# Resetting with stochastic return through linear confining potential

Deepak Gupta1, Arnab Pal2, Anupam Kundu3<EMAIL_ADDRESS>1Dipartimento di Fisica ‘G. Galilei’, INFN, Università di Padova, Via Marzolo
8, 35131 Padova, Italy  2School of Chemistry, The Center for Physics and
Chemistry of Living Systems, Tel Aviv University, Tel Aviv 6997801, Israel
3International Centre for Theoretical Sciences, Tata Institute of Fundamental
Research, Bengaluru 560089, India

###### Abstract

We consider motion of an overdamped Brownian particle subject to stochastic
resetting in one dimension. In contrast to the usual setting where the
particle is instantaneously reset to a preferred location (say, the origin),
here we consider a finite time resetting process facilitated by an external
linear potential $V(x)=\lambda|x|~{}(\lambda>0)$. When resetting occurs, the
trap is switched on and the particle experiences a force $-\partial_{x}V(x)$
which helps the particle to return to the resetting location. The trap is
switched off as soon as the particle makes a first passage to the origin.
Subsequently, the particle resumes its free diffusion motion and the process
keeps repeating. In this set-up, the system attains a non-equilibrium steady
state. We study the relaxation to this steady state by analytically computing
the position distribution of the particle at all time and then analysing this
distribution using the spectral properties of the corresponding Fokker-Planck
operator. As seen for the instantaneous resetting problem, we observe a ‘cone
spreading’ relaxation with travelling fronts such that there is an inner core
region around the resetting point that reaches the steady state, while the
region outside the core still grows ballistically with time. In addition to
the unusual relaxation phenomena, we compute the large deviation functions
associated to the corresponding probability density and find that the large
deviation functions describe a dynamical transition similar to what is seen
previously in case of instantaneous resetting. Notably, our method, based on
spectral properties, complements the existing renewal formalism and reveals
the intricate mathematical structure responsible for such relaxation
phenomena. We verify our analytical results against extensive numerical
simulations.

Keywords: Stochastic resetting, Brownian motion, Fokker-Planck equation

##  1 Introduction

Stochastic resetting is a mechanism that interrupts an ongoing process at
random times and resets it to some pre-determined state only to recommence all
over again [1]. Such tendencies are only natural and appear in many search
processes e.g., animal foraging [2, 3], randomized computer search algorithms
[4, 5], facilitated diffusion [6], and single molecular search processes such
as backtracking by RNA polymerases [7], motor driven intracellular transport
[8], cytoneme based search of morphogen [9] and enzymatic reactions [10, 11].
Beyond search strategies, the notion of resetting has also been found
ubiquitously in several other contexts e.g., in population dynamics [12],
queuing system [13], extreme and catastrophic events [14, 15, 16].

Systematic theoretical study of resetting in the physics literature started
with the canonical Evans-Majumdar model for diffusion with stochastic
resetting [17] and since then the subject has been a focal point of recent
studies. Simply put, resetting brings a particle back to a preferred location
persistently thus putting it into an effective confinement. This generates a
non-zero probability current in the system and eventually a non-equilibrium
steady state is attained [18, 19, 20, 21, 22, 23, 24, 25]. Steady state
properties under resetting have also been studied in a wide spectrum of
stochastic processes namely anomalous diffusion [26, 27], underdamped process
[28], random acceleration process [29], scaled Brownian motion [30],
continuous time random walk [31, 32], Lévy flights [33], telegraphic process
[34], run and tumble motion [35, 36] and others [37, 38, 39, 40, 41]. Steady
state properties under resetting were also studied in many-body interacting
systems such as fluctuating interface [42], spin system [43] and exclusion
process [44, 45].

Another hallmark property of resetting is to expedite the completion of a
first passage process which otherwise would hinder e.g., resetting renders the
mean first passage time finite for a diffusive search process [17, 18, 23].
Regularization by resetting happens by taking advantage of the large
stochastic fluctuations in the underlying first passage processes and this
property was extensively studied in diffusive systems [46, 47, 48, 49, 50, 51,
52, 53, 54], generic stochastic processes [55, 56, 57, 58, 59] and nonlinear
dynamical systems such as chaotic Lorenz system [60]. Resetting has also been
studied in the context of stochastic thermodynamics [61, 62, 63, 64, 65, 66],
large deviation [38, 67] and in quantum systems [68].

Majority of the studies to date in the field assumed the resetting process to
take zero time [1]. However, in practice, getting from one place to another
for the particle always takes non-zero time. This idea has recently been
augmented with a series of theoretical studies which attempted to account for
such realistic situations. To be more specific, in the set-up of diffusion
with resetting, the particle is reset/returned to the origin at a finite time
making use of various deterministic protocols e.g., returns with a constant or
space dependent velocity [3, 69, 70, 71, 72, 73], constant or space dependent
acceleration [72, 73], and returns with finite time [74]. This intuition also
stems from the first experimental case study of the resetting phenomena [75]
(also see [76] which followed after) where it was shown why it is only natural
to consider resetting processes with non-instantaneous returns.

Our motivation to the current paper complements the idea of deterministic
return protocols with the stochastic ones since in reality, a deterministic
driving is never perfect and will always be accompanied by uncontrollable
random fluctuations. Taking these facts into account, we recently have
proposed the idea of stochastic resetting followed by stochastic return in
[77] (also see another recent study [78] that proposes similar ideas of
diffusion with stochastic returns using intermittent resetting potentials). To
illustrate the set-up, let us again consider the Brownian particle which
diffuses in one dimension until a random resetting time. At that very moment,
an external potential centered at the resetting location, is turned on and the
particle starts experiencing a force towards the potential minimum (i.e., the
resetting location). Once the particle hits the resetting location for the
first time, the trap is switched off immediately, and the particle resumes its
diffusing motion. Thus, motion of the particle can viewed as a compound
process which is a superposition of the exploration and return phase. In Ref.
[77], we focused on the steady state and discussed shapes and forms of the
position density in each phase analytically for different choices of external
confining potentials. In this paper, we go beyond the steady state analysis to
study the time dependent problem in whole generality. We provide a
comprehensive method to compute the exact time dependent density, and then
apply the framework to a set-up where a Brownian particle is returned to the
origin using a linear trap potential i.e., $V(x)=\lambda|x|$ ( $\lambda>0$).
We first provide the exact form of the position density and then study its
large time relaxation behaviour.

Often relaxation of a system to a stationary state is described by the
exponential relaxation through spectral representation of the Fokker-Planck
operator where the relaxation occurs homogeneously over the full available
space with uniform rate. In contrast, for the canonical diffusion with
instantaneous resetting model, it was demonstrated using the renewal structure
that the relaxation to the steady state occurs in-homogeneously [21]. More
elaborately, it was shown that as time progresses an inner core region around
the resetting point reaches the steady state, while the region outside the
core is still transient. The boundaries of the core region are seen to grow
with time linearly [21]. In this paper we show that a similar ‘cone spreading’
type relaxation with travelling fronts also occurs in the case of resetting
with stochastic return using a linear trap. We find that details of the core
e.g., the speed of its boundaries depends on the relative strength between the
resetting rate $r$ and the potential $\lambda$. We solve the Fokker-Planck
equation using Laplace transform method which reveals the spectral properties
of the operator. Analysing the spectral properties through inverse Laplace
transform we study the relaxation to the steady state at large time. This
approach does not only complement the existing renewal formalism but also
proves to be effective where a renewal framework is less tractable. The
current problem at our hand exactly depicts such a scenario.

Finally, Finally, we analyse the large deviation property of the particle’s
position density. For the instantaneous case, the large deviation function
(LDF) was studied in [21] where it has been described to demonstrate a
dynamical transition in the density with time when looked at a fixed position.
In the present case of non-instantaneous resetting, we also find a similar
dynamical transition described by appropriate LDFs. In particular, we find
that the position density satisfies a large deviation form with an explicit
LDF. The form of the LDF depends on the relative strength between the
resetting rate $r$ and the potential strength $\lambda$. In each case, the LDF
can be interpreted to describe a dynamical transition occurring at a fixed
position similar to what one observes in the instantaneous case [21].

The structure of the paper is organized as follows. In Sec. 2, we briefly
introduce the model and discuss the governing Fokker-Planck equations for the
process. In Sec. 3, we sketch out the method to solve these equations in
Laplace space. Sec. 4 contains the exact results for the probability density
function of the particle at all times. Next in Sec. 5, we compute the
probability for the system to be in the two phases. Sec. 6 is dedicated to the
detailed computation of the relaxation of the total density to its steady
state. We summarise our paper in Sec. 7. Some details of the calculations are
reserved for the Appendix.

##  2 Model

We consider motion of a Brownian particle interrupted by stochastic resetting
in one-dimension. We refer the uninterrupted motion of the particle as the
exploration or diffusive phase. Resetting occurs at a rate $r$ and at each
resetting event, we switch on a trap that creates a potential $V(x)$ centered
(i.e., has its global minimum) at the resetting location $x_{r}$. Thus, after
every resetting event the particle experiences a force
$F(x)=-\partial_{x}V(x)$ which facilitates the particle to return to the
resetting location. This motion will be referred as the return phase. The
return phase persists until the particle hits the resetting location for the
first time and at that moment the trap is switched off. This marks the end of
the return phase, and the particle resumes its motion in the (diffusive)
exploration phase. In what follows, we will set the resetting location as the
origin $x_{r}=0$ without loss of any generality.

Let $x(t)$ be the position of the particle at time $t$. The evolution of the
particle in the subsequent time interval $dt$ depends on which phase the
particle is in. If the particle is in the exploration phase then within time
interval $dt$, a resetting event can occur with probability $rdt$ and if so
happens the external potential is switched on, and the particle starts moving
in the force field. Otherwise, the particle continues to perform the free
diffusion with probability $(1-rdt)$. On the other hand if at time $t$ the
particle is in the return phase then it moves in force field until it reaches
the origin where the free diffusion starts again. The evolution can be written
as following:

$\displaystyle\begin{split}&\text{In~{}exploration~{}phase}\\\
&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}x(t+dt)=\left[\begin{array}[]{@{}l@{\quad}l@{}}x(t)+\sqrt{2D}~{}dW(t),{}{}&\text{with~{}prob.}~{}(1-rdt)\\\
x(t)+F(x(t))dt+\sqrt{2D}~{}dW(t),{}{}&\text{with~{}prob.}~{}rdt\end{array}\right.\\\
&\text{In~{}return~{}phase}\\\
&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}x(t+dt)=x(t)+F(x(t))dt+\sqrt{2D}~{}dW(t),\end{split}$
(1)

where $dW(t)=\int_{t}^{t+dt}dt^{\prime}\eta(t^{\prime})$ with $\eta(t)$ being
a zero-mean Gaussian white noise and $D$ is the diffusion constant.

In each realization, the particle starts from the origin in the exploration
phase, and has multiple switching between the two phases along a long
trajectory of duration $t$. Thus at any observation time, the particle can be
found at random in either of the two phases. Let $\rho_{D}(x,t)$ and
$\rho_{R}(x,t)$ denote the probability density functions of finding the
particle at position $x$ at time $t$ in the exploration and return phase
respectively. Thus, the probability density function for the compound process
at time $t$ is given by [77]

$\displaystyle\rho(x,t)=\rho_{D}(x,t)+\rho_{R}(x,t),$ (2)

which is normalized $\int^{+\infty}_{-\infty}~{}dx~{}\rho(x,t)=1$. On the
other hand, the individual phase densities are normalized to the time
dependent probabilities to be in each of the phases. To be specific,
$p_{D}(t)=\int^{+\infty}_{-\infty}~{}dx~{}\rho_{D}(x,t)$ and
$p_{R}(t)=\int^{+\infty}_{-\infty}~{}dx~{}\rho_{R}(x,t)$ are the time
dependent probabilities to find the particle in the exploration and return
phases respectively, with the normalization $p_{D}(t)+p_{R}(t)=1$.

