# Elastic drifted Brownian motions and non-local boundary conditions

Mirko D’Ovidio  Department of Basic and Applied Sciences for Engineering
Sapienza University of Rome
via A. Scarpa 10, Rome, Italy  and  Francesco Iafrate  Department of Basic and
Applied Sciences for Engineering
Sapienza University of Rome
via A. Scarpa 10, Rome, Italy<EMAIL_ADDRESS>
###### Abstract.

We provide a deep connection between elastic drifted Brownian motions and
inverses to tempered subordinators. Based on this connection, we establish a
link between multiplicative functionals and dynamical boundary conditions
given in terms of non-local equations in time. Indeed, we show that the
multiplicative functional associated to the elastic Brownian motion with drift
is equivalent to a multiplicative functional associated with fractional
boundary conditions of tempered type. By exploiting such connection we write
some functionals in terms of a simple (positive and non-decreasing) process.
In our view, such a representation is useful in many applications.

##  1\. Introduction

In this paper we focus on elastic drifted Brownian motions and their governing
equations equipped with fractional boundary conditions of the form

$\displaystyle\mathfrak{D}^{\Phi}_{t}\varphi(t,0)+c_{1}\,\varphi(t,0)=c_{2}$
(1.1)

where the constants $c_{1},c_{2}$ will be better specified and
$\mathfrak{D}^{\Phi}_{t}$ is a non-local operator characterized by the
Bernstein symbol $\Phi$. The constants $c_{1},c_{2}$ and the symbol $\Phi$
depend on the drift. In particular, for $\lambda\geq 0$,
$\Phi(\lambda)=\sqrt{\lambda+\eta}-\sqrt{\eta}$ where $\eta\geq 0$ will be
written in terms of the drift of the elastic Brownian motion. This symbol
$\Phi$ introduces the so-called tempered fractional derivative with tempering
parameter $\eta$ (see Section 3).

A first relevant fact is that the tempered fractional derivative turns out to
be strictly related with the infinitesimal generator of the drifted Brownian
motion. However, the condition (1.1) is more than a surprising relation
involving this generator. Indeed, we provide a deep connection between the
time-dependent boundary condition and the multiplicative functional associated
with the elastic drifted Brownian motion. In particular, (1.1) is associated
with an equivalent multiplicative functional which is written in terms of
tempered subordinators and their inverses. This entails a deep relation also
between subordinators and functionals of the Brownian motion.

A family $M=\\{M_{t}\\}_{t\geq 0}$ of real-valued random variable is called
multiplicative functional (of a given Markov process) provided: $i)$ $M_{t}$
is progressively measurable; $ii)$
$M_{t+s}=M_{t}(M_{s}\circ\theta_{t})=M_{s}(M_{t}\circ\theta_{s})$ a.s. for
each $t,s\geq 0$ ($\theta_{\alpha}$ is the translation operator); $iii)$
$0\leq M_{t}\leq 1$ for all $t\geq 0$ ([6, Chapter III]). It is well known
that two multiplicative functionals are equivalent if and only if they
generate the same semigroup ([6, Proposition 1.9]). In particular, the
multiplicative functional uniquely characterizes the semigroup ([6, Theorem
3.3]).

Let us consider the drift $\pm\mu$ with $\mu\geq 0$. For the elastic drifted
Brownian motion
$\widetilde{X}^{\pm\mu}=\\{\widetilde{X}^{\pm\mu}_{t}\\}_{t\geq 0}$ on
$[0,\infty)$ we can write

$\displaystyle\mathbf{E}_{x}[f(\widetilde{X}^{\pm\mu}_{t})]=\mathbf{E}_{x}[f(X^{\pm\mu}_{t})\,M^{\pm\mu}_{t}]$
(1.2)

where $X^{\pm\mu}=\\{X^{\pm\mu}_{t}\\}_{t\geq 0}$ is a reflecting Brownian
motion on $[0,\infty)$ and $M^{\pm\mu}_{t}$ is the multiplicative functional
associated with the elastic condition. Formula (1.2) gives the probabilistic
representation of the solution to

$\begin{cases}\displaystyle\frac{\partial u}{\partial t}=\pm\mu\frac{\partial
u}{\partial x}+\frac{\partial^{2}u}{\partial x^{2}},&x\geq 0,\,t>0\\\
\displaystyle u(0,x)=f(x),&x\geq 0\end{cases},\quad\mu\geq 0$ (1.3)

with (elastic) boundary condition

$\displaystyle\frac{\partial u}{\partial x}(t,0)=c_{\pm\mu}\,u(t,0),\quad t>0$
(1.4)

where

$\displaystyle c_{\mu}:=c_{0}-\frac{\mu}{2},\quad c_{0}\geq 0.$

We are interested in the equivalence between (1.1) and (1.4) for the Cauchy
problem (1.3). Thus, we focus on the lifetime of $\widetilde{X}^{\pm\mu}$ and
we provide some connections between the fractional boundary condition and the
corresponding process, that is a non negative and non decreasing process which
is an inverse to a tempered subordinator with symbol $\Phi$ with tempering
parameter $\eta=(\pm\mu/2)^{2}$.

