# Fractional Boundary Value Problems and Elastic Sticky Brownian Motions, I:
The half line

Mirko D’Ovidio

###### Abstract

We extend the results obtained in [14] by introducing a new class of boundary
value problems involving non-local dynamic boundary conditions. We focus on
the problem to find a solution to a local problem on a domain $\Omega$ with
non-local dynamic conditions on the boundary $\partial\Omega$. Due to the
pioneering nature of the present research, we propose here the apparently
simple case of $\Omega=(0,\infty)$ with boundary $\\{0\\}$ of zero Lebesgue
measure. Our results turn out to be instructive for the general case of
boundary with positive (finite) Borel measures. Moreover, in our view, we
bring new light to dynamic boundary value problems and the probabilistic
description of the associated models.

###### Contents

  1. 1 Introduction
    1. 1.1 Presentation of the results
    2. 1.2 Plan of the work
    3. 1.3 Motivations
  2. 2 Preliminaries
    1. 2.1 The Caputo-Dzherbashian fractional derivative
    2. 2.2 The process $X$
    3. 2.3 The random times $L$ and $H$
  3. 3 The fractional initial value problem
    1. 3.1 Time changes for FIVPs
  4. 4 The fractional boundary value problem
    1. 4.1 The process $\bar{X}$ and the FBVP
    2. 4.2 On the occupation times for $\bar{X}$
    3. 4.3 On the holding times of $\bar{X}$
    4. 4.4 On the lifetime of $\bar{X}$

Keywords and phrases: Sticky Brownian motions, dynamic boundary conditions,
fractional boundary value problems, fractional derivatives, time changes

MCS2020 subject classifications: 60J50 ; 60J55 ; 35C05 ; 26A33

##  1 Introduction

###  1.1 Presentation of the results

We study the solution to the (local) heat equation subjected to a non-local
dynamic boundary condition. The non-local condition is written in terms of the
Caputo-Dzherbashian fractional derivative defined as the convolution operator

$\displaystyle
D^{\alpha}_{t}\varphi(t)=(\varphi^{\prime}*\kappa)(t),\quad\alpha\in(0,1)$

involving the singular kernel $\kappa$ described below (see Section 2.1). Our
results can be generalized by considering a general kernel $\kappa$ and
therefore a non-local operator (see $D^{\Phi}_{t}$ below) as defined in [22,
31, 11, 23]. Here we maintain our focus on $D^{\alpha}_{t}$ and refer to [5]
for general non-local boundary conditions. The Caputo-Dzherbashian derivative
has been introduced in [7, 8, 9] by the first author and separately in a
series of works starting from [17, 18] by the second author.

The fractional Boundary condition introduces an anomalous behaviour of the
associated process only on the lower dimensional space given by the boundary.
In the present paper we extend the result given in [14] and provide a
discussion in case a Brownian motion on $(0,\infty)$ has an anomalous
behaviour only at the boundary point $\\{0\\}$. An interesting formulation of
the problem would be given for an arbitrary domain
$\overline{\Omega}=\Omega\sqcup\partial\Omega$ as the problem to find a
solution $u$ such that

$\left\\{\begin{array}[]{l}\displaystyle\frac{\partial u}{\partial
t}=Au\quad\textrm{on }\Omega,\\\ \displaystyle\mathfrak{u}=Tu,\\\
\displaystyle D^{\alpha}_{t}\mathfrak{u}=Bu\quad\textrm{on }\partial\Omega,\\\
\displaystyle u_{0}=f\end{array}\right.$ (1)

where the trace operator $T$ must be well-defined on $L^{2}(m_{\partial})$ for
a suitable finite Borel measure $m_{\partial}$ supported on $\partial\Omega$.
Thus, we may focus on the problem to find a continuous kernel for $u$ on
$L^{2}(\overline{\Omega},m)$ where $m$ is obtained as the sum of the Lebesgue
measure $dx$ and the surface measure $m_{\partial}$, that is
$L^{2}(\overline{\Omega},m)=L^{2}(\Omega,dx)\oplus
L^{2}(\partial\Omega,m_{\partial})$. Notice that the non-local boundary
condition says that we can not obtain a semigroup for the problem (1).
Although the problem (1) is interesting, there are no results on this problem
and we are lead to approach the problem by first considering the instructive
case $\overline{\Omega}=[0,\infty)$ with $\partial\Omega=\\{0\\}$. For a
smooth domain $\Omega\subset\mathbb{R}^{d}$ with $d>1$ we refer to the work
[15]. For $\Omega=[0,\infty)$ we refer to [14] where the fractional dynamic
boundary condition has been introduced for a problem with constant initial
datum. In order to prepare our discussion we first fix below some notation and
ideas.

As usual $\mathbf{E}_{x}$ is the expectation with respect to a probability
measure $\mathbf{P}_{x}$ with $x$ denoting the starting point for the process
we are dealing with. We also use the following notation:

$\displaystyle\dot{\varphi}=\frac{\partial\varphi}{\partial
t},\quad\varphi^{\prime}=\frac{\partial\varphi}{\partial
x},\quad\varphi^{\prime\prime}=\frac{\partial^{2}\varphi}{\partial x^{2}}.$

Let $\eta,\sigma,c$ be positive constants. We consider the following
processes:

  * i)

$X^{+}=\\{X^{+}_{t}\\}_{t\geq 0}$ is a reflecting Brownian motion on
$[0,\infty)$ with boundary local time $\gamma^{+}=\\{\gamma^{+}_{t}\\}_{t\geq
0}$;

  * ii)

$X^{\dagger}=\\{X^{\dagger}_{t}\\}_{t\geq 0}$ is a Brownian motion on
$(0,\infty)$ killed upon reaching the boundary $\\{0\\}$ for which

$\displaystyle\mathbf{E}_{x}[f(X^{\dagger}_{t})]=\mathbf{E}_{x}[f(X^{+}_{t}),t<\tau_{0}],\quad
t>0,\;x\in(0,\infty)$

where $\tau_{0}=\inf\\{t\geq 0\,:\,X^{+}_{t}=0\\}$;

  * iii)

$X=\\{X_{t}\\}_{t\geq 0}$ is an elastic sticky Brownian motion on $[0,\infty)$
with boundary local time $\gamma=\\{\gamma_{t}\\}_{t\geq 0}$. The term
“elastic sticky ”is referred to the fact that, for $c\geq 0$,

$\displaystyle\mathbf{E}_{x}[f(X^{el}_{t})]=\mathbf{E}_{x}[f(X^{+}_{t})M_{t}],\quad
M_{t}:=\exp(-c\gamma^{+}_{t})$

is the well-known representation of the elastic Brownian motion
$\\{X^{el}_{t}\\}_{t\geq 0}$ in terms of $X^{+}$ and the multiplicative
functional $M^{c}_{t}$. Passing through the random time change
$V_{t}=t+(\eta/\sigma)\gamma^{+}_{t}$ we therefore use the representation (a
well-known result given in [25])

$\displaystyle\mathbf{E}_{x}[f(X_{t})]=\mathbf{E}_{x}[f(X^{el}\circ
V^{-1}_{t})]$

where $V^{-1}_{t}:=\inf\\{s\,:\,V_{s}>t\\}$ is the inverse of $V_{t}$. We
recall the associated boundary condition

$\displaystyle\eta\varphi^{\prime\prime}=\sigma\varphi^{\prime}-c\varphi$ (2)

which is termed Wentzell-Robin or Feller-Robin boundary condition and includes
a (pure) sticky condition obtained as $\eta\to\infty$ and an elastic condition
obtained as $\eta\to 0$. We say that $X$ has generator $(G,D(G))$ where
$G\varphi=\varphi^{\prime\prime}$ and

$\displaystyle D(G)=\\{\varphi\in C^{2}([0,\infty))\,:\,\varphi\textrm{
satisfies \eqref{BCintro} at the boundary point $\\{0\\}$}\\}.$

Let us write $\gamma^{+}_{t}=\gamma^{+}_{t,0}$ where $\gamma^{+}_{t,z}$ is the
jointly continuous local time of $X^{+}$ for the level $z\in[0,\infty)$ and
time $t\geq 0$ for which, given a bounded measurable function
$f:[0,\infty)\mapsto[0,\infty)$, the following occupation time formula holds

$\displaystyle\int_{0}^{t}f(X^{+}_{s})ds=\int_{0}^{\infty}f(z)\gamma^{+}_{t,z}dz,\quad
t\geq 0.$

The boundary point $\\{0\\}$ must be carefully treated. We therefore consider
the measure

$\displaystyle m(dz)=dz+(\eta/\sigma)\delta_{0}(dz)$

on $\overline{\Omega}=[0,\infty)$ given by the sum of the Lebesgue measure on
$(0,\infty)$ and the Dirac measure at $\\{0\\}$. We observe that

$\displaystyle\int_{[0,\infty)}\gamma^{+}_{t,z}m(dz)=\int_{0}^{\infty}\gamma^{+}_{t,z}dz+(\eta/\sigma)\gamma^{+}_{t,0}=t+(\eta/\sigma)\gamma^{+}_{t}=:V_{t}$
(3)

gives the random time introduced above;

  * iv)

$H=\\{H_{t}\\}_{t\geq 0}$ is a stable subordinator of order $\alpha\in(0,1)$;

  * v)

$L=\\{L_{t}\\}_{t\geq 0}$ is the inverse $L_{t}:=\inf\\{s\geq
0\,:\,H_{s}>t\\}$ to $H$;

  * vi)

$\bar{X}=\\{\bar{X}_{t}\\}_{t\geq 0}$ with boundary local time
$\bar{\gamma}=\\{\bar{\gamma}_{t}\\}_{t\geq 0}$ introduced and discussed below
in the present section.

We are now ready to discuss our results.

Let $\eta,\sigma,c$ be positive constants. We consider the following problems:

$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
D^{\alpha}_{t}v(t,x)=v^{\prime\prime}(t,x),\quad t>0,\,x>0\\\ \\\
\displaystyle\eta v^{\prime\prime}(t,0)=\sigma v^{\prime}(t,0)-c\,v(t,0),\quad
t>0\\\ \\\ \displaystyle v(0,x)=f(x),\quad x\geq 0\end{array}\right.$ (9)
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
D^{\alpha}_{t}w(t,x)=w^{\prime\prime}(t,x),\quad t>0,\,x>0\\\ \\\
\displaystyle\eta D^{\alpha}_{t}w(t,0)=\sigma w^{\prime}(t,0)-c\,w(t,0),\quad
t>0\\\ \\\ \displaystyle w(0,x)=f(x),\quad x\geq 0\end{array}\right.$ (15)
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\dot{u}(t,x)=u^{\prime\prime}(t,x),\quad
t>0,\,x>0\\\ \\\ \displaystyle\eta D^{\alpha}_{t}u(t,0)=\sigma
u^{\prime}(t,0)-c\,u(t,0),\quad t>0\\\ \\\ \displaystyle u(0,x)=f(x),\quad
x\geq 0\end{array}\right.$ (21)

The problem (9) is a fractional initial value problem (FIVP) also called
fractional Cauchy problem (FCP). It has been investigated by many researchers,
see for example [21, 20, 29, 28, 2, 1, 13, 19, 16, 27, 26]. All these works
basically consider the semigroup associated with $\alpha=1$ as a base object
to deal with. Then, for $\alpha\in(0,1)$, the solution to the FCP comes out,
via integration of that semigroup with a suitable kernel. From the
probabilistic point of view we have a time change for a Markov (base) process
where the new random time is given by $L$. The composition is not Markov and
the associated operator is not a semigroup.

Our results have some impact on occupation measures, then we think the
following discussion will help the readers. Let $Y=\\{Y_{t}\\}_{t\geq 0}$ be a
Markov process with generator $(A_{Y},D(A_{Y}))$. Then, the solution to the
FCP

$\displaystyle D^{\alpha}_{t}\varphi=A_{Y}\varphi,\quad\varphi_{0}=f\in
D(A_{Y})$ (22)

has the probabilistic representation

$\displaystyle\varphi(t,x)=\mathbf{E}_{x}[f(Y^{L}_{t})]=\int_{0}^{\infty}\mathbf{E}_{x}[f(Y_{s})]\,\mathbf{P}_{0}(L_{t}\in
ds)$

where $Y^{L}=\\{Y^{L}_{t}\\}_{t\geq 0}$ defined as $Y^{L}_{t}=Y\circ L_{t}$ is
obtained via time change from the base process $Y$. For $Y$ on a bounded
domain $\Omega$ and $\forall\,\Lambda\subset\Omega$ we introduce
$\tau_{Y}(\Lambda)=\inf\\{t\,:\,Y_{t}\notin\Lambda\\}$ and
$\tau^{L}_{Y}(\Lambda)=\inf\\{t\,:\,Y^{L}_{t}\notin\Lambda\\}$ as the first
exit time from the subset $\Lambda$ respectively of $Y$ and $Y^{L}$. For
$x\in\Lambda$, $S\subset\Lambda$, we have ([11])

$\displaystyle\mu_{Y}(x,S):=\mathbf{E}_{x}\left[\int_{0}^{\tau_{Y}(\Lambda)}\mathbf{1}_{S}(Y_{s})ds\right]<\infty$

and

$\displaystyle\mu_{Y}^{L}(x,S):=\mathbf{E}_{x}\left[\int_{0}^{\tau^{L}_{Y}(\Lambda)}\mathbf{1}_{S}(Y^{L}_{s})ds\right]=\infty,$

for the occupation measures $\mu_{Y}$ and $\mu_{Y}^{L}$ of a subset $S$. This
result well agrees with the fact that ([6, 11])

$\displaystyle\mathbf{E}_{x}[\zeta]<\infty,\quad\mathbf{E}_{x}[\zeta^{L}]=\infty$

where $\zeta$ and $\zeta^{L}=H\circ\zeta$ respectively are the lifetimes of
$Y$ and $Y^{L}$ on $\Omega$ if we allow a kill on $\partial\Omega$ (Dirichlet
condition on the boundary). Thus, we have infinite average amount of time
spent by the process $Y^{L}_{t}$ in $\Omega$ and every subsets $\Lambda$ of
$\Omega$. We say that the process is delayed by $L$.

