# Optimization and Growth in First-Passage Resetting

B. De Bruyne  Perimeter Institute, 31 Caroline Street North, Waterloo, ON, N2L
2Y5, Canada  LPTMS, CNRS, Univ. Paris-Sud, Université Paris-Saclay, 91405
Orsay, France  J. Randon-Furling  SAMM, Université Paris 1 – FP2M (FR2036)
CNRS, 75013 Paris, France  Department of Mathematics, Columbia University, New
York, NY 10027, USA  MSDA, Mohammed VI Polytechnic University, Ben Guerir
43150, Morocco  S. Redner  Santa Fe Institute, 1399 Hyde Park Rd., Santa Fe,
New Mexico 87501, USA

###### Abstract

We combine the processes of resetting and first-passage to define _first-
passage resetting_ , where the resetting of a random walk to a fixed position
is triggered by a first-passage event of the walk itself. In an infinite
domain, first-passage resetting of isotropic diffusion is non-stationary, with
the number of resetting events growing with time as $\sqrt{t}$. We calculate
the resulting spatial probability distribution of the particle analytically,
and also obtain this distribution by a geometric path decomposition. In a
finite interval, we define an optimization problem that is controlled by
first-passage resetting; this scenario is motivated by reliability theory. The
goal is to operate a system close to its maximum capacity without experiencing
too many breakdowns. However, when a breakdown occurs the system is reset to
its minimal operating point. We define and optimize an objective function that
maximizes the reward (being close to maximum operation) minus a penalty for
each breakdown. We also investigate extensions of this basic model to include
delay after each reset and to two dimensions. Finally, we study the growth
dynamics of a domain in which the domain boundary recedes by a specified
amount whenever the diffusing particle reaches the boundary after which a
resetting event occurs. We determine the growth rate of the domain for the
semi-infinite line and the finite interval and find a wide range of behaviors
that depend on how much the recession occurs when the particle hits the
boundary.

##  1 Introduction

Random walks are ubiquitous in phenomena across a wide range of fields, such
as physics, chemistry, finance and social sciences [1, 2, 3, 4]. In addition
to the applications of the random walk, many useful extensions of the basic
model have been developed (see, e.g., [5, 6, 7, 8]). Almost a decade ago, the
notion of resetting of a random walk was introduced [9, 10, 11]. The basic
idea of resetting is simplicity itself: at a given rate, a random walk is
reset to its starting point. The rich phenomenology induced by this extension
of the random walk has sparked much interest (see, e.g., [9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19]). In the context of search strategies, where the
walker is searching for a target at a fixed location, resetting changes the
average search time from being infinite (in an infinite domain) to finite [20,
21, 22, 23]. Moreover, there exists an optimal resetting rate that minimizes
the search time. A very different, but also fruitful concept in the theory of
random walks is the notion of a first-passage process [24, 2, 25, 26]. Of
particular importance is first-passage probability, which is defined as the
probability that a walker reaches a specified location for the first time.
This notion has many applications where a particular event happens when a
threshold is first reached. One such example is a limit order for a stock.
When the price of a stock, whose evolution is often modeled as a geometric
random walk, first reaches a limit price, this event triggers the sale or the
purchase of the stock.

In this work, we combine these disparate notions of first passage and
resetting into _first-passage resetting_ , in which the particle is reset
whenever it reaches a specified threshold. Contrary to standard resetting, the
time at which first-passage resetting occurs is defined by the motion of the
diffusing particle itself rather than being imposed externally [9, 10, 11].
Feller showed that such a process is well defined mathematically and provided
existence and uniqueness theorems [27], while similar ideas were pursued in
[28]. First-passage resetting was initially treated in the physics literature
for the situation in which two Brownian particles are biased toward each other
and the particles are reset to their initial positions when they encounter
each other [29].

In our work, we first focus on the related situation of a diffusing particle
on the semi-infinite line $x\leq L$ that is reset to the origin whenever the
particle hits the boundary $x=L$ (Fig. 1). The model studied in [29]
corresponds to a drift toward the boundary in our semi-infinite geometry; this
setting leads to a stationary state. In contrast, the absence of drift in our
model leads to a variety of new phenomena. In particular, the probability
distribution for the position of the particle is non-stationary. We also
construct two simple path decompositions for first-passage resetting, in which
the trajectory of the resetting particle is mapped onto a free diffusion
process. This approach provides useful geometrical insights, as well as simple
ways to derive the average number of reset events and the spatial probability
distribution with essentially no calculation.

We then treat first-passage resetting on a finite interval, which has a
natural application to reliability theory. Here the particle is restricted to
the interval $[0,L]$ where $x=L$ is again the boundary where resetting occurs
and the particle is immediately reinjected at $x=0$ when it reaches $x=L$. We
may view this mechanism as characterizing the performance of a driven
mechanical system [30, 31, 32, 33], with the coordinates $x=0$ and $x=L$
indicating poor and maximal performance, respectively. While one ideally wants
to operate the system close to its maximum performance level ($x=L$), there is
a risk of overuse, leading to breakdowns whenever $x=L$ is reached.
Subsequently, the system has to be repaired and then restarted from $x=0$.
This dynamics corresponds to resetting that is induced by a first passage to
the boundary $x=L$. We will find the optimal bias velocity that optimizes the
performance of the system. We will also investigate additional features of
this first-passage resetting, such as a random maintenance delay at each
breakdown and resetting in higher dimensions. A preliminary account of some of
these results was given in [34].

Finally, we investigate a very different aspect of first-passage resetting
where the domain boundary at which resetting occurs moves by a specified
amount each time the diffusing particle reaches this boundary. Many features
of this moving boundary problem can be readily calculated because of the
renewal structure of the theory. For both the semi-infinite and finite
interval geometries, we find a variety of scaling behaviors for the motion of
the resetting boundary. These behaviors depend on the initial geometry and by
how much the boundary moves at each resetting event.

##  2 First-Passage Resetting in the Semi-Infinite Geometry

In the standard resetting process, reset events occur at a fixed rate $r$ that
are uncorrelated with the position of the diffusing particle. In contrast,
first-passage resetting directly couples the times at which resetting occurs
and the particle position. For first-passage resetting in the semi-infinite
line geometry, the particle starts at $x(0)=0$ and freely diffuses in the
range $x\leq L$ (with $L>0$). Each time $L$ is reached, the particle is
instantaneously reset to the origin (Fig. 1). We are interested in two basic
characteristics of the particle motion: the spatial probability distribution
of the particle and the time dependence of the number of reset events. To
compute these quantities, we rely on the renewal structure of the process,
which allows us to first compute the probability for $n$ reset events in a
direct way. With this result, as well as the propagator for free diffusion in
the presence of an absorbing boundary, we can readily obtain the spatial
distribution of the particle and the average number of resets up to a given
time.

Figure 1: Schematic illustration of first-passage resetting for diffusion on
the semi-infinite line $x\leq L$. Each time the particle reaches the threshold
$L$, it is reset to the origin. The times of the resetting events are denoted
by $t_{1},t_{2},\ldots$.

###  2.1 The $n^{\text{t}h}$ reset probability distribution

Define $F_{n}(L,t)$ as the probability that the particle resets for the
$n^{\text{t}h}$ time at time $t$. When $n=1$, this quantity is the standard
first-passage probability for a freely diffusing particle that starts at the
origin, to first reach $L$ [25, 26]:

$\displaystyle F_{1}(L,t)\equiv F(L,t)=\frac{L}{\sqrt{4\pi
Dt^{3}}}\,e^{-L^{2}/4Dt}\,.$

For the particle to reset for the $n^{\text{th}}$ time at time $t$ with $n>1$,
the particle must reset for the $(n-1)^{\text{th}}$ time at some earlier time
$t^{\prime}<t$, and reset one more time at time $t$. Because the process is
renewed at each reset, $F_{n}(L,t)$ is formally given by the renewal equation

$\displaystyle
F_{n}(L,t)=\int_{0}^{t}dt^{\prime}\,F_{n-1}(L,t^{\prime})\,F_{1}(L,t-t^{\prime}),\qquad
n>1\,.$ (1a) The convolution structure of Eq. (1a) lends itself to a Laplace
transform analysis. The corresponding equation in the Laplace domain is
simply:
$\displaystyle\widetilde{F}_{n}(L,s)=\widetilde{F}_{n-1}(L,s)\widetilde{F}_{1}(L,s)=\widetilde{F}_{1}(L,s)^{n},$
(1b)

where quantities with tildes denote Laplace transforms. Using the Laplace
transform of the first-passage probability:

$\displaystyle\widetilde{F}_{1}(L,s)=\int_{0}^{\infty}dt\,F_{1}(L,t)\,e^{-s\,t}=e^{\sqrt{sL^{2}/D}}\equiv
e^{-\ell}\,,$

where we define the scaled coordinate $\ell\equiv\sqrt{s/D}\,L$ henceforth,
then Eq. (1b) becomes

$\displaystyle\widetilde{F}_{n}(L,s)=e^{-n\ell}\,.$

Notice that $\widetilde{F}_{n}(L,s)$ has the same form as
$\widetilde{F}_{1}(L,s)$ with $L\to nL$. That is, the time for a diffusing
particle to reset $n$ times at a fixed boundary $x=L$ is the same as the time
for a freely diffusing particle to first reach $x=nL$.

Figure 2: Relation between a first-passage path to $x=3L$ and a third-passage
path to $x=L$, with resetting each time $x=L$ is reached. The green and blue
paths in (b) have merely been shifted vertically downward by $L$ and $2L$
compared to (a), respectively.

In hindsight, this equivalence between the first-passage probability to $x=nL$
and the $n^{\text{t}h}$-passage probability to $x=L$ with resetting at $x=L$
is self evident. As indicated in Fig. 2 for the case $n=3$, a first-passage
path from 0 to $3L$ is composed of a first-passage path from 0 to $L$,
followed by a first-passage path from $L$ to $2L$, and finally a first-passage
path from $2L$ to $3L$. Resetting causes each of these three segments to
(re)start from the origin. Thus the point $x=L$ is first reached for the third
time after resetting by these displaced paths.