###  1.1. Main results and plan of the work

First we provide some deep relations between elastic drifted Brownian motion
and an inverse to a tempered subordinator. Then, we prove equivalence between
the multiplicative functionals $M$ and $\overline{M}$ where the latter is
written in terms of an inverse to a tempered subordinator. This permits a very
fruitful change between the elastic drifted Brownian motion and a non-
decreasing process (the inverse to a subordinator) in studying the problem
(1.3) - (1.4).

In Section 2 and Section 3 we introduce the tempered subordinator $H$ and its
inverse $L$ together with non-local operators in time. In section 4 we
introduce the elastic drifted Brownian motion and the following equivalences
in law (Theorem 4.1 and Theorem 4.2)

$\displaystyle\max_{0\leq s\leq t}X^{\mu}_{s}\stackrel{{\scriptstyle
d}}{{=}}L_{t},\quad t\geq 0\quad\textrm{and}\quad\max_{0\leq s\leq
t}X^{-\mu}_{s}\stackrel{{\scriptstyle d}}{{=}}L_{t}\wedge T_{\mu},\quad t\geq
0$ (1.5)

where $T_{\mu}$ is an independent exponential random variable.

In Section 5 we discuss an intuitive example in case of zero drift. This
corresponds to the case of stable subordinator (indeed $\eta=0$) and
therefore, the Caputo derivative is involved.

In Section 6 we confirm the relations discussed above in (1.5) in terms of
boundary value problems. Indeed, the process $L_{t}$ (associated to the
problem (2.17)) is related with $X^{\pm\mu}$ (in terms of formulas (1.5)) as
well as the condition (1.4) is related with (1.1) in the domain
$D(G_{\pm\mu})$. In particular, we are concerned with the solution

$\displaystyle u(t,x)=\mathbf{E}_{x}[\overline{M}_{t}^{\pm\mu}]$

to the problem

$\begin{cases}\displaystyle\frac{\partial u}{\partial t}=\pm\mu\frac{\partial
u}{\partial x}+\frac{\partial^{2}u}{\partial x^{2}},&x\geq 0,\,t>0\\\
\displaystyle u(0,x)=\mathbf{1}(x\geq 0),&x\geq 0\end{cases},\qquad\mu\geq 0$

with the boundary condition (1.1) where $c_{1}$ characterizes an inverse to a
subordinator and $c_{2}$ characterizes the exponential random variable
$T_{\mu}$. The non-local boundary condition (1.1) takes the following forms:

  * -

In case of positive drift $\mu$ (Theorem 6.1, Theorem 6.2 and Theorem 6.3),

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+(\mu+c_{\mu})\,u(t,0)=\mu,\quad
t>0;$ (1.6)

  * -

In case of negative drift $-\mu$ (Theorem 6.4 and Theorem 6.5),

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+c_{-\mu}\,u(t,0)=0,\quad
t>0;$ (1.7)

  * -

In case of zero drift $\mu=0$ (as a by-product of the previous theorems),

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+c_{0}\,u(t,0)=0,\quad
t>0,$ (1.8)

where $\mathfrak{D}^{\frac{1}{2},\mu}_{t}$ denotes the tempered Caputo
derivative defined in Section 3. First we show that (1.4) is equivalent to
(1.6) and (1.7). That is, $\overline{M}^{\pm\mu}_{t}$ is equivalent to the
multiplicative functional $M^{\pm\mu}_{t}$ associated with the elastic
condition for a drifted Brownian motion. Moreover, we show that
$\overline{M}^{\pm\mu}_{t}$ can be written in terms of inverses to tempered
stable subordinators $L_{t}$ and the exponential r.v. $T_{\mu}$ for which
$\mathbf{P}(T_{\mu}>x)=e^{-\mu x}$ and $T_{0}=\infty$ with probability $1$. We
show that $c_{2}=\mu$ in (1.6) introduces $T_{\mu}$ with $\mu>0$ whereas,
$c_{2}=0$ in (1.7) introduces $T_{0}$. If $\mu=0$, then we obtain the elastic
Brownian motion with elastic coefficient $c_{0}$. The corresponding boundary
condition is therefore given by (1.8).

Such results highlight the following facts.