######  Remark 1.1.

This can be better explained by considering the general operator

$\displaystyle
D^{\Phi}_{t}\varphi=\varphi^{\prime}*\kappa,\quad\kappa(z)=\Pi(z,\infty)$

where

$\displaystyle\Phi(\lambda)=\int_{0}^{\infty}(1-e^{-\lambda
z})\Pi(dz),\quad\textrm{s.t.}\quad\int_{0}^{\infty}(1\wedge z)\Pi(dz)<\infty$
(23)

is the symbol of a subordinator with Lévy measure $\Pi$ and $\kappa$ is the
tail of $\Pi$ with

$\displaystyle\int_{0}^{\infty}e^{-\lambda
z}\kappa(z)dz=\frac{\Phi(\lambda)}{\lambda}.$

The fact that $E_{x}[\zeta^{L}]<\infty$ can be related with the elliptic
problem associated to the FCP and the integrability of $D^{\Phi}_{t}$. In
particular, from the problem (22), via integration with respect to time, we
get

$\displaystyle\int_{0}^{\infty}D^{\alpha}_{t}\varphi=A_{Y}\int_{0}^{\infty}\varphi
dt,\quad\varphi_{0}=1$

and

$\displaystyle\int_{0}^{\infty}\varphi
dt=\int_{0}^{\infty}\mathbf{P}_{x}(\zeta>t)dt$

gives the mean lifetime of $Y^{L}$. Notice that $\zeta$ coincides with a
killing time if we assume Dirichlet boundary condition. The elliptic problem
associated with (22) makes sense according with

$\displaystyle\|D^{\Phi}_{t}\varphi\|^{p}_{p}\leq\|\varphi\|^{p}_{p}\left(\lim_{\lambda\to\infty}\frac{\Phi(\lambda)}{\lambda}\right)^{p},\quad
p\in[1,\infty)$

for $\varphi\in L^{p}(0,\infty)$ and
$\lim_{\lambda\to\infty}\Phi(\lambda)/\lambda<\infty$. The derivative
$D^{\alpha}_{t}$ is the special case with $\Phi(\lambda)=\lambda^{\alpha}$ for
which the limit above is infinite. In [6] we discuss this property as a
delaying effect for the base process $Y$.

The problem (15) can be associated with (9). Indeed, it can be regarded as a
simple extension to the well-known results about fractional PDE in case the
following assumption turns out to be verified:

$\displaystyle\forall\,t>0\quad\lim_{x\to
0}\big{(}D^{\alpha}_{t}w(t,x)-w^{\prime\prime}(t,x)\big{)}=0.$ (A1)

Thus, under (A1) the identities

$\displaystyle\eta D^{\alpha}_{t}w(t,0)\stackrel{{\scriptstyle(A1)}}{{=}}\eta
w^{\prime\prime}(t,0)\stackrel{{\scriptstyle\eqref{BCintro}}}{{=}}\sigma
w^{\prime}(t,0)-c\,w(t,0),\quad t>0$

give the equivalence between (9) and (15). We observe that (A1) can be
considered also in a weaker sense. However, the key role here is played by the
continuity of the second derivative at zero. We observe that an equivalent
formulation of these problems can be given as

$\displaystyle D^{\alpha}_{t}\left(\begin{matrix}\varphi\\\
\varphi|_{0}\end{matrix}\right)=\mathcal{A}\left(\begin{matrix}\varphi\\\
\varphi|_{0}\end{matrix}\right)$

with the suitable definition of the operator matrix $\mathcal{A}$ on
$C(\overline{\Omega})$. This does not hold for the next case.

The problem (21) involves a non-local equation on the boundary which can be
written as

$\displaystyle\eta D^{\alpha}_{t}u(t,0)=$ $\displaystyle\lim_{x\downarrow
0}\left(\eta\frac{1}{\Gamma(1-\alpha)}\int_{0}^{\infty}u^{\prime\prime}(s,x)(t-s)^{-\alpha}ds\right)$
$\displaystyle=$
$\displaystyle\eta\frac{1}{\Gamma(1-\alpha)}\int_{0}^{\infty}u^{\prime\prime}(s,0)(t-s)^{-\alpha}ds,\quad
t>0.$

by exploiting the definition of $D^{\alpha}_{t}$. Our main goal in the present
work is to study this problem and obtain a probabilistic description of the
associated models. In some sense we extend the results obtained in [25] and
[32] on the second order boundary condition. In probability, we refer to the
sticky condition. A bridge between second order boundary conditions and
dynamic boundary conditions can be given for the heat equation under the fact
that $\dot{u}(t,x)|_{x=0}=\dot{u}(t,0)$ as in the Assumption (A1). Thus, a
Feller-Wentzell boundary condition can be treated as a dynamic boundary
condition for the heat equation.

The probabilistic representation of the problem (9) is concerned with the
process $\bar{X}$. We show (Theorem 4.1) that $\bar{X}$ admits the
representation $\bar{X}_{t}=X^{el}\circ\bar{V}^{-1}_{t}$ where
$\bar{V}^{-1}_{t}$ is the inverse to the process

$\displaystyle\bar{V}_{t}=t+H\circ(\eta/\sigma)\gamma^{+}_{t},\quad t\geq 0.$
(24)

Notice that, for $\alpha=1$ $H_{t}=t$ a.s. and therefore $\bar{V}_{t}=V_{t}$
a.s., thus $\bar{X}$ coincides with $X$ for $\alpha=1$. Let us define
$\tau_{\bar{X}}(\Lambda)=\inf\\{t\,:\,\bar{X}_{t}\notin\Lambda\\}$ and
$\tau_{X}(\Lambda)=\inf\\{t\,:\,X_{t}\notin\Lambda\\}$. For $\alpha\in(0,1]$,
we have finite occupation measures, that is, for $S\subset\Lambda$,

$\displaystyle\forall
x\in\Lambda\subset(0,\infty),\quad\mu_{\bar{X}}(x,S):=\mathbf{E}_{x}\left[\int_{0}^{\tau_{\bar{X}}(\Lambda)}\mathbf{1}_{S}(\bar{X}_{s})ds\right]<\infty.$

This is related with the fact that $\bar{X}$ (and $X$ for $\alpha=1$) behaves
like a Brownian motion on $(0,\infty)$. It moves on finite intervals with
finite times. For the time the process $\bar{X}$ spends on the boundary point
we refer, as usual, to the holding times. Here we have different behaviours
depending on $\alpha$. It is well-known that $X$ spends an exponential time on
the boundary with each visit, we denote by $\\{e_{i}\\}_{i}$ the sequence of
i.i.d. holding times for $X$. The average amount of time the process $X$
spends at $\\{0\\}$ is therefore given by

$\displaystyle\mathbf{E}[e_{i}]=(\eta/\sigma),\quad\forall\,i.$

Concerning the process $\bar{X}$ we refer to the sequence
$\\{\bar{e}_{i}\\}_{i}$ of i.i.d. holding times, that is $\bar{e}_{i}$ for the
visit $i$ is a random time such that, for $\bar{X}_{0}=0$ and
$\bar{X}_{\bar{e}_{i}}>0$, it holds (Theorem 4.3)

$\displaystyle\mathbf{P}_{0}(\bar{e}_{i}>t|\bar{X}_{\bar{e}_{i}}>0)=E_{\alpha}(-(\sigma/\eta)t^{\alpha}),\quad
t\geq 0$

where $E_{\alpha}$ is the Mittag-Leffer function. It is well-known that
$E_{\alpha}\notin L^{1}(0,\infty)$. The time the process $\bar{X}$ spends on
the boundary point $\\{0\\}$ with each visit, is a Mittag-Leffler random
variable with

$\displaystyle\forall\,\alpha\in(0,1),\quad\mathbf{E}[\bar{e}_{i}]=\infty,\quad\forall\,i.$

This special behaviour of $\bar{X}$ for $\alpha\in(0,1)$ entails a new role
for the boundary point $\\{0\\}$. This point can be regarded as a trap for the
Brownian motion on $(0,\infty)$.

###  1.2 Plan of the work

We begin with some preliminaries. In Section 3 we compare the problems (9) and
(15). In Section 4 we present our main results on the problem (21).

###  1.3 Motivations

The present work brings new light to the boundary value problems with dynamic
conditions. Although the main interest would be apparently given by motions on
bounded domains $\Omega\subset\mathbb{R}^{d}$ with $d>1$ our problem could be
very attractive for a number of applications. Indeed, the study of non-local
dynamic problems on higher dimension introduces motions on the lower
dimensional space $\partial\Omega\subset\mathbb{R}^{d-1}$. This is obviously
of interest and reasonably leads to pure jump processes on $\partial\Omega$,
the associated trace process. This is motivated by our construction given in
terms of $X^{+}$, instantaneous reflections may only give holding times (we do
not have boundary motions). On the other hand, the fact that we deal with a
zero Lebesgue measure boundary, the point $\\{0\\}$, it does not seem to be
restrictive. Indeed, the independent holding times at $\\{0\\}$ (or a general
point $x>0$) for a Brownian motion find interesting applications in different
fields of applied sciences. We mention the financial models in which investors
update their beliefs too slowly or traffic models in which we construct
motions on metric graphs with (independent) delay on a vertex. In general, the
sticky point for the process $X$ becomes a trap point for the process
$\bar{X}$. In our analysis, the boundary point $\\{0\\}$ can be regarded as a
trap point for the Brownian motion $X^{+}$ on $[0,\infty)$.

##  2 Preliminaries

###  2.1 The Caputo-Dzherbashian fractional derivative

For a function $\varphi(\cdot,x):(0,\infty)\mapsto(0,\infty)$,
$\forall\,x\in[0,\infty)$ we consider the convolution-type operator defined as

$\displaystyle
D^{\alpha}_{t}\varphi(t,x)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{\partial\varphi}{\partial
s}(s,x)\,(t-s)^{-\alpha}ds$

for $\alpha\in(0,1)$.

We notice that the Caputo-Dzherbashian derivative is well-defined for
functions which are exponentially bounded together with their first
derivative. In particular, $D^{\alpha}_{t}\varphi$ is well defined for
$\varphi(\cdot,x)\in W^{1,\infty}(0,\infty)$, $\forall\,x\in[0,\infty)$ and
this ensures existence of the Laplace transform

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}D^{\alpha}_{t}\varphi(t,x)\,dt=$
$\displaystyle\left(\int_{0}^{\infty}e^{-\lambda
t}\frac{t^{-\alpha}}{\Gamma(1-\alpha)}\right)\left(\int_{0}^{\infty}e^{-\lambda
t}\frac{\partial\varphi}{\partial t}(t,x)\,dt\right)$ $\displaystyle=$
$\displaystyle\frac{\lambda^{\alpha}}{\lambda}\big{(}\lambda\tilde{\varphi}(\lambda,x)-u(0,x)\big{)},\quad\lambda>0$
(25)

where

$\displaystyle\tilde{\varphi}(\lambda,x)=\int_{0}^{\infty}e^{-\lambda
t}\varphi(t,x)dt,\quad\lambda>0.$

Formula (25) can be obtained by observing that $D^{\alpha}_{t}$ is a
convolution operator. Notice that we do not require $\varphi(\cdot,x)\in
L^{1}(0,\infty)$ for some $x$.

###  2.2 The process $X$

Our discussion is mainly based on the well-known book [4] and the pioneering
work [25]. Let us consider the natural filtration
$\mathcal{F}_{t}=\sigma\\{X_{s},\,0\leq s<t\\}$ and a good function $f$ for
which $\mathbf{E}[f(X_{s})|\mathcal{F}_{t}]=\mathbf{E}[f(X_{s})|X_{t}]$,
$t\leq s$, and
$\mathbf{E}_{x}[f(X_{t+s})|\mathcal{F}_{t}]=\mathbf{E}_{X_{t}}[f(X_{s})]$,
$s,t>0$. We say that $X$ is an elastic process meaning that
$\mathbf{E}_{x}[f(X_{t})]=\mathbf{E}_{x}[f(X_{t}),t<\zeta]$ is written in
terms of the multiplicative functional $M_{t}=\mathbf{1}_{(t<\zeta)}$ where
the lifetime $\zeta$ well accords with an elastic kill. There exists an
independent exponential random variable (with parameter $c/\eta$) for which

$\displaystyle\mathbf{E}[M_{t}|X_{t}]=e^{-(c/\eta)\gamma_{t}}.$ (26)

On the other hand, from the sticky condition,
$\\{t\,:\,X_{t}\in\partial\Omega\\}$ is a Lebesgue measurable set obtained
from the holding times of $X$ on $\partial\Omega$. In particular, we may
consider a sequence $\\{e_{i}\\}_{i}$ of (identically distributed) independent
exponential random variables (with parameter $\sigma/\eta$) for which
$\mathbf{P}_{x}(e_{1}>t,X_{e_{1}}\in
dy)=e^{-(\sigma/\eta)t}\mathbf{P}_{x}(X_{e_{1}}\in dy)$ for
$x\in\partial\Omega$ and

$\displaystyle\mathbf{P}_{x}(e_{1}>t|X_{e_{1}})=e^{-(\sigma/\eta)t},\quad
x\in\partial\Omega.$ (27)

As announced we consider $\overline{\Omega}=[0,\infty)$ with the boundary
point $\\{0\\}$. Since $X$ is an elastic sticky Brownian motion we have the
representation ([25, Section 10]

$\displaystyle\mathbf{E}_{x}[f(X_{t})]=\mathbf{E}_{x}\left[f(X^{+}\circ
V^{-1}_{t})\exp\left(-c/\sigma\,\gamma^{+}\circ V^{-1}_{t}\right)\right]$ (28)

where

$\displaystyle V^{-1}_{t}=\inf\\{s\geq 0\,:\,V_{s}>t\\}$

is the inverse of

$\displaystyle V_{t}=t+(\eta/\sigma)\gamma^{+}_{t}.$

Notice that $V_{t}$ and $V^{-1}_{t}$ are both continuous and strictly
increasing. Moreover,

$\displaystyle V\circ V^{-1}_{t}=t,\quad t\geq 0.$

Under the representation (28) for $X$, the boundary local time $\gamma$ of $X$
can be given as the composition (see [25], formula 18, Section 10)

$\displaystyle\gamma_{t}=(\eta/\sigma)\gamma^{+}\circ V^{-1}_{t},\quad t>0.$
(29)

We recall the resolvent

$\displaystyle
R_{\lambda}f(x)=e^{-x\sqrt{\lambda}}\frac{\sigma\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy+\eta
f(0)}{c+\eta\lambda+\sigma\sqrt{\lambda}}+R^{\dagger}_{\lambda}f(x),\quad
x\in[0,\infty),\;\lambda>0$ (30)

where $R^{\dagger}_{\lambda}f=\int_{0}^{\infty}e^{-\lambda
t}Q^{\dagger}_{t}fdt$ is the resolvent for the Dirichlet semigroup
$Q^{\dagger}_{t}$. For the explicit calculation we refer to [25] and [14].
Moreover, we write

$\displaystyle\mathbf{P}_{x}(X_{t}\in dy)=p(t,x,y)dy$

where $p$ is the continuous kernel of $X$.