In a recent work [77], a general framework has been developed to describe
stochastic processes under stochastic returns. Following this formalism, and
translating to our set-up, we get the following coupled Fokker-Planck (FP)
equations for the densities

$\displaystyle\partial_{t}\rho_{D}(x,t)=-\partial_{x}J_{D}(x,t)-r\rho_{D}(x,t)+\delta(x)\mathscr{J}_{R}(0,t),$
(3)
$\displaystyle\partial_{t}\rho_{R}(x,t)=-\partial_{x}J_{R}(x,t)+r\rho_{D}(x,t)-\delta(x)\mathscr{J}_{R}(0,t),$
(4)

where $J_{D}(x,t)=-D\partial_{x}\rho_{D}(x,t)$ and
$J_{R}(x,t)=F(x)\rho_{R}(x,t)-D\partial_{x}\rho_{R}(x,t)$ are the fluxes in
the exploration and return phases respectively. Note here,
$\mathscr{J}_{R}(0,t)=J_{R}(0^{-},t)-J_{R}(0^{+},t)$ is the net flux of
returning particles those enter at the origin from the negative and positive
side, and immediately switch their motion to the exploration phase. Since all
the particles initially start their motion in exploration phase (which is free
diffusion), we have

$\displaystyle\rho_{D}(x,0)=\delta(x)~{},~{}~{}\rho_{R}(x,0)=0.$ (5)

Interestingly, the return phase density can be written as a sum of the
densities $\rho_{R}^{+}(x,t)$ and $\rho_{R}^{-}(x,t)$ which describe the
particles returning from $x>0$ and $x<0$, respectively, such that [77]

$\displaystyle\rho_{R}(x,t)=\rho_{R}^{-}(x,t)\Theta(-x)+\rho_{R}^{+}(x,t)\Theta(x),$
(6)

where $\Theta(x)$ represents the Heaviside step function. This decomposition
is only natural since the returning particles can either be in the positive or
negative $x$ and switching between them is possible only via the exploration
phase. Thus, the origin acts as an effective absorbing boundary for the return
probability densities such that

$\displaystyle\rho_{R}^{\pm}(0,t)=0.$ (7)

Furthermore, since the probability can not get accumulated at infinity at any
time for motions in both phases, we have the natural boundary conditions

$\displaystyle\lim_{x\to\pm\infty}\rho_{D}(x,t)=0,$ (8a)
$\displaystyle\quad\lim_{x\to\pm\infty}\rho_{R}^{\pm}(x,t)=0.$ (8b)

Finally, the exploration phase must be continuous at the origin implying

$\displaystyle\rho_{D}(0^{+},t)=\rho_{D}(0^{-},t).$ (9)

We now proceed to solve the set of time dependent equations (namely Eqs. (3)
and (4)) along with the initial conditions in Eq. (5) and boundary conditions
in Eqs. (7)-(9) discussed above.

##  3 Solution of the master equations

In the previous section, we have presented the FP Eqs. (3) and (4) that govern
the motion of particles in the exploration and the return phase. To solve
them, we first perform the following Laplace transforms
$\tilde{\rho}_{D,R}(x,s)=\int_{0}^{\infty}~{}dt~{}e^{-st}~{}\rho_{D,R}(x,t)$
and $\tilde{j}(s)=\int_{0}^{\infty}~{}dt~{}e^{-st}~{}\mathscr{J}_{R}(0,t)$ on
both sides of Eqs. (3) and (4) and get

$\displaystyle D\dfrac{\partial^{2}\tilde{\rho}_{D}(x,s)}{\partial x^{2}}$
$\displaystyle-(s+r)\tilde{\rho}_{D}(x,s)+\delta(x)\left[1+\tilde{j}(s)\right]=0,$
(10) $\displaystyle D\dfrac{\partial^{2}\tilde{\rho}_{R}(x,s)}{\partial
x^{2}}$ $\displaystyle-\dfrac{\partial[F(x)\tilde{\rho}_{R}(x,s)]}{\partial
x}-s\tilde{\rho}_{R}(x,s)+r\tilde{\rho}_{D}(x,s)-\delta(x)\tilde{j}(s)=0,$
(11)

where we have used the initial conditions $\rho_{D}(x,0)=\delta(x)$ and
$\rho_{R}(x,0)=0$. Using the boundary conditions from Eq. (8a), the solution
of Eq. (10) can be easily found as

$\displaystyle\tilde{\rho}_{D}(x,s)=A(s)~{}\exp\left[-\sqrt{\frac{s+r}{D}}|x|\right],~{}~{}\text{for}~{}~{}-\infty<x<\infty,$
(12)

where $A(s)$ is a $s$-dependent constant. To compute $A(s)$, we integrate both
sides of Eq. (10) over $[-\epsilon,\epsilon]$ around $x=0$ and finally take
the $\epsilon\to 0$ limit. This provides us the following jump condition

$\displaystyle D\left[\dfrac{\partial\tilde{\rho}^{+}_{D}(x,s)}{\partial
x}\Big{|}_{x\to 0^{+}}-\dfrac{\partial\tilde{\rho}^{-}_{D}(x,s)}{\partial
x}\Big{|}_{x\to 0^{-}}\right]=-[1+\tilde{j}(s)],$ (13)

where the superscripts ± indicate the solutions in the positive and negative
sides of the origin. Substituting $\tilde{\rho}^{\pm}_{D}(x,s)$ from Eq. (12)
into Eq. (13), we find

$\displaystyle A(s)=\dfrac{1+\tilde{j}(s)}{2\sqrt{D(r+s)}},$ (14)

where $\tilde{j}(s)$ is still unknown and has to be determined from the return
phase density as we will show now. Integrating both sides of Eq. (11) over a
small region across the origin $x=0$, we arrive at the second discontinuity
condition

$\displaystyle D\left[\dfrac{\partial\tilde{\rho}^{+}_{R}(x,s)}{\partial
x}-\dfrac{\partial\tilde{\rho}^{-}_{R}(x,s)}{\partial
x}\right]_{x=0}=\tilde{j}(s).$ (15)

Hence, using Eqs. (11) and (15), we can now solve for $\tilde{j}(s)$ which, in
turn, gives $A(s)$, although we still need to find
$\tilde{\rho}_{R}^{\pm}(x,s)$.

The formalism presented so far provides a framework to compute the densities
in Laplace space. Naturally, this framework can easily be applied to any set-
up concerning an arbitrary confining potential $V(x)$ exerted by the trap.
Here, we consider the trap to be linear so that $V(x)=\lambda|x|$ with
$\lambda>0$. This results in an external force
$F(x)=-\lambda~{}\text{sgn}(x)$, where $\text{sgn}(x)=1$ for $x>0$,
$\text{sgn}(x)=-1$ for $x<0$ and equal to $0$ for $x=0$. Using this expression
of force in Eq. (11), we write the following two differential equation for
positive and negative $x$ regions

$\displaystyle D\dfrac{\partial^{2}\tilde{\rho}^{+}_{R}(x,s)}{\partial
x^{2}}+\lambda\dfrac{\partial[\tilde{\rho}^{+}_{R}(x,s)]}{\partial
x}-s\tilde{\rho}^{+}_{R}(x,s)+rA(s)~{}e^{-\sqrt{\frac{s+r}{D}}x}$
$\displaystyle=0,~{}{\text{for}~{}x>0,}$ (16) $\displaystyle
D\dfrac{\partial^{2}\tilde{\rho}^{-}_{R}(x,s)}{\partial
x^{2}}-\lambda\dfrac{\partial[\tilde{\rho}^{-}_{R}(x,s)]}{\partial
x}-s\tilde{\rho}^{-}_{R}(x,s)+rA(s)~{}e^{\sqrt{\frac{s+r}{D}}x}$
$\displaystyle=0,~{}{\text{for}~{}x<0.}$ (17)

Using the boundary conditions $\tilde{\rho}_{R}^{\pm}(\pm\infty,s)\to 0$ (from
Eq. (8b)) and $\tilde{\rho}_{R}^{\pm}(0,s)=0$ (from Eq. (7)), we arrive at the
full solution for the returning density

$\displaystyle\tilde{\rho}_{R}(x,s)=\dfrac{rA(s)}{\lambda\alpha_{s}-r}\left(e^{-\alpha_{s}|x|}-e^{-\mu_{s}|x|}\right),~{}~{}\text{for}~{}~{}-\infty<x<\infty,$
(18)

with

$\displaystyle\mu_{s}=\dfrac{\lambda+\sqrt{\lambda^{2}+4sD}}{2D}~{}\text{and}~{}\alpha_{s}=\sqrt{\dfrac{s+r}{D}}~{}.$
(19)

We now substitute $\tilde{\rho}_{R}(x,s)$ and $A(s)$ from Eqs. (18) and (14),
respectively, into Eq. (15) to find

$\displaystyle\tilde{j}(s)=\frac{r(\alpha_{s}-\mu_{s})}{\mu_{s}r-\alpha_{s}^{2}\lambda},$
(20)

which, in turn, results in

$\displaystyle
A(s)=\frac{\lambda\alpha_{s}-r}{\lambda(r+2s)-r\sqrt{4Ds+\lambda^{2}}}{=\frac{\lambda\sqrt{s+r}-r\sqrt{D}}{\sqrt{D}(\lambda(r+2s)-r\sqrt{4Ds+\lambda^{2}})}}.$
(21)

Therefore, Eqs. (12) and (18) along with the explicit expression for $A(s)$
provides us the full solutions for the master equations (3) and (4) in the
Laplace space. The normalization condition for the joint density
$\int_{-\infty}^{+\infty}~{}dx~{}\tilde{\rho}(x,s)=1/s$ also follows
trivially.

Having fully solved the FP equations in Laplace space, the steady state
density can be readily obtained by making use of the final value theorem which
states $\rho^{\text{ss}}(x)=\lim_{s\to 0}[s\tilde{\rho}(x,s)]$. Applying this
to Eqs. (12) and (18) along with Eq. (21), we find

$\displaystyle\rho_{D}^{\text{ss}}(x)$
$\displaystyle=\dfrac{\alpha_{0}p_{D}}{2}~{}e^{-\alpha_{0}|x|},$ (22)
$\displaystyle\rho_{R}^{\text{ss}}(x)$
$\displaystyle=\dfrac{rp_{D}}{2(\lambda-\alpha_{0}D)}\left(e^{-\alpha_{0}|x|}-e^{-\frac{\lambda}{D}|x|}\right),$
(23)

for $-\infty\leq x\leq\infty$ where
$p_{D}=\lim_{t\to\infty}p_{D}(t)=\frac{\alpha_{0}\lambda}{\alpha_{0}\lambda+r}$
is the steady state probability for the system to be in the exploration phase
and recall that $\alpha_{0}\equiv\alpha_{s=0}=\sqrt{\frac{r}{D}}$ is the
typical inverse distance traveled by the particle in the exploration phase
[77]. Note that the functions in Eqs. (22) and (23) correspond to the right
eigenfunctions of the Fokker-Planck operator with the largest eigenvalue which
is zero. The steady state flux at the origin can be computed similarly and
this gives

$\displaystyle\mathscr{J}_{R}(0,\infty)=\lim_{s\to
0}[s\tilde{j}(s)]=\dfrac{r\lambda}{\alpha_{0}D+\lambda},$ (24)

which is consistent with
$\mathscr{J}_{R}(0,\infty)=J_{R}(0^{-},\infty)-J_{R}(0^{+},\infty)$ as defined
in Sec. 2. For $\lambda\gg\alpha_{0}D$, one essentially reaches the limit of
instantaneous resetting where flux due to resetting is
$\mathscr{J}_{R}(0,\infty)=r$, as expected [17]. For more details on the
properties of steady states we refer the readers to Ref. [77]. In the
following, we go beyond steady states and study time dependent solutions of
the FP equations in (3) and (4) to understand the approach to the steady
state. First, we present exact solutions for the time dependent propagators in
the next section.