  * i)

Equivalence between boundary conditions: the generator of a drifted Brownian
motion appears to be intimately connected with the (time) tempered fractional
derivative;

  * ii)

Equivalence between multiplicative functionals: the elastic drifted Brownian
motion and the tempered subordinator are intimately related. In particular,
some functionals of $\widetilde{X}^{\pm\mu}$ can be written in terms of
$L_{t}$ with tempering parameter $\eta=(\pm\mu/2)^{2}$ and $T_{\mu}$ with
$\mu\geq 0$.

###  1.2. Motivations and discussion of the results

Our aim is to consider an alternative formulation of the boundary conditions
based on the time-derivatives. Such a formulation relies on the fact that the
multiplicative functional charactering the semigroup can be also described by
an operator in time. Actually, we obtain that time-dependent (or dynamical)
boundary conditions characterize uniquely such a class of functionals. The
present work has been inspired by [35, 36] where the authors have obtained a
beautiful characterization of the sticky Brownian motion in terms of a time-
dependent boundary condition. We have been also moved by the fundamental
awareness that fractional powers of operators (and therefore non-local
operators) are strictly related with their local higher-order counterparts,
when they exist (as discussed in [14]). A recent result concerning the Caputo-
Dzherbashian derivative has been given in [13]. The case
$\Phi(\lambda)=\lambda$ corresponds to the ordinary derivative, in this case
the dynamical boundary condition have a clear physical interpretation (see
[20]). As far as we known, in the literature there is no mention about the
physical meaning of non-local time-dependent boundary conditions. This is far
from our scope at this stage of the research. Moreover, to the best of our
knowledge, there are no works dealing with non-local dynamical boundary
conditions. From the probabilistic point of view, beside the interesting
connection between processes ($X^{\pm\mu}_{t}$, $L_{t}$ and $L_{t}\wedge
T_{\mu}$), the arguments in points i), ii) above turn out to be evidently
useful in applications, simulation and numerical methods.

##  2\. Non-local operators and random times

We introduce the processes $L_{t}$ with $\lambda$-potentials

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\,f(L_{t})\,dt\right]=\frac{\Phi(\lambda)}{\lambda}\int_{0}^{\infty}e^{-s\,\Phi(\lambda)}\,f(s)\,ds$
(2.1)

where the symbol $\Phi$ is a Bernstein function uniquely characterized by the
measure $\Pi$ as follows

$\displaystyle\Phi(\lambda)=\int_{0}^{\infty}(1-e^{-s\lambda})\,\Pi(ds).$
(2.2)

It is well-known that the process $L_{t}$ can be regarded as the inverse to a
subordinator with symbol $\Phi$, say $H_{t}$, for which we have

$\displaystyle\mathbf{E}[e^{-\lambda
H_{t}}]=e^{-t\Phi(\lambda)},\quad\lambda>0,\quad t>0$ (2.3)

and the relation $\mathbf{P}_{0}(L_{t}<s)=\mathbf{P}_{0}(H_{s}>t)$ holds true.
In this context, the measure $\Pi$ is termed Lévy measure of $H_{t}$. Both
processes are random times in the sense that they are non-negative and non-
decreasing. The subordinator $H_{t}$ may have jumps, thus the inverse $L_{t}$
defined as

$\displaystyle L_{t}:=\inf\\{s>0\,:\,H_{s}>t\\},\quad t>0$ (2.4)

is a continuous process with non-decreasing paths. We also assume that
$H_{0}=0$ and $L_{0}=0$. We denote by $\ell$ and $h$ the density of $L_{t}$
and $H_{t}$ respectively, that is

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

As usual, $\mathbf{P}_{x}$ denotes the probability measure of the process
started at $x$. We notice that, by definition of inverse process, $L_{t}$ is
the first exit time of $H_{t}$ from the interval $(0,t)$. Since $H_{t}$ has
strictly increasing paths with jumps (we are not including the case
$\Pi((0,\infty))<\infty$, the Poisson case for instance) the process $L_{t}$
has continuous paths with plateaus. This is an interesting aspects introducing
the concept of delayed and rushed motions for time-changed processes ([8]).

We provide the following result which will be useful further on.

######  Proposition 2.1.

Let $\theta>0$ be fixed. Let $\Phi$ be the symbol defined in (2.2). Then, for
$x\geq 0$,

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\frac{1-e^{-\theta(L_{t}-x)}}{\theta}\,\mathbf{1}_{(L_{t}\geq
x)}\right]dt=\frac{1}{\lambda}\frac{1}{\theta+\Phi(\lambda)}e^{-x\,\Phi(\lambda)},\quad\lambda>0$
(2.5)

and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}[e^{-\theta(L_{t}-x)}\mathbf{1}_{(L_{t}\geq
x)}]\,dt=\frac{\Phi(\lambda)}{\lambda}\frac{1}{\theta+\Phi(\lambda)}e^{-x\,\Phi(\lambda)},\quad\lambda>0$
(2.6)

hold true.