###  2.3 The random times $L$ and $H$

The process $L$ is the inverse to the $\alpha$-stable subordinator $H$ defined
by $L_{t}=\inf\\{s\geq 0\,:\,H_{s}>t\\}$ and for which
$\mathbf{P}_{0}(L_{t}<s)=\mathbf{P}_{0}(t<H_{s})$, $t,s>0$. We assume that
$H_{0}=0=L_{0}$ and write $\mathbf{P}_{0}$ for the associated probability
measure. Denote by $l$ and $h$ the corresponding probability densities for
which

$\displaystyle\mathbf{P}_{0}(H_{t}\in
ds)=h(t,s)ds,\quad\mathbf{P}_{0}(L_{t}\in ds)=l(t,s)ds.$

Then

$\displaystyle\int_{0}^{\infty}e^{-\xi
s}h(t,s)ds=e^{-t\xi^{\alpha}},\quad\int_{0}^{\infty}e^{-\lambda
t}l(t,s)dt=\frac{\lambda^{\alpha}}{\lambda}e^{-s\lambda^{\alpha}},\quad\xi,\lambda>0.$
(31)

We recall that

$\displaystyle\int_{0}^{\infty}e^{-\xi s}l(t,s)ds=E_{\alpha}(-\xi
t^{\alpha})\quad\textrm{with}\quad\int_{0}^{\infty}e^{-\lambda
t}E_{\alpha}(-\xi
t^{\alpha})dt=\frac{\lambda^{\alpha-1}}{\lambda^{\alpha}+\xi},\quad\lambda>0$
(32)

where the Mittag-Leffler function $E_{\alpha}$ is analytic and such that

$\displaystyle|E_{\alpha}(-\mu t^{\alpha})|\leq\frac{C}{1+\mu
t^{\alpha}},\quad t\geq 0,\;\mu>0\quad\textrm{for a constant }C>0$ (33)

(see [3, 24]). We underline that $E_{\alpha}\notin L^{1}(0,\infty)$ for
$\alpha\in(0,1)$.

We recall that $\lambda^{\alpha}$ is the so-called symbol of $H$ such that
$\mathbf{E}_{0}[\exp-\lambda H_{t}]=\exp-t\lambda^{\alpha}$, $\lambda>0$. For
the reader’s convenience we also recall that $\lambda^{\alpha}$ is a Bernstein
function with representation (according with (23))

$\displaystyle\lambda^{\alpha}=\int_{0}^{\infty}\left(1-e^{-\lambda
z}\right)\frac{\alpha}{\Gamma(1-\alpha)}z^{-\alpha-1}dz,\quad\lambda\geq 0.$

Moreover,

$\displaystyle\overline{\Pi}(z):=\int_{z}^{\infty}\frac{\alpha}{\Gamma(1-\alpha)}y^{-\alpha-1}dy=l(z,0),\quad
z>0$

as the Laplace transform of both sides entails.

##  3 The fractional initial value problem

###  3.1 Time changes for FIVPs

######  Theorem 3.1.

Let us consider the solution $v$ to the problem (9) and the solution $w$ to
the problem (15):

  * i)

$v,w\in D(G)$;

  * ii)

$\forall\,t\geq 0,\;\forall\,x\in[0,\infty),\;v(t,x)=w(t,x)$.

Moreover,

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}v(t,x)dt=\frac{\lambda^{\alpha}}{\lambda}R_{\lambda^{\alpha}}f(x),\quad\lambda>0,\;x\in[0,\infty)$

where $R_{\lambda}$ has been defined in (30) and the following probabilistic
representation holds true

$\displaystyle v(t,x)=\mathbf{E}_{x}[f(X\circ L_{t})],\quad t\geq
0,\;x\in[0,\infty).$ (34)

###### Proof.

Let us consider the problem (9). The problem

$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle
D^{\alpha}_{t}\varphi(t,x)=\varphi^{\prime\prime}(t,x),\quad t>0,\,x>0\\\ \\\
\displaystyle\eta\varphi^{\prime\prime}(t,0)=\sigma\varphi^{\prime}(t,0)-c\,\varphi(t,0),\quad
t>0\\\ \\\ \displaystyle\varphi(0,x)=f(x),\quad x\geq 0,\quad f\in
C_{b}[0,\infty)\end{array}\right.$ (40)

has a unique solution with probabilistic representation

$\displaystyle\varphi(t,x)=\mathbf{E}_{x}[f(X\circ
L_{t})]=\mathbf{E}_{x}[f(X^{+}\circ V^{-1}\circ L_{t})\,M\circ V^{-1}\circ
L_{t}].$ (41)

To prove our statement we first assume that the representation (41) holds
true. Then we consider the $\lambda$-potential

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\varphi(t,x)dt=$
$\displaystyle\frac{\lambda^{\alpha}}{\lambda}\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda^{\alpha}t}f(X^{+}\circ
V^{-1}_{t})M\circ
V^{-1}_{t}\,dt\right]=\frac{\lambda^{\alpha}}{\lambda}R_{\lambda^{\alpha}}f(x)$

where $R_{\lambda}f$ has been introduced in (30). Thus

$\displaystyle\lim_{x\to 0}\int_{0}^{\infty}e^{-\lambda
t}\varphi(t,x)dt=\frac{\lambda^{\alpha}}{\lambda}\frac{\sigma\int_{0}^{\infty}e^{-y\sqrt{\lambda^{\alpha}}}f(y)dy+\eta
f(0)}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda^{\alpha}}}=:\widetilde{\varphi}(\lambda,0)$

We get

$\displaystyle\eta\lambda^{\alpha}\widetilde{\varphi}(\lambda,0)-\frac{\lambda^{\alpha}}{\lambda}\eta
f(0)=\sigma\frac{\lambda^{\alpha}}{\lambda}\int_{0}^{\infty}e^{-y\sqrt{\lambda^{\alpha}}}f(y)dy-\sigma\sqrt{\lambda^{\alpha}}\widetilde{\varphi}(\lambda,0)-c\widetilde{\varphi}(\lambda,0)$

where

$\displaystyle\eta\lambda^{\alpha}\widetilde{\varphi}(\lambda,0)-\frac{\lambda^{\alpha}}{\lambda}\eta
f(0)=\int_{0}^{\infty}e^{-\lambda t}\eta D^{\alpha}_{t}\varphi(t,0)dt$

and

$\displaystyle\sigma\frac{\lambda^{\alpha}}{\lambda}\int_{0}^{\infty}e^{-y\sqrt{\lambda^{\alpha}}}f(y)dy-\sigma\sqrt{\lambda^{\alpha}}\widetilde{\varphi}(\lambda,0)-c\widetilde{\varphi}(\lambda,0)$

equals

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\left(\sigma\frac{\partial\varphi}{\partial
x}(t,0)-c\,\varphi(t,0)\right)dt.$

We conclude that $\varphi=w$, the solution of (15).

On the other hand,

$\displaystyle\lim_{x\to 0}\int_{0}^{\infty}e^{-\lambda
t}\varphi^{\prime\prime}(t,x)dt=$ $\displaystyle\lim_{x\to
0}\left(\frac{\lambda^{\alpha}}{\lambda}R_{\lambda^{\alpha}}f(x)\right)^{\prime\prime}$
(42) $\displaystyle=$
$\displaystyle\frac{\lambda^{\alpha}}{\lambda}\left(\lambda^{\alpha}R_{\lambda^{\alpha}}f(0)-f(0)\right)$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\dot{\varphi}(t,0)\,dt.$

Notice that

$\displaystyle\frac{\lambda^{\alpha}}{\lambda}\left(\lambda\frac{\lambda^{\alpha}}{\lambda}R_{\lambda^{\alpha}}f(0)-f(0)\right)=\left(\int_{0}^{\infty}e^{-\lambda
t}\overline{\Pi}(t)\,dt\right)\left(\int_{0}^{\infty}e^{-\lambda
t}\varphi^{\prime}(t,0)dt\right)$

and therefore

$\displaystyle\lim_{x\to 0}\int_{0}^{\infty}e^{-\lambda
t}\varphi^{\prime\prime}(t,x)dt=(\varphi^{\prime}*\overline{\Pi})(t,0)$

which is the fractional derivative $D^{\alpha}_{t}\varphi$ at $x=0$. We
conclude that $\varphi=v$ is the solution of (9).

In both cases we have a continuous inverse of the Laplace transform, then
there exist at most one inverse. The uniqueness of the continuous solutions
for the two problems says that $w=v$ pointwise on
$[0,\infty)\times[0,\infty)$.

∎

##  4 The fractional boundary value problem

We recall that by FBVP we mean a local equation (the heat equation) on
$(0,\infty)$ equipped with non-local condition at the boundary point
$\\{0\\}$. In the literature many authors use to refer to a FBVP in case of
non-local equations with local boundary conditions, that is, in our paper, a
FIVP.

###  4.1 The process $\bar{X}$ and the FBVP

We state here the main result of the work.

Further on we denote by $\bar{p}=\bar{p}(t,x,y)$ and $p^{+}=p^{+}(t,x,y)$ the
continuous kernels associated with $\bar{X}$ and $X^{+}$. It will be clear
that $\bar{X}$ becomes a Markov process only in case $\alpha=1$, that is the
case $\bar{X}=X$. For $\alpha\in(0,1)$, $\bar{p}$ can not be associated with a
semigroup operator. We also underline that $X$ does not enjoy the strong
Markov property because of the holding times at $\\{0\\}$. However, the
holding times for $X$ are exponential random variables and this maintains the
Markov property.

We recall that $X^{+}$ is a reflected Brownian motion on $[0,\infty)$ with
local time

$\displaystyle\gamma^{+}_{t}=\int_{0}^{t}\mathbf{1}_{\\{0\\}}(X^{+}_{s})ds$

properly defined as a limit on the interval $[0-\epsilon,0+\epsilon]$ as
$\epsilon\to 0$. We also recall that a positive continuous additive functional
$A=\\{A_{t}\\}_{t\geq 0}$ for $X^{+}$ is in Revuz correspondence with a
measure $\mu_{A}$, that is

$\displaystyle\lim_{t\to
0}\mathbf{E}_{m}\left[\frac{1}{t}\int_{0}^{t}f(X^{+}_{s})dA_{s}\right]=\int_{\Omega}f(x)\mu_{A}(dx)$

where $f$ is Borel measurable on $\Omega$ and
$\mathbf{E}_{m}=\int\mathbf{E}_{x}m(dx)$. Moreover, for
$x\in\overline{\Omega}=[0,\infty)$,

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{t}f(X^{+}_{s})dA_{s}\right]=\int_{0}^{t}\int_{\Omega}f(y)p^{+}(s,x,y)\mu_{A}(dy)ds.$

In particular, we observe that given $A$ for $X$ with
$supp[\mu_{A}]=\Lambda\subset[0,\infty)$ we have

$\displaystyle\lim_{t\to
0}\mathbf{E}_{m}\left[\frac{1}{t}\int_{0}^{t}f(X_{s})dA_{s}\right]=\int_{\Lambda}f(x)dx+(\eta/\sigma)\int_{\Lambda}f(x)\delta_{0}(dx).$

We provide the following result for the process $\bar{X}$ on $[0,\infty)$,
that is an elastic Brownian motion on $[0,\infty)$ running with the new clock
$\bar{V}_{t}^{-1}$. Let us introduce the space

$\displaystyle D_{L}=$
$\displaystyle\bigg{\\{}u:(0,\infty)\times[0,\infty)\to\mathbb{R}\textrm{ such
that }\dot{u}(s,0)(t-s)^{-\alpha}\in L^{1}(0,t),\,t>s>0\bigg{\\}}.$

######  Theorem 4.1.

The solution $u\in C^{1,2}((0,\infty)\times[0,\infty))\cap D_{L}$ to

$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle\dot{u}(t,x)=u^{\prime\prime}(t,x),\quad
t>0,\,x>0\\\ \\\ \displaystyle\eta D^{\alpha}_{t}u(t,0)=\sigma
u^{\prime}(t,0)-c\,u(t,0),\quad t>0\\\ \\\ \displaystyle u(0,x)=f(x),\quad
x\geq 0,\quad f\in C_{b}[0,\infty)\end{array}\right.$ (48)

has the probabilistic representation

$\displaystyle
u(t,x)=\mathbf{E}_{x}[f(\bar{X}_{t})]=\mathbf{E}_{x}\left[f(X^{+}\circ\bar{V}^{-1}_{t})\exp\left(-c/\sigma\,\gamma^{+}\circ\bar{V}^{-1}_{t}\right)\right]$

where $\bar{V}^{-1}_{t}$ is the inverse of the process

$\displaystyle\bar{V}_{t}=t+(\eta/\sigma)^{1/\alpha}H\circ\gamma^{+}_{t}.$

Moreover,

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}u(t,x)dt=R^{\dagger}_{\lambda}f(x)+\frac{e^{-x\sqrt{\lambda}}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}\left(\sigma\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy+\eta\frac{\lambda^{\alpha}}{\lambda}f(0)\right)$

with $\lambda>0$, where $R^{\dagger}_{\lambda}$ has been defined in (30).