###  2.2 Spatial probability distribution

We now compute the probability distribution of the diffusing particle at time
$t$, $P(x,t)$, on the semi-infinite line $x\leq L$ under the influence of
first-passage resetting. This distribution can be obtained in several ways.
Here we make use of the path transformation discussed above for the derivation
of the $n^{\rm th}$ passage probability (see Fig. 2). (A calculational
approach based on Laplace transforms that relies on the renewal structure of
the process is given in A.) When the walker is at position $x\in[0,L]$ at time
$t$ and has experienced exactly $n$ resets, this is equivalent to a free
particle being at position $x+nL$ without having reached level $(n+1)L$. As a
result, the corresponding probability is that of a free particle with position
$x(t)=x+nL$ and running maximum position $M(t)<(n+1)L$; the latter is defined
by $M(t)=\max_{t^{\prime}\leq t}x(t^{\prime})$. The joint distribution of the
position and maximum, $x(t)=x$ and $M(t)=m$, is given by

$\Pi(x,m,t)=\frac{2m-x}{\sqrt{4\pi D^{3}t^{3}}}\,\,e^{-(2m-x)^{2}/4Dt}\,.$ (2)

This formula was established by Lévy [35, 36] and it can be derived by using
the reflection property of Brownian motion. From this joint probability, we
find that

$\displaystyle P(x,t)$ $\displaystyle=\sum_{n\geq
0}\mathrm{Prob}\big{(}M(t)<(n+1)L\text{ and }x(t)=x+nL\big{)}$
$\displaystyle=\sum_{n\geq 0}\,\int_{x+nL}^{(n+1)L}\,dm\,\Pi(x+nL,m,t)$
$\displaystyle=\frac{1}{\sqrt{4\pi Dt}}\,\sum_{n\geq
0}\left[e^{-(x+nL)^{2}/4Dt}-e^{-[x-(n+2)L]^{2}/4Dt}\right],\qquad 0<x\leq
L\,.$ (3)

While this expression is exact, it is not in a convenient form to determine
its long-time behavior. However, the long-time limit of $P(x,t)$ is simple to
obtain by expanding its Laplace transform (see [34]) for small $s$,

$\displaystyle\widetilde{P}(x,s)\simeq\frac{1}{\sqrt{Ds}}\,\frac{L-x}{L}\qquad
0\leq x\leq L\,,\quad s\rightarrow 0\,,$ (4a) from which the inverse Laplace
transform is $\displaystyle P(x,t)\simeq\frac{1}{\sqrt{\pi
Dt}}\,\frac{L-x}{L}\qquad 0\leq x\leq L\,,\quad t\rightarrow\infty\,.$ (4b)

The linear $x$ dependence arises from the balance between the diffusive flux
that exits at the reset point $x=L$ and this same flux being re-injected at
$x=0$.

For $x<0$, the integral in the second line of (2.2) ranges from $nL$ to
$(n+1)L$ rather than from $x+nL$. This leads to

$\displaystyle P(x,t)$ $\displaystyle=\frac{1}{\sqrt{4\pi Dt}}\,\sum_{n\geq
0}\left[e^{-(x-nL)^{2}/4Dt}-e^{-[x-(n+2)L]^{2}/4Dt}\right]$
$\displaystyle=\frac{1}{\sqrt{4\pi
Dt}}\,\left[e^{-x^{2}/4Dt}+e^{-(x-L)^{2}/4Dt}\right],\qquad x<0\,.$ (5)

Thus the probability distribution is merely the sum of two Gaussians. In the
long-time limit and for $|x|/\sqrt{4Dt}\gg 1$, the factor $L$ in the second
term becomes irrelevant and the distribution reduces to that of diffusion on
the half line in the presence of a reflecting boundary.

Figure 3: Schematic space-time trajectory of diffusion with first-passage
resetting on a semi-infinite line. (a) A path with an odd number of resets is
equivalent to (c) a freely diffusing path that starts at $x(t\\!=\\!0)=L$. (b)
A path with an even number of resets is equivalent to (d) a freely diffusing
path that starts from $x(t\\!=\\!0)=0$. This equivalence underlies the spatial
probability distribution for $x<0$ in Eq. (2.2).

An appealing way to obtain the probability distribution (2.2) is by a path
decomposition construction. Consider the original resetting problem and
partition all trajectories into those that undergo either an odd or an even
number of resets. In the former case, we invert the segments _before_ each
odd-numbered resetting about the origin and then translate each such segment
by a distance $+L$ (blue arrows in Fig. 3(a) & (c)). As shown in this portion
of the figure, the resulting trajectory is simply a Brownian path that starts
at $x=L$ and propagates freely to its final position $x$. For a path that
consists of an even number of resetting events, we perform this same inversion
and translation on the segments _after_ each odd-numbered resetting (green
arrows in Fig. 3(b) & (d)). The resulting trajectory is now a Brownian path
that starts at the origin and propagates freely to its final position $x$. It
is worth emphasizing that this decomposition applies for any symmetric and
continuous stochastic process (provided it is homogeneous and stationary).

###  2.3 Average number of resets

To find the average number of resets that occur up to time $t$, we first
compute the probability that $n$ resets have occurred during this time. By
relying on the path transformation shown in Fig. 2, we find that the
probability for exactly $n$ resets to occur by time $t$ equals the probability
for a freely diffusing particle to have a running maximum $M(t)$ greater than
$nL$ but less than $(n+1)L$, that is:

$\displaystyle P(N(t)=n)$ $\displaystyle=\mathrm{Prob}\big{(}nL\leq
M(t)<(n+1)L\big{)}\,.$ (6)

The distribution of $M(t)$ is known [37, 38, 36] and may be readily rederived
from (2),

$\displaystyle P(M(t)=m)=\frac{1}{\sqrt{\pi Dt}}\,\,e^{-m^{2}/4Dt}\,,$

from which it follows that

$P\big{(}N(t)\\!=\\!n\big{)}=\mathrm{erf}\left(\frac{(n+1)L}{\sqrt{4Dt}}\right)-\mathrm{erf}\left(\frac{nL}{\sqrt{4Dt}}\right),$
(7)

where $\mathrm{erf}$ is the Gauss error function.

We can compute the average number of reset events,
$\mathcal{N}(t)\equiv\left\langle N(t)\right\rangle$, from (7), but it is
quicker to use again the mapping with the running maximum of free diffusion.
Indeed, writing $\mathcal{M}(t)$ for the average maximum position of a freely
diffusing particle up to time $t$, one has

$\mathcal{N}(t)\,L\,\leq\,\mathcal{M}(t)\,<\,\left[\mathcal{N}(t)+1\right]\,L.$
(8)

Since $\mathcal{M}(t)=\sqrt{4Dt/\pi}$, we find that the long-time behavior of
$\mathcal{N}(t)$ is given by

$\displaystyle\mathcal{N}(t)\simeq\sqrt{4Dt/\pi L^{2}}\,.$ (9)

The resetting process is non-stationary, as the number of reset events grows
as $\sqrt{t}$.

###  2.4 Biased diffusion

The case where the diffusing particle is biased towards the resetting boundary
is equivalent to the model studied by Falcao and Evans [29]. Here we briefly
discuss the complementary situation in which the particle is biased away from
the resetting boundary, with drift velocity $v<0$. In the absence of
resetting, a particle that starts at the origin eventually reaches $x=L$ with
probability $H=e^{-\text{Pe}}$ and escapes to $x\\!=\\!-\infty$ with
probability $1-H$ [24, 25], where $\text{Pe}\equiv vL/2D$ is the Péclet number
(the dimensionless bias velocity). When resetting can occur, $H$ now becomes
the probability that a resetting event actually happens. Consequently, the
probability that the particle resets exactly $n$ times is given by
$R_{n}=H^{n}(1-H)$. The average number of resetting events before ultimate
escape therefore is [35]

$\displaystyle\mathcal{N}(t)=\sum_{n}nR_{n}=\frac{H}{1-H}=\frac{1}{e^{\text{Pe}}-1}\,.$
(10)

In the limit of $v\to 0$, the number of resetting events diverges as
$\mathcal{N}(t)\simeq 2D/(vL)$. The time between resetting events is known to
be $L/v$ [25, 39]. Thus after typically $1/(e^{\text{Pe}}-1)$ resetting
events, each of which requires a time of $L/v$, the particle escapes to
$-\infty$.

We can also compute the spatial probability distribution of the particle when
it undergoes biased diffusion with bias velocity of magnitude $v$. We make use
again of the path transformation shown in Fig. 2. For a Brownian particle with
drift $v$, the analog of Eq. (2) is

$\Pi(x,m,t)=\frac{2m-x}{\sqrt{4\pi
D^{3}t^{3}}}\,\,e^{-(2m-x)^{2}/4Dt}\,e^{\text{Pe}\,\left(x/L-\text{Pe}\,Dt/L^{2}\right)}\,,$
(11)

which leads to

$\displaystyle P(x,t)$ $\displaystyle=\frac{1}{\sqrt{4\pi Dt}}\,\sum_{n\geq
0}\left[e^{-(x+nL)^{2}/4Dt}-e^{-[x-(n+2)L]^{2}/4Dt}\right]\,e^{\text{Pe}\,\left(x/L+n-\text{Pe}\,Dt/L^{2}\right)},\quad
0<x\leq L\,$ (12a) and $\displaystyle P(x,t)$
$\displaystyle=\frac{1}{\sqrt{4\pi Dt}}\,\sum_{n\geq
0}\left[e^{-(x-nL)^{2}/4Dt}-e^{-[x-(n+2)L]^{2}/4Dt}\right]\,e^{\text{Pe}\,\left(x/L+n-\text{Pe}\,Dt/L^{2}\right)},\quad
x<0\,.$ (12b)

In contrast to the driftless case, there is no simplification for the
probability distribution when $x<0$. In particular, the path transformation of
Fig. 3 requires symmetry and thus does not hold in the presence of drift.

##  3 Optimization in First-Passage Resetting

###  3.1 The finite interval

We now introduce an optimization problem that is induced by first-passage
resetting. We envisage a mechanical system whose operating coordinate $x(t)$
lies in the range $[0,L]$. Increasing the value of $x$ corresponds to
increasing its level of operation, and it is desirable to run the system as
close as possible to its maximum capacity $L$. However, the system breaks down
whenever $x$ reaches $L$, after which repairs have to be made before the
system can restart its operation from $x=0$. While the dynamics of the
operating coordinate is typically complicated and dependent on multiple
parameters, we view the coordinate $x$ has undergoing a drift-diffusion
process for the sake of parsimonious modeling. For the system to be close to
$x=L$, the drift should be positive. On the other hand, breakdowns of the
system are to be avoided because a cost is incurred with each breakdown. This
suggests that the drift velocity should be negative. The goal is to determine
the optimal operation that maximizes the performance of the system as a
function of the cost for each breakdown and the drift velocity. Although the
analogy between first-passage resetting and a mechanical system is naive, this
formulation allows us to determine the optimal operation in a concrete way.

The basic control parameter is the magnitude of the drift velocity. If the
velocity is large and negative, the system is under-exploited because it
operates far from its maximum capacity. Conversely, if the velocity is large
and positive, the system breaks down often. We seek the optimal operation by
maximizing an objective function $\mathcal{F}$ that rewards high performance
and penalizes breakdowns. A natural choice for $\mathcal{F}$ is

$\displaystyle\mathcal{F}=\lim_{T\to\infty}\frac{1}{T}\left[\frac{1}{L}\int_{0}^{T}x(t)\,dt-C\,\mathcal{N}(T)\right]\,,$
(13)

where $T$ is the total operation time, $\mathcal{N}(T)$ is the average number
of breakdowns within a time $T$, with $T$ much longer than the mean breakdown
time, and $C$ is the cost of each breakdown. As defined, this objective
function rewards operation close to the maximum point $L$ and penalizes
breakdowns.