##  4 Time dependent density $\rho(x,t)$

The position distribution functions in time domain can be obtained by
performing the inverse Laplace transform so that

$\displaystyle\rho_{D,R}(x,t)=\dfrac{1}{2\pi
i}\int_{\Gamma-i\infty}^{\Gamma+i\infty}~{}ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s),$
(25)

which is expressed in terms of the Bromwich integral [80]. Here, the
integration is done along the imaginary axis (vertical contour through
$\text{Re}(s)=\Gamma$) in the complex-$s$ plane such that all the
singularities lie on the left of it. We first recall the densities in the
exploration and return phases from Eqs. (12) and (18)

$\displaystyle\tilde{\rho}_{D}(x,s)$
$\displaystyle=\dfrac{(\lambda\sqrt{s+r}-r\sqrt{D})[\lambda(r+2s)+2r\sqrt{D}\sqrt{s+\beta}]}{4\lambda^{2}\sqrt{D}s(s-\gamma)}~{}e^{-\sqrt{s+r}|z|},$
(26) $\displaystyle\tilde{\rho}_{R}(x,s)$
$\displaystyle=\dfrac{r[\lambda(r+2s)+2r\sqrt{D}\sqrt{s+\beta}]}{4\lambda^{2}s(s-\gamma)}~{}\left(e^{-\sqrt{s+r}|z|}-e^{-\sqrt{\beta}|z|}~{}e^{-\sqrt{s+\beta}|z|}\right),$
(27)

where we have used Eq. (21) and defined

$\displaystyle{z=\frac{x}{\sqrt{D}},~{}~{}\beta=\frac{\lambda^{2}}{4D},~{}~{}\gamma=\frac{r^{2}}{4\beta}-r.}$
(28)

It is easy to see that $\tilde{\rho}_{D,R}(x,s)$ have three singularities for
$r\neq\beta$: one pole at $s=0$ and two branch points at $s=-r$ and
$s=-\beta$. Naturally when $r=\beta$, we have one pole at $s=0$ and one branch
point at $s=-r=-\beta$. Moreover, substituting $s$ by $1/u$, one can see that
there is one more branch point at infinity which we choose to be at
$s=-\infty$ for both cases. Looking at the expressions in (26)-(27) at a first
glance, it seems that there is a pole also at $s=\gamma$ for both
$\tilde{\rho}_{D}$ and $\tilde{\rho}_{R}$, however it turns out to be a
removable singularity. To see this, we first note that
$s-\gamma=\frac{D}{\lambda^{2}}(\lambda\alpha_{s}-r)(\lambda\alpha_{s}+r)$.
The factor $(\lambda\alpha_{s}-r)$ gets cancelled with the first factor term
in the numerator in Eq. (26). On the other hand, for $\tilde{\rho}_{R}(x,s)$,
the cancellation happens differently for $r>\beta$ and $r<\beta$. In the
former case, note from Eq. (27) that
$e^{-\sqrt{s+r}|z|}-e^{-\sqrt{\beta}|z|}~{}e^{-\sqrt{s+\beta}|z|}={(s-\gamma)g_{1}(\gamma)+O((s-\gamma)^{2})}$,
which essentially eliminates the factor $s-\gamma$ in the denominator and thus
removing the pole at $s=\gamma$. In the latter case, this cancellation is done
by the factor term in the numerator in Eq. (27). Taking into account the
structure of these singularities in details, we now proceed to perform the
integrals in Eq. (61) using Bromwich contours (see Fig. 6) for $r\neq\beta$
and $r=\beta$ cases separately in the next subsections.

It is worth noticing that the parameters $r$ and $\beta$ are associated to two
different time scales in the system. The first time scale $\tau_{res}=r^{-1}$
comes from the resetting events which provides the relaxation time for the
density to reach steady state in case of instantaneous resetting [1]. On the
other hand, $\beta=\frac{\lambda^{2}}{4D}$ provides the relaxation time scale
$\tau_{pot}=\beta^{-1}$ for the particle to reach steady state in absence of
any resetting events. In presence of both, these two times scales compete and
naturally we get the following two cases: $r=\beta$ and $r\neq\beta$. In the
next sections we will observe relaxation terms with prefactors $e^{-rt}$ and
$e^{-\beta t}$ appearing naturally in the expressions of in $\rho_{D,R}(x,t)$.

###  4.1 Case 1: $r=\beta$

In this case, the branch points $s=-r$ and $s=-\beta$ coalesce with each
other, and thus, we now have one pole at $s=0$ and two branch points at $s=-r$
and $s=-\infty$ (hence one branch cut). We construct a contour as shown in
Fig. 6 and perform the integral by considering the contributions from the
branch cuts and the pole. We follow the steps described in A and collect the
contributions from the pole and the branch cuts. After making some
simplifications we get the following expressions for the densities in the
exploration and return phases:

$\displaystyle\rho_{D}(x,t)=\rho^{ss}_{D}(x)+\frac{1}{4\pi\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(r+y)}}{(r+y)(r+\gamma+y)}\bigg{[}\lambda
r^{2}\sin\left(\sqrt{y}\left|z\right|\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+~{}\sqrt{\frac{y}{D}}\left(2r^{2}D+\lambda^{2}r+2\lambda^{2}y\right)\cos\left(\sqrt{y}\left|z\right|\right)\bigg{]},$
(29)
$\displaystyle\rho_{R}(x,t)=\rho^{ss}_{R}(x)-\frac{r}{4\pi\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(r+y)}(1-e^{-\sqrt{r}|z|})}{(r+y)(r+\gamma+y)}\bigg{[}2r\sqrt{Dy}\cos\left(\sqrt{y}\left|z\right|\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\lambda(r+2y)\sin\left(\sqrt{y}\left|z\right|\right)\bigg{]},$
(30)

where $z=x/\sqrt{D}$ and steady state densities $\rho^{\rm ss}_{D,R}(x,t)$ are
given in Eqs. (22) and (23). We verify these expressions against numerical
simulations in Fig. 1. Although the results presented above are exact, they do
not provide much insights. Moreover, it is difficult to quantify from these
expressions how the densities relax to their corresponding steady states. In
Sec. 6, we will analyze these expressions to understand the relaxation
behaviour.

###  4.2 Case 2: $r\neq\beta$

In this case, there are three branch points: $s=-r,s=-\beta,s=-\infty$, and
one pole at $s=0$. We refer to A for details of these Laplace inversions to
obtain expressions of the densities. Since these expressions are quite
lengthy, we prefer to present them in the Appendix A (see Eqs. (65-68)). In
Fig. 1, we compare the expressions for the densities in the exploration and
the return phases given in Eqs. (65-68), against numerical simulations for
both $r>\beta$ and $r<\beta$. We observe excellent agreement between the two
in each case.

Figure 1: Comparison of analytical results for density functions
$\rho_{D}(x,t)$, $\rho_{R}(x,t)$ and $\rho(x,t)=\rho_{D}(x,t)+\rho_{R}(x,t)$,
with the same obtained from numerical simulation. First, second, and third-
column, respectively, correspond to three different cases: $r<\beta$,
$r=\beta$, $r>\beta$. Solid curves are the analytical results given in Eqs.
(65-68) whereas the circles are obtained from numerical simulation for
$10^{8}$ realizations. In each plot, dashed curves indicates the steady state
densities given in Eqs. (22) and (23). Common parameters used in these plots
are $D=0.5$, $dt=10^{-3}$, $t=15$.

##  5 Probability to be in the exploration and return phases

Before we get into the relaxation of the density $\rho(x,t)$, it is imperative
to study simple quantities to get some insights. To this end, in this section,
we study the probability for the particle to be in the exploration phase. This
quantity was formally defined in Sec. 2. We recall here for brevity

$\displaystyle p_{D}(t)=\int^{+\infty}_{-\infty}~{}dx~{}\rho_{D}(x,t).$ (31)

Since the particle starts in the exploration phase, we have $p_{D}(0)=1$ and
then the probability decreases with time and finally saturates to a steady
state value ($<1$) at large time. In this section, we obtain an exact
expression for $p_{D}(t)$. For simplicity, we will consider the case when
$r=\beta$ but the other cases also can be computed following the same
procedure. To begin with, we integrate $\tilde{\rho}_{D}(x,s)$ given in Eq.
(26) over the entire real space, and we obtain

$\displaystyle\tilde{p}_{D}(s)=\dfrac{2\sqrt{s+r}-\sqrt{r}}{(2s+r)\sqrt{s+r}-\sqrt{r}(s+r)},$
(32)

where we have used the condition $\lambda=\sqrt{4Dr}$ that comes from
$r=\beta$ limit. The probability $p_{D}(t)$ can be obtained using the inverse
Laplace transform

$\displaystyle p_{D}(t)=\dfrac{1}{2\pi
i}\int_{\Gamma-i\infty}^{\Gamma+\infty}~{}ds~{}e^{st}~{}\tilde{p}_{D}(s).$
(33)

Clearly, we can see from Eq. (32) that the integrand of the above equation has
one pole at $s=0$ and one branch point at $s=-r$. Therefore, we consider the
contour shown in Fig. 6 with the branch points $s_{b}^{(1)}$ and $s_{b}^{(2)}$
coalescing at $s=-r$. Following similar steps of calculation as shown in A, we
finally get

$\displaystyle
p_{D}(t)=\dfrac{2}{3}+\dfrac{r^{3/2}e^{-rt}}{\pi}\int_{0}^{\infty}~{}dy~{}\dfrac{e^{-yt}}{\sqrt{y}(r+y)(r+4y)}.$
(34)

where the integral can be computed easily. The full expression for $p_{D}(t)$
is given by

$\displaystyle
p_{D}(t)=\dfrac{2}{3}+\frac{1}{3}\left[2e^{-\frac{3rt}{4}}\text{erfc}\left(\frac{\sqrt{rt}}{2}\right)-\text{erfc}\left(\sqrt{rt}\right)\right],$
(35)

where $\text{erfc}(u)=\frac{2}{\sqrt{\pi}}\int_{u}^{\infty}~{}dt~{}e^{-t^{2}}$
is the complementary error function. The probability to find the particle in
the return phase at time $t$ is then simply given by $p_{R}(t)=1-p_{D}(t)$.

At large time, we see that $p_{D}(t)$, relaxes exponentially as $e^{-rt}$ to
its steady state value $2/3$ for $r=\beta$. Note here that $p_{D}(\infty)$ is
nothing but the fraction of time spent in the exploration phase in steady
state such that

$\displaystyle
p_{D}(\infty)=\frac{1/r}{1/r+\langle\overline{\tau}(x)\rangle},$ (36)

where $1/r$ and $\langle\overline{\tau}(x)\rangle$ are the mean times spent in
the exploration and return phases respectively. In [77], it was shown that
$\langle\overline{\tau}(x)\rangle=\frac{1}{\alpha\lambda}$, which upon
plugging to Eq. (36) yields the value as expected. In Fig. 2, we show the
comparison of the analytical result Eq. (35) with the numerical simulation for
$r=\beta=0.1$ and they show a very good agreement.

Figure 2: Probability of the system in the exploration phase with time. Solid
line is the analytical result shown in Eq. (35) while circles are obtained
using numerical simulation for $10^{5}$ realizations. The parameters for the
simulation are $r=\beta=0.1$, $D=0.5$, and $dt=10^{-3}$.

Setting the stage, we now delve deeper to understand the relaxation of the
probability densities. This is done in the next section.

