the response in the continuous limit by first replacing
$B_{i+1,i}=B(x_{i})$, and then

$\displaystyle\lim_{\Delta x\to 0}\sum_{i=a}^{b-1}\frac{\partial\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial B_{i+1,i}}=\lim_{\Delta x\to
0}\sum_{i=a}^{b-1}\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial B(x_{i})}=\int_{x_{a}}^{x_{b}}\frac{\delta\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\delta B(x)}dx,$ (73)

where we have utilized (71), with ${\mathcal{I}}=\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)$, $f(x)=B(x)$, and $g(x)=I_{[a,b]}(x)$ is
the indicator function on the set $x\in[a,b]$. Substituting into (62), noting
the change of variables $\delta\ln\mu(x)=-\delta B(x)$ (with $E(x)$ fixed)
from (70), and that the sole thermodynamic force is $F_{C}=fL$ we arrive at
Eq. (7) [main text].

_Aymmetric edge parameters_ : Lastly, for asymmetric edge perturbations, we
link the $f$-perturbations via

$\left|\frac{\partial\ln(\langle Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial
f}\right|=\left|\lim_{\Delta x\rightarrow
0}\sum_{i=1}^{N}\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial(F_{i+1,i}/\Delta x)}\right|\leq\lim_{\Delta x\to
0}N\Delta x=L,$ (74)

where the inequality is due to (63), and the desired result Eq. (8) [main
text] follows.

###  VI.4 Failure of bounds in the continuous limit for higher dimensions

It turns out that the results known for discrete Markov process Owen et al.
(2020) are not sufficient to constrain the steady-state response of diffusion
processes in higher dimensions.

To demonstrate this possibility, we focus here on a two-dimensional diffusion
process with positions $(x,y)$ on a torus whose circumferences in both
directions are $L$. As before, we discretize the dynamics by placing the
evolution on a square lattice with lattice spacing $l$, and discretized
positions $(x_{i},y_{j})=(il,jl)$. The transition rates are only nonzero for
nearest neighbor hops in the positive and negative $x$ and $y$ directions.
Motivated by our previous discussion we introduce the smooth functions defined
on the torus, $B_{x}(x,y)$, $B_{y}(x,y)$, $E(x,y)$, $f_{x}$ and $f_{y}$,
allowing us to specify the transition rates

$\displaystyle W_{i+1,i}^{j}$
$\displaystyle=e^{-{\bm{(}}B_{x}(x_{i},y_{j})-E(x_{i},y_{j})-f_{x}l/2{\bm{)}})}$
(75) $\displaystyle W_{i-1,i}^{j}$
$\displaystyle=e^{-{\bm{(}}B_{x}(x_{i-1},y_{j})-E(x_{i},y_{j})+f_{x}l/2{\bm{)}}}$
(76) $\displaystyle W_{i}^{j+1,j}$
$\displaystyle=e^{-{\bm{(}}B_{y}(x_{i},y_{j})-E(x_{i},y_{j})-f_{y}l/2{\bm{)}}}$
(77) $\displaystyle W_{i}^{j-1,j}$
$\displaystyle=e^{-{\bm{(}}B_{y}(x_{i},y_{j-1})-E(x_{i},y_{j})+f_{y}l/2{\bm{)}}}.$
(78)

For similar reasons as above, these rates limit to a diffusion process as
$l\to 0$

Now imagine we perturb all the symmetric edge parameters in a square region
from $x_{a}=al$ to $x_{b}=bl$ and from $y_{a^{\prime}}=a^{\prime}l$ to
$y_{b^{\prime}}=b^{\prime}l$, totaling $N_{\rm
e}=(b-a)(b^{\prime}-a^{\prime}-1)+(b-a-1)(b^{\prime}-a^{\prime})$ edges. Now
Eq. (20) of Ref. Owen et al. (2020) predicts that the response is no worse
than the number of vertices on the perimeter of this region $N_{\rm
p}=2(b-a+b^{\prime}-a^{\prime})$ as

$\left|\sum_{i=a}^{b-1}\sum_{j=a^{\prime}}^{b^{\prime}-1}\left(\frac{\partial\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial
B_{x}(x_{i},y_{j})}+\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial B_{y}(x_{i},y_{j})}\right)\right|\leq N_{\rm p}-1.$
(79)

In the continuous limit $l\to 0$, the left hand side tends to a finite value
given by the functional derivative

$\lim_{l\to
0}\sum_{i=a}^{b-1}\sum_{j=a^{\prime}}^{b^{\prime}-1}\left(\frac{\partial\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\partial
B_{x}(x_{i},y_{j})}+\frac{\partial\ln(\langle Q_{1}\rangle/\langle
Q_{2}\rangle)}{\partial
B_{y}(x_{i},y_{j})}\right)=\int_{x_{a}}^{x_{b}}\int_{y_{a^{\prime}}}^{y_{b^{\prime}}}\frac{\delta\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\delta B_{x}(x,y)}+\frac{\delta\ln(\langle
Q_{1}\rangle/\langle Q_{2}\rangle)}{\delta B_{y}(x,y)}dydx.$ (80)

However, the right hand side tends to infinity, since the number of vertices
on the perimeter grows without bound as the spacing tends to zero. Thus, the
inequalities derived in Ref. Owen et al. (2020) for discrete Markov processes
are uninformative in the continuous limit in dimensions above one.

## References

  * Owen et al. (2020) J. A. Owen, T. R. Gingrich, and J. M. Horowitz, Phys. Rev. X 10, 011066 (2020).
  * Gardiner (2004) C. W. Gardiner, _Handbook of Stochastic Methods for Physics, Chemistry and the Natural Sciences_ (Springer-Verlag, New York, 2004), 3rd ed.
  * (3) M. Aleandri, M. Colangeli, and D. Gabrielli, arXiv:2002.00654v1.