We now determine this objective function when the operating coordinate $x(t)$
evolves according to drift-diffusion, with the additional constraint that
$x(t)$ is reset to zero whenever $x$ reaches $L$. Mathematically, we need to
solve the convection-diffusion equation with the following additional
conditions: (i) a $\delta$-function source at the origin whose magnitude is
determined by the outgoing flux $j(x)=-D\partial_{x}c+vc$ at $x=L$, (ii) a
reflecting boundary condition at $x=0$, and (iii) the initial condition
$x(t=0)=0$. That is, we want to solve

$\displaystyle\partial_{t}c+v\partial_{x}c=D\partial_{xx}c+\delta(x)(-D\partial_{x}c+vc)\big{|}_{x=L}\,,$
(14a) subject to
$\displaystyle\begin{cases}\left(D\partial_{x}c-vc\right)\rvert_{x=0}=\delta(t)\,&\\\
c(L,t)=0\,&\\\ c(x,0)=0\,&\end{cases}.$ Here, $c\equiv c(x,t)$ is the
probability density for the operating coordinate, the subscripts denote
partial differentiation, $D$ is the diffusion coefficient, and $v$ is the
drift velocity. Notice that the reflecting boundary condition at $x=0$ holds
_except_ at the start of the process to account for the unit input of flux at
$t=0$. This construction allows one to take the initial condition to be
$c(x,t\\!=\\!0)=0$, which greatly simplifies all calculations. Effectively,
this flux initial condition corresponds to starting the system with the
particle at $x=0$.

As in Sec. 2, we first solve the free theory, where the delta-function term in
(14a) is absent, and then use renewal equations to solve the full problem. In
the free case, Eq. (14a) becomes, in the Laplace domain:

$\displaystyle
s\widetilde{c_{0}}+v\partial_{x}\widetilde{c_{0}}=D\partial_{xx}\widetilde{c_{0}}\,,$
(14b)

and is subject to the boundary conditions

$\displaystyle\begin{cases}\left(D\partial_{x}\widetilde{c_{0}}-v\widetilde{c_{0}}\right)\rvert_{x=0}=1\,\\\
\widetilde{c_{0}}(L,s)=0\,,\end{cases}$

where the subscript $0$ denotes the concentration without flux re-injection.
The solution to (14b) is standard and the result is (see B for details):

$\displaystyle\widetilde{c}_{0}(x,s)$
$\displaystyle=\frac{2\,e^{\text{P}}\sinh\left[w(L-x)\right]}{\mathcal{W}}\,,$
(15a) where $\text{P}\equiv vx/2D$, $w=\sqrt{v^{2}+4Ds}/2D$ and
$\mathcal{W}=2Dw\cosh(Lw)+v\sinh(Lw)$. In terms of $\widetilde{c}_{0}$, the
Laplace transform of the first-passage probability to $x=L$ is
$\displaystyle\widetilde{F}_{1}(L,s)$
$\displaystyle=\left(-D\partial_{x}\widetilde{c}_{0}+v\widetilde{c}_{0}\right)\rvert_{x=L}=\frac{2\,D\,w\,e^{\text{Pe}}}{\mathcal{W}}\,,$
(15b)

where again $\text{Pe}=vL/2D$ is the Péclet number. With re-injection of the
outgoing flux, the concentration obeys the renewal equations. In the Laplace
domain and using $\widetilde{c}_{0}$ above from the free theory, we obtain

$\displaystyle\widetilde{c}(x,s)=\frac{\widetilde{c}_{0}(x,s)}{1-F_{1}(L,s)}=\frac{2\,e^{\text{P}}\sinh[w(L-x)]}{\mathcal{W}-2D\,w\,e^{\text{Pe}}}\,,$
(16)

where we substitute in the results from Eqs. (15) to obtain the final result.

Figure 4: The stationary distribution for the first-passage resetting process
on the interval $[0,1]$ for different Péclet numbers.

Contrary to the semi-infinite case, a stationary distribution is attained on
the finite interval. To determine this steady state, we use the duality
between the limits $s\to 0$ in the Laplace domain and $t\to\infty$ in the time
domain. With this approach, the coefficient of the term proportional to $1/s$
in $\widetilde{c}(x,s)$ gives the steady-state concentration, $c_{\text{ss}}$,
in the time domain:

$\displaystyle c_{\text{ss}}(x)$
$\displaystyle\simeq\frac{1}{L}\times\frac{1\,-\,e^{-2(\text{Pe}-\text{P})}}{1\,-\,\text{Pe}^{-1}\,e^{-\text{Pe}}\,\sinh\left(\text{Pe}\right)}\,,$
(17)

from which the normalized first moment in the steady state is

$\displaystyle\frac{\langle
x\rangle}{L}=\frac{1}{L}\int_{0}^{L}\\!\\!x\,c(x)\,dx=\frac{\left(2\text{Pe}^{2}-2\text{Pe}+1\right)\,e^{2\text{Pe}}-1}{2\text{Pe}\left[(2\text{Pe}-1)\,e^{2\text{Pe}}+1\right]}\,.$
(18)

Representative plots of the stationary-state concentration for different
Péclet numbers are given in Fig. 4. As one might anticipate, the density
profile is concentrated near $x=0$ for negative drift velocity, while for
positive drift there is a constant cycling of outgoing flux that is reinjected
at $x=0$, which leads to a nearly constant density profile.

Figure 5: The objective function versus Péclet number Pe for different
normalized cost values $C^{\prime}\equiv C/(L^{2}/D)$ for: (a) no delay upon
resetting, and (b) a dimensionless delay time of $\overline{\tau}=0.1$ at each
resetting. Indicated on each curve is the optimal operating point.

The average number of reset events $\mathcal{N}$ satisfies the renewal
equation (54a), and substituting in $\widetilde{F}_{1}$ from (15b), we obtain

$\displaystyle\widetilde{\mathcal{N}}(s)$
$\displaystyle=\frac{2D\,w\,e^{\text{Pe}}}{s\,\left[\mathcal{W}-2Dw\,e^{\text{Pe}}\right]}\,.$
(19a) We now extract the long-time behavior for the average number of times
that $x=L$ is reached by taking the limit $s\rightarrow 0$ of
$\widetilde{\mathcal{N}}(s)$ to give $\displaystyle\mathcal{N}(T)$
$\displaystyle\simeq\frac{4\text{Pe}^{2}}{2\text{Pe}-1+e^{-2\text{Pe}}}\,\,\frac{T}{L^{2}/D}\,.$
(19b)

Substituting these expressions for $\langle x\rangle/L$ and $\mathcal{N}$ into
(13) immediately gives the objective function, and representative plots are
shown in Fig. 5(a). For a given cost of a breakdown, there is an optimal drift
velocity or optimal Péclet number. The higher this cost, the smaller the
optimal bias and the value of $\mathcal{F}$. Moreover, the optimal bias is not
necessarily negative. Indeed, if the cost of a breakdown is relatively small,
then it is advantageous to operate the system close to its limit $L$ and
absorb the (small) cost of many breakdowns. On the contrary, if the cost of a
breakdown is high, it is better to run the system at low level and with a
negative bias to avoid breakdowns.

###  3.2 Time delay for repair

When a mechanical system breaks down, there is usually some downtime during
which repairs are made before the system can be restarted. Such a downtime can
naturally be incorporated into our model by including a random delay time
after each resetting event. Thus when the particle reaches $x=L$ and is
returned to $x=0$, we posit that the particle waits at the origin for a random
time $\tau$ that is drawn from the exponential distribution
$\sigma^{-1}e^{-\tau/\sigma}$ before the particle starts moving again. We now
determine the role of this delay on the optimal operation of the system.

The governing renewal equations can be readily extended to incorporate this
delay. This delay mechanism can also be viewed as the so-called “sticky”
Brownian motion [40, 41, 42] that is then combined with first-passage
resetting. When we include this delay, the renewal equation for the
probability distribution becomes:

$\displaystyle P(x,t)=G(x,L,t)+\int_{0}^{t}\\!\\!dt^{\prime}$ $\displaystyle
F_{1}(t^{\prime})\bigg{[}\delta_{0}(x)e^{-(t-t^{\prime})/\sigma}+\int_{0}^{t-t^{\prime}}\\!\\!\frac{d\tau}{\sigma}\,e^{-\tau/\sigma}P(x,t\\!-\\!t^{\prime}\\!-\\!\tau)\bigg{]}.$
(20a) This equation encapsulates the two possibilities for the subsequent
behavior of the particle when it first reaches $x=L$ at time $t^{\prime}$.
Either the particle remains at $x=0$ for the remaining time $t-t^{\prime}$ or
the particle waits for a time $\tau<t-t^{\prime}$ and then the process starts
anew from $(x,t)=(0,t^{\prime}+\tau)$ for the remaining time
$t-t^{\prime}-\tau$.

In a similar fashion, the renewal equation for the average number of resetting
events is

$\displaystyle\begin{split}\mathcal{N}(t)=\int_{0}^{t}dt^{\prime}&F_{1}(t^{\prime})\bigg{\\{}e^{-(t-t^{\prime})/\sigma}+\int_{0}^{t-t^{\prime}}\frac{d\tau}{\sigma}e^{-\tau/\sigma}\big{[}1+\mathcal{N}(t-t^{\prime}-\tau)\big{]}\bigg{\\}}.\end{split}$
(20b)

Equation (20b) accounts for the particle first hitting $L$ at time
$t^{\prime}$ and either waiting at the origin for the entire remaining time
$t-t^{\prime}$ or waiting there for a time $\tau<t-t^{\prime}$ and then
renewing the process for the remaining time. For this latter possibility,
there will be, on average, $1+\mathcal{N}(t-t^{\prime}-\tau)$ resetting
events.