###### Proof.

Since $u$ solves the heat equation, then $u$ can be written as

$\displaystyle u(t,x)=a(t,x)+\int_{0}^{t}b(t-s,x)u(s,0)ds$ (49)

with Laplace transform

$\displaystyle\tilde{u}(\lambda,x)=\tilde{a}(\lambda,x)+\tilde{b}(\lambda,x)\,\tilde{\upsilon}(\lambda,0),\quad\lambda>0$

for some sufficiently regular and (piecewise continuous) integrable functions
$a,b$. After some calculation we arrive at $\tilde{a}(\lambda,0)=0$ and
$\tilde{b}(\lambda,0)=1$. Moreover, we must have $u(0,x)=a(0,x)$ and
$\tilde{u}^{\prime\prime}=\lambda\tilde{u}-f$. In particular, it turns out
that $a,b$ are given by

$\displaystyle
u(t,x)=Q^{\dagger}_{t}f(x)+\int_{0}^{t}\frac{x}{s}g(s,x)\,u(t-s,0)\,ds$ (50)

where

$\displaystyle
Q^{\dagger}_{t}f(x)=\int_{0}^{\infty}\big{(}g(t,x-y)-g(t,x+y)\big{)}\,f(y)\,dy$
(51)

is the Dirichlet semigroup and $g(t,z)=e^{-z^{2}/4t}/\sqrt{4\pi t}$ is the
Gaussian kernel. The associated $\lambda$-potential is given by

$\displaystyle\tilde{u}(\lambda,x)=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda t}u(t,x)dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}Q^{\dagger}_{t}f(x)dt+e^{-x\sqrt{\lambda}}\tilde{u}(\lambda,0),\quad\lambda>0$

where the potential of $Q^{\dagger}_{t}$ can be written as

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}Q^{\dagger}_{t}f(x)\,dt=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}\left(\frac{e^{-|x-y|\sqrt{\lambda}}}{\sqrt{\lambda}}-\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}\right)f(y)\,dy$
(52) $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}\left(\frac{e^{(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}-\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}\right)f(y)\,dy$
$\displaystyle-\frac{1}{2}\int_{0}^{x}\left(\frac{e^{(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}-\frac{e^{-(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}\right)f(y)\,dy,\quad\lambda>0$

by exploiting the fact that

$\displaystyle\int_{0}^{\infty}e^{-|x-y|}f(y)\,dy$ $\displaystyle=$
$\displaystyle\int_{0}^{x}e^{-(x-y)}f(y)\,dy+\int_{x}^{\infty}e^{-(y-x)}f(y)\,dy$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-(y-x)}f(y)\,dy+\int_{0}^{x}e^{-(x-y)}f(y)\,dy-\int_{0}^{x}e^{-(y-x)}f(y)\,dy.$

We can immediately see that

$\displaystyle\frac{\partial^{2}\tilde{u}}{\partial
x^{2}}(\lambda,x)=\lambda\tilde{u}(\lambda,x)-f(x)$

and

$\displaystyle\frac{\partial\tilde{u}}{\partial
x}(\lambda,x)\bigg{|}_{x=0}=\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)-\sqrt{\lambda}\tilde{u}(\lambda,x).$
(53)

Let us recall that $\bar{\tau}_{0}=\tau_{0}$ is the first time the process
$\bar{X}$ hits the boundary point $x=0$. The representation (50) can be
obtained from the fact that $\bar{X}=X$ and therefore enjoys the Markov
property up to the time $\bar{\tau}_{0}$. In particular, we have that

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}f(\bar{X}_{t})dt\right]$ $\displaystyle=$
$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\bar{\tau}_{0}}e^{-\lambda
t}f(\bar{X}_{t})dt\right]+\mathbf{E}_{x}[e^{-\lambda\bar{\tau}_{0}}]\,\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(\bar{X}_{t})dt\right],\quad\lambda>0.$

Let us focus on

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(\bar{X}_{t})dt\right]=$
$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}\circ\bar{V}^{-1}_{t})\exp\left(-c/\sigma\,\gamma^{+}\circ\bar{V}^{-1}_{t}\right)dt\right]$
$\displaystyle=$
$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda\bar{V}_{t}}f(X^{+}_{t})\exp\left(-c/\sigma\,\gamma^{+}_{t}\right)d\bar{V}_{t}\right],\quad\lambda>0$
(54)

where

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\mathbf{P}_{0}(X^{+}_{t}\in
dy,0<\gamma^{+}_{t}\in dw)=e^{-(y+w)\sqrt{\lambda}}dy\,dw,\quad\lambda>0.$

Based on the definition of $\bar{V}_{t}$, we write (54) as the sum of

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda(\eta/\sigma)^{1/\alpha}H\circ\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)dt\right]$

and

$\displaystyle(\eta/\sigma)^{1/\alpha}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda(\eta/\sigma)^{1/\alpha}H\circ\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)d(H\circ\gamma^{+}_{t})\right].$

From the first integral we get

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda(\eta/\sigma)^{1/\alpha}H\circ\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)dt\right]$
$\displaystyle=$
$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda^{\alpha}\eta/\sigma\,\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)dt\right]$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}f(y)\,e^{-w\,c/\sigma-w\lambda^{\alpha}\eta/\sigma}e^{-(y+w)\sqrt{\lambda}}dw\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{c/\sigma+\lambda^{\alpha}\eta/\sigma+\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy$
$\displaystyle=$
$\displaystyle\frac{\sigma}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy,\quad\lambda>0.$

Set

$\displaystyle
I_{3}:=\frac{\sigma}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy,\quad\lambda>0.$

Now, observe that

$\displaystyle\mathbf{E}\left[e^{-\theta
H_{t}}dH_{t}\right]=-\frac{1}{\theta}d\mathbf{E}\left[e^{-\theta
H_{t}}\right]=\frac{\theta^{\alpha}}{\theta}e^{-\theta^{\alpha}t}dt,\quad\theta>0$

and write (recall that $\gamma^{+}\circ\gamma^{+,-1}_{t}=t$ a.s.)

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda(\eta/\sigma)^{1/\alpha}H\circ\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)d(H\circ\gamma^{+}_{t})\right]$
$\displaystyle=$
$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda\gamma^{-1}_{t}}f(X^{+}\circ\gamma^{+,-1}_{t})\exp\left(-\lambda(\eta/\sigma)^{1/\alpha}H_{t}-c/\sigma\,t\right)dH_{t}\right]$
$\displaystyle=$
$\displaystyle\frac{\sigma^{1/\alpha}}{\eta^{1/\alpha}}\frac{\lambda^{\alpha}\eta}{\lambda\sigma}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-t\lambda^{\alpha}\eta/\sigma}e^{-\lambda\gamma^{+,-1}_{t}-c/\sigma\,t}f(X^{+}\circ\gamma^{+,-1}_{t})dt\right]$
$\displaystyle=$
$\displaystyle\frac{\sigma^{1/\alpha}}{\eta^{1/\alpha}}\frac{\lambda^{\alpha}\eta}{\lambda\sigma}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda^{\alpha}\eta/\sigma\,\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)d\gamma^{+}_{t}\right],\quad\lambda>0.$

Set

$\displaystyle
I_{4}:=\frac{\sigma^{1/\alpha}}{\eta^{1/\alpha}}\frac{\lambda^{\alpha}\eta}{\lambda\sigma}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda^{\alpha}\eta/\sigma\,\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)d\gamma^{+}_{t}\right],\quad\lambda>0$

where

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(X^{+}_{t})\exp\left(-\lambda^{\alpha}\eta/\sigma\,\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)d\gamma^{+}_{t}\right]$
$\displaystyle=$
$\displaystyle\frac{f(0)}{c/\sigma+\lambda^{\alpha}\eta/\sigma}\mathbf{E}_{0}\left[1-\lambda\int_{0}^{\infty}e^{-\lambda
t}\exp\left(-\lambda^{\alpha}\eta/\sigma\,\gamma^{+}_{t}-c/\sigma\,\gamma^{+}_{t}\right)dt\right]$
$\displaystyle=$
$\displaystyle\frac{f(0)}{c/\sigma+\lambda^{\alpha}\eta/\sigma}\left(1-\sqrt{\lambda}\int_{0}^{\infty}e^{-w\,c/\sigma-w\,\lambda^{\alpha}\eta/\sigma-w\sqrt{\lambda}}dw\right)$
$\displaystyle=$
$\displaystyle\frac{f(0)}{c/\sigma+\lambda^{\alpha}\eta/\sigma+\sqrt{\lambda}}$
$\displaystyle=$
$\displaystyle\frac{\sigma\,f(0)}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}},\quad\lambda>0.$

That is,

$\displaystyle
I_{4}=\frac{\sigma^{1/\alpha}}{\eta^{1/\alpha}}\frac{\lambda^{\alpha}}{\lambda}\frac{\eta
f(0)}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}},\quad\lambda>0.$

Then, we obtain

$\displaystyle\tilde{u}(\lambda,0)=$ $\displaystyle
I_{3}+(\eta/\sigma)^{1/\alpha}I_{4}$ $\displaystyle=$
$\displaystyle\frac{\sigma}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy+\frac{\lambda^{\alpha}}{\lambda}\frac{\eta
f(0)}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}$ (55)

from which we get that

$\displaystyle\eta\lambda^{\alpha}\tilde{u}-\eta\frac{\lambda^{\alpha}}{\lambda}f(0)=\sigma\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy-\sigma\sqrt{\lambda}\tilde{u}-c\,\tilde{u}.$

By observing that

$\displaystyle\eta\lambda^{\alpha}\tilde{u}-\eta\frac{\lambda^{\alpha}}{\lambda}f(0)=\eta\int_{0}^{\infty}e^{-\lambda
t}D^{\alpha}_{t}u(t,0)\,dt,$

from (53), we obtain

$\displaystyle\eta D^{\alpha}_{t}u(t,0)=\sigma\frac{\partial u}{\partial
x}(t,0)-c\,u(t,0)$

which is the claimed boundary condition.

The requirement $u\in D_{L}$ ensures existence of $D^{\alpha}_{t}u(t,0)$.
Uniqueness follows from the continuity of $u$ and the Laplace transform
machinery (Lerch’s Theorem).

∎

For the reader’s convenience we recall that

$\displaystyle\tilde{u}(\lambda,0)=\int_{0}^{\infty}e^{-\lambda t}u(t,0)dt$

and $u$ is the solution to the problem (48). Now we study

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)=\int_{0}^{\infty}\lambda
e^{-\lambda
t}\mathbf{E}_{0}[f(\bar{X}_{t})-f(\bar{X}_{0})]dt=\mathbf{E}_{0}[f(\bar{X}_{\chi})-f(\bar{X}_{0})]$

where $\chi$ is an exponential r.v. with $\mathbf{P}(\chi>t)=e^{-\lambda t}$,
$\lambda>0$. Denote by $H^{1/2}$ the stable subordinator of order $1/2$
independent from $H$. Recall that $H$ is a stable subordinator of order
$\alpha\in(0,1)$.

######  Lemma 1.

If $\|f\|_{\infty}=f(0)$, then

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq-f(0)\,\mathbf{E}\left[\exp\left(-\lambda
H^{1/2}_{\sigma\chi}-\lambda H_{\eta\chi}\right)\right],\quad\lambda\geq 0$
(56)

with $\mathbf{P}(\chi>t)=e^{-ct}$ and

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\geq-f(0)\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda\geq
0.$ (57)

Otherwise,

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq\frac{1}{2}\big{(}\|f\|_{\infty}-f(0)\big{)}\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda>0$
(58)

and

$\displaystyle\lim_{\lambda\to
0}\lambda\tilde{u}(\lambda,0)-f(0)\leq\big{(}\|f\|_{\infty}-f(0)\big{)}.$ (59)

Moreover,

$\displaystyle|\lambda\tilde{u}(\lambda,0)-f(0)|\leq\|f\|_{\infty}\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda>0.$
(60)

###### Proof.

We use the fact that, for $\lambda>0$,

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\frac{\partial u}{\partial
t}(t,0)dt\leq\int_{0}^{\infty}e^{-\lambda
t}\varphi(t)dt\quad\textrm{iff}\quad\frac{\partial u}{\partial
t}(t,0)\leq\varphi(t)$

for a given function $\varphi:(0,\infty)\mapsto(0,\infty)$ for which the
Laplace transform exists. We observe that, from (4.1),

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq$
$\displaystyle\frac{\sigma\sqrt{\lambda}}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\|f\|_{\infty}+\frac{\eta\lambda^{\alpha}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)-f(0)$
(61) $\displaystyle=$
$\displaystyle\frac{\sigma\sqrt{\lambda}}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\|f\|_{\infty}-\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)$
$\displaystyle\leq$
$\displaystyle\big{(}\|f\|_{\infty}-f(0)\big{)}\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}},\quad\lambda\geq
0.$

This leads to (59). By exploiting the Young’s inequality $a+b\geq 2\sqrt{ab}$
for $a=c+\sigma\sqrt{\lambda}$ and $b=\eta\lambda^{\alpha}$, we obtain

$\displaystyle\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}\leq\frac{1}{2}\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}}$
(62)

and this proves (58). Since $-f\leq|f|$ we also get

$\displaystyle\big{(}\|f\|_{\infty}-f(0)\big{)}\leq\big{(}\|f\|_{\infty}+|f(0)|\big{)}\leq
2\|f\|_{\infty}$

and

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq\|f\|_{\infty}\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda\geq
0.$

Moreover,

$\displaystyle f(0)-\lambda\tilde{u}(\lambda,0)=$
$\displaystyle\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)-\frac{\sigma\lambda}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}f(y)dy$
$\displaystyle\leq$
$\displaystyle\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)+\frac{\sigma\lambda}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}|f(y)|dy$
$\displaystyle\leq$
$\displaystyle\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)+\frac{\sigma\sqrt{\lambda}}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\|f\|_{\infty}$
$\displaystyle\leq$
$\displaystyle\frac{c+\sigma\sqrt{\lambda}}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}\big{(}f(0)+\|f\|_{\infty}\big{)}$

where

$\displaystyle\big{(}f(0)+\|f\|_{\infty}\big{)}\leq 2\|f\|_{\infty}.$

Thus,

$\displaystyle
f(0)-\lambda\tilde{u}(\lambda,0)\leq\|f\|_{\infty}\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda>0$

says that

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\geq-\|f\|_{\infty}\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda>0$

which is (57). We conclude that

$\displaystyle|\lambda\tilde{u}(\lambda,0)-f(0)|\leq\|f\|_{\infty}\,\sqrt{(c/\eta)\frac{1}{\lambda^{\alpha}}+(\sigma/\eta)\frac{\sqrt{\lambda}}{\lambda^{\alpha}}},\quad\lambda>0$

as stated in (60).