##  6 Relaxation to the steady state

Presence of resetting affects the stochastic dynamics of a particle in many
different ways. For example, a free Brownian particle under resetting
mechanism reaches a steady state which in absence of resetting is described by
a Gaussian propagator with variance growing linearly with time. Moreover, the
relaxation properties to the steady state is also unusual compared to the
equilibration of the same particle in a confining potential in the absence of
resetting. In the context of instantaneous resetting of a free Brownian
particle, it has been shown that at a given large time $t$ the probability
density $\rho(x,t)$ as a function of $x$ exhibits two markedly different
phases: an inner core which has already reached the steady state and an outer
core which is still explicitly time dependent [1, 21]. The boundaries that
separate these two regions grow linearly with time. Physically, these two
regions are created by two groups of trajectories: ones that have encountered
zero (almost zero) resetting events up to the observation time $t$ and the
ones that have undergone many resetting events. The former group contributes
to the outer region and the latter group contributes to the inner core region
of the density function. Such ‘cone spreading’ relaxation with travelling
fronts is studied and observed in few other stochastic dynamics with
instantaneous resetting e.g., underdamped Brownian particle [28], interface
dynamics [21], random acceleration process [29] and resetting occurring with
time dependent rate [23].

Notably, in all these resetting systems the relaxation properties are studied
using the renewal approach. However, in many problems it might be difficult to
follow the renewal approach analytically. In such cases one would require to
rely upon the master equation approach where the relaxation is usually
described through spectral decomposition. In this section, we demonstrate how
the above mentioned ‘cone spreading’/‘traveling boundary’ type relaxation
behavior can be understood also from the master equation approach. We first
discuss the limit $\lambda\to\infty$ (more precisely, $\lambda\gg\alpha_{0}D$)
which corresponds to the well studied instantaneous resetting scenario and
recover the existing results. Subsequently, we extend our analysis to the non-
instantaneous resetting case, (i.e., finite $\lambda$) and provide new
insights that emerge due to the finite duration of return events.

###  6.1 Instantaneous resetting

We start with the case of instantaneous resetting. Taking the limit
$\lambda\to\infty$ in Eqs. (26) and (27), we find

$\displaystyle\tilde{\rho}_{R}(x,s)=$ $\displaystyle 0,$ (37)
$\displaystyle\tilde{\rho}_{D}(x,s)=$
$\displaystyle\dfrac{1}{2s}\sqrt{\dfrac{s+r}{D}}~{}e^{-\sqrt{s+r}|z|}.$ (38)

Since the return events are instantaneous, $\rho_{D}(x,t)$ is essentially the
total density function. Hence, in the following we drop its subscript $D$. To
get $\rho(x,t)$ we perform the inverse Laplace transform using the following
Bromwich integral

$\displaystyle\rho(x,t)=\dfrac{1}{2\pi
i}\int_{\Gamma-i\infty}^{\Gamma+i\infty}~{}ds~{}e^{st}~{}\tilde{\rho}(x,s),$
(39)

where $\Gamma$ is the vertical contour passing through $\Gamma=\mathrm{Re}(s)$
of the complex $s$-plane (see Fig. 6 in A). We notice that $\tilde{\rho}(x,s)$
in Eq. (38) has two branch points at $s=-r$ and $s=-\infty$, and a pole at
$s=0$ and they all lie to the left of the vertical contour $\Gamma$. Following
the steps sketched in A, we finally obtain

$\displaystyle\rho(x,t)=\frac{\alpha_{0}}{2}~{}e^{-\sqrt{r}|z|}+\frac{e^{-rt}}{2\pi\sqrt{D}}\int_{0}^{\infty}~{}dy~{}\dfrac{\sqrt{y}~{}e^{-yt}}{y+r}\cos\left(|z|\sqrt{y}\right),~{}\text{with}~{}z=\frac{x}{\sqrt{D}},$
(40)

where the first term is the contribution from the pole and the second term
gets contribution from the branch cut $(-\infty,-r]$. Further, we make a
change of variable $y=w^{2}$ and extend the limit of integration on the entire
real line by writing $\cos(w)=\frac{e^{iw}-e^{-iw}}{2}$ which yields

$\displaystyle\rho(x,t)$
$\displaystyle=\sqrt{\frac{r}{D}}\frac{e^{-\sqrt{r}|z|}}{2}+\frac{e^{-rt}}{2\pi\sqrt{D}}\mathcal{K}_{\sqrt{r}}^{\mathcal{A}^{(0)}}(x,t),~{}~{}~{}\text{where,}$
(41) $\displaystyle\mathcal{K}_{a}^{\mathcal{A}}(x,t)$
$\displaystyle=\int_{-\infty}^{+\infty}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|},~{}~{}~{}{\text{with}~{}z=\frac{x}{\sqrt{D}}}~{}~{}\text{and},$
(42) $\displaystyle\mathcal{A}^{(0)}(w)$
$\displaystyle=\frac{w^{2}}{w+i\sqrt{r}}.$ (43)

Note that Eq. (41) provides the exact density function at all time. Also note
that the first term in Eq. (41) is the steady state density $\rho^{\rm ss}(x)$
of the Brownian particle under instantaneous resetting [17]. The second term
provides the relaxation to this steady state. To understand this relaxation
better and in detail, we now analyze the integral in Eq. (42) at large $t$.
For that we extend this integral in the complex $w$ plane and perform the
integral on the closed contour $\mathcal{C}$ as shown in Fig. 7. Performing
the saddle point integral as sketched in B, we evaluate
$\mathcal{K}_{\sqrt{r}}^{\mathcal{A}}(x,t)$ and obtain the following
approximate expression of the second term in Eq. (41) for large $t$:

$\displaystyle\frac{e^{-rt}}{2\pi\sqrt{D}}\mathcal{K}_{\sqrt{r}}^{\mathcal{A}^{(0)}}(x,t)$
$\displaystyle=-\frac{e^{-\sqrt{r}|z|}}{2}\sqrt{\frac{r}{D}}\Theta(|z|-\sqrt{4r}t)+\frac{e^{-rt-\frac{z^{2}}{{4}t}}}{2\pi\sqrt{D}}~{}K_{z}^{\mathcal{A}^{(0)}}\left(\sqrt{r}-\frac{|z|}{2t},t\right),$
(44) $\displaystyle\text{where,}~{}K_{z}^{\mathcal{A}^{(0)}}(b,t)$
$\displaystyle\approx\mathcal{A}^{(0)}\left(ib+i\frac{|z|}{2t}\right)e^{tb^{2}}\left[i\pi~{}\text{sgn}(b)~{}\text{erfc}(|b|\sqrt{t})\right]+\psi(0)\sqrt{\frac{\pi}{t}}~{},$
(45)

with
$\psi(0)=\frac{i}{b}\left[\mathcal{A}^{(0)}\left(i{|z|}/{2t}\right)-\mathcal{A}^{(0)}\left(ib+i{|z|}/{2t}\right)\right]$.
The first term in Eq. (44) appears for $|z|>\sqrt{4r}t$ and interestingly, it
is exactly same as the first term in Eq. (41) but with opposite sign. As a
result the steady state form for $\rho(x,t)$ serves as the dominant
contribution for $|x|\leq\sqrt{4Dr}t$ and thus creating a core region which
expands in time with a speed $\sqrt{4Dr}$ on both sides of $x=0$ [21]. Outside
this region, $\rho(x,t)$ is explicitly time dependent. Performing further
simplifications and keeping the dominant order terms in the large $t$ limit,
we get the following simpler and more explicit asymptotic form for the density
as

$\displaystyle\begin{split}\rho(x,t)\approx\frac{1}{2}\sqrt{\frac{r}{D}}e^{-\sqrt{r}|z|}&\Theta(\sqrt{4r}t-|z|)+\dfrac{e^{-rt}e^{-z^{2}/4t}}{\sqrt{4\pi
Dt}}\left[\frac{z^{2}}{z^{2}-4rt^{2}}-\frac{\sqrt{r}t}{|z|-\sqrt{4r}t}\right]\\\
&+\frac{\alpha_{0}}{4}\text{sgn}(|z|-2\sqrt{r}t)~{}e^{-\sqrt{r}|z|}~{}\text{erfc}\left(\frac{||z|-2\sqrt{r}t|}{\sqrt{4t}}\right),\end{split}$
(46)

where recall $z=x/\sqrt{D}$. This expression clearly describes the ‘cone
spreading’ relaxation behaviour with travelling fronts, first observed using
renewal approach in Ref. [21]. In this reference the relaxation to the steady
state is interpreted as a dynamical transition occurring in the system when
observed at a fixed spatial point. The transition was further characterised by
non-analytic properties of an appropriate large deviation function (LDF) [21].
Looking at the large $t$ asymptotic of the different terms in Eq. (46) it is
easy to see that the density in the large $t$ limit satisfies a large
deviation principle (for a review on large deviation principle, see Ref. [79])

$\displaystyle\rho(x,t)$ $\displaystyle\asymp
e^{-t\Phi(x/t)},~{}\textit{i.e.}~{}\Phi\left(v=\frac{x}{t}\right)=-\lim_{t\to\infty}\frac{\ln\rho(x,t)}{t}~{}\text{where},~{}$
(47) $\displaystyle\Phi\left(v\right)$
$\displaystyle=\begin{cases}\sqrt{\frac{r}{D}}|v|,~{}&\text{for}~{}|v|<v^{*}\\\
r+\frac{v^{2}}{4D},~{}&\text{for}~{}|v|\geq
v^{*},\end{cases}~{}\text{with}~{}v^{*}=\sqrt{4Dr}.$ (48)

The non-analytic structure of the large deviation function $\Phi(v)$ at
$v=v^{*}$ is interpreted as a dynamical transition. This transition mentioned
above is manifested as follows: at a fixed spatial point $x$, the LDF is
quadratic at short time when $|v|>v^{*}$. On the other hand, as time
increases, $v$ decreases and eventually crosses $v^{*}$ from above. In this
case, the LDF changes from the quadratic to a linear form as in [21].

While renewal approach clearly depicts the physical mechanism behind this
unusual relaxation behaviour, our analysis reveals that mathematically this
transition happens exactly when the saddle point hits the pole, i.e., when
$w^{*}=w_{+}$ which gives $|z^{*}|=t\sqrt{4r}$ i.e., $|x^{*}|=t\sqrt{4Dr}$
[see Fig. 7]. In a broader sense, our approach identifies a connection between
the structure of the singularities in propagator in Laplace space and the
relaxation behaviour of the particle.

###  6.2 Non-instantaneous resetting

We now turn our attention to the non-instantaneous resetting. Recalling from
Sec. 4, there are three different regimes $r<\beta$, $r=\beta$ and $r>\beta$
for which the exact time dependent densities were obtained in Sec. 4. In the
following, we discuss the relaxation properties for these cases separately
using the saddle point method applied in the previous section. In each case,
we will write the total density $\rho(x,t)=\rho_{D}(x,t)+\rho_{R}(x,t)$ in the
form of Eq. (41) i.e., the steady state part $\rho^{\rm ss}(x)$ plus
relaxation parts which are in the form of integral in Eq. (42).

####  6.2.1 Case I ($r=\beta$):

In this case, the densities in the exploration and return phases were already
obtained in Eqs. (29) and (30). Adding these two densities and simplifying, we
find that the we total density $\rho(x,t)$ can be written in the form of Eq.
(41) as

$\displaystyle\rho(x,t)$
$\displaystyle=\rho^{ss}(x)+\frac{e^{-rt}}{8\pi\sqrt{D}}\mathcal{K}_{\sqrt{r}}^{\mathcal{A}^{(1)}}(x,t),~{}\text{with}$
(49) $\displaystyle\mathcal{A}^{(1)}(w)$
$\displaystyle=\dfrac{w\bigg{(}3rw+4w^{3}-ir^{3/2}+\sqrt{r}(e^{-\sqrt{r}|z|}-1)\left[\sqrt{r}w-i(r+2w^{2})\right]\bigg{)}}{(w+i\sqrt{r})\big{(}\frac{r}{4}+w^{2}\big{)}},$
(50)

and $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$ is defined in Eq. (42). As done for
the instantaneous case we follow the calculations in B to compute
$\mathcal{K}_{\sqrt{r}}^{\mathcal{A}^{(1)}}(x,t)$ and find an approximate
expression of $\rho(x,t)$ at large $t$ given by

$\displaystyle\rho(x,t)\approx\rho_{ss}(x)\Theta(t\sqrt{4r}-|z|)+\dfrac{e^{-rt-\frac{z^{2}}{4t}}}{8\pi\sqrt{D}}K^{\mathcal{A}^{(1)}}_{z}\left(\sqrt{r}-\frac{|z|}{2t},t\right),$
(51)

where $K_{z}^{\mathcal{A}}(b,t)$ is given in Eq. (45). A simple observation of
the above equation conforms a similar ‘cone spreading’ type relaxation
behavior of the density to the steady state.