Solving Eqs. (20) in the Laplace domain yields:

$\displaystyle\begin{split}\widetilde{P}(x,s)&=\frac{\delta(x)\,\sigma\widetilde{F}_{1}(s)+\widetilde{G}(x,s)(1+\sigma
s)}{1-F_{1}(s)+\sigma s}\,,\\\
\widetilde{\mathcal{N}}(s)&=\frac{\widetilde{F}_{1}(s)(\sigma+1/s)}{1+s\sigma-\widetilde{F}_{1}(s)}\,.\end{split}$
(21a) We now use the results from Sec. 3.1 for the optimization problem on the
interval with no delay. Namely, we substitute in Eqs. (21a) the first-passage
probability $\widetilde{F}_{1}(s)$ from Eq. (15b) and the probability
distribution in Eq. (15a) for $\widetilde{G}(x,s)$ to obtain
$\displaystyle\begin{split}\widetilde{P}(x,s)&=\frac{\delta(x)\,\sigma
2Dw+2\sinh(w(L-x))(1+\sigma s)}{\mathcal{W}e^{-\text{Pe}}(1+\sigma s)-2Dw},\\\
\widetilde{\mathcal{N}}(s)&=\frac{2Dw(\sigma+1/s)}{\mathcal{W}e^{-\text{Pe}}(1+s\sigma)-2Dw}\,.\end{split}$
(21b)

From the Laplace transform of the spatial probability density, we compute its
stationary distribution by taking the $s\rightarrow 0$ limit and obtain

$\displaystyle
P(x)\simeq\frac{e^{\text{Pe}}\,\text{Pe}\,(2\overline{\tau}\,\text{Pe}\,L\,\delta_{0}(x)+1)-\text{Pe}\,e^{\text{P}}}{e^{\text{Pe}}\left(\overline{\tau}\,\text{Pe}^{2}+\text{Pe}-1\right)+1}\frac{1}{L},$
(22a) where $\overline{\tau}=D\sigma/L^{2}$ is the dimensionless delay time.
From this distribution, the average position of the particle is
$\displaystyle\frac{\langle
x\rangle}{L}=\frac{\left[(\text{Pe}-2)\text{Pe}+2\right]e^{\text{Pe}}-2}{2\left[\text{Pe}(\overline{\tau}\,\text{Pe}^{2}+\text{Pe}-1)e^{\text{Pe}}+\text{Pe}\right]}\,.$
(22b)

Similarly, the average number of resetting events, or equivalently, the
average number of breakdowns in the long-time limit is

$\displaystyle\mathcal{N}=\frac{\text{Pe}^{2}}{\text{Pe}-1+\overline{\tau}\,\text{Pe}^{2}+e^{-\text{Pe}}}\,\frac{T}{L^{2}/D}\,.$
(23)

These two results, when substituted into Eq. (13), give an objective function
$\mathcal{F}$ whose qualitative features are similar to the case of no delay
(Fig. 5(b)). This behavior is what might anticipate, since delay may be viewed
as an additional form of cost.

The primary difference with the no-delay case is that the optimal Péclet
number and the corresponding optimal objective function $\mathcal{F}$ both
decrease as the delay time is increased (Fig. 5). Indeed, delay reduces the
number of resetting events/breakdowns, but also induces the coordinate to
remain closer to the origin. In the limit where the delay is extremely long,
the optimal Péclet number will be small. Moreover this optimal value will be
nearly independent of the cost per breakdown, as the particle will almost
never hit the resetting boundary.

###  3.3 Two dimensions

It is natural to extend the optimization problem on the interval to higher-
dimensional domains. Here, we treat the case where the domain is an annulus of
outer radius $L$, inner radius $a<L$, and the diffusing particle is reset to
$r=a$ whenever the outer domain boundary is reached. In analogy with the one-
dimensional problem, the particle also experiences drift velocity
$v(r)=v_{0}/r$. As we shall see, the choice of a potential flow field is
convenient because the velocity can be combined with the centrifugal term in
the Laplacian, which simplifies the form of the solution. The finite inner
radius is needed to eliminate the infinite-velocity singularity that would
occur if the inner radius was zero.

In close analogy with the finite-interval system, Eq. (14a), the equation of
motion for the particle is

$\displaystyle\partial_{t}c+\frac{v}{r}\partial_{r}c=D\left(\partial_{rr}c+\frac{1}{r}\partial_{r}c\right)+\delta_{a}(r)\big{[}2\pi
r(-D\partial_{r}c+vc)\big{]}\big{|}_{r=L}\,.$ (24)

Here, the flux term has a factor $2\pi r$ due to an integration over all
angles. In this geometry, the probability density of finding a particle at a
radius $r$ is $2\pi r\,c(r,t)$. We now introduce the dimensionless variables
$x=r/L$, $x_{0}=a/L$, the Fourier number $\text{Fo}=Dt/L^{2}$, and
$\text{Pe}=v_{0}/D$111Note that $v_{0}$ has units of velocity times length, so
this definition of the Péclet number is dimensionally correct. to transform
Eq. (24) into

$\displaystyle\partial_{\text{Fo}}c(x,\text{Fo})$
$\displaystyle=\partial_{xx}c(x,\text{Fo})+\frac{1-\text{Pe}}{x}\,\partial_{x}c(x,\text{Fo})+\delta_{x_{0}}(x)\big{[}2\pi
x(-\partial_{x}c+\text{Pe}\,c)\big{]}\big{\rvert}_{x=1},$ (25)

and the appropriate boundary conditions for this equation are

$\displaystyle\begin{cases}\big{[}\text{Pe}\,c(x,\text{Fo})-x\partial_{x}c(x,\text{Fo})\big{]}\big{\rvert}_{x=x_{0}}=\delta(\text{Fo})/(2\pi)\\\
c(1,\text{Fo})=0\\\ c(x,0)=0\,.\end{cases}$

By performing similar calculations as in the one-dimensional case we find the
following expression for the steady-state probability density in the time
domain (see C for the details):

$\displaystyle 2\pi x\,c(x)$
$\displaystyle\simeq\frac{2(\text{Pe}+2)x\left(x^{\text{Pe}}-1\right)}{\text{Pe}\left(x_{0}^{2}-1\right)-2x_{0}^{2}\left(x_{0}^{\text{Pe}}-1\right)}\,.$
(26)

From this expression, the average radial displacement is

$\displaystyle\langle
x\rangle=\int_{x_{0}}^{1}\\!\\!x\,\,2\pi\,x\,c(x)\,dx=\frac{2(\text{Pe}+2)\left[\text{Pe}\left(x_{0}^{3}-1\right)-3x_{0}^{3}\left(x_{0}^{\text{Pe}}-1\right)\right]}{3(\text{Pe}+3)\left[\text{Pe}\left(x_{0}^{2}-1\right)-2x_{0}^{2}\left(x_{0}^{\text{Pe}}-1\right)\right]}\,.$
(27)

Figure 6: The objective function versus Péclet number Pe for different
normalized cost values $C^{\prime}\equiv C/(L^{2}/D)$ in two dimensions.

The average number of reset events $\mathcal{N}$ satisfies a renewal equation
and using $\widetilde{F}_{1}$ from Eq. (C) we find

$\displaystyle\widetilde{\mathcal{N}}(s)$
$\displaystyle=\frac{1}{s}\frac{1}{\,\mathcal{W}-1}\,,$ (28a) where
$\displaystyle\mathcal{W}=x_{0}^{1+\text{Pe}/2}\sqrt{s}\left[K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)I_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)+I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)\right]\,,$
and $I_{\nu}(x)$ and $K_{\nu}(x)$ are the modified Bessel functions of the
first and second kind, respectively. We now extract the long-time behavior for
the average number of times that $x=L$ is reached by taking the limit
$s\rightarrow 0$ of $\widetilde{\mathcal{N}}(s)$. We find
$\displaystyle\mathcal{N}(t)$
$\displaystyle\simeq\frac{2\text{Pe}(\text{Pe}+2)}{\text{Pe}+2x_{0}^{2}\left(x_{0}^{\text{Pe}}-1\right)-\text{Pe}\,x_{0}^{2}}\,\,\text{Fo}\,.$
(28b)

From Eqs. (27) and (28b), we immediately obtain the objective function and
representative results are given in Fig. 6. Overall, the two-dimensional
system has the same qualitative behavior as in one dimension. In the limit
$x_{0}\rightarrow 0$ the average particle position $\langle x\rangle$ and the
average number of resetting events $\mathcal{N}$ take an even simpler form
than in one dimension:

$\displaystyle\langle x\rangle$
$\displaystyle\simeq\frac{2(\text{Pe}+2)}{3(\text{Pe}+3)}\,\Theta(\text{Pe}+2)\,,$
(29) $\displaystyle\mathcal{N}(T)$ $\displaystyle\simeq
2(2+\text{Pe})\,\Theta(\text{Pe}+2)\,\text{Fo}\,,$ (30)

where $\Theta(x)$ is the Heaviside step function. The step function arises
because of the curious feature that when $\text{Pe}<-2$, the flow field
$\text{Pe}/r$ at the origin is so strong that the particle remains trapped
there forever.

##  4 Domain Growth by First-Passage Resetting

We now turn to a different aspect of first-passage resetting—the growth of a
domain as a result of a diffusing particle that reaches the resetting boundary
and causes this boundary to recede by a specified amount at each resetting
event. Moving boundaries typically arise at the interface between two
thermodynamic phases that undergo a first-order phase transition [43, 44]. In
this case, the interface moves continuously as the stable phase grows into the
unstable phase. A simple example is water freezing at the interface between
water and air, when the air temperature is held below $0^{\circ}$C. A layer of
ice grows on top of the water as heat is transported away from the ice-water
interface. In these types of systems, the temperature field evolves by
diffusion and the movement of the interface is determined by the heat flow at
the interface.

In contrast, for a growth process that is induced by first-passage resetting,
a single diffusing particle is discontinuously reset to a distant location
when the boundary is reached. Concomitantly, the motion of the interface is
intermittent and discontinuous when the interface recedes by a finite distance
upon hitting. A related behavior also occurs in the absence of resetting:
returning to the situation depicted in Fig. 2 (a), a boundary that is
initially at $L$ moves to $2L$ when it is hit, and then to $3L$, etc. This
interface position clearly moves discontinuously and its position moves as
$\sqrt{4Dt/\pi}$.

In the next paragraphs, we study the interface motion and related properties
in the presence of first-passage resetting, when the domain of interest is
either the semi-infinite line or the finite interval. We find a variety of
growth laws that depend on how far the boundary recedes at each resetting
event.

###  4.1 Expanding Semi-Infinite Geometry

Suppose that the diffusing particle starts at the origin and diffuses in the
range $[-\infty,L_{n}]$, with $L_{n}>0$. Each time the particle reaches
$L_{n}$, the particle is reset to the origin and the interface moves forward
by a specified amount $\delta L_{n}$ so that $L_{n+1}=L_{n}+\delta L_{n}$.
Since the resetting events occur at separated discrete times, it is convenient
to index the position of the interface by $n$, the number of resetting events.
We consider two natural cases: additive and multiplicative interface growth.