Now we use (4.1) once again and write

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq$
$\displaystyle\frac{\sigma\sqrt{\lambda}}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\|f\|_{\infty}-\frac{(c+\sigma\sqrt{\lambda})}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0)$
$\displaystyle=$
$\displaystyle\big{(}\|f\|_{\infty}-f(0)\big{)}\frac{\sigma\sqrt{\lambda}}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}-\frac{c}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0).$
(63)

In case $\|f\|_{\infty}=f(0)$ we get

$\displaystyle\lambda\tilde{u}(\lambda,0)-f(0)\leq-\frac{c}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}f(0).$
(64)

Thus, by observing that

$\displaystyle\mathbf{E}e^{-\theta H_{t}}=e^{-t\theta^{\alpha}},$

from

$\displaystyle\frac{c}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}=c\int_{0}^{\infty}e^{-ct}e^{-t\sigma\sqrt{\lambda}-t\eta\lambda^{\alpha}}dt$

we prove (56). ∎

######  Remark 4.1.

We argue on $u\in D_{L}$. This ensures that

$\displaystyle\big{|}D^{\alpha}_{t}u(t,0)\big{|}\leq\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\left|\frac{\partial
u}{\partial s}(s,0)\right|(t-s)^{-\alpha}ds<\infty.$

If $\sigma=0$ in (58), then

$\displaystyle\sqrt{c/\eta}\frac{1}{\lambda^{\alpha/2}}=\sqrt{c/\eta}\frac{1}{\Gamma(\alpha/2)}\int_{0}^{\infty}e^{-\lambda
t}t^{\alpha/2-1}dt$

and $\forall\,\alpha/2>0$,

$\displaystyle\int_{0}^{t}\frac{\partial u}{\partial
s}(s,0)(t-s)^{-\alpha}ds\leq(\|f\|_{\infty}-f(0))\int_{0}^{t}t^{\alpha/2-1}(t-s)^{-\alpha}ds<\infty.$

Moreover, from (60) and the fact that

$\displaystyle\bigg{|}\frac{\partial u}{\partial
t}(t,0)\bigg{|}<\infty\;\;a.e.$

we conclude that

$\displaystyle\exists\,\mathcal{C}>0\,:\,\bigg{|}\frac{\partial u}{\partial
t}(t,0)\bigg{|}\leq\mathcal{C}\,\|f\|_{\infty}\,\frac{\sqrt{(c/\eta)}}{2}t^{\alpha/2-1},\quad
t>0.$ (65)

We notice that $\sigma=0$ is strictly related with $f(0)=\|f\|_{\infty}$ as
formula (63) entails. In this regard, we notice that $f\geq 0$ and formula
(56) says also that

$\displaystyle\tilde{u}(\lambda,0)\leq
f(0)\frac{1}{\lambda}\mathbf{E}\left[1-\exp(-\lambda
H^{1/2}_{\sigma\chi}-\lambda H_{\eta\chi})\right],\quad
0\leq\sigma<\infty,\;0\leq\eta<\infty$

where $\chi$ is an independent exponential r.v. with parameter $c>0$. Observe
that $\sigma=\eta=0$ gives the Dirichlet boundary condition. For $\sigma=0$ we
have

$\displaystyle\tilde{u}(\lambda,0)\leq
f(0)\frac{1}{\lambda}\mathbf{E}\left[1-\exp(-\lambda H\circ
e_{0})\right],\quad 0\leq\eta<\infty$

where $e_{0}$ is the holding time (at zero) for $X$ (see also Theorem 4.3
below).

If $c=0$, from (58) we write

$\displaystyle\frac{\partial u}{\partial
t}(t,0)\leq\varphi(t)\quad\textrm{with}\quad\int_{0}^{\infty}e^{-\lambda
t}\varphi(t)dt=C\,\lambda^{1/4-\alpha/2},\quad\lambda>0$

for some constant $C\leq(\|f\|_{\infty}-f(0))\sqrt{\sigma/\eta}$. Thus,

$\displaystyle
D^{\alpha}_{t}u(t,0)\leq\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\varphi(s)(t-s)^{-\alpha}ds$

and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}D^{\alpha}_{t}u(t,0)\,dt\leq\frac{\lambda^{\alpha}}{\lambda}\,C\,\lambda^{1/4-\alpha/2}=C\,\lambda^{1/4+\alpha/2-1}.$

We conclude that

$\displaystyle D^{\alpha}_{t}u(t,0)\leq C\,t^{(1-1/4-\alpha/2)-1},\quad t>0$

and $u\in D_{L}$.

######  Remark 4.2.

Finally we focus on (61) with $\sigma=0$. The inequality

$\displaystyle\lambda\tilde{u}(\lambda,0)\leq$ $\displaystyle
f(0)\,\frac{\eta\lambda^{\alpha}}{c+\eta\lambda^{\alpha}}$

implies that

$\displaystyle\tilde{u}(\lambda,0)\leq$ $\displaystyle
f(0)\int_{0}^{\infty}e^{-\lambda
t}E_{\alpha}(-(c/\eta)t^{\alpha})\,dt,\quad\lambda>0.$

That is,

$\displaystyle u(t,0)\leq
f(0)\,E_{\alpha}(-(c/\eta)t^{\alpha})=f(0)\,\mathbf{P}(\chi>t)$

where $\chi$ denotes a Mittag-Leffler random variable with parameters $\alpha$
and $(c/\eta)$.

Figure 1: A representation of $\bar{X}$ on $[0,\infty)$ as a motion on the
path of $X^{+}$ (a Brownian motion reflected at zero). The plateau is given by
the inverse of $\bar{V}_{t}=t+H\circ(\eta/\sigma)\gamma^{+}_{t}$. As the local
time at zero $\gamma^{+}$ increases the jump of $H$ produces a plateau for
$\bar{V}^{-1}_{t}$. According with this plateau, the process
$X^{+}\circ\bar{V}^{-1}_{t}$ spends more time on the boundary point $\\{0\\}$.
The path exhibits a delayed reflection. The delay is given by $H$ which is
independent from $X^{+}$. This delay is the holding time with Mittag-Leffler
distribution.

###  4.2 On the occupation times for $\bar{X}$

We recall that $\\{t\,:\,X^{+}_{t}=0\\}$ is a perfect closed set of zero
Lebesgue measure. The definition of $\gamma^{+}$ can be therefore given with
some care as a proper limit which holds a.s. and $L^{2}(0,\infty)$. On the
other hand $\\{t\,:\,X_{t}=0\\}$ has positive Lebesgue measure as well as
$\\{t\,:\,\bar{X}_{t}=0\\}$. For $\alpha\in(0,1]$ and $\epsilon>0$ we can
write

$\displaystyle\mathbf{P}_{0}(0\leq\bar{X}_{t}\leq\epsilon)=\int_{0}^{\epsilon}\bar{p}(t,0,y)dy+(\eta/\sigma)\bar{p}(t,0,0)$
(66)

whose potential writes

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{[0,\epsilon]}(\bar{X}_{t})dt\right]=$
$\displaystyle\frac{\sigma}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\frac{1-e^{-\epsilon\sqrt{\lambda}}}{\sqrt{\lambda}}+\frac{\lambda^{\alpha}}{\lambda}\frac{\eta}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}$
(67)

as obtained in (4.1). Notice that both limits $\epsilon\to 0$ and
$\epsilon\to\infty$ make sense and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\bar{p}(t,0,0)dt=\frac{\lambda^{\alpha}}{\lambda}\frac{\sigma}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}},\quad\lambda>0$

becomes

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}p(t,0,0)dt=\frac{\sigma}{c+\eta\lambda+\sigma\sqrt{\lambda}},\quad\lambda>0$

for $\alpha=1$. We immediately see that

$\displaystyle\int_{0}^{\infty}p(t,0,0)dt=\sigma/c\quad\textrm{and}\quad(\eta/\sigma)\int_{0}^{\infty}p(t,0,0)dt=\eta/c$

give interesting readings of the parameters $\eta,\sigma,c$ and their
asymptotic analysis.

We also observe that simple manipulation leads to

$\displaystyle\bar{p}(t,0,0)=\int_{0}^{t}\frac{\sigma s}{t-z}g(t-z,\sigma
s)\,\eta\,l(z,\eta s)dz$

where $g$ and $l$ have been respectively defined in pages 51 and 33. Explicit
representations of $l$ are well-known for some values of $\alpha$. For
example:

  * i)

$\alpha=1/2$,

$\displaystyle l(t,x)=2g(t,x)=2\frac{e^{-\frac{x^{2}}{4t}}}{\sqrt{4\pi
t}},\quad t>0,\,x\in(0,\infty),$

  * ii)

$\alpha=1/3$,

$\displaystyle
l(t,x)=\frac{3}{\sqrt[3]{3t}}Ai\left(\frac{x}{\sqrt[3]{3t}}\right),\quad
t>0,\,x\in(0,\infty)$

where $Ai$ is the Airy function.

The reader can consult [30] and the references therein for further details.

Now we recall that

$\displaystyle
f(\bar{X}_{t})=f(X^{+}\circ\bar{V}^{-1}_{t})\,M\circ\bar{V}^{-1}_{t}$

where $M\leq 1$ is the multiplicative functional associated with the Robin
boundary condition. If we assume $c=0$, then $\bar{X}$ has no elastic kill and
we write

$\displaystyle f(\bar{X}_{t})=f(X^{+}\circ\bar{V}^{-1}_{t}).$

We present the following result concerning the processes
$\bar{V}_{t}=t+H\circ(\eta/\sigma)\gamma^{+}_{t}$, $t\geq 0$ and the right-
inverse $\bar{V}^{-1}_{t}=\inf\\{s\,:\,\bar{V}_{s}>t\\}$, $t\geq 0$. Recall
that a.s. $\bar{V}_{t}\geq t$ and $\bar{V}^{-1}_{t}\leq t$. In particular,

$\displaystyle\mathbf{P}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\geq
s)>0\quad\textrm{for}\quad s\geq t.$

For $\alpha=1$, $\bar{V}_{t}=V_{t}:=t+(\eta/\sigma)\gamma^{+}_{t}$ is a
continuous process. We introduce the processes

$\displaystyle
T^{H}_{t}=H\circ(\eta/\sigma)\gamma^{+}_{t}\quad\textrm{and}\quad
T^{L}_{t}=\gamma^{+,-1}\circ(\sigma/\eta)L_{t}$

such that, for $t>0$, $s>0$,

$\displaystyle\mathbf{P}_{0}(T^{H}_{s}\geq t)=\mathbf{P}_{0}(s\geq T^{L}_{t})$

where $\gamma^{+,-1}=\\{\gamma^{+,-1}_{t}\\}_{t\geq 0}$ is the inverse of
$\gamma^{+}$. Then, we write

$\displaystyle\bar{V}_{t}=t+T^{H}_{t}\quad\textrm{and}\quad\bar{\mathcal{V}}_{t}=t+T^{L}_{t}$

######  Lemma 2.

The following holds true:

  * i)

$\mathbf{P}_{0}(\gamma^{+}_{t}\geq s)=\mathbf{P}_{0}(\gamma^{+,-1}_{s}\leq t)$
where $\gamma^{+,-1}$ is a $1/2$-stable subordinator and $s,t>0$;

  * ii)

$\mathbf{P}_{0}(\bar{V}_{t}\geq s)=\mathbf{P}_{0}(t\geq T^{L}_{s-t}),\quad
0\leq t\leq s$;

  * iii)

$\mathbf{P}_{0}(\bar{\mathcal{V}}_{t}\geq s)=\mathbf{P}_{0}(t\geq
T^{H}_{s-t}),\quad 0\leq t\leq s$.

###### Proof.

Point $i)$ can be easily verified. Thus we move on point $ii)$. We prove,
$\forall\,s>0$, the identity

$\displaystyle\int_{0}^{s}e^{-\xi
t}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\geq s)dt=$
$\displaystyle\int_{0}^{s}e^{-\xi
t}\mathbf{P}_{0}(t\geq\gamma^{+,-1}\circ(\sigma/\eta)L_{s-t})dt$

with $\xi>0$ from which $\mathbf{P}_{0}(\bar{V}_{t}\geq
s)=\mathbf{P}_{0}(t\geq\bar{\mathcal{V}}^{-1}_{s,t})$ for $s\geq t\geq 0$. The
left-hand side of the integral above gives

$\displaystyle\int_{0}^{\infty}e^{-\lambda s}\int_{0}^{s}e^{-\xi
t}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\geq s)dt\,ds$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\xi
t}\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\geq s)ds\,dt$
$\displaystyle=$ $\displaystyle\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}\left(e^{-\lambda t}-\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\in ds)\right)dt$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\lambda+\xi)}-\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\in ds)dt$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\lambda+\xi)}-\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}\mathbf{E}_{0}\left[e^{-\lambda t-\lambda
H\circ(\eta/\sigma)\gamma^{+}_{t}}\right]dt$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda(\lambda+\xi)}-\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}e^{-\lambda
t}\mathbf{E}_{0}[e^{-\lambda^{\alpha}(\eta/\sigma)\gamma^{+}_{t}}]dt$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\lambda+\xi)}-\frac{1}{\lambda}\int_{0}^{\infty}(\xi+\lambda)^{-1/2}e^{-w(\xi+\lambda)^{1/2}}e^{-\lambda^{\alpha}(\eta/\sigma)w}dw$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\lambda+\xi)}-\frac{1}{\lambda}\frac{(\xi+\lambda)^{-1/2}}{(\xi+\lambda)^{1/2}+\lambda^{\alpha}(\eta/\sigma)}$
$\displaystyle=$
$\displaystyle\frac{(\xi+\lambda)^{-1/2}}{\lambda}\left(\frac{1}{(\xi+\lambda)^{1/2}}-\frac{1}{(\xi+\lambda)^{1/2}+\lambda^{\alpha}(\eta/\sigma)}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}\frac{\lambda^{\alpha}(\eta/\sigma)}{(\xi+\lambda)^{1/2}+\lambda^{\alpha}(\eta/\sigma)},\quad\xi>0,\quad\lambda>0$

which equals

$\displaystyle\int_{0}^{\infty}e^{-\lambda s}\int_{0}^{s}e^{-\xi
t}\mathbf{P}_{0}(t\geq\gamma^{+,-1}\circ(\sigma/\eta)L_{s-t})dt\,ds$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\xi
t}\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t\geq\gamma^{+,-1}\circ(\sigma/\eta)L_{s-t})ds\,dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\xi
t}\int_{0}^{\infty}e^{-\lambda t}e^{-\lambda
z}\mathbf{P}_{0}(t\geq\gamma^{+,-1}\circ(\sigma/\eta)L_{z})dz\,dt$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-(\xi+\lambda)t}\int_{0}^{\infty}e^{-\lambda
z}\mathbf{P}_{0}(t\geq\gamma^{+,-1}\circ(\sigma/\eta)L_{z})dz\,dt$
$\displaystyle=$
$\displaystyle\frac{1}{\xi+\lambda}\int_{0}^{\infty}e^{-\lambda
z}\mathbf{E}_{0}[e^{-(\xi+\lambda)\gamma^{+,-1}\circ(\sigma/\eta)L_{z}}]dz$
$\displaystyle=$
$\displaystyle\frac{1}{\xi+\lambda}\int_{0}^{\infty}\mathbf{E}_{0}[e^{-(\xi+\lambda)\gamma^{+,-1}\circ(\sigma/\eta)w}]\lambda^{\alpha-1}e^{-\lambda^{\alpha}w}dw$
$\displaystyle=$
$\displaystyle\frac{1}{\xi+\lambda}\int_{0}^{\infty}e^{-(\xi+\lambda)^{1/2}(\sigma/\eta)w}\lambda^{\alpha-1}e^{-\lambda^{\alpha}w}dw$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}\frac{\lambda^{\alpha}}{(\xi+\lambda)^{1/2}(\sigma/\eta)+\lambda^{\alpha}},\quad\xi>0,\quad\lambda>0$

from the right-hand side. Concerning the point $iii)$ we simply follow the
same arguments as before and we evaluate