After analysing the different terms in Eq. (51), in this case also, one
observes that the total density $\rho(x,t)$ satisfies a large deviation form.
Interestingly, in this case, the LDF is same as that of the instantaneous
resetting given in Eq. (48).

Figure 3: Comparison of the densities $\rho(x,t)$ obtained using saddle point
calculation in Eq. (54) and computing the integrals numerically in Eq. (52)
for $r<\beta$. We observe that the distribution within an inner core region
(inside the dashed vertical lines given by $x=\pm\sqrt{4Dr}t$) has reached
steady state while the outer region is still in the transient state. These two
regions are separated by a travelling front which moves ballistically.
Parameters used in this plot are: $D=0.5,~{}r=0.3$ and $\beta=0.8$ i.e.,
$\lambda=1.2649$.

####  6.2.2 Case II ($r<\beta$):

Adding the right hand sides of Eqs. (65) and (66), and performing some tedious
manipulations we write the total density
$\rho(x,t)=\rho_{D}(x,t)+\rho_{R}(x,t)$ in the following form

$\displaystyle\begin{split}\rho(x,t)=\rho^{ss}(x)&+\frac{ie^{-rt}}{4\pi\sqrt{D}(\sqrt{r+\gamma}-\sqrt{r})}\left[\mathcal{K}_{\sqrt{r}}^{\mathcal{A}^{(2)}}(x,t)-\mathcal{K}_{\sqrt{r+\gamma}}^{\mathcal{A}^{(2)}}(x,t)\right]\\\
&+\frac{ire^{-\sqrt{\beta}|z|-\beta
t}}{4\pi\lambda^{2}(\sqrt{\beta+\gamma}-\sqrt{\beta})}\left[\mathcal{K}_{\sqrt{\beta}}^{\mathcal{A}^{(4)}}(x,t)-\mathcal{K}_{\sqrt{\beta+\gamma}}^{\mathcal{A}^{(4)}}(x,t)\right]\\\
&-\frac{re^{-rt}}{\pi\lambda(\sqrt{r+\gamma}-\sqrt{r})}\text{Im}\left[i\left(\mathcal{J}_{\sqrt{r}}^{\mathcal{A}^{(3)}}(x,t)-\mathcal{J}_{\sqrt{r+\gamma}}^{\mathcal{A}^{(3)}}(x,t)\right)\right]~{},\end{split}$
(52)

where $\gamma\neq 0$ and the $\mathcal{A}^{(i)}(w)$ functions are given in C
and $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$ is defined in Eq. (42). In this case
we have encountered a new type of integral of the form

$\displaystyle\mathcal{J}^{\mathcal{A}}_{a}(x,t)=\int_{0}^{\infty}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|},~{}~{}~{}\text{with}~{}~{}z=\frac{x}{\sqrt{D}}.$
(53)

This integral is similar to $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$ except that
the integration range is now restricted to $[0,\infty]$. One can follow a
similar procedure as done for $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$ to compute
$\mathcal{J}^{\mathcal{A}}_{a}(x,t)$ for large $t$. In D we evaluate this
integral using saddle point method and get an approximate expression for large
$t$ given by Eq. (96). Evaluating this expression for $\mathcal{A}^{(3)}(w)$
and $\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ from Eq. (96) for
$\mathcal{A}^{(2)}(w)$ and $\mathcal{A}^{(4)}(w)$, we insert them in Eq. (52)
to get an approximate expression for the density $\rho(x,t)$ in the large $t$
limit

$\rho(x,t)\approx~{}~{}\boxed{\begin{aligned} &\text{Eq.~{}\eqref{r-lthan-
beta}~{}with~{}large~{}}t~{}\text{approximations~{}of~{}}\mathcal{K}_{a}^{\mathcal{A}}(x,t)~{}\text{and}~{}\mathcal{J}_{a}^{\mathcal{A}}(x,t)\\\
&\text{are~{}inserted~{}from~{}Eqs.~{}\eqref{K_a^A-F}~{}and~{}\eqref{J_a^A-F}~{}respectively.}\end{aligned}}$
(54)

From the presence of the $\Theta$ functions in the expressions of
$\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ and $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$
in Eqs. (79) and (96) respectively, one can see the appearance of many
travelling fronts. However, after some manipulations one can see from the
relaxation part of the density in Eq. (52) that one gets an equal and opposite
contributions of the steady state part from terms with
$\Theta(|x|-\sqrt{4Dr}t)$, which gets manifested as ‘cone spreading’
relaxation as seen in case of $r=\beta$ in Sec. 6.2.1. Such a relaxation is
demonstrated in Fig. 3 where we provide a comparison between the saddle point
result in Eq. (54) and the one from direct numerical evaluation of the
integrals in Eq. (52). We observe nice agreement between them.

To understand the large deviation behaviour of the density $\rho(x,t)$ in this
case, one should look at the large $t$ asymptotic of the integrals of the form
$\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ and $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$
in Eq. (52). Before proceeding, we first note that the total contribution
coming from these integrals with $a=\sqrt{r+\gamma}$ and
$a=\sqrt{\beta+\gamma}$ [i.e. the total contributions from the second terms in
the square brackets in Eq. (52)] is very small because the contributions from
the poles at $w=i\sqrt{r+\gamma}$ and $w=i\sqrt{\beta+\gamma}$ should cancel
each other. Only the contributions from saddle points survive. This can also
be verified by evaluating them in Mathematica. This cancellation is expected
due to the following observation: recall that these extra poles, in fact,
originate from the pole $s=\gamma$, which appears while performing the inverse
Laplace transform of $\tilde{\rho}_{D}(x,s)$ and $\tilde{\rho}_{D}(x,s)$ given
in Eqs. (26) and (27). Moreover recall from sec. 4 (see the paragraph after
Eq. (28)) that $s=\gamma$ is a removable singularity. Hence, the total
contributions of these poles at $w=i\sqrt{r+\gamma}$ and
$w=i\sqrt{\beta+\gamma}$ from the integrals of the form
$\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ and $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$
in Eq. (52) should cancel each other and the net sum would be zero. Keeping
this in mind and finding the leading contributions of the other three
integrals (with poles at $w=i\sqrt{r}$ and $w=i\sqrt{\beta}$) and the saddle
point contributions from all the integrals in Eq. (52), we find that for large
$t$, the total density $\rho(x,t)$ satisfies a large deviation form with a LDF
which is, interestingly, once again same as that of the instantaneous case
given in Eq. (48). Here too, one observes the same dynamical transition
similar to the previous cases.

Figure 4: Comparison of the densities $\rho(x,t)$ obtained using saddle point
calculation in Eq. (57) and computing the integrals numerically in Eq. (55)
for $r>\beta$. We observe that the distribution within an inner core region
(inside the dashed vertical lines given by $x=\pm\sqrt{4D\beta}t$)) has
reached steady state while the outer region is still in the transient state.
This two regions are separated by a travelling front which moves
ballistically. Parameters used in this plot are: $D=0.5,~{}r=0.6$ and
$\beta=0.4$ i.e., $\lambda=0.8944$.

####  6.2.3 Case II: $r>\beta$

In this case, the total density is obtained by combining Eqs. (67)-(68) and
after some manipulations and rearrangements, we get

$\displaystyle\begin{split}\rho(x,t)=\rho^{ss}(x)&+\frac{ie^{-rt}}{4\pi\sqrt{D}(\sqrt{r+\gamma}-\sqrt{r})}\left[\mathcal{K}^{\mathcal{A}^{(2)}}_{\sqrt{r}}(x,t)-\mathcal{K}^{\mathcal{A}^{(2)}}_{\sqrt{r+\gamma}}(x,t)\right]\\\
&+\frac{ire^{-\sqrt{\beta}\left|z\right|}e^{-\beta
t}}{4\pi\lambda^{2}(\sqrt{\beta+\gamma}-\sqrt{\beta})}\left[\mathcal{K}^{\mathcal{A}^{(4)}}_{\sqrt{\beta}}(x,t)-\mathcal{K}^{\mathcal{A}^{(4)}}_{\sqrt{\beta+\gamma}}(x,t)\right]\\\
&-\frac{re^{-rt}}{\pi\lambda(\sqrt{r+\gamma}-\sqrt{r})}\text{Im}\left[i\left(\mathcal{J}_{\sqrt{r}}^{\mathcal{A}^{(3)}}(x,t)-\mathcal{J}_{\sqrt{r+\gamma}}^{\mathcal{A}^{(3)}}(x,t)\right)\right]\\\
&+\frac{re^{-rt}}{\pi\lambda(\sqrt{r+\gamma}-\sqrt{r})}\left[\mathbb{J}_{\sqrt{r},\sqrt{r-\beta}}^{\mathcal{A}^{(3)}}(x,t)-\mathbb{J}_{\sqrt{r+\gamma},\sqrt{r-\beta}}^{\mathcal{A}^{(3)}}(x,t)\right]~{},\end{split}$
(55)

for $\gamma\neq 0$, where the $\mathcal{A}^{(i)}(w)$ functions are given in C.
The functions $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$ and
$\mathcal{J}^{\mathcal{A}}_{a}(x,t)$ are defined in Eqs. (42) and (53)
respectively. We also define the following integral on real line segment
$[0,\zeta]$ for $\zeta>0$ as

$\displaystyle\mathbb{J}_{a,\zeta}^{\mathcal{A}}(x,t)=\int_{0}^{\zeta}dv~{}\frac{\mathcal{A}(iv)}{v-a}e^{tv^{2}-v|z|},~{}\text{with}~{}z=\frac{x}{\sqrt{D}}.$
(56)

This integral naturally appears while evaluating the integral
$\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ in D. For
$\mathcal{A}(w)=\mathcal{A}^{(3)}(w)$ we analyse this integral in E where it
can be seen using Eq. (98) that, a part of the contribution from third line of
Eq. (55) is same to the terms coming from the fourth line but of opposite sign
yielding some cancellations and hence further simplifications [see discussions
around Eqs. (98) and (99)]. Note that the above calculations are done assuming
$\gamma\neq 0$. One can do a similar calculation for $\gamma=0$ i.e.
$r=4\beta$ and evaluate the integrals using the saddle point method as used in
other cases.

Accumulating all the terms we, for large $t$, get

$\rho(x,t)\approx~{}~{}\boxed{\begin{aligned} &\text{Eq.~{}\eqref{r-g-
beta}~{}with~{}large~{}}t~{}\text{approximations~{}of~{}}\mathcal{K}_{a}^{\mathcal{A}}(x,t),~{}\mathcal{J}_{a}^{\mathcal{A}}(x,t)~{}\text{and}\\\
&\mathbb{J}_{a,\zeta}^{\mathcal{A}}(x,t)~{}\text{are~{}given~{}in~{}Eqs.~{}\eqref{K_a^A-F},~{}\eqref{J_a^A-F}~{}and~{}\eqref{mbbJ_a^A>}~{}respectively.}\end{aligned}}$
(57)

In this case also, many travelling fronts appear due the presence of the
$\Theta$ functions in the expressions of $\mathcal{K}_{a}^{\mathcal{A}}(x,t)$
and $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ in Eqs. (79) and (96) respectively.
Once again, after some manipulations one can see from the relaxation part of
the density in Eq. (55) that one gets an equal and opposite contributions of
the steady state part from terms with $\Theta(|x|-\sqrt{4D\beta}t)$, which
gets manifested as ‘cone spreading’ relaxation as seen in case of $r<\beta$ in
Sec. 6.2.2. Such a relaxation is demonstrated in Fig. 4 where we provide a
comparison between the saddle point result in Eq. (57) and the one from direct
numerical evaluation of the integrals in Eq. (55). Here too we observe a nice
agreement.