####  4.1.1 Additive growth: $L_{n}=L_{n-1}+L$.

In this case, the right boundary starts at $L$ and then moves to $2L$ at the
first reset event, then to $3L$, etc. We make use of the simple relation
between diffusion with resetting and free diffusion as shown in Fig. 2. By
this equivalence, the probability for $n$ reset events to occur in the time
range $[0,t]$ equals the probability that free diffusion travels further than
$L_{n}=n(n+1)L/2$ but no further than $L_{n+1}=(n+1)(n+2)L/2$ in $[0,t]$; that
is, the maximum of the freely diffusing particle is located in the range
$[L_{n},L_{n+1}]$. So, writing again $\mathcal{M}(t)$ for the average of the
maximum $M(t)$, one has

$\displaystyle\sum_{n\geq 0}\,L_{n}\,P\left(N(t)=n\right)\;\leq\;\sum_{n\geq
0}\,\int_{L_{n}}^{L_{n+1}}\,dm$
$\displaystyle\,m\,P\left(M(t)=m\right)\;\leq\;\sum_{n\geq
0}\,L_{n+1}\,P\left(N(t)=n\right)\,,$

that is

$\displaystyle\sum_{n\geq 0}\,L_{n}\,P\left(N(t)=n\right)\;\leq\;$
$\displaystyle\mathcal{M}(t)\;\leq\;\sum_{n\geq
0}\,L_{n+1}\,P\left(N(t)=n\right)\,.$ (31)

This leads to

$\displaystyle\frac{L}{2}\left[\left\langle N^{2}\right\rangle+\left\langle
N\right\rangle\right]\;\leq\;$
$\displaystyle\mathcal{M}(t)\;\leq\;\frac{L}{2}\left[\left\langle
N^{2}\right\rangle+3\left\langle N\right\rangle+2\right]\,,$ (32)

where we write $N$ for $N(t)$ to simplify the notation, from which it follows
that

$\displaystyle\left\langle N^{2}\right\rangle\simeq 4\sqrt{\frac{Dt}{\pi
L^{2}}}\quad\text{or}\quad\sqrt{\left\langle N^{2}\right\rangle}\simeq
2\left[\frac{Dt}{\pi L^{2}}\right]^{\frac{1}{4}}$ (33)

Note that we do not obtain directly the average number of reset events
$\mathcal{N}(t)\equiv\left\langle N\right\rangle$ from (33), but only that it
scales as $t^{1/4}$. However, we can derive $\mathcal{N}(t)$ by exploiting the
renewal structure of the problem in the Laplace domain (see D) and find

$\displaystyle\mathcal{N}(t)\simeq\sqrt{\frac{\pi}{2}}\,\frac{1}{\Gamma(5/4)}\times\left(\frac{t}{\tau}\right)^{1/4}\,,$
(34)

where $\tau=L^{2}/D$ is the diffusion time.

The $t^{1/4}$ scaling is to be compared with the $t^{1/2}$ scaling when the
boundary is moving through first-passage dynamics but without resetting. When
resetting occurs, the number of encounters with the boundary is reduced
because after each reset, the boundary is further away. As one might expect,
this boundary recession leads to an anomalously slow interface growth.

####  4.1.2 Multiplicative growth: $L_{n}=\alpha L_{n-1}$, $\alpha>1$.

The approach given above can be applied to multiplicative interface recession.
That is, upon the first resetting, the initial boundary at $x=L$ moves to
$x=\alpha L$. In the next resetting, the boundary moves from $x=\alpha L$ to
$x=\alpha^{2}L$, etc. For this recession rule, Eq. (31) remains valid, with
$L_{n}=\alpha^{n}L$. Thus, we have

$\displaystyle\sum_{n\geq 0}\,\alpha^{n}L\,P\left(N(t)=n\right)\;\leq\;$
$\displaystyle\mathcal{M}(t)\;\leq\;\sum_{n\geq
0}\,\alpha^{n+1}L\,P\left(N(t)=n\right)\,,$ (35)

from which

$\displaystyle\left\langle\alpha^{N}\right\rangle L\;\leq\;$
$\displaystyle\mathcal{M}(t)\;\leq\;\left\langle\alpha^{N+1}\right\rangle
L\,.$ (36)

These inequalities suggest that $\mathcal{N}(t)$ scales as
$\ln(t/\tau)\,/\,2\ln\alpha$. From the exact Laplace transform approach (see
D), we also find the same prefactor in the scaling of $\mathcal{N}(t)$ with
$t$. Thus we conclude that

$\displaystyle\mathcal{N}(t)\simeq\ln(t/\tau)\,/\,2\ln\alpha\,.$ (37)

After $n$ resets, the boundary is located at $\alpha^{n}L$. We also checked
numerically that the distribution of $N(t)$ is concentrated sufficiently
tightly around its average value $\mathcal{N}(t)$ so that
$\langle\alpha^{N}\rangle\sim\alpha^{\langle N\rangle}=\alpha^{\mathcal{N}}$.
As a result, the average position of the boundary at time $t$ scales as

$\displaystyle\alpha^{\mathcal{N}(t)}\,L\simeq\sqrt{t/\tau}\,L\,.$ (38)

Thus the boundary moves as $t^{1/2}$, which is faster than in the additive
case, where the boundary moves as $t^{1/4}$. Despite a smaller number of reset
events, each of these moves the boundary far enough for the overall motion to
be almost as fast as in the case of first-passage growth without resetting,
with the difference being only a factor of 2.

###  4.2 Expanding Interval

We now study the case where a diffusing particle is confined to a finite and
growing interval $[0,L_{n}]$, with a reflecting boundary condition at $x=0$.
Each time the particle reaches the right boundary at $x=L_{n}$, the particle
is instantaneously reset to $x=0$, while the position of the boundary recedes
by a specified amount. We want to understand how the interval grows with time
and related statistical properties of this process. We first give the formal
result for an arbitrary dependence of $L_{n}$ on $n$ and then specialize to
the additive case where $L_{n}=nL$.

We again start with the analog of Eq. (67b) for the finite domain, namely, the
Laplace transform of the probability to reset for the $n^{\text{t}h}$ time at
$t$:

$\displaystyle\widetilde{R}_{n}(s)$
$\displaystyle=\widetilde{R}_{n-1}(s)\sech\big(\sqrt{s/D}\,L_{n}\big{missing})=\prod_{m=1}^{n}\sech\big(\sqrt{s/D}\,L_{m}\big{missing})\,.$
(39)

Here $\sech\big(\sqrt{s/D}\,L_{n}\big{missing})$ is the Laplace transform of
the first-passage probability to the right boundary of the finite interval
$[0,L_{n}]$ [25]. Similarly, the average number of resetting events obeys the
renewal equation

$\displaystyle\mathcal{N}(t)=0\times
Q(L,t)+\sum_{n=1}^{\infty}n\int_{0}^{t}dt^{\prime}\,Q(L_{n+1},t-t^{\prime})R_{n}(t^{\prime})\,,$
(40a) where $Q(L,t)$ is now the survival probability of a diffusing particle
in the finite interval $[0,L_{n}]$, with reflection at $x=0$ and absorption at
$x=L_{n}$. In the Laplace domain Eq. (40a) becomes
$\displaystyle\widetilde{\mathcal{N}}(s)=\sum_{n=0}^{\infty}n\widetilde{Q}(L_{n+1},s)\widetilde{R}_{n}(s)\,.$
(40b)

Substituting in $\widetilde{Q}(L_{n},s)=[1-\widetilde{F}(L_{n},s)]/s$ and Eq.
(67b) into the above equation gives

$\displaystyle\widetilde{\mathcal{N}}(s)=$
$\displaystyle\sum_{n=0}^{\infty}\frac{n}{s}\,\left[1-\sech\big(\sqrt{s/D}L_{n+1}\big{missing})\right]\prod_{m=1}^{n}\sech\big(\sqrt{s/D}L_{m}\big{missing})$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{n}{s}\left[\widetilde{R}_{n}(s)-\widetilde{R}_{n+1}(s)\right]=\frac{1}{s}\sum_{n=1}^{\infty}\widetilde{R}_{n}(s)$
$\displaystyle=$
$\displaystyle\frac{1}{s}\sum_{n=1}^{\infty}\prod_{m=1}^{n}\sech\big(\sqrt{s/D}L_{m}\big{missing})$
(41)

To extract the asymptotic behavior of $\mathcal{N}(t)$, we now focus on the
additive case where $L_{n}=nL$; that is, the boundary recedes by a fixed
distance $L$ after each resetting event. Using the dimensionless coordinate
$\ell=\sqrt{sL^{2}/D}$, Eq. (41) now gives

$\displaystyle\widetilde{\mathcal{N}}(s)$
$\displaystyle=\frac{1}{s}\sum_{n=1}^{\infty}\prod_{m=1}^{n}\sech\big(\sqrt{s/D}\,\,mL\big{missing})=\frac{1}{s}\sum_{n=1}^{\infty}\prod_{m=1}^{n}\sech(m\,\ell)$
$\displaystyle\approx\frac{1}{s}\sum_{n=1}^{\infty}\exp\left\\{\int_{0}^{n}dm\ln[\sech(m\,\ell)\big{]}\right\\}$
$\displaystyle\approx\frac{1}{s}\sum_{n=1}^{\infty}\exp\left\\{-\frac{1}{\ell}\left[-\frac{1}{2}\,\ell^{2}n^{2}-\frac{1}{2}\text{Li}_{2}\left(-e^{2\ell
n}\right)-\ell n\ln 2-\frac{\pi^{2}}{24}\right]\right\\}$
$\displaystyle\approx\frac{1}{s\ell}\int_{0}^{\infty}du\,\exp\left\\{-\frac{1}{\ell}\left[-\frac{1}{2}u^{2}-\frac{1}{2}\text{Li}_{2}\left(-e^{2u}\right)-u\ln
2-\frac{\pi^{2}}{24}\right]\right\\}\,.$

where $\text{Li}_{2}(x)$ is the polylogarithm function of order $2$. This
integral can now be computed in the small $s$ limit using the saddle-point
approximation (see E) and the final result is:

$\displaystyle\widetilde{\mathcal{N}}(s)$
$\displaystyle\approx\frac{1}{s\ell}\int_{0}^{\infty}du\,\exp\left(-\frac{u^{3}}{6}\right)$
$\displaystyle\approx\Gamma\left(\frac{4}{3}\right)\,\frac{1}{s}\,\left(\frac{6}{\ell^{2}}\right)^{1/3}+o\left(\frac{1}{s^{4/3}}\right)\,.$
(42)

The Laplace inversion of the above expression gives the average number of
resetting events up to time $t$ in the $t\to\infty$ limit:

$\displaystyle\mathcal{N}(t)\simeq\left(\frac{6t}{\tau}\right)^{1/3}+o\left(t^{1/3}\right)\,.$
(43)

This result implies that the length of the interval also grows as $t^{1/3}$.
For determining the standard deviation (see below), we also need the next
correction to the asymptotic behavior. Numerically, we find that
$\mathcal{N}\simeq(6t/\tau)^{1/3}+C_{1}$ where $C_{1}=-0.8$.

The $t^{1/3}$ dependence of $\mathcal{N}(t)$ can be understood in a simple
way. The mean time for a particle, which starts at $x=0$, to reach the
boundary at $x=L_{n}$ is $L_{n}^{2}/2D=(nL)^{2}/2D\equiv n^{2}\tau/2$ [25]. If
the particle is immediately reset to the origin each time the boundary is
reached, then the time required for $\mathcal{N}$ reset events is
$\sum^{\mathcal{N}}n^{2}\tau/2\simeq\mathcal{N}^{3}\tau/6$. This gives
$\mathcal{N}\simeq(6t/\tau)^{1/3}$.