$\displaystyle\int_{0}^{\infty}e^{-\xi t}\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t+\gamma^{+,-1}\circ(\sigma/\eta)L_{t}\geq s)ds\,dt$
$\displaystyle=$ $\displaystyle\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}\left(e^{-\lambda t}-\mathbf{E}_{0}[e^{-\lambda
t-\lambda\gamma^{+,-1}\circ(\sigma/\eta)L_{t}}]\right)dt$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}-\frac{1}{\lambda}\int_{0}^{\infty}e^{-\xi
t}\mathbf{E}_{0}[e^{-\lambda t-\lambda^{1/2}(\sigma/\eta)L_{t}}]dt$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}-\frac{1}{\lambda}\int_{0}^{\infty}e^{-\lambda^{1/2}(\sigma/\eta)w}(\xi+\lambda)^{\alpha-1}e^{-(\xi+\lambda)^{\alpha}w}dw$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}-\frac{1}{\lambda(\xi+\lambda)}\frac{(\xi+\lambda)^{\alpha}}{(\xi+\lambda)^{\alpha}+\lambda^{1/2}(\sigma/\eta)}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda(\xi+\lambda)}\frac{\lambda^{1/2}(\sigma/\eta)}{(\xi+\lambda)^{\alpha}+\lambda^{1/2}(\sigma/\eta)},\quad\xi>0,\quad\lambda>0.$

Since

$\displaystyle\int_{0}^{\infty}e^{-\xi t}\int_{t}^{\infty}e^{-\lambda
s}\mathbf{P}_{0}(t\geq T^{H}_{s-t})\,ds\,dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-(\xi+\lambda)t}\int_{0}^{\infty}e^{-\lambda
z}\mathbf{P}_{0}(t\geq T^{H}_{z})\,dz\,dt$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
z}\frac{1}{\xi+\lambda}\mathbf{E}_{0}[e^{-(\xi+\lambda)H\circ(\eta/\sigma)\gamma^{+}_{z}}]dz$
$\displaystyle=$
$\displaystyle\frac{1}{\xi+\lambda}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-(\xi+\lambda)z}e^{-(\xi+\lambda)^{\alpha}(\eta/\sigma)\gamma^{+}_{z}}dz\right]$
$\displaystyle=$
$\displaystyle\frac{1}{\xi+\lambda}\int_{0}^{\infty}e^{-(\xi+\lambda)^{\alpha}(\eta/\sigma)w}\lambda^{-1/2}e^{-w\lambda^{1/2}}dw$
$\displaystyle=$
$\displaystyle\frac{\lambda^{-1/2}}{\xi+\lambda}\frac{1}{(\xi+\lambda)^{\alpha}(\eta/\sigma)+\lambda^{1/2}}$

we conclude that $iii)$ holds true. ∎

For the reader’s convenience we report the following identities which are
obtained from the potentials above:

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\int_{0}^{\infty}\mathbf{P}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\geq
s)dt\,ds$ $\displaystyle=$ $\displaystyle\lim_{\xi\to
0}\frac{1}{\lambda(\xi+\lambda)}\frac{\lambda^{\alpha}(\eta/\sigma)}{(\xi+\lambda)^{1/2}+\lambda^{\alpha}(\eta/\sigma)}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{\lambda^{\alpha}}{\lambda}\frac{\eta}{\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}},\quad\lambda>0$
(68)

and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\int_{0}^{\infty}\mathbf{P}(t+\gamma^{+,-1}\circ(\sigma/\eta)L_{t}\geq
s)dt\,ds$ $\displaystyle=$ $\displaystyle\lim_{\xi\to
0}\frac{1}{\lambda(\xi+\lambda)}\frac{\lambda^{1/2}(\sigma/\eta)}{(\xi+\lambda)^{\alpha}+\lambda^{1/2}(\sigma/\eta)},\quad\lambda>0$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{\sigma}{\sigma\sqrt{\lambda}+\eta\lambda^{\alpha}}\frac{1}{\sqrt{\lambda}}$
(69) $\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{\sigma}{\sigma\sqrt{\lambda}+\eta\lambda^{\alpha}}\int_{0}^{\infty}e^{-y\sqrt{\lambda}}dy.$

We can recognize (68) and (69) in formula (4.1) for $c=0$ and suitable initial
data. These formulas will be useful further on. We also observe that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\int_{0}^{s}\mathbf{P}_{0}(t+H\circ(\eta/\sigma)\gamma^{+}_{t}\leq s)dt\,ds$
$\displaystyle=$ $\displaystyle\lim_{\xi\to
0}\frac{1}{\lambda}\frac{(\xi+\lambda)^{-1/2}}{(\xi+\lambda)^{1/2}+\lambda^{\alpha}(\eta/\sigma)},\quad\lambda>0$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{\sigma}{\sigma\sqrt{\lambda}+\eta\lambda^{\alpha}}\frac{1}{\sqrt{\lambda}}$

coincides with (69) and can be regarded as

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\int_{0}^{s}\mathbf{P}_{0}(\bar{V}_{t}\leq
s)dt\,ds=\int_{0}^{\infty}e^{-\lambda
s}\int_{0}^{s}\mathbf{P}_{0}(\bar{V}^{-1}_{s}\geq t)dt\,ds$

where

$\displaystyle\int_{0}^{s}\mathbf{P}_{0}(\bar{V}_{s}^{-1}\geq
t)dt=\int_{0}^{\infty}\mathbf{P}_{0}(\bar{V}_{s}^{-1}\geq
t)dt=\mathbf{E}_{0}[\bar{V}_{s}^{-1}],\quad s\geq 0.$

Formula (69) leads to

$\displaystyle\int_{0}^{\infty}e^{-\lambda s}\int_{0}^{s}e^{-\xi
t}\mathbf{P}_{0}(t+\gamma^{+,-1}\circ(\sigma/\eta)L_{t}\leq s)dt\,ds$
$\displaystyle=$ $\displaystyle\lim_{\xi\to
0}\frac{1}{\lambda(\xi+\lambda)}\frac{(\xi+\lambda)^{\alpha}}{(\xi+\lambda)^{\alpha}+\lambda^{1/2}(\sigma/\eta)},\quad\lambda>0$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{\lambda^{\alpha}}{\lambda}\frac{\eta}{\lambda^{\alpha}+(\sigma/\eta)\sqrt{\lambda}}$

which coincides with (68) and is the Laplace transform of

$\displaystyle\int_{0}^{s}\mathbf{P}(\bar{\mathcal{V}}^{-1}_{s}\geq
t)dt=\int_{0}^{\infty}\mathbf{P}(\bar{\mathcal{V}}^{-1}_{s}\geq
t)dt=\mathbf{E}_{0}[\bar{\mathcal{V}}^{-1}_{s}],\quad s\geq 0.$

We are now ready for the next result.

######  Theorem 4.2.

For $\lambda>0$,

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\bar{V}^{-1}_{t}\,dt\right]=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}\mathbf{1}_{(0,\infty)}(\bar{X}_{s})ds\,dt\right]$

and

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\bar{\mathcal{V}}^{-1}_{t}\,dt\right]=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}\mathbf{1}_{\\{0\\}}(\bar{X}_{s})ds\right].$

###### Proof.

Let us take $f=\mathbf{1}_{(0,\infty)}$ in (4.1) with $c=0$. Then,

$\displaystyle\frac{1}{\lambda}\tilde{u}(\lambda,0)=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}\mathbf{1}_{(0,\infty)}(\bar{X}_{s})ds\,dt\right]$

which coincides with (69), that is with

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}[\bar{V}^{-1}_{t}]dt.$

On the other hand, by taking

$\displaystyle f(x)=\left\\{\begin{array}[]{ll}0,&x\in(0,\infty)\\\
1,&x=0\end{array}\right.$

we obtain

$\displaystyle u(t,0)=$ $\displaystyle\mathbf{E}_{0}[f(\bar{X}_{t})]$
$\displaystyle=$ $\displaystyle\int_{[0,\infty)}f(y)\bar{p}(t,0,y)m(dy)$
$\displaystyle=$
$\displaystyle(\eta/\sigma)\int_{[0,\infty)}f(y)\bar{p}(t,0,y)\delta_{0}(dy)$
$\displaystyle=$ $\displaystyle(\eta/\sigma)\bar{p}(t,0,0).$

From (4.1) and the fact that

$\displaystyle\frac{1}{\lambda}\tilde{u}(\lambda,0)=\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}u(s,0)ds\,dt=\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{t}\mathbf{E}_{0}[f(\bar{X}_{s})]ds\,dt$

we write

$\displaystyle\frac{1}{\lambda}\tilde{u}(\lambda,0)=\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\int_{0}^{t}\mathbf{1}_{\\{0\\}}(\bar{X}_{s})ds\right]dt$

which coincides with (68) and therefore with

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}[\bar{\mathcal{V}}^{-1}_{t}]dt.$

∎

We study the occupation measure $\mu_{\bar{X}}$ defined as

$\displaystyle\mu_{\bar{X}}(x,S):=\mathbf{E}_{x}\left[\int_{0}^{\tau_{\bar{X}}(\Lambda)}\mathbf{1}_{S}(\bar{X}_{s})ds\right],\quad
x\in\Lambda\subseteq[0,\infty),\quad S\subset\Lambda,\quad\alpha\in(0,1]$

and for which

$\displaystyle\mu_{\bar{X}}(x,\Lambda)=\mathbf{E}_{x}[\tau_{\bar{X}}(\Lambda)],\quad
x\in\Lambda,\quad\alpha\in(0,1].$

######  Lemma 3.

Let $\Lambda$ be an interval with $m(\Lambda)<\infty$.

  * i)

If $\Lambda\subset(0,\infty)$, then

$\displaystyle\forall\,\alpha\in(0,1],\quad\forall\,x\in\Lambda,\quad\mathbf{E}_{x}[\tau_{\bar{X}}(\Lambda)]<\infty.$

  * ii)

Otherwise,

$\displaystyle\forall\,\alpha\in(0,1),\quad\forall\,x\in\Lambda,\quad\mathbf{E}_{x}[\tau_{\bar{X}}(\Lambda)]=\infty$

and

$\displaystyle\quad\forall\,x\in\Lambda,\quad\mathbf{E}_{x}[\tau_{X}(\Lambda)]<\infty.$

###### Proof.

For the solution $u$ to the problem (21) we observe that

$\displaystyle\lim_{\lambda\to
0}\tilde{u}(\lambda,x)=\int_{0}^{\infty}Q^{\dagger}_{t}f(x)\,dt+\lim_{\lambda\to
0}\tilde{u}(\lambda,0),\quad x\in[0,\infty)$

where

$\displaystyle\int_{0}^{\infty}Q^{\dagger}_{t}f(x)\,dt=\mathbf{E}_{x}\left[\int_{0}^{\tau_{0}}f(\bar{X}_{t})dt\right]=\mathbf{E}_{x}\left[\int_{0}^{\tau_{0}}f(X^{\dagger}_{t})dt\right]$

and $\tilde{u}(\lambda,0)$ is given in (4.1). We consider
$\Lambda=(0,\epsilon)$ with $\epsilon>0$ and $f=\mathbf{1}_{\Lambda}$. From
(52) we obtain

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{0}}\mathbf{1}_{\Lambda}(X^{\dagger}_{t})dt\right]=x\int_{0}^{\epsilon}dy+2x\int_{0}^{\epsilon}y\,dy=x\epsilon+x\epsilon^{2},\quad
x\in(0,\epsilon),\quad\forall\epsilon>0.$

From (67) we get

$\displaystyle\lim_{\lambda\to
0}\tilde{u}(\lambda,0)=(\sigma/c)\epsilon,\quad\forall\epsilon>0$

and therefore, we conclude that

$\displaystyle\mathbf{E}_{x}[\tau_{\bar{X}}(\Lambda)]=x\epsilon+x\epsilon^{2}+(\sigma/c)\epsilon<\infty\quad\forall\,x\in\Lambda\subset(0,\infty).$

For $\Lambda=(a,b)\subset(0,\infty)$ we use the fact that $\bar{X}$ behaves
like $X^{\dagger}$ and the invariance of $X^{\dagger}$ with respect to
translations. Thus, we write
$\mathbf{E}_{x-a}[\tau_{\bar{X}}(\Lambda^{\prime})]$ where
$\Lambda^{\prime}=(0,b-a)$. This concludes the proof of $i)$.