Figure 5: Plots of the LDF $\Phi(v)$ of the density as given in Eq. (58)
(dashed lines) for three choices of $r$ and $\beta$. The symbols are obtained
from the (saddle point) approximate expressions of the density $\rho(x,t)$
presented in Eqs. (54) and (57). The diffusion constant used to prepare these
plots is $D=0.5$.

Finally we discuss the large deviation behaviour of $\rho(x,t)$ in this case.
To get the LDF in this case we follow the same procedure as done in the
previous section 6.2.2. We find that, unlike the previous cases, in this case
the LDF can be different from the instantaneous case. In particular, for
$\beta<r<4\beta$ we find the same LDF as in Eq. (48). But for $r\geq 4\beta$
we get a different LDF.

Below we summarize the explicit forms of LDF for all values of $r$ and
$\beta$:

$\displaystyle\Phi\left(v\right)$
$\displaystyle=\begin{cases}&\left[\begin{array}[]{@{}l@{\quad}l@{}}\sqrt{\frac{r}{D}}|v|,{}&~{}~{}~{}~{}~{}\text{for}~{}|v|<v^{*}\\\
r+\frac{v^{2}}{4D},{}&~{}~{}~{}~{}~{}\text{for}~{}|v|\geq
v^{*},\end{array}\right.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\text{for}~{}0<r\leq
4\beta,\\\ &\\\
&\left[\begin{array}[]{@{}l@{\quad}l@{}}2\sqrt{\frac{\beta}{D}}|v|,{}&\text{for}~{}|v|<v_{1}^{*}\\\
\frac{(2\sqrt{\beta
D}+|v|)^{2}}{4D},{}&\text{for}~{}v_{1}^{*}\leq|v|<v_{2}^{*}\\\
\sqrt{\frac{r}{D}}|v|,{}&\text{for}~{}v_{2}^{*}\leq|v|<v^{*}\\\
r+\frac{v^{2}}{4D},{}&\text{for}~{}|v|\geq
v^{*}\end{array}\right.~{}~{}~{}~{}~{}~{}~{}~{}\text{for}~{}4\beta\leq
r\leq(3+2\sqrt{2})\beta,\\\ &\\\
&\left[\begin{array}[]{@{}l@{\quad}l@{}}2\sqrt{\frac{\beta}{D}}|v|,{}&\text{for}~{}|v|<v_{1}^{*}\\\
\frac{(2\sqrt{\beta
D}+|v|)^{2}}{4D},{}&\text{for}~{}v_{1}^{*}\leq|v|<v_{3}^{*}\\\
r+\frac{v^{2}}{4D},{}&\text{for}~{}|v|\geq
v_{3}^{*}\end{array}\right.~{}~{}~{}~{}~{}~{}~{}~{}\text{for}~{}r\geq(3+2\sqrt{2})\beta>0.\\\
\end{cases}$ (58)

where $v^{*}=\sqrt{4Dr}$, $v_{1}^{*}=\sqrt{4D\beta}$,
$v_{2}^{*}=\sqrt{4D}(\sqrt{r}-\sqrt{\beta}+\sqrt{r-\sqrt{4r\beta}})$ and
$v_{3}^{*}=\sqrt{D}(r-\beta)/\sqrt{\beta}$. Note that for $r=4\beta$ the
points $v_{1}^{*}$ and $v_{2}^{*}$ coincide. On the other hand, for
$r=(3+2\sqrt{2})\beta$, the three points $v^{*},~{}v_{2}^{*}$ and $v_{3}^{*}$
coincide. The above expressions for the LDF in different ranges are plotted in
Fig. 5. Similar to the instantaneous case, the LDF associated to the density
$\rho(x,t)$ for finite $\lambda$ also exhibits dynamical phase transition at a
fixed spatial point as described in Ref. [21].

##  7 Summary

In this paper, we have studied motion of a Brownian particle under stochastic
resetting with finite time return. The return dynamics is facilitated by a
linear trap whose minimum is located at the resetting position. The return
time is stochastic and it also depends on the position of the particle at the
time of resetting. Realizing the fact that motion of the particle can be
decomposed into two phases namely the exploration and return phases, we
construct Fokker-Planck equations for each phase with suitable boundary
conditions. We go beyond the steady state limit, and compute exact time
dependent forms of the position density. This also allows us to study the
temporal relaxation of the densities at large time, and we show that depending
on the relative strength between the resetting rate and potential, different
variants of the relaxation forms are found. We have shown that the density has
travelling boundaries which separate two regions: an inner core which has
already relaxed to the steady state, and an outer core which is still at
transient. An important thing to note that the speed at which the boundary
between the cores moves depends on the ratio of the resetting rate and
strength of the potential.

It is worth stressing that in this paper we have taken the master equation
approach to discuss the relaxation properties which, to the best of our
knowledge, was only shown before using the renewal formalism for the
instantaneous resetting. While the latter approach in the current set-up is
still a far-off, the master equation formalism allows us to make a connection
between the singularities in the densities in Laplace space and their
relaxation properties. We benchmark this approach first for the instantaneous
case, and then employ it in the case of non-instantaneous stochastic returns.

Computation of time dependent distributions for a stochastic process in the
presence of resetting is rather limited unlike that of the non-equilibrium
steady state. In this paper, we have provided a systematic way to compute
these probability densities. Various physical limits can be reached from our
formulation – thus, we believe that the methods shown here can be very useful
to compute propagators and study relaxation in other resetting systems.
Finally, we mention that one interesting future direction would be to study
the effect of non-instantaneous stochastic return on the first passage
problems.

##  8 Acknowledgement

We thank Carlos A. Plata for many fruitful discussions. DG acknowledges the
support from the University of Padova through “Excellence Project 2018” of the
Fondazione Cassa di Risparmio di Padova e Rovigo. AP gratefully acknowledges
Raymond and Beverly Sackler postdoctoral fellowship and Ratner Center for
Single Molecule Science at the Tel Aviv University for funding. AK would like
to acknowledge support of the Department of Atomic Energy, Government of
India, under project no.12-R&D-TFR-5.10-1100 and the support from DST,
Government of India grant under project No. ECR/2017/000634.

##  Appendix A Inverse Laplace transform of Eqs. (26) and (27)

Figure 6: Bromwich contour for inverse Laplace transform in the complex
$s-$plane. Two branch point singularities at $s_{b}^{(1)}$ and
$s_{b}^{(2)}<s_{b}^{(1)}<0$ and a pole at $s=0$. The pole is indicated by
cross and branch points are shown by filled circles. Zigzag curves represent
the branch cuts that connect the branch points to the branch point at
$s=-\infty$.

In this section, we provide details of inverse Laplace transforms of Eqs. (26)
and (27) to get the densities in the exploration and return phases as given in
Eqs. (65-30). We start by recalling $\tilde{\rho}_{D,R}(x,s)$ below as

$\displaystyle\tilde{\rho}_{D}(x,s)$
$\displaystyle=\dfrac{\lambda(r+2s)+r\sqrt{4D}\sqrt{s+\beta}}{4Ds(\lambda\sqrt{(s+r)/D}+r)}~{}e^{-\sqrt{s+r}|z|},$
(59) $\displaystyle\tilde{\rho}_{R}(x,s)$
$\displaystyle=r\dfrac{\lambda(r+2s)+r\sqrt{4D}\sqrt{s+\beta}}{4\lambda^{2}s(s-\gamma)}~{}\big{(}e^{-\sqrt{s+r}|z|}-e^{-\sqrt{\beta}|z|}~{}e^{-\sqrt{s+\beta}|z|}\big{)},$
(60)

where $z=\frac{x}{\sqrt{D}}$, $\beta=\frac{\lambda^{2}}{4D}$, and
$\gamma=-r+\frac{r^{2}}{4\beta}$. Following the Bromwich integral formula and
the residue theorem, the inverse Laplace transform of
$\tilde{\rho}_{D,R}(x,s)$ can be formally written as

$\displaystyle\rho_{D,R}(x,t)=\dfrac{1}{2\pi
i}\int_{\Gamma-i\infty}^{\Gamma+i\infty}~{}ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s),$
(61)

where the vertical contour passes through $\text{Re}(s)=\Gamma$ in the complex
$s$-plane such that all the singularities lie to left of it (see e.g., Fig.
6). Both the functions $\tilde{\rho}_{D,R}(x,s)$ have a simple pole at $s=0$
and three branch points: $s=-r,s=-\beta$ and $s=-\infty$.

To compute the Bromwich integral in (61) in this case, we consider the contour
integral $\dfrac{1}{2\pi
i}\oint_{\mathcal{C}}~{}ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s)$ along the
contour $\mathcal{C}$ shown in Fig. 6. It is straightforward to see that it
has the value

$\displaystyle\dfrac{1}{2\pi
i}\oint_{\mathcal{C}}~{}ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s)$
$\displaystyle=\text{Residue}[s=0],$ (62)

The integral along the contour $\mathcal{C}$ in Fig. 6 can be decomposed in
the following manner

$\displaystyle\begin{split}\dfrac{1}{2\pi
i}\int_{\mathcal{C}}~{}ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s)&=\\\
&=\dfrac{1}{2\pi
i}\left[\int_{AB}+\int_{BC}+\int_{CD}+\int_{DE}+\int_{EF}\right.\\\
&~{}~{}~{}~{}~{}\left.+\int_{FG}+\int_{GH}+\int_{HI}+\int_{IJ}+\int_{JA}\right]ds~{}e^{st}~{}\tilde{\rho}_{D,R}(x,s).\\\
&=\text{Residue}[s=0]\end{split}$ (63)

Note that the inverse Laplace transform in Eq. (61) is actually the integral
over the part ‘AB’ on the contour. In the limits of $R\to\infty$ and
$\epsilon_{1,2}\to 0$, one can show that the integrals along the chords ‘BC’,
‘DE’, ‘FG’, ‘HI’ and ‘JA’ vanish. Hence, we get

$\displaystyle\rho_{D,R}(x,t)=\frac{1}{2\pi
i}\int_{A}B=\text{Residue}[s=0]-\frac{1}{2\pi
i}\left[\int_{CD}+\int_{EF}+\int_{GH}+\int_{IJ}\right],$ (64)

where the integrals along ‘CD’, and ‘EF’ are performed on the negative real
axis above the branch cut whereas the integrals along ‘GH’, and ‘IJ’ are
performed below the branch cut. Writing these integrals in terms of real
integration variables and performing some manipulations, we get the following
expressions for $\rho_{D}(x,t)$ and $\rho_{R}(x,t)$ for different choices of
$r$ and $\lambda$ (equivalently $\beta$).