The second moment of the probability distribution for the number of encounters
is:

$\displaystyle\langle\widetilde{{N}}^{2}(s)\rangle=\sum_{n=0}^{\infty}\frac{n^{2}}{s}\left[\widetilde{R}_{n}(s)-\widetilde{R}_{n+1}(s)\right]=\sum_{n=1}^{\infty}\frac{2n-1}{s}\widetilde{R}_{n}(s)$
(44)

By following similar steps as those to compute $\mathcal{N}$, we obtain the
following result for fixed $n$ and small $s$:

$\displaystyle\langle\widetilde{{N}}^{2}(s)\rangle$
$\displaystyle\approx\frac{1}{s\ell}\int_{0}^{\infty}du\,\left(\frac{2u}{\ell}-1\right)\exp\left\\{-\frac{1}{\ell}\left[-\frac{1}{2}u^{2}-\frac{1}{2}\text{Li}_{2}\left(-e^{2u}\right)-u\ln
2-\frac{\pi^{2}}{24}\right]\right\\}\,.$

This integral can now be computed in the small-$s$ limit using the saddle-
point approximation:

$\displaystyle\langle\widetilde{{N}}^{2}(s)\rangle$
$\displaystyle\approx\frac{1}{s\ell}\int_{0}^{\infty}du\,\left(\frac{2u}{\ell}-1\right)\,\exp\left(-\frac{\ell\,u^{3}}{6}\right)$
$\displaystyle\approx\Gamma\left(\frac{5}{3}\right)\frac{1}{s}\left(\frac{6}{\ell^{2}}\right)^{2/3}+o\left(\frac{1}{s^{5/3}}\right)\,.$
(45)

Performing a Laplace inversion of the above expression gives:

$\displaystyle\langle\mathcal{N}^{2}(t)\rangle\simeq\left(\frac{6\,t}{\tau}\right)^{2/3}+o\left(t^{2/3}\right)\,.$
(46)

Numerically, we find that the next correction is $C_{2}\,(6\,t/\tau)^{1/3}$
with $C_{2}\approx-1.47$. Hence, the standard deviation grows as
$\sqrt{\langle{N}^{2}(t)\rangle-\langle{N}(t)\rangle^{2}}\approx\sqrt{C_{2}-2\,C_{1}}\,(6\,t/\tau)^{1/6}$.

Figure 7: Numerical simulation results for $P(N\\!=\\!n,t)$ for the expanding
interval. The initial interval length $L=1$ and the length grows by 1 after
each resetting event. The distribution is shown at $t=10^{9}$ for $1000$
walkers. The diffusion constant was set to $\Delta x^{2}/(2\Delta t)=5\times
10^{-3}$.

Numerical simulations of this growth process (Fig. 7) show that the
distribution of $\mathcal{N}(t)$ is highly localized around its average value
and decreases rapidly as one moves away from the maximum. While we do not know
how to compute the full distribution analytically, we can determine the tails
of the distribution by a simple extremal argument [45, 46]. For notational
simplicity we take $L=1$ and $D=1$. From the time dependence of the average
value of $\mathcal{N}(t)$ (Eq. (43)), we posit that the natural scaling
variable is $z\equiv n/\mathcal{N}(t)\simeq n/t^{1/3}$. We further assume that
the distribution can be expressed in the scaling form $P({N}=n,t)\propto f(z)$
that decays as a stretched exponential for both $z\to\infty$ and $z\to 0$.
That is, $f(z)=\exp(-z^{a})$, where $a>0$, for $z\to\infty$ and
$f(z)=\exp(-z^{b})$, where $b<0$, for $z\to 0$.

Consider now the extreme event in which the particle always moves towards the
resetting boundary up to time $t$. This event occurs with probability
$2^{-t}\sim e^{-t}$. For this directed motion, the particle requires 1 time
step to first reach the boundary, 2 time steps to reach it for a second time,
3 time steps for a third time, etc. This leads to the total number $n$ of
encounters with the boundary that is determined by $\sum_{k=1}^{n}k=t$. Hence,
$n\simeq\sqrt{2t}$. In terms of our scaling function, the probability to reach
the boundary $\sqrt{2t}$ times occurs with probability $e^{-t^{a/6}}$.
Equating this with $e^{-t}$ gives $a=6$.

For the small-$z$ tail, we focus on the situation where the boundary is
encountered as little as possible. This extremal event is achieved by a random
walk that alternately and deterministically moves one step left, then one step
right, etc. In the case the boundary is encountered once and only once. This
event again occurs with probability $2^{-t}\simeq e^{-t}$. On the other hand,
this event of a single boundary encounter corresponds to the scaling variable
$z=1/t^{1/3}\to 0$, and thus occurs with probability $e^{-t^{-b/3}}$. Equating
these two asymptotic forms of the distribution gives $b=-3$. In summary, we
find the following asymptotics:

$\displaystyle P(N=n,t)\simeq\begin{cases}e^{-(n/\mathcal{N}(t))^{6}}\qquad\
n\rightarrow\infty\,,\\\ e^{-(n/\mathcal{N}(t))^{-3}}\qquad n\rightarrow
0\,.\end{cases}$ (47)

Because the exponent values in the scaling forms are fairly large, it does not
seem possible to verify these asymptotic behaviors numerically.

##  5 Summary and Discussion

We presented the concept of first-passage resetting, in which a random walk is
reset to its starting point whenever it reaches a specified location. This
situation contrasts with constant-rate resetting in which a random walk is
reset to its starting point at a fixed rate. In the simple case of a semi-
infinite line, $[-\infty,L]$ with $L>0$, the particle diffuses freely and is
reset to the origin whenever it reaches $L$. The resulting probability
distribution has dramatically different behavior depending on whether $0<x<L$
or $x<0$. In the former case, the distribution has a simple linear profile
that arises from the balance between flux leaving at the reset point and the
flux being reinjected at $x=0$. In the latter case, the probability
distribution reduces to free diffusion in the presence of a reflecting
boundary. We derived this result analytically and also via a path
decomposition that is reminiscent of the image method.

In the finite interval geometry, we defined an optimization problem that
describes, in a schematic way, aspects of the repeated breakdown of a driven
mechanical system. The operation domain of the system is a finite interval;
this interval could be interpreted as the RPM range of an engine. The
resetting boundary corresponds the system reaching its operating limit or
maximum RPMs, after which a breakdown occurs and the system has to be
restarted from scratch. The control parameter is the bias velocity (not to be
confused with the RPM of the engine), which may either drive the system
towards breakdown or towards minimal-level operation. We showed that there
exists an optimal bias velocity that optimizes the performance of the system.
This optimum balances the gain by operating close to $x=L$ while minimizing
the number of breakdowns. A similar physical picture arises if breakdown is
accompanied by a random delay before restarting the system or by extending to
a two-dimensional geometry.

We also studied a variety domain growth phenomena that are driven by first-
passage dynamics with resetting. When each resetting event moves the boundary
by a fixed amount, the boundary recedes as $t^{1/4}$ and as $t^{1/3}$ for the
semi-infinite geometry and the finite interval, respectively. In the semi
infinite geometry, if the boundary position grows by a factor $\alpha>1$ with
each resetting event, then the interface moves much more quickly, as
$t^{1/2}$. The case where the boundary moves by a fixed amount at each
resetting is actually a version of the internal diffusion-limited aggregation
problem for which there is extensive literature that has focused on the
geometrical properties of the growing domain (see, e.g., [47, 48, 49, 50]). We
instead focused on the rich dynamical aspects of the model and we suggest,
based on the correspondence with internal diffusion-limited aggregation, that
it will be worthwhile to treat our first–passage resetting in a finite two-
dimensional domain.

There are a variety of extensions of the optimization problem may be worth
exploring. First, the control strategy could be finer than simply a uniform
bias velocity [51]. More realistically, one could also associate a cost to a
strategy that becomes more expensive as the control mechanism becomes more
sophisticated. For instance, it would be natural to turn on a bias velocity
_away_ from the breakdown point when the system is very close to breakdown. It
would also be useful to identify the optimal region over which the particle
experiences a bias (both toward and away from the breakdown point). In
addition to a more refined control strategy, other simple geometries may be
worthwhile to study. One such example is a one-dimensional interval with a
first-passage resetting mechanism at each end of the interval and with a
different cost in reaching each end. First-passage resetting in a bounded
planar geometry with a cost function that depends on the hitting angle of the
boundary might be another geometry that will be worthwhile to study.

Given the rich behavior exhibited by first-passage resetting, it should also
be worthwhile to investigate both extensions of the basic model and
applications. An example of the former is the Fleming-Viot branching process,
in which there are $N+1$ particles and when one of them resets, it resets to
one of the positions of the remaining $N$ particles [52, 53, 54, 55, 56]. More
generally, the first-passage resetting in the presence of multiple diffusing
particles could lead to new phenomenology. On another note, applications also
exist in cash flow management: cash levels in a large firm are sometimes
modeled as a diffusion process in which one wishes to have cash fully invested
in profitable ventures, while at the same time keeping enough cash available
so as to avoid being indebted [57, 58]. These types of problems seem to be
ripe for further exploration.

##  6 Acknowledgments

BBs research at the Perimeter Institute was supported in part by the
Government of Canada through the Department of Innovation, Science and
Economic Development Canada and by the Province of Ontario through the
Ministry of Colleges and Universities. JRFs research at Columbia University
was supported by the Alliance Program. SR thanks Paul Hines for helpful
conversations and financial support from NSF grant DMR-1910736. We thank one
of the referees for providing helpful suggestions about possible extensions of
the basic optimization problem.

##  Appendix A Laplace transform approach in the semi-infinite geometry

###  A.1 Spatial probability distributions

The probability distribution $P(x,t)$ of the diffusing particle at time $t$ on
the semi-infinite line $x\leq L$, can be obtained in several ways. We
presented in the main text a path transformation approach and we detail here
the Laplace transform approach (see also [34]). We first partition the
trajectory according to the number of reset events up to time $t$. Between
consecutive resets, the particle undergoes free diffusion with an absorbing
boundary at $x\\!=\\!L$. This part of the motion is described by the free
propagator

$\displaystyle
G(x,L,t)=\big{[}e^{-x^{2}/4Dt}-e^{-(x-2L)^{2}/4Dt}\big{]}/\sqrt{4\pi Dt}\,,$
(48)

which can be computed, for example, by the image method [24, 25]. Summing over
all numbers of reset events, the spatial probability is determined by

$\displaystyle P(x,t)=G(x,L,t)+\sum_{n\geq
1}\,\int_{0}^{t}dt^{\prime}\,F_{n}(L,t^{\prime})\,G(x,L,t\\!-\\!t^{\prime})\,.$
(49a) Equation (49a) states that for the particle to be at $x$ at time $t$, it
either: (i) must never hit $L$, in which case its probability distribution is
just $G(x,L,t)$, or (ii), the particle first hits $L$ for the $n^{\rm th}$
time at $t^{\prime}<t$, after which the particle restarts at the origin and
then propagates to $x$ in the remaining time $t-t^{\prime}$ without hitting
$L$ again. The equivalent way of writing Eq. (49a) in a renewal fashion is:
$\displaystyle
P(x,t)=G(x,L,t)+\int_{0}^{t}dt^{\prime}\,F_{1}(L,t^{\prime})P(x,t-t^{\prime})\,.$
(49b)

The first term accounts for the particle never reaching $x=L$, while the
second term accounts for the particle reaching $x=L$ at time $t^{\prime}$,
after which the process starts anew from $x(t^{\prime})=0$ for the remaining
time $t-t^{\prime}$. Note that this is a renewal equation in the sense that
the second term contains the full propagator $P(x,t-t^{\prime})$ and not the
free propagator $G(x,t-t^{\prime})$, thereby accounting for any number of
resetting events in the time interval $[t^{\prime},t]$.