From (67) we also get, for $\Lambda=[0,\epsilon)$,

$\displaystyle\lim_{\lambda\to
0}\tilde{u}(\lambda,0)=(\sigma/c)\epsilon+(\eta/c)\lim_{\lambda\to
0}\frac{\lambda^{\alpha}}{\lambda},\quad\alpha\in(0,1]$

which is finite only in case $\alpha=1$. That is the case $\bar{X}=X$ and

$\displaystyle\lim_{\lambda\to
0}\tilde{u}(\lambda,0)=(\sigma/c)\,\int_{[0,\epsilon)}m(dx).$

This holds $\forall\,\Lambda$ bounded such that $\Lambda\ni\\{0\\}$ and proves
that $ii)$ holds true. ∎

######  Remark 4.3.

Consider $\Lambda=[0,\epsilon)$ with $\epsilon>0$. We observe that, for
$\tau_{X^{+}}(\Lambda)=\inf\\{t\,:\,X^{+}_{t}\notin\Lambda\\}$ and $\tau_{X}$
as above, after simple calculation,

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{X^{+}}(\Lambda)}\mathbf{1}_{\Lambda}(X^{+}_{t})dt\right]=\frac{\epsilon^{2}-x^{2}}{2}$

and

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{X}(\Lambda)}\mathbf{1}_{\Lambda}(X_{t})dt\right]=\frac{\epsilon^{2}-x^{2}}{2}+(\eta/\sigma)(\epsilon-x).$

Thus, the (average) extra time the process $X$ spends on $\\{0\\}$ is given by

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{X}(\Lambda)}\mathbf{1}_{\Lambda}(X_{t})dt\right]-\mathbf{E}_{x}\left[\int_{0}^{\tau_{X^{+}}(\Lambda)}\mathbf{1}_{\Lambda}(X^{+}_{t})dt\right]=(\eta/\sigma)(\epsilon-x),\quad
x\in\Lambda.$

We observe that

$\displaystyle\mathbf{E}[e_{0}]=\eta/\sigma$

and, for the local time $\gamma^{+}$ accumulated up to time
$\tau_{\epsilon}=\tau_{X^{+}}(\Lambda)$, from [12, Theorem 7.7],

$\displaystyle\mathbf{E}_{x}[\gamma^{+}\circ\tau_{\epsilon}]=\epsilon-x.$

We recall that $\gamma^{+}$ is the local time at zero of the process $X^{+}$
started at $X^{+}_{0}=x$. Thus, the extra time above is given by the holding
time of $X$ and satisfies the Wald identity

$\displaystyle\mathbf{E}_{x}[Holding\
time]=\mathbf{E}[e_{0}]\,\mathbf{E}_{x}[\gamma^{+}\circ\tau_{\epsilon}]=\mathbf{E}[e_{0}]\,\mathbf{E}_{x}\left[\int_{0}^{\tau_{\epsilon}}d\gamma^{+}_{t}\right],\quad
x\in\Lambda.$

######  Lemma 4.

Let $\Lambda$ be an interval with $m(\Lambda)<\infty$.

  * i)

If $\Lambda\subset(0,\infty)$, then

$\displaystyle\forall\,\alpha\in(0,1],\quad\forall\,x\in\Lambda,\quad\mu_{\bar{X}}(x,S)<\infty.$

  * ii)

Otherwise,

$\displaystyle\quad\forall\,x\in\Lambda,\quad\mu_{X}(x,S)<\infty\quad\textrm{and}\quad\forall\,\alpha\in(0,1),\quad\forall\,x\in\Lambda,\quad\mu_{\bar{X}}(x,S)\leq\infty.$

###### Proof.

$\forall x\in\Lambda$ with $\Lambda$ a bounded subsect of $(0,\infty)$,
$S\subset\Lambda$,

$\displaystyle\mu_{X}(x,S)=$
$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{X}(\Lambda)}\mathbf{1}_{S}(X_{s})ds\right]$
$\displaystyle\leq$
$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\infty}\mathbf{1}_{S}(X_{s})ds\right]$
$\displaystyle=$ $\displaystyle\lim_{\lambda\to
0}\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{S}(X_{s})ds\right]$

where (see formula (4.1))

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{S}(X_{s})ds\right]=:$ $\displaystyle
R_{\lambda}\mathbf{1}_{S}(x)$ $\displaystyle\leq$
$\displaystyle\frac{\sigma\,m(S)}{c+\eta\lambda+\sigma\sqrt{\lambda}}\to(\sigma/c)m(S)<\infty\quad\textrm{as}\quad\lambda\to
0$

where $m(S)\leq m(\Lambda)$. With the same arguments we obtain

$\displaystyle\mu_{\bar{X}}(x,S)=$
$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{\bar{X}}(\Lambda)}\mathbf{1}_{S}(\bar{X}_{s})ds\right]$
$\displaystyle\leq$ $\displaystyle\frac{\sigma
m(S)}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}\to(\sigma/c)m(S)<\infty\quad\textrm{as}\quad\lambda\to
0.$

Both limits above are obtained uniformly in $x\in[0,\infty)$ and this
concludes the proof of $i)$.

For $\Lambda\ni\\{0\\}$, for example $\Lambda=[0,\epsilon)$ with $\epsilon>0$,
we have that

$\displaystyle\mu_{\bar{X}}(x,\Lambda)\leq\mathbf{E}_{x}\left[\int_{0}^{\infty}\mathbf{1}_{\Lambda}(\bar{X}_{s})ds\right]=\lim_{\lambda\to
0}\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{\Lambda}(\bar{X}_{s})ds\right]$

where (see formula (66))

$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{\Lambda}(\bar{X}_{s})ds\right]=\frac{\sigma}{c+\lambda^{\alpha}\eta+\sigma\sqrt{\lambda}}\frac{1-e^{-\epsilon\sqrt{\lambda}}}{\sqrt{\lambda}}+\frac{\lambda^{\alpha}}{\lambda}\frac{\eta}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}$

and

$\displaystyle\lim_{\lambda\to
0}\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}\mathbf{1}_{\Lambda}(\bar{X}_{s})ds\right]=\lim_{\lambda\to
0}\frac{\lambda^{\alpha}}{\lambda}\frac{\eta}{c+\eta\lambda^{\alpha}+\sigma\sqrt{\lambda}}=\infty$

only in case $\alpha<1$ concluding the proof. ∎

######  Remark 4.4.

We notice that

$\displaystyle\forall\,S\subset\Lambda,\quad\mathbf{E}_{x}\left[\int_{0}^{\tau_{\bar{X}}(\Lambda)}\mathbf{1}_{S}(\bar{X}_{s})ds\right]\leq\mathbf{E}_{x}\left[\int_{0}^{\tau_{\bar{X}}(\Lambda)}\mathbf{1}_{\Lambda}(\bar{X}_{s})ds\right].$

The processes $X$ and $\bar{X}$ move along the path of $X^{+}$ according with
their clocks $V$ and $\bar{V}$. Assume $c=0$. By definition
$\bar{V}^{-1}_{t}\leq t$ and $V^{-1}_{t}\leq t$ a.s. and
$\forall\,x\in\Lambda\subset(0,\infty)$, $\bar{V}^{-1}_{t}=t$ and
$V^{-1}_{t}=t$ for $t<\tau_{0}$, thus

$\displaystyle\tau_{\bar{X}}(\Lambda)=\tau_{X}(\Lambda)=\inf\\{t\,:X^{+}_{t}\notin\Lambda\\}.$

Consider $\Lambda=(a,b)\ni x=X^{+}_{0}$ with $b>a>0$ and write

$\displaystyle\tau_{X}(\Lambda)=\inf\\{t\,:X^{+}_{t}<a\\}\wedge\inf\\{t\,:X^{+}_{t}>b\\}.$

Since

$\displaystyle\inf\\{t\,:X^{+}_{t}<a\\}=\inf\\{t\,:X^{\dagger}_{t}=a\\}=:\tau_{a}$

and

$\displaystyle\inf\\{t\,:X^{+}_{t}>b\\}=\inf\\{t\,:X^{\dagger}_{t}=b\\}=:\tau_{b},$

then we get

$\displaystyle\mu_{\bar{X}}(x,\Lambda)=$
$\displaystyle\mathbf{E}_{x}\left[\int_{0}^{\tau_{a}\wedge\tau_{b}}\mathbf{1}_{(a,b)}(X^{\dagger}_{t})dt\right]$
$\displaystyle=$ $\displaystyle\mathbf{E}_{x}[\tau_{a}\wedge\tau_{b}]$
$\displaystyle=$ $\displaystyle-\frac{(x-a)^{2}}{2}+\frac{(b-a)(x-a)}{2}.$

This provides an alternative proof of the first statement of the previous
Lemma.

###  4.3 On the holding times of $\bar{X}$

We say that $\\{e_{i}\\}_{i}$ is a sequence of exponential holding times for
$X$ meaning that, for $X_{0}=0$,

$\displaystyle\mathbf{P}_{0}(e_{i}>t|X_{e_{i}}>0)=e^{-(\sigma/\eta)t}.$

Despite the fact that $X$ is not strong Markov, the exponential law is
directly related with the fact that $X$ is Markov. Let $\\{\bar{e}_{i}\\}_{i}$
be the sequence of holding time for $\bar{X}$. The process $\bar{X}$ is
defined via time-change along the path of $X^{+}$. We have instantaneous
reflection at zero for $X^{+}$ and a delayed reflection for $\bar{X}$. Indeed,
the non-local boundary condition introduces a delaying effect due to the
process $L$.

A sequence of holding times (at the boundary point) for $\bar{X}$ is a
sequence of random variables, say $\\{\bar{e}_{i}\\}_{i}$, with distribution
given by

$\displaystyle\forall\,i,\quad\mathbf{P}_{0}(\bar{e}_{i}>t|\bar{X}_{\bar{e}_{i}}>0),\quad
t>0.$

Thus the process started from $\bar{X}_{0}=0$ leaves the boundary point after
an holding time $\bar{e}_{i}$. We have that $\bar{X}$ moves along the path of
$X^{+}$ which exhibits instantaneous reflection. The time change given by
$\bar{V}_{t}=t+H\circ(\eta/\sigma)\gamma^{+}_{t}$ is the clock with the extra
time $H\circ(\eta/\sigma)\gamma^{+}_{t}$ depending on the time the process
$X^{+}$ spend at zero up to time $t$. In particular $\bar{V}$ plays the same
role of $V$ for which the extra time $(\eta/\sigma)\gamma^{+}_{t}$ leads to
the holding time $e_{i}$, $i\in\mathbb{N}$ for $X$.

######  Theorem 4.3.

The sequence $\\{\bar{e}_{i}\\}_{i}$ of holding times for $\bar{X}$ is given
by i.i.d. Mittag-Leffler random variables.

###### Proof.

By definition, $\bar{e}_{0}$ is such that $\bar{X}_{t}=0$ for $0\leq
t<\bar{e}_{0}$ if $\bar{X}_{0}=0$. Focus on the case $\alpha=1$. Assume
$X_{0}=0$, then $X_{t}=0$ for $0\leq t<e_{0}$ by definition of holding time
$e_{0}$ for $X$. The process $V_{t}=t+(\eta/\sigma)\gamma^{+}_{t}$ has
continuous paths as well as the inverse process $V^{-1}_{t}$. As $\eta=0$ we
get $V_{t}=t=V^{-1}_{t}$ and the process $X=X^{+}$ has instantaneous
reflection. For $\eta>0$, the extra time $(\eta/\sigma)\gamma^{+}_{t}$ of
$V_{t}$ introduces the holding time $e_{0}$ via time change with $V^{-1}$. In
particular,

$\displaystyle\forall\,t\geq 0,\;V_{t}-t\geq 0\quad\textrm{and}\quad
V_{t}-t>0\quad\textrm{if}\quad 0<t<e_{0}$

and

$\displaystyle\mathbf{P}_{0}(0<V_{t}-t<e_{0})=\mathbf{P}_{0}((\eta/\sigma)\gamma^{+}_{t}<e_{0}).$

With $\tau_{0}=0$, let $\tau_{i}=\inf\\{t>\tau_{i-1}\,:\,X_{t}=0\\}$,
$i\in\mathbb{N}$ be the sequence of return times at zero for $X$ and define

$\displaystyle N_{t}=\max\\{i\,:\,\tau_{i}<t\\}.$

The time $t$ is a clock for $X^{+}$ as well as the time $V_{t}$ is a clock for
$\bar{X}$. The difference $V_{t}-t$ is given by the amount of time the process
$\bar{X}$ has been stopped at zero up to time $t$. That is,

$\displaystyle V_{t}-t=\sum_{i=0}^{N_{t}}e_{i}=\sum_{i\geq
0}e_{i}\,\mathbf{1}_{(i\leq N_{t})}=\sum_{i\geq
0}e_{i}\,\mathbf{1}_{(\tau_{i}<t)}$

and the first difference $e_{0}$ between the two clocks is obtained at the
first instantaneous visit of $X^{+}$ at zero. Moreover, we can write

$\displaystyle\bar{V}_{t}-t=\sum_{i\geq
0}\bar{e}_{i}\,\mathbf{1}_{(\bar{\tau}_{i}<t)}$

which gives $\bar{e}_{0}$ according with the previous case. Notice that
$\bar{X}_{0}=0$ and $\bar{\tau}_{0}=0$, $X$ and $\bar{X}$ move along the path
of $X^{+}$ and $\bar{X}=X$ for $\alpha=1$. Now we observe that

$\displaystyle V_{t}-t=(\eta/\sigma)\gamma^{+}_{t}=e_{0},\quad 0\leq
t<\tau_{1}$

and therefore

$\displaystyle\bar{V}_{t}-t=H\circ(\eta/\sigma)\gamma^{+}_{t}=\bar{e}_{0},\quad
0\leq t<\tau_{1}\quad\Rightarrow\quad\bar{V}_{t}-t=H\circ e_{0},\quad 0\leq
t<\tau_{1},$

that is, $\bar{e}_{0}=H\circ e_{0}$. Observe that $\tau_{1}$ is the excursion
time on $(0,\infty)$ for a Brownian motion, thus both processes $V_{t}-t$ and
$\bar{V}_{t}-t$ run over $[0,\tau_{1})$ respectively taking the values $e_{0}$
and $\bar{e}_{0}$. Since $e_{0}$ is and exponential r.v. we conclude that

$\displaystyle\mathbf{E}_{0}[e^{-\lambda H\circ
e_{0}}]=\int_{0}^{\infty}(\sigma/\eta)e^{-(\sigma/\eta)s}e^{-\lambda^{\alpha}s}ds=\frac{(\sigma/\eta)}{(\sigma/\eta)+\lambda^{\alpha}}$

and $H\circ e_{0}$ has a Mittag-Leffler distribution. Indeed,

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}E_{\alpha}(-(\sigma/\eta)t^{\alpha})dt=\frac{\lambda^{\alpha}}{\lambda}\frac{1}{(\sigma/\eta)+\lambda^{\alpha}},\quad\lambda>0$

and

$\displaystyle\mathbf{P}_{0}(H\circ
e_{0}>t)=E_{\alpha}(-(\sigma/\eta)t^{\alpha}).$

Since $e_{i}\sim e_{0}$ are i.i.d. random variables, then $H\circ e_{i}$ are
i.i.d. random variables. Indeed, $H$ is independent from $X$. Moreover, $H$ is
a Markov process for which

$\displaystyle H\circ e_{0}\perp H\circ e_{1}=H\circ(e_{0}+e_{1})-H\circ
e_{0}$

as well as $H\circ e_{i}\perp H\circ e_{k}$ $\forall\,i\neq k$. Thus, we
obtain the claim. ∎

######  Corollary 1.

The process $\bar{X}$ spends an infinite average amount of time at $\\{0\\}$
with each visit.