###  A.1 $r<\beta$

For the density $\rho_{D}(x,t)$ in the exploration phase, we get

$\displaystyle\begin{split}&\rho_{D}(x,t)=\rho^{ss}_{D}(x)+\frac{1}{4\pi\lambda^{2}\sqrt{D}}\int_{0}^{\beta-r}~{}dy~{}\frac{e^{-t(r+y)}}{(r+y)(r+\gamma+y)}\\\
&~{}\times\left[r\left(\lambda-2\sqrt{D(\kappa-y)}\right)+2\lambda
y\right]\left[\lambda\sqrt{y}\cos\left(\sqrt{y}\left|z\right|\right)+r\sqrt{D}\sin\left(\sqrt{y}\left|z\right|\right)\right]\\\
&~{}+\frac{1}{4\pi\sqrt{D}\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(\beta+y)}}{(\beta+y)(\beta+\gamma+y)}\bigg{[}\left\\{2Dr^{2}\sqrt{y}+\lambda^{2}(2(\beta+y)-r)\sqrt{\kappa+y}\right\\}\\\
&~{}\times\cos\left(\left|z\right|\sqrt{\kappa+y}\right)-r\lambda\sqrt{D}\left\\{2\sqrt{y(\kappa+y)}+r-2(\beta+y)\right\\}\sin\left(\left|z\right|\sqrt{\kappa+y}\right)\bigg{]},\end{split}$
(65)

where recall $z=x/\sqrt{D}$ and $\kappa=\beta-r>0$. On the other hand for the
density in the return phase, we get

$\displaystyle\rho_{R}(x,t)=\rho^{ss}_{R}(x)$
$\displaystyle+\frac{r}{4\pi\lambda^{2}}\int_{0}^{\beta-r}~{}dy~{}\frac{e^{-(r+y)t}}{(r+y)(r+\gamma+y)}\bigg{[}\left\\{2r\sqrt{D(\kappa-y)}-\lambda
r-2\lambda y\right\\}\sin\left(\sqrt{y}\left|z\right|\right)\bigg{]}$
$\displaystyle+\frac{r}{4\pi\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(\beta+y)}}{(\beta+y)(\beta+\gamma+y)}\bigg{[}2r\sqrt{Dy}\left\\{e^{-\sqrt{\beta}|z|}\cos\left(\sqrt{y}\left|z\right|\right)\right.$
(66)
$\displaystyle~{}-\left.\cos\left(\left|z\right|\sqrt{\kappa+y}\right)\right\\}-\lambda[r-2(\beta+y)]\left\\{e^{-\sqrt{\beta}|z|}\sin\left(\sqrt{y}\left|z\right|\right)-\sin\left(\left|z\right|\sqrt{\kappa+y}\right)\right\\}\bigg{]},$

where $z=x/\sqrt{D}$ and the densities in the steady state are given in Eqs.
(22) and (23).

###  A.2 $r>\beta$

Similar prescription also applies for $r>\beta$. Following the steps described
above, we find

$\displaystyle\rho_{D}(x,t)=\rho^{ss}_{D}(x)+\frac{1}{2\pi\lambda^{2}}\int_{0}^{\Lambda}~{}dy~{}\frac{r\sqrt{Dy}~{}e^{-t(\beta+y)}e^{-\left|z\right|\sqrt{\Lambda-y}}}{(\beta+y)(\beta+\gamma+y)}\left(r-\lambda\sqrt{\frac{\Lambda-y}{D}}\right)$
$\displaystyle~{}~{}~{}~{}+\frac{1}{4\pi\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(r+y)}}{(r+y)(r+\gamma+y)}\bigg{[}\cos\left(\sqrt{y}\left|z\right|\right)\left(2r^{2}\sqrt{D(\Lambda+y)}+\lambda^{2}r\sqrt{\frac{y}{D}}\right.$
(67)
$\displaystyle~{}~{}~{}~{}\left.+~{}2\lambda^{2}y\sqrt{\frac{y}{D}}\right)+\lambda
r\sin\left(\sqrt{y}\left|z\right|\right)\\{-2\sqrt{y(\Lambda+y)}+r+2y\\}\bigg{]},$
$\displaystyle\rho_{R}(x,t)=\rho^{ss}_{R}(x)++\frac{r}{4\pi\lambda^{2}}\int_{0}^{\Lambda}~{}dy~{}\frac{e^{-t(\beta+y)}}{(\beta+y)(\beta+\gamma+y)}\bigg{[}2r\sqrt{Dy}\left\\{e^{-\sqrt{\beta}|z|}\cos\left(\sqrt{y}\left|z\right|\right)\right.$
$\displaystyle\left.-\cosh\left(\left|z\right|\sqrt{\Lambda-y}\right)\right\\}+e^{-\sqrt{\beta}|z|}\lambda[2(\beta+y)-r]\sin\left(\sqrt{y}\left|z\right|\right)+2r\sqrt{Dy}\sinh\left(\left|z\right|\sqrt{\Lambda-y}\right)\bigg{]}$
$\displaystyle-\frac{r}{4\pi\lambda^{2}}\int_{0}^{\infty}~{}dy~{}\frac{e^{-t(r+y)}}{(r+y)(r+\gamma+y)}\bigg{[}2r\sqrt{D(\Lambda+y)}\left\\{\cos\left(\sqrt{y}\left|z\right|\right)\right.$
(68)
$\displaystyle\left.-e^{-\sqrt{\beta}|z|}\cos\left(\left|z\right|\sqrt{\Lambda+y}\right)\right\\}+\lambda(r+2y)\left\\{\sin\left(\sqrt{y}\left|z\right|\right)-e^{-\sqrt{\beta}|z|}\sin\left(\left|z\right|\sqrt{\Lambda+y}\right)\right\\}\bigg{]},$

where $z=x/\sqrt{D}$ and $\Lambda=-\kappa=r-\beta>0$.

##  Appendix B Evaluation of the integral $\mathcal{K}^{\mathcal{A}}_{a}(x,t)$
appearing in Eq. (42)

Figure 7: Closed contour $\mathcal{C}$ for the integral (70). $w^{*}$ and
$w_{+}$, respectively, are the saddle point and pole, and both of them are
function of a variable $z$. The horizontal path through the saddle point
$w^{*}$ is the steepest decent path.

In this section, we evaluate the integral of the form

$\displaystyle\mathcal{K}_{a}^{\mathcal{A}}(x,t)$
$\displaystyle=\int_{-\infty}^{+\infty}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|},~{}~{}a>0,$
(69)

where $z=x/\sqrt{D}$. To compute the above integral, let us consider the
following integral over a closed contour $\mathcal{C}$ as shown in Fig. 7 in
the complex plane:

$\displaystyle\begin{split}\oint_{\mathcal{C}}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|}&=2\pi
i~{}\text{Residue}[ia]~{}\Theta(|z|-2at)\\\ &=2\pi
i~{}\mathcal{A}(ia)~{}e^{a^{2}t-a|z|}\Theta(|z|-2at),\end{split}$ (70)

where we have assumed that $\mathcal{A}(w)$ does not have any singularity
inside the contour $\mathcal{C}$. We choose this particular contour because,
as can be easily seen that, the horizontal line parallel to the real axis
passes through the saddle point $w^{*}=\frac{i|z|}{2t}$ and is actually the
steepest-descent path. It is easy to identify that the desired integral
$\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ in Eq. (69) corresponds to the integral
$\int_{AB}$. Since, in the $R\to\infty$, the integrals $\int_{BC}$ and
$\int_{DA}$ go to zero, we get
$\mathcal{K}_{a}^{\mathcal{A}}(x,t)=\int_{AB}=2\pi
i~{}\text{Residue}[ia]~{}\Theta(|z|-2at)-\int_{CD}$. Hence

$\displaystyle\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ $\displaystyle=2\pi
i~{}\mathcal{A}(ia)~{}e^{a^{2}t-a|z|}~{}\Theta(|z|-2at)+e^{-\frac{z^{2}}{4t}}~{}K_{z}^{\mathcal{A}}\left(a-\frac{|z|}{2t},t\right),$
(71) where
$\displaystyle~{}K_{z}^{\mathcal{A}}(b,t)=\int_{-\infty}^{\infty}~{}dw~{}\frac{\mathcal{A}\left(w+i{|z|}/{2t}\right)}{w-ib}~{}e^{-tw^{2}}.$
(72)

The integral $K_{z}^{\mathcal{A}}(b,t)$ in the above equation can be evaluated
for large $t$ using saddle point method [80, 81]. However, we need to be
careful when the saddle point $w^{*}=0$ is close to the pole at $w_{+}=ib$. To
proceed, one needs to separate the singular and non-singular part of the
integrand. For this we rewrite

$\displaystyle K_{z}^{\mathcal{A}}(b,t)$
$\displaystyle=\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)\int_{-\infty}^{\infty}~{}dw~{}\frac{e^{-tw^{2}}}{w-ib}+\int_{-\infty}^{\infty}~{}dw~{}\psi(w)~{}e^{-tw^{2}},$
(73) where
$\displaystyle~{}\psi(w)=\frac{\mathcal{A}\left(w+i{|z|}/{2t}\right)-\mathcal{A}\left(ib+i{|z|}/{2t}\right)}{w-ib}.$
(74)

Note that now the function $\psi(w)$ does not have any singularity since
$\mathcal{A}(w)$ is an analytic function on the upper half plane. Following
[80, 81], we evaluate the first integral by defining

$\displaystyle L_{f}(b,t)$
$\displaystyle=\int_{-\infty}^{\infty}~{}dw~{}\frac{e^{-tw^{2}}}{w-ib},~{}~{}\text{such~{}that,}$
(75) $\displaystyle K_{z}^{\mathcal{A}}(b,t)$
$\displaystyle=\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)~{}L_{f}(b,t)+\int_{-\infty}^{\infty}~{}dw~{}\psi(w)~{}e^{-tw^{2}}.$
(76)

Note that the function $L_{f}(b,t)$ satisfies the following differential
equation

$\displaystyle\frac{dL_{f}}{dt}=b^{2}L_{f}(t)-ib\sqrt{\frac{\pi}{t}},$ (77)

which can be solved with boundary condition $L_{f}(\infty)=0$. We get,

$\displaystyle L_{f}(b,t)$
$\displaystyle=ibe^{tb^{2}}~{}\int_{t}^{\infty}~{}d\tau~{}e^{-\tau
b^{2}}~{}\sqrt{\frac{\pi}{\tau}}=i\pi
e^{tb^{2}}\text{sgn}(b)~{}\text{erfc}(|b|\sqrt{t}).$ (78)

On the other hand the integral in Eq. (76) for large $t$ can be evaluated
straightforwardly using Laplace method and one gets a contribution
$\psi(0)\sqrt{\frac{\pi}{t}}$ in the leading order. The next order terms can
be found easily by expanding $\psi(w)$ in a Taylor series around $w=0$.
Collecting both the contributions, we finally get an explicit expression of
$K_{z}^{\mathcal{A}}(b,t)$ at large $t$. Inserting this expression in Eq. (71)
one gets an explicit expression of $\mathcal{K}_{a}^{\mathcal{A}}(x,t)$ as

$\displaystyle\begin{split}\mathcal{K}_{a}^{\mathcal{A}}(x,t)&=2\pi
i~{}\mathcal{A}(ia)~{}e^{a^{2}t-a|z|}~{}\Theta(|z|-2at)+e^{-\frac{z^{2}}{4t}}~{}K_{z}^{\mathcal{A}}\left(a-\frac{|z|}{2t},t\right),\\\
\text{where,}~{}K_{z}^{\mathcal{A}}(b,t)&\approx\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)e^{tb^{2}}\left[i\pi~{}\text{sgn}(b)~{}\text{erfc}(|b|\sqrt{t})\right]+\psi(0)\sqrt{\frac{\pi}{t}}~{},\end{split}$
(79)

with
$\psi(0)=\frac{i}{b}\left[\mathcal{A}\left(i{|z|}/{2t}\right)-\mathcal{A}\left(ib+i{|z|}/{2t}\right)\right]$.