To solve for $P(x,t)$ we again treat the problem in the Laplace domain. While
we can find the solution from the Laplace transform of Eq. (49a), the solution
is simpler and more direct from the Laplace transform of (49b):

$\displaystyle\widetilde{P}(y,s)$
$\displaystyle=\widetilde{G}(y,\ell,s)+\widetilde{F}_{1}(\ell,s)\widetilde{P}(y,s),$
(50a) with
$\displaystyle\widetilde{G}(y,\ell,s)=\left[e^{-|y|}-e^{-|y-2\ell|}\right]/\sqrt{4Ds}\,,$
the Laplace transform of $G(x,L,t)$, where we have introduced the scaled
coordinates $y\equiv x\sqrt{s/D}$ and $\ell\equiv L\sqrt{s/D}$. Solving for
$\widetilde{P}(y,s)$ yields: $\displaystyle\widetilde{P}(y,s)$
$\displaystyle=\frac{\widetilde{G}(y,\ell,s)}{1-\widetilde{F}_{1}(\ell,s)}=\frac{1}{\sqrt{4Ds}}\,\frac{\left[e^{-|y|}-e^{-|y-2\ell|}\right]}{1-e^{-\ell}}\,.$
(50b)

To invert this Laplace transform, we separately consider the cases $0\leq
y\leq\ell$ and $y<0$. In the former, we expand the denominator in a Taylor
series to give

$\displaystyle\widetilde{P}(y,s)$
$\displaystyle=\frac{1}{\sqrt{4Ds}}\,\left[e^{-y}-e^{-(2\ell-y)}\right]\,\sum_{n\geq
0}e^{-n\ell}$ $\displaystyle=\frac{1}{\sqrt{4Ds}}\,\sum_{n\geq
0}\left[e^{-(y+n\ell)}-e^{-[(n+2)\ell-y]}\right]\,,\qquad 0<y\leq\ell\,,$
(51a) from which $\displaystyle P(x,t)$ $\displaystyle=\frac{1}{\sqrt{4\pi
Dt}}\,\sum_{n\geq
0}\left[e^{-(x+nL)^{2}/4Dt}-e^{-[x-(n+2)L]^{2}/4Dt}\right],\qquad 0<x\leq
L\,.$ (51b)

In the case of $y<0$, $\widetilde{P}(y,s)$ in Eq. (50b) is factorizable:

$\displaystyle\widetilde{P}(y,s)$
$\displaystyle=\frac{1}{\sqrt{4Ds}}\left[\frac{e^{y}-e^{y-2\ell}}{1-e^{-\ell}}\right]=\frac{1}{\sqrt{4Ds}}\left[e^{y}+e^{(y-\ell)}\right],$
(52a) and this latter form can be readily inverted to give: $\displaystyle
P(x,t)=\frac{1}{\sqrt{4\pi
Dt}}\,\left[e^{-x^{2}/4Dt}+e^{-(x-L)^{2}/4Dt}\right]\quad x<0\,.$ (52b)

###  A.2 Average number of resets

The average number of resets that occur up to time $t$ may also be derived by
using the Laplace transform approach. The probability for $n$ resets to occur
by time $t$ equals the probability to have at least $n$ resets minus the
probability to have at least $n+1$ resets:

$\displaystyle\begin{split}P\left(N(t)\\!=\\!n\right)&=P\left(N(t)\\!\geq\\!n\right)-P\left(N(t)\\!\geq\\!n+1\right)\,,\\\
&=\int_{0}^{t}\\!dt^{\prime}\,F_{n}(L,t^{\prime})\,-\int_{0}^{t}\\!dt^{\prime}\,F_{n+1}(L,t^{\prime})\,.\end{split}$
(53a) Using our earlier result that $F_{n}(L,t)=F_{1}(nL,t)$, we have
$P\left(N(t)\\!=\\!n\right)=\mathrm{erf}\left(\frac{(n+1)L}{\sqrt{4Dt}}\right)-\mathrm{erf}\left(\frac{nL}{\sqrt{4Dt}}\right)\,,$
(53b)

where $\mathrm{erf}$ is the Gauss error function.

We can compute the average number of reset events,
$\mathcal{N}(t)\equiv\left\langle N(t)\right\rangle$, from (7), but it is
quicker to use a renewal equation approach. Here we can write

$\displaystyle\mathcal{N}(t)=\int_{0}^{t}dt^{\prime}\,F_{1}(L,t-t^{\prime})\big{[}1+\mathcal{N}(t^{\prime})\big{]}\,.$
(54a) Equation (54a) states that to have $\mathcal{N}$ reset events up to time
$t$, $\mathcal{N}-1$ reset events must have occurred at some earlier time
$t^{\prime}<t$ and then one more reset event occurs exactly at time $t$.
Taking the Laplace transform of (54a) gives
$\displaystyle\widetilde{\mathcal{N}}(s)$
$\displaystyle=\frac{\widetilde{F}_{1}(L,s)}{s\,(1-\widetilde{F}_{1}(L,s))}=\frac{e^{-\ell}}{s(1-e^{-\ell})}\,,$
(54b)

where again $\ell=L\sqrt{s/D}$. We extract the long-time behavior of the
average number of reset events by taking the $s\rightarrow 0$ limit and then
Laplace inverting this limiting expression. We thus find

$\displaystyle\mathcal{N}(t)\simeq\sqrt{4Dt/\pi L^{2}}\,.$ (55)

##  Appendix B  Solution to the convection-diffusion equation

The general solution to Eq. (14b) is

$\displaystyle\widetilde{c_{0}}(x,s)=e^{\text{P}}(Ae^{wx}+Be^{-wx})\,,$ (56)

where $\text{P}=vx/2D$, $w=\sqrt{v^{2}+4Ds}/2D$, and $A,B$ are integration
constants. To determine $A$ and $B$, we apply the boundary conditions that
accompany Eq. (56) to give the linear system

$\displaystyle\begin{cases}v(A+B)-\frac{1}{2}v(A+B)-ADw+BDw=1\,,\\\
e^{vL/(2D)}\left(Ae^{Lw}+Be^{-Lw}\right)=0\,,\end{cases}$ (57)

whose solution is

$\displaystyle\begin{cases}A=-\dfrac{2}{2Dwe^{2Lw}+2Dw+ve^{2Lw}-v}=-\dfrac{e^{-Lw}}{2Dw\cosh(Lw)+v\sinh(Lw)}\,,\\\\[6.99997pt]
B=\dfrac{2e^{2Lw}}{2Dwe^{2Lw}+2Dw+ve^{2Lw}-v}=\dfrac{e^{Lw}}{2Dw\cosh(Lw)+v\sinh(Lw)}\,.\end{cases}$
(58)

We define $\mathcal{W}\equiv 2Dw\cosh(Lw)+v\sinh(Lw)$, from which
$A=-e^{-Lw}/\mathcal{W}$ and $B=e^{Lw}/\mathcal{W}$. Substituting these
constants back into the general solution Eq. (56) leads to Eq. (15a).

##  Appendix C First-passage resetting in the annular geometry

For the convection-diffusion equation in two dimensions with a radial drift
$v/r$; that is, Eq. (25) without the delta-function term, the general solution
in the Laplace domain is [59]

$\displaystyle\widetilde{c}_{0}(x,s)$
$\displaystyle=x^{\text{Pe}/2}\big{[}A\,\text{I}_{\text{Pe}/2}(\sqrt{s}x)+B\,\text{K}_{\text{Pe}/2}(\sqrt{s}x)\big{]}\,,$
(59)

where $A$, and $B$ are integration constants, and $I_{\nu}(x)$ and
$K_{\nu}(x)$ are the modified Bessel functions of the first and second kind,
respectively. The subscript $0$ refers to the concentration without flux re-
injection. Imposing the boundary conditions that accompany Eq. (25) leads to a
linear system to solve for $A$ and $B$:

$\displaystyle\begin{cases}AI_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)+BK_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)&=0\,,\\\
\sqrt{s}x_{0}^{1+{\text{Pe}}/{2}}\left(BK_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)-AI_{1+\text{Pe}/2}\left(\sqrt{s}x\right)\right)&=1/(2\pi)\,.\end{cases}$
(60)

whose solution is:

$\displaystyle\begin{cases}A=-\dfrac{x_{0}^{-(1+\text{Pe}/2)}K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)}{2\pi\sqrt{s}\left(K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)I_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)+I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)\right)}\,,\\\\[6.99997pt]
B=\dfrac{x_{0}^{-(1+\text{Pe}/2)}I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)}{2\pi\sqrt{s}\left(K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)I_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)+I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)\right)}\,.\end{cases}$
(61)

We now define $\mathcal{W}\equiv
x_{0}^{1+\text{Pe}/2}\sqrt{s}\left[K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)I_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)+I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{1+\text{Pe}/2}\left(\sqrt{s}x_{0}\right)\right]$.
In terms of this function, we have
$A=K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)/(2\pi\mathcal{W})$ and
$B=I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)/(2\pi\mathcal{W})$. Substituting
these constants back into Eq. (59) yields:

$\displaystyle\widetilde{c}_{0}(x,s)=\frac{x^{\text{Pe}/2}\left[I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)-K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)I_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)\right]}{2\pi\mathcal{W}}\,,$
(62)

The first-passage probability is obtained by computing the outlet flux at
$x=1$ from the concentration in Eq. (62):

$\displaystyle\widetilde{F}_{1}(s)$
$\displaystyle=2\pi\left(-x\partial_{x}\widetilde{c}_{0}(x,s)+\text{Pe}\,\widetilde{c}_{0}(x,s)\right)\rvert_{x=1}$
$\displaystyle=2\pi\left(-\partial_{x}\widetilde{c}_{0}(x,s)\right)\rvert_{x=1}$
$\displaystyle=\frac{\sqrt{s}I_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{1-{\text{Pe}}/{2}}\left(\sqrt{s}\right)+\sqrt{s}I_{1-{\text{Pe}}/{2}}\left(\sqrt{s}\right)K_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)}{\mathcal{W}}=\frac{1}{\mathcal{W}}\,,$
(63)

where we used the absorbing boundary condition to go to the second line.