###### Proof.

It follows from the fact that $E_{\alpha}\notin L^{1}(0,\infty)$ and

$\displaystyle\forall\,i\in\mathbb{N}_{0},\quad\mathbf{E}[\bar{e}_{i}]=\int_{0}^{\infty}E_{\alpha}(-(\sigma/\eta)t^{\alpha})dt.$

∎

###  4.4 On the lifetime of $\bar{X}$

We provide some connection between the results obtained in [14] and the
results obtained in the present paper. Let us introduce

$\displaystyle\bar{H}_{t}=\frac{\sigma}{\eta}t+H^{\beta}_{t},\;t\geq
0\quad\textrm{with inverse}\quad\bar{L}_{t},\;t\geq 0$

where $H^{\beta}=\\{H^{\beta}_{t}\\}_{t\geq 0}$ is a $\beta$-stable
subordinator with inverse $L^{\beta}=\\{L^{\beta}_{t}\\}_{t\geq 0}$. We still
denote by $H$ and $L$ the $\alpha$-stable subordinator and its inverse. Let
$\chi$ be an exponential random variable such that
$\mathbf{P}(\chi>x)=e^{-cx}$, $x>0$. We also introduce $\chi(\beta)$ which is
a Mittag-Leffler random variable with
$\mathbf{P}(\chi(\beta)>x)=E_{\beta}(-cx^{\beta})$, $c>0$. It is well-know
that $\chi(1)=\chi$.

In [14] the FBVP (48) has been investigated for $\alpha=\beta/2\in(0,1/2]$ and
an alternative probabilistic representation has been considered. Such a
representation has been related to the fractional telegraph process. Let us
consider the lifetime

$\displaystyle\zeta\stackrel{{\scriptstyle law}}{{=}}\inf\\{t\geq
0\,:\,\gamma^{+}_{t}<\chi(\beta)\\}\quad\textrm{with }\beta/2=\alpha\leq 1/2.$

For $f=\mathbf{1}$ and $\sigma=0$, from [14] we known that

$\displaystyle
u(t,x)=\mathbf{P}_{x}(\zeta>t)=\mathbf{E}_{x}\left[\exp\left(-(c/\eta)L^{\beta}\circ\gamma^{+}_{t}\right)\right]=\mathbf{E}_{x}\left[E_{\beta}\big{(}-(c/\eta)(\gamma^{+}_{t})^{\beta}\big{)}\right]$

which is the case of the elastic sticky fractional condition. Moreover, the
special case $\alpha=1/2$ gives

$\displaystyle
u(t,x)=\mathbf{E}_{x}\left[\exp\left(-\frac{c}{\eta+\sigma}\gamma^{+}_{t}\right)\right]$

which is the case of the elastic condition. Thus, for $\alpha=1/2$ and
$\sigma=0$, we have that

$\displaystyle\zeta\stackrel{{\scriptstyle law}}{{=}}\inf\\{t\geq
0\,:\,\gamma^{+}_{t}<\chi\\}$

coincides with the lifetime of an elastic Brownian motion. This relates the
Neumann boundary condition to the fractional dynamic boundary condition with
$\alpha=1/2$.

######  Remark 4.5.

Focus on the case $\sigma=0$. We have

$\displaystyle u(t,x)=$
$\displaystyle\mathbf{E}_{x}\left[\exp\left(-(c/\eta)\,L^{\beta}\circ\gamma^{+}_{t}\right)\right]$
$\displaystyle=$
$\displaystyle\mathbf{E}_{x}\left[E_{\beta}\big{(}-(c/\eta)(\gamma^{+}_{t})^{\beta}\big{)}\right]$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}E_{\beta}(-(c/\eta)s^{\beta})\mathbf{P}_{x}(\gamma^{+}_{t}\in
ds)$

for which

$\displaystyle\frac{\partial u}{\partial t}(t,x)=$
$\displaystyle\int_{0}^{\infty}E_{\beta}(-(c/\eta)z^{\beta})\frac{\partial^{2}}{\partial
z^{2}}\frac{e^{-(x+z)^{2}/4t}}{\sqrt{4\pi t}}dz$ $\displaystyle=$
$\displaystyle-
E_{\beta}(-(c/\eta)z^{\beta})\frac{(x+z)}{2t}\frac{e^{-(x+z)^{2}/4t}}{\sqrt{4\pi
t}}\bigg{|}_{0}^{\infty}$
$\displaystyle+\int_{0}^{\infty}\frac{\partial}{\partial
z}E_{\beta}(-(c/\eta)z^{\beta})\frac{(x+z)}{2t}\frac{e^{-(x+z)^{2}/4t}}{\sqrt{4\pi
t}}dz$ $\displaystyle=$
$\displaystyle\frac{x}{2t}\frac{e^{-x^{2}/4t}}{\sqrt{4\pi
t}}+\int_{0}^{\infty}\frac{\partial}{\partial
z}E_{\beta}(-(c/\eta)z^{\beta})\frac{(x+z)}{2t}\frac{e^{-(x+z)^{2}/4t}}{\sqrt{4\pi
t}}dz.$

Thus,

$\displaystyle\bigg{|}\frac{\partial u}{\partial t}(t,0)\bigg{|}=$
$\displaystyle\bigg{|}\int_{0}^{\infty}\frac{\partial}{\partial
z}E_{\beta}(-(c/\eta)z^{\beta})\frac{z}{2t}\frac{e^{-z^{2}/4t}}{\sqrt{4\pi
t}}dz\bigg{|}$ $\displaystyle\leq$
$\displaystyle\int_{0}^{\infty}\bigg{|}\frac{\partial}{\partial
z}E_{\beta}(-(c/\eta)z^{\beta})\bigg{|}\frac{z}{2t}\frac{e^{-z^{2}/4t}}{\sqrt{4\pi
t}}dz.$

According with [24, equation (17)], we have that, for $\mathtt{C}>0$,

$\displaystyle\bigg{|}\frac{\partial u}{\partial t}(t,0)\bigg{|}\leq$
$\displaystyle(c/\eta)\int_{0}^{\infty}\mathtt{C}\,z^{\beta-1}\frac{z}{2t}\frac{e^{-z^{2}/4t}}{\sqrt{4\pi
t}}dz$ $\displaystyle=$
$\displaystyle[z=\sqrt{y}]=(c/\eta)\,\mathtt{C}\frac{1}{2}\frac{1}{2t}\frac{1}{\sqrt{4\pi
t}}\int_{0}^{\infty}y^{\beta/2+1/2-1}e^{-y/4t}dy$ $\displaystyle=$
$\displaystyle(c/\eta)\,\mathtt{C}\frac{1}{2}\frac{1}{2t}\frac{1}{\sqrt{4\pi
t}}(4t)^{\beta/2+1/2}\Gamma(\beta/2+1/2)$ $\displaystyle=$
$\displaystyle(c/\eta)\,\mathtt{C}\frac{1}{2t}\frac{1}{\sqrt{4\pi
t}}(4t)^{\beta/2+1/2}2^{1-\beta}\frac{\Gamma(\beta)}{\Gamma(\beta/2)}\sqrt{\pi}$
$\displaystyle=$
$\displaystyle(c/\eta)\,\mathtt{C}\,2^{-\beta}\frac{\Gamma(\beta)}{\Gamma(\beta/2)}t^{\beta/2-1}$

with $\beta/2\in(0,1/2]$. This agrees with the formula (65) which writes

$\displaystyle\bigg{|}\frac{\partial u}{\partial
t}(t,0)\bigg{|}\leq\mathcal{C}\,\|f\|_{\infty}\,\frac{\sqrt{(c/\eta)}}{2}t^{\alpha/2-1},\quad
t>0.$ (70)

Let us denote by $\bar{\zeta}$ the lifetime of $\bar{X}$. For the process
$\bar{V}_{t}=t+H\circ(\eta/\sigma)\gamma^{+}_{t}$ defined as above we provide
the following result.

######  Theorem 4.4.

For $\alpha=\beta/2\in(0,1/2)$,

$\displaystyle\mathbf{P}_{x}(\bar{\zeta}>t)=\mathbf{P}_{x}(\chi>\bar{L}\circ\gamma^{+}_{t})=\mathbf{P}_{x}(\chi>\gamma^{+}\circ\bar{V}^{-1}_{t}),\quad
t>0,\quad x\geq 0$ (71)

and

$\displaystyle\bar{\zeta}\stackrel{{\scriptstyle law}}{{=}}\inf\\{t\geq
0\,:\,\gamma^{+}_{t}<\bar{H}_{\chi}\\}.$ (72)

For $\alpha\in(0,1)$

$\displaystyle\mathbf{P}_{x}(\bar{\zeta}>t)=\mathbf{P}_{0}(H^{1/2}_{x}+H^{1/2}_{\chi}+(\eta/\sigma)^{1/\alpha}H^{\alpha}_{\chi}>t),\quad
t>0,\quad x\geq 0$ (73)

where $H^{1/2}_{x}$, $H^{1/2}_{\chi}$, $H^{\alpha}_{\chi}$, $\bar{X}$ are
independent.

###### Proof.

Focus on

$\displaystyle\mathbf{E}_{x}\left[\exp\left(-c\gamma^{+}\circ\bar{V}^{-1}_{t}\right)\right]=\int_{0}^{\infty}e^{-cs}\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}\in
ds).$

Let us consider

$\displaystyle\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}>s)=\mathbf{P}_{x}(\bar{V}^{-1}_{t}>\gamma^{+,-1}_{s})=\mathbf{P}_{x}(t>\bar{V}\circ\gamma^{+,-1}_{s}).$

Notice that $\gamma^{+}_{t}$ is non decreasing and therefore the process
$\gamma^{+,-1}$ may have jumps. Thus, $\gamma^{+,-1}\circ\gamma^{+}_{t}$ is a
little tricky whereas $\gamma^{+}\circ\gamma^{+,-1}_{t}=t$ almost surely. We
can therefore write

$\displaystyle\mathbf{P}_{x}(t>\bar{V}\circ\gamma^{+,-1}_{s})=$
$\displaystyle\mathbf{P}_{x}(t>\gamma^{+,-1}_{s}+(\eta/\sigma)^{1/\alpha}H_{s})$
$\displaystyle=$
$\displaystyle\mathbf{P}_{x}(t-(\eta/\sigma)^{1/\alpha}H_{s}>\gamma^{+,-1}_{s})$
$\displaystyle\displaystyle=$
$\displaystyle\int_{0}^{t}\mathbf{P}_{x}(t-z>\gamma^{+,-1}_{s})\mathbf{P}_{0}((\eta/\sigma)^{1/\alpha}H_{s}\in
dz)$

which is a convolution with Laplace transform

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}>s)dt$ (74) $\displaystyle=$
$\displaystyle\left(\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{x}(t>\gamma^{+,-1}_{s})dt\right)\left(\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}((\eta/\sigma)^{1/\alpha}H_{s}\in dt)\right)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{x}(\gamma^{+}_{t}>s)dt\,e^{-s\lambda^{\alpha}\eta/\sigma}$
$\displaystyle=$
$\displaystyle\int_{s}^{\infty}\frac{\sqrt{\lambda}}{\lambda}e^{-w\sqrt{\lambda}}dw\,e^{-s\lambda^{\alpha}\eta/\sigma}\,e^{-x\sqrt{\lambda}}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}e^{-s\sqrt{\lambda}-s\lambda^{\alpha}\eta/\sigma}\,e^{-x\sqrt{\lambda}},\quad\lambda>0.$
(75)

This implies that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}\in
ds)\,dt=\frac{\sqrt{\lambda}+\lambda^{\alpha}\eta/\sigma}{\lambda}e^{-s\sqrt{\lambda}-s\lambda^{\alpha}\eta/\sigma}\,e^{-x\sqrt{\lambda}}ds,\quad\lambda>0.$

Since, for the independent subordinators $H^{1/2}_{x}$, $H^{1/2}_{s}$,
$H^{\alpha}_{s}$,

$\displaystyle\frac{1}{\lambda}\mathbf{E}_{0}e^{-\lambda H^{1/2}_{x}-\lambda
H^{1/2}_{s}-\lambda(\eta/\sigma)^{1/\alpha}H^{\alpha}_{s}}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}(H^{1/2}_{x}+H^{1/2}_{s}+(\eta/\sigma)^{1/\alpha}H^{\alpha}_{s}<t)dt$

from (75), we get that

$\displaystyle\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}>s)=\mathbf{P}_{0}(H^{1/2}_{x}+H^{1/2}_{s}+(\eta/\sigma)^{1/\alpha}H^{\alpha}_{s}<t)$

and

$\displaystyle\mathbf{P}_{x}(\gamma^{+}\circ\bar{V}^{-1}_{t}<\chi)=$
$\displaystyle\mathbf{P}_{0}(H^{1/2}_{x}+H^{1/2}_{\chi}+(\eta/\sigma)^{1/\alpha}H^{\alpha}_{\chi}>t),\quad
t>0,\;x\geq 0.$

This concludes the proof. ∎

######  Remark 4.6.

From (73) we see that

$\displaystyle\mathbf{E}_{x}[\bar{\zeta}]=\mathbf{E}_{0}[H^{1/2}_{x}]+\mathbf{E}\Big{[}\mathbf{E}_{0}[H^{1/2}_{\chi}|\chi]+(\eta/\sigma)^{1/\alpha}\mathbf{E}_{0}[H^{\alpha}_{\chi}|\chi]\Big{]}$

is infinite for $\alpha\in(0,1]$ also in case $c>0$, that is the case of
elastic kill.

Acknowledgements The author thanks his institution for the support under the
Ateneo Grant 2020 and MUR for the support under PRIN 2022 - 2022XZSAFN:
Anomalous Phenomena on Regular and Irregular Domains: Approximating Complexity
for the Applied Sciences - CUP B53D23009540006.
Web Site: https://www.sbai.uniroma1.it/~mirko.dovidio/prinSite/index.html.

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