##  Appendix C Explicit forms of $\mathcal{A}^{(i)}(w)$ functions

Here, we present the explicit forms for the $\mathcal{A}^{(i)}$ functions
which were introduced in the main text

$\displaystyle\mathcal{A}^{(2)}(w)$
$\displaystyle=\frac{w^{2}(2w^{2}+r)}{(w+i\sqrt{r})(w+i\sqrt{r+\gamma})},$
(80) $\displaystyle\mathcal{A}^{(3)}(w)$
$\displaystyle=\frac{w^{2}\sqrt{w^{2}+r-\beta}}{(w+i\sqrt{r})(w+i\sqrt{r+\gamma})},~{}$
(81) $\displaystyle\mathcal{A}^{(4)}(w)$
$\displaystyle=\frac{\sqrt{4r^{2}D}w^{2}-i\lambda
w(2\beta+2w^{2}-r)}{(w+i\sqrt{\beta})(w+i\sqrt{\beta+\gamma})}.~{}$ (82)

Figure 8: Closed contour $\mathcal{C}$ for the integral (83). The saddle point
is located at $\frac{i|z|}{2t}$, while the semi-circular integral contributes
for $|z|\geq 2at$.

##  Appendix D Evaluation of the integral $\mathcal{J}^{\mathcal{A}}_{a}(x,t)$
appearing in Eq. (53)

In this section, we evaluate the integral of the form

$\displaystyle\mathcal{J}^{\mathcal{A}}_{a}(x,t)=\int_{0}^{\infty}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|},~{}~{}~{}\text{with}~{}~{}z=\frac{x}{\sqrt{D}},$
(83)

and $a>0$. To compute the above integral, we consider the following integral
over a closed contour $\mathcal{C}$ as shown in Fig. 8 in the complex plane:

$\displaystyle\oint_{\mathcal{C}}~{}dw~{}\frac{\mathcal{A}(w)}{w-ia}~{}e^{-tw^{2}+iw|z|}=0,$
(84)

where we have assumed that $\mathcal{A}(w)$ does not have any singularity
inside the contour $\mathcal{C}$. It is easy to identify that the desired
integral $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ in Eq. (83) corresponds to the
integral $\int_{AB}$. Since, in the $R\to\infty$, the integral $\int_{BC}$
goes to zero, we get

$\displaystyle\mathcal{J}_{a}^{\mathcal{A}}(x,t)$
$\displaystyle=\int_{AB}=-\int_{\gamma}-\int_{DA}-\int_{CD}=\frac{1}{2}\ointctrclockwise_{w=ia}+\int_{AD}+\int_{DC},$
(85)

where the integral $\int_{AD}$ represents integral along $AD$. The integrals
on the small circle around $w=ia$ can be evaluated easily and one gets

$\displaystyle\frac{1}{2}\ointctrclockwise_{w=ia}=i\pi\mathcal{A}(ia)e^{a^{2}t-a|z|}~{}\Theta(|z|-2at).$
(86)

On the other hand, the integral $\int_{AD}$ on AD is exactly the integral
$\mathbb{J}_{a,\zeta}^{\mathcal{A}}(x,t)$ defined in Eq. (56) with
$\zeta=|z|/2t$ and is written explicitly as

$\displaystyle\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}\left(x,t\right)=\int_{0}^{\frac{|z|}{2t}}dv~{}\frac{\mathcal{A}(iv)}{v-a}e^{tv^{2}-v|z|}.$
(87)

Note that this is an integral over $v$ within the finite range $[0,|z|/2t]$ on
the real line. Collecting all the contributions in Eq. (85) explicitly, we get

$\displaystyle\begin{split}\mathcal{J}_{a}^{\mathcal{A}}(x,t)&=i\pi\mathcal{A}(ia)e^{a^{2}t-a|z|}~{}\Theta(|z|-2at)+\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}\left(x,t\right)+e^{-\frac{z^{2}}{4t}}~{}J_{z}^{\mathcal{A}}\left(a-\frac{|z|}{2t},t\right),\end{split}$
(88) where,
$\displaystyle~{}J_{z}^{\mathcal{A}}(b,t)=\int_{0}^{\infty}~{}dw~{}\frac{\mathcal{A}\left(w+i\frac{|z|}{2t}\right)}{w-ib}~{}e^{-tw^{2}}.$
(89)

We now proceed to evaluate the integral $J_{z}^{\mathcal{A}}(b,t)$ in the
above equation for large $t$ using saddle point method in similar way as done
in B. Once again we need to be careful when the saddle point $w^{*}=0$ is
close to the pole at $w_{+}=ib$. To proceed, one needs to separate the
singular and non-singular part of the integrand. For this we rewrite

$\displaystyle J_{z}^{\mathcal{A}}(b,t)$
$\displaystyle=\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)\int_{0}^{\infty}~{}dw~{}\frac{e^{-tw^{2}}}{w-ib}+\int_{0}^{\infty}~{}dw~{}\psi(w)~{}e^{-tw^{2}},$
(90) $\displaystyle\text{where,}~{}~{}\psi(w)$
$\displaystyle=\frac{\mathcal{A}\left(w+i{|z|}/{2t}\right)-\mathcal{A}\left(ib+i{|z|}/{2t}\right)}{w-ib}.$
(91)

We evaluate the first integral by defining

$\displaystyle L_{h}(b,t)$
$\displaystyle=\int_{0}^{\infty}~{}dw~{}\frac{e^{-tw^{2}}}{w-ib},~{}~{}\text{such~{}that,}$
(92) $\displaystyle J_{z}^{\mathcal{A}}(b,t)$
$\displaystyle=\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)L_{h}(b,t)+\int_{0}^{\infty}~{}dw~{}\psi(w)~{}e^{-tw^{2}}.$
(93)

Note that the function $L_{h}(b,t)$ satisfies the following differential
equation

$\displaystyle\begin{split}\frac{dL_{h}}{dt}&=b^{2}L_{h}(t)-\int_{0}^{\infty}~{}dw~{}(w+ib)~{}e^{-tw^{2}}\\\
&=b^{2}L_{h}(t)-\left(\frac{1}{2t}+\frac{ib}{2}\sqrt{\frac{\pi}{t}}\right),\end{split}$
(94)

which can be solved with boundary condition $L_{f}(\infty)=0$. We get,

$\displaystyle L_{h}(b,t)$
$\displaystyle=e^{tb^{2}}~{}\int_{t}^{\infty}~{}d\tau~{}e^{-\tau
b^{2}}~{}\left(\frac{1}{2\tau}+\frac{ib}{2}\sqrt{\frac{\pi}{\tau}}\right)$
(95)
$\displaystyle=\frac{e^{tb^{2}}}{2}\left[\Gamma(0,tb^{2})+i\pi\text{sgn}(b)~{}\text{erfc}(|b|\sqrt{t})\right],$

where $\Gamma(0,z)=\int_{z}^{\infty}~{}du~{}\frac{e^{-u}}{u}$ is an incomplete
Gamma function. On the other hand the second integral in Eq. (93) for large
$t$ can be evaluated straightforwardly using Laplace method because the
function $\psi(w)$ in Eq. (91) does not have any singularity for
$\mathcal{A}(w)$ being analytic function on the upper half plane. Hence from
the second integral one gets a contribution
$\frac{\psi(0)}{2}\sqrt{\frac{\pi}{t}}$ at large $t$. Now, collecting both the
contributions, we finally get an explicit expression of
$J_{z}^{\mathcal{A}}(b,t)$ at large $t$. Inserting this expression in Eq. (88)
one gets an explicit expression of $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ as

$\displaystyle\begin{split}\mathcal{J}_{a}^{\mathcal{A}}(x,t)&=i\pi\mathcal{A}(ia)e^{a^{2}t-a|z|}~{}\Theta(|z|-2at)+\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}\left(x,t\right)+e^{-\frac{z^{2}}{4t}}~{}J_{z}^{\mathcal{A}}\left(a-\frac{|z|}{2t},t\right),\\\
\text{where,}&\\\
J_{z}^{\mathcal{A}}(b,t)&\approx\mathcal{A}\left(ib+i\frac{|z|}{2t}\right)\frac{e^{tb^{2}}}{2}\left[\Gamma(0,tb^{2})+i\pi~{}\text{sgn}(b)~{}\text{erfc}(|b|\sqrt{t})\right]+\frac{\psi(0)}{2}\sqrt{\frac{\pi}{t}}~{},\end{split}$
(96)

with
$\psi(0))=\frac{i}{b}\left[\mathcal{A}\left(i{|z|}/{2t}\right)-\mathcal{A}\left(ib+i{|z|}/{2t}\right)\right]$
and $\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}(x,t)$ is defined in Eq. (87).

##  Appendix E Remarks on the integral
$\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}^{(3)}}\left(x,t\right)$ defined in
Eq. (87)

Since in Eqs. (52) and (55), we require to evaluate
$\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ for
$\mathcal{A}(w)=\mathcal{A}^{(3)}(w)$, we here rewrite the function
$\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}(x,t)$ [given in Eq. (87)] with
$\mathcal{A}=\mathcal{A}^{(3)}(w)$ for $r<\beta$ and $r>\beta$. For $r<\beta$,
the integral $\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}^{(3)}}(x,t)$ reads

$\displaystyle\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}^{(3)}}\left(x,t\right)=$
$\displaystyle~{}~{}i\int_{0}^{\frac{|z|}{2t}}dv~{}\frac{v^{2}\sqrt{v^{2}+\beta-r}}{(v-a)(v+\sqrt{r})(v+\sqrt{r+\gamma})}e^{tv^{2}-v|z|},$
(97)

where we have used the expression of $\mathcal{A}^{(3)}(w)$ in Eq. (81). Note
that in this case $\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}}(x,t)$ is purely
imaginary and hence will not finally contribute in Eq. (52).

On the other hand for $r>\beta$, we rewrite the integral as

$\displaystyle\begin{split}\mathbb{J}_{a,\frac{|z|}{2t}}^{\mathcal{A}^{(3)}}\left(x,t\right)=&~{}~{}i\int_{\min(\sqrt{r-\beta},\frac{|z|}{2t})}^{\frac{|z|}{2t}}dv~{}\frac{v^{2}\sqrt{v^{2}+\beta-r}}{(v-a)(v+\sqrt{r})(v+\sqrt{r+\gamma})}e^{tv^{2}-v|z|}\\\
&~{}~{}+\int_{0}^{\min(\sqrt{r-\beta},\frac{|z|}{2t})}dv~{}\frac{v^{2}\sqrt{r-\beta-v^{2}}}{(v-a)(v+\sqrt{r})(v+\sqrt{r+\gamma})}e^{tv^{2}-v|z|}.\end{split}$
(98)

It is clear from the above equation and Eq. (96) that the contribution of the
form $\mathbb{J}_{a,\zeta}^{\mathcal{A}^{(3)}}\left(x,t\right)$ from terms
like $\mathcal{J}_{a}^{\mathcal{A}}(x,t)$ in the third line of Eq. (55) gets
partly cancelled with the fourth line of Eq. (55) and the remaining
contribution in the form
$\mathbb{J}_{a,\zeta}^{\mathcal{A}^{(3)}}\left(x,t\right)$ is given by:

Combined
$\displaystyle\text{contribution~{}of~{}the~{}form~{}}\mathbb{J}_{a,\zeta}^{\mathcal{A}^{(3)}}\left(x,t\right)~{}\text{from~{}the~{}$3^{rd}$~{}and~{}$4^{th}$~{}line~{}in~{}Eq.~{}\eqref{r-g-
beta}}$
$\displaystyle=\int_{\min(\sqrt{r-\beta},\frac{|z|}{2t})}^{\sqrt{r-\beta}}dv~{}\frac{v^{2}\sqrt{r-\beta-v^{2}}}{(v-a)(v+\sqrt{r})(v+\sqrt{r+\gamma})}e^{tv^{2}-v|z|}.$
(99)

Note that the above integral has to be performed on a real line segment and is
non zero only for $|z|<2t\sqrt{r-\beta}$. This integral can again be computed
using saddle point method as discussed in previous sections because
$\beta+\gamma>0$ for $r>\beta$ (recall $r+\gamma=r^{2}/(4\beta)$ from Eq.
(28)).

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