On the other hand, the survival probability ${Q}(t)$ is defined as the
integral of $2\pi x\,\widetilde{c}_{0}(x,t)$ over the interval $[x_{0},1]$.
Alternatively, it can be computed as the probability of not having hit the
absorbing boundary until time $t$:
$Q(t)=1-\int_{0}^{t}dt^{\prime}F_{1}(t^{\prime})$, which in the Laplace domain
translates to $\widetilde{Q}(s)=(1-\widetilde{F}_{1}(s))/s$. Using Eq. (C), we
obtain:

$\displaystyle\widetilde{Q}(s)=\frac{1}{s}\left(1-\frac{1}{\mathcal{W}}\right)\,.$
(64)

When there is re-injection of the outgoing flux, the concentration obeys the
renewal equation below. In the Laplace domain and using the form for
$\widetilde{c}_{0}$ obtained above, we find:

$\displaystyle\widetilde{c}(x,s)=\frac{\widetilde{c}_{0}(x,s)}{1-\widetilde{F}_{1}(s)}=\frac{x^{\text{Pe}/2}\left[\text{I}_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)\text{K}_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)-\text{K}_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)\text{I}_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)\right]}{2\pi\left(\,\mathcal{W}-1\right)}\,.$
(65)

In the $s\to 0$ limit, we find that:

$\displaystyle\begin{split}&x^{\text{Pe}/2}[\text{I}_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)\text{K}_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)-\text{K}_{{\text{Pe}}/{2}}\left(\sqrt{s}\right)\text{I}_{{\text{Pe}}/{2}}\left(\sqrt{s}x\right)]\simeq\frac{1-x^{\text{Pe}}}{\text{Pe}}\,,\\\
&\mathcal{W}-1\simeq\frac{\text{Pe}-\text{Pe}\,x_{0}^{2}+2x_{0}^{2}\,(-1+x_{0}^{\text{Pe}})}{2\text{Pe}\,(2+\text{Pe})}\,s\,.\end{split}$
(66)

Substituting these asymptotic expressions in Eq. (65), the coefficient of the
term proportional to $1/s$ in $\widetilde{c}(x,s)$ gives the steady-state
probability density in the time domain that is quoted in Eq. (26).

##  Appendix D Expanding semi-infinite geometry: Laplace transforms

We derive here with Laplace transforms the properties for domain growth in the
semi-infinite geometry that was obtained in the main text using a path
transformation. Suppose that the diffusing particle starts at the origin and
diffuses in the range $[-\infty,L_{n}]$, with $L_{n}>0$. Each time the
particle reaches $L_{n}$, the particle is reset to the origin and the
interface moves forward by a specified amount $\delta L_{n}$ so that
$L_{n+1}=L_{n}+\delta L_{n}$. Since the resetting events occur at separated
discrete times, it is convenient to index the position of the interface by
$n$, the number of resetting events. We consider two natural cases: additive
and multiplicative interface growth.

###  D.1 Additive growth: $L_{n}=L_{n-1}+L$.

In this case, the right boundary starts at $L$ and then moves to $2L$ at the
first reset event, then to $3L$, etc. The probability that the $n^{\text{t}h}$
reset occurs at time $t$, $R_{n}(t)$, is given by the renewal equation

$\displaystyle R_{n}(t)$
$\displaystyle=\int_{0}^{t}dt^{\prime}\,R_{n-1}(t-t^{\prime})\,F(L_{n}\\!=\\!nL,t^{\prime})\,,$
(67a) where $F(L_{1},t)$ is the standard first-passage probability to reach
$L_{1}$ when the particle starts from the origin. In the Laplace domain Eq.
(67a) becomes $\displaystyle\widetilde{R}_{n}(s)$
$\displaystyle=\widetilde{R}_{n-1}(s)\,e^{-n\ell}=e^{-n(n+1)\ell/2}\,,$ (67b)

with $\ell\equiv\sqrt{sL^{2}/D}$.

By similar considerations, the average number of resetting events is

$\displaystyle\mathcal{N}(t)=0\times
Q(L,T)+\sum_{n=1}^{\infty}n\int_{0}^{t}dt^{\prime}\,Q\big{(}(n\\!+\\!1)L,t-t^{\prime}\big{)}\,R_{n}(t^{\prime})\,,$
(68a) which, in the Laplace domain, becomes
$\displaystyle\widetilde{\mathcal{N}}(s)=\sum_{n=0}^{\infty}n\,\widetilde{Q}\big{(}(n+1)L,s\big{)}\,\widetilde{R}_{n}(s)\,.$
(68b)

Here $Q(L,t)=1-\int_{0}^{T}dtF(L,t)$ is the survival probability for a
diffusing particle that starts at the origin to not reach an absorbing
boundary at $L$ within time $T$.

In the Laplace domain, this relation becomes
$\widetilde{Q}(nL,s)=\big{[}1-\widetilde{F}(nL,s)\big{]}/s$. We now substitute
this expression for $\widetilde{Q}$ and the above expression for
$\widetilde{R}_{n}(s)$ into (68b), and then convert the sum to an integral to
give

$\displaystyle\widetilde{\mathcal{N}}(s)\approx$
$\displaystyle\int_{0}^{\infty}dn\,\,\frac{n}{s}\,\left[1-e^{-(n+1)\ell}\right]\,\,e^{-n(n+1)\ell/2}$
$\displaystyle=$
$\displaystyle\frac{e^{-\ell}}{4\,s\,\ell}\Big{\\{}\sqrt{2\pi\ell}\,\,e^{9\,\ell/8}\big{[}2+\text{erf}(\sqrt{\ell/8})-3\,\text{erf}(3\sqrt{\ell/8})\big{]}+4\,e^{\ell}-4\Big{\\}}$
$\displaystyle\to$ $\displaystyle\,\frac{1}{s}\sqrt{\frac{\pi}{2\,\ell}}\qquad
s\rightarrow 0\,.$ (69a) Laplace inverting this expression, the average number
of reset events asymptotically scales as
$\displaystyle\mathcal{N}(t)\simeq\sqrt{\frac{\pi}{2}}\,\frac{1}{\Gamma(5/4)}\times\left(\frac{t}{\tau}\right)^{1/4}\,,$
(69b)

where $\tau=L^{2}/D$ is the diffusion time.

###  D.2 Multiplicative growth: $L_{n}=\alpha\,L_{n-1},\alpha>1$.

The approach given above can be applied to multiplicative interface recession.
That is, upon the first resetting, the initial boundary at $x=L$ moves to
$x=\alpha L$. In the next resetting, the boundary moves from $x=\alpha L$ to
$x=\alpha^{2}L$, etc. For this recession rule, the probability for the
$n^{\text{t}h}$ reset event to occur at time at $t$, $R_{n}(t)$, is

$\displaystyle R_{n}(t)$
$\displaystyle=\int_{0}^{t}dt^{\prime}\,R_{n-1}(t-t^{\prime})\,F(\alpha^{n-1}L,t^{\prime})\,,$
(70a) which, in the Laplace domain, becomes
$\displaystyle\widetilde{R}_{n}(s)$
$\displaystyle=\widetilde{R}_{n-1}(s)\,e^{-\sqrt{s/D}\,\alpha^{n-1}L}\equiv\widetilde{R}_{n-1}(s)e^{-\ell\,\alpha^{n-1}}=e^{-\ell\,(1-\alpha^{n})/(1-\alpha)}\,.$
(70b)

The average number of reset events up to time $t$ is

$\displaystyle\mathcal{N}(t)=0\times
Q(L,T)+\sum_{n=1}^{\infty}n\int_{0}^{t}dt^{\prime}\,Q(\alpha^{n}L,t-t^{\prime})\,R_{n}(t^{\prime})\,,$
(71a) which becomes, in the Laplace domain,
$\displaystyle\widetilde{\mathcal{N}}(s)=\sum_{n=0}^{\infty}n\,\widetilde{Q}(\alpha^{n}L,s)\,\widetilde{R}_{n}(s)\,.$
(71b)

Substituting in
$\widetilde{Q}(\alpha^{n}L,s)=[1-\widetilde{F}(\alpha^{n}L,s)]/s$ and (70b)
into the above equation, we obtain

$\displaystyle\widetilde{\mathcal{N}}(s)=$
$\displaystyle\sum_{n=0}^{\infty}\frac{n}{s}\,\left(1-e^{-\ell\alpha^{n}}\right)e^{-\ell\,(1-\alpha^{n})/(1-\alpha)}$
$\displaystyle\approx$
$\displaystyle\int_{0}^{\infty}dn\,\frac{n}{s}\,\left(1-e^{-\ell\alpha^{n}}\right)e^{-\ell\,(1-\alpha^{n})/(1-\alpha)}$
$\displaystyle=$
$\displaystyle\frac{1}{s}\,\,e^{-\ell/(1-\alpha)}\int_{0}^{\infty}dn\,n\,(1-e^{-\ell\,\alpha^{n}})\,e^{\ell\,\alpha^{n}/(1-\alpha)}\,.$
(72)

To evaluate the above integral, we make the variable change
$z=\ell\,\alpha^{n-1}$, from which $n=\ln(z\,\alpha/\ell)/\ln\alpha$ and
$dn=dz/(z\ln\alpha)$. The above integral now becomes:

$\displaystyle\widetilde{\mathcal{N}}(s)=$
$\displaystyle\,\frac{1}{s}\,\,\frac{e^{-\ell/(1-\alpha)}}{(\ln\alpha)^{2}}\int_{\ell/\alpha}^{\infty}\,\,dz\,\,\frac{\ln(z\,\alpha/\ell)}{z}\,(1-e^{-\alpha\,z})\,e^{\alpha\,z/(1-\alpha)}\,,$

For $s\to 0$, we compute the above integral by first splitting it into two
terms, by using $\ln(z\alpha/\ell)=\ln z+\ln(\alpha/\ell)$. The contribution
from the first term is while the second one diverges. Dropping the finite
term, we obtain

$\displaystyle\widetilde{\mathcal{N}}(s)\simeq\,$
$\displaystyle\frac{\ln(\alpha/\ell)}{s}\,\,\frac{e^{-\ell/(1-\alpha)}}{(\ln\alpha)^{2}}\int_{\ell/\alpha}^{\infty}\,\,dz\,\frac{1-e^{-\alpha
z}}{z}\,e^{\alpha z/(1-\alpha)}\,\simeq-\frac{\ln(s\tau)}{2\,s\ln\alpha},\quad
s\rightarrow 0\,.$ (73)

Note that we introduce the factor $\tau$ inside the logarithm, so that this
term is manifestly dimensionless. Thus in the long-time limit, the average
number of resetting events scales as

$\displaystyle\mathcal{N}(t)\simeq\frac{\ln(t/\tau)}{2\ln\alpha}\,.$ (74)

##  Appendix E Expanding interval: Saddle point approximation

We want to compute the following integral in the small $\ell$ limit:

$\displaystyle I(\ell)=\int_{0}^{\infty}du\,\exp[-\frac{1}{\ell}\,f(u)]\,,$
(75)

where
$f(u)=-\frac{1}{2}u^{2}-\frac{1}{2}\text{Li}_{2}\left(-e^{2u}\right)-u\log(2)-\frac{\pi^{2}}{24}$.
Because the negative exponential is rapidly decreasing as $\ell\rightarrow 0$,
the main contribution will occur when $f(u)$ is minimum. This function has a
global minimum located at $u=0$ and can be locally approximated by $f(u)\simeq
u^{3}/6$. Substituting this into Eq. (75), we recover Eq. (42).

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