###### Proof.

First we notice that

$\displaystyle\int_{0}^{t}h(s,x)dx=\mathbf{P}_{0}(H_{s}\leq
t)=\mathbf{P}_{0}(L_{t}\geq s)=\int_{s}^{\infty}\ell(t,x)dx,\qquad t>0,\;s>0.$

Formula (2.5) can be obtained by considering the Tonelli’s theorem and the
fact that

$\displaystyle\frac{1}{\lambda}\frac{1}{\theta+\Phi(\lambda)}e^{-x\,\Phi(\lambda)}=$
$\displaystyle\frac{1}{\lambda}\int_{0}^{\infty}e^{-w(\theta+\Phi(\lambda))-x\Phi(\lambda)}\,dw$
$\displaystyle=$ $\displaystyle\left[\textrm{by \eqref{symbH}}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\int_{0}^{\infty}e^{-w\theta}\mathbf{E}_{0}[e^{-\lambda
H_{w+x}}]\,dw$ $\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{\infty}e^{-w\theta}\left[\int_{0}^{t}h(w+x,s)ds\right]\,dw\,dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{\infty}e^{-w\theta}\,\mathbf{P}_{0}(H_{w+x}\leq t)\,dw\,dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{\infty}e^{-w\theta}\,\mathbf{P}_{0}(L_{t}\geq w+x)\,dw\,dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-w\theta}\,\mathbf{1}_{(L_{t}-x\geq
w)}\,dw\right]dt$ $\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\left(\int_{0}^{L_{t}-x}e^{-w\theta}dw\right)\mathbf{1}_{(L_{t}\geq
x)}\right]dt$ $\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\frac{1-e^{-\theta(L^{\Phi}_{t}-x)}}{\theta}\,\mathbf{1}_{(L_{t}\geq
x)}\right]dt,\quad\lambda>0.$

Formula (2.6) immediately follows from (2.1). ∎

The non-local operator associated with $H_{t}$ is given by (Bochner-Phillips)

$\displaystyle-\Phi(-\partial_{x})\varphi(x):=\int_{0}^{\infty}\left(\psi(x)-\psi(x-s)\right)\Pi(ds),\quad
x\geq 0.$ (2.7)

Indeed, from (2.2), the Laplace transform of the right-hand side of (2.7)
gives

$\displaystyle\left(\int_{0}^{\infty}(1-e^{-\lambda
s})\,\Pi(ds)\right)\widetilde{\psi}(\lambda)=\Phi(\lambda)\,\widetilde{\psi}(\lambda)$

for a function $\psi$ compactly supported on the positive real line. Thus, in
the Laplace analysis, the symbol $\Phi$ turns out to be the multiplier of the
operator (2.7). Formula (2.7) recall the definition of fractional derivative
given by Marchaud, thus we may refer to (2.7) as a Marchaud (type) operator
(the definition coincides in case of stable subordinator, that is for
$\Phi(\lambda)=\lambda^{\alpha}$). An interesting discussion about the
comparison between fractional derivatives has been given in [18]. The Riemann-
Liouville (type) operator is therefore written for a general symbol $\Phi$ as

$\displaystyle\mathcal{D}^{\Phi}_{x}\psi(x):=\frac{d}{dx}\int_{0}^{x}\psi(x-s)\,\overline{\Pi}(s)\,ds$

where $\overline{\Pi}(s)=\Pi((s,\infty))$ is the tail of $\Pi$. We can check
that the symbol $\Phi$ still plays the role of multiplier for this operator,
that is

$\displaystyle\int_{0}^{\infty}e^{-\lambda
x}\mathcal{D}^{\Phi}_{x}\psi(x)\,dx=\Phi(\lambda)\,\widetilde{\psi}(\lambda).$
(2.8)

We now introduce the time fractional operator we will deal with further on.
Let $N>0$ and $n\geq 0$. Let $\mathcal{N}_{\omega}$ be the set of (piecewise)
continuous function on $[0,\infty)$ of exponential order $\omega$ such that
$|\psi(t)|\leq Ne^{\omega t}$. Denote by $\widetilde{\psi}$ the Laplace
transform of $\psi$. Then, we define the operator
$\mathfrak{D}^{\Phi}_{t}:\mathcal{N}_{\omega}\mapsto\mathcal{N}_{\omega}$ as
the Caputo (type) operator for which

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}^{\Phi}_{t}\psi(t)\,dt=\Phi(\lambda)\,\widetilde{\psi}(\lambda)-\frac{\Phi(\lambda)}{\lambda}\psi_{0},\quad\lambda>\omega.$
(2.9)

This immediately introduces the definition

$\displaystyle\mathfrak{D}^{\Phi}_{t}\psi:=\mathcal{D}^{\Phi}_{t}\psi-\Pi((t,\infty))\psi_{0}=\mathcal{D}^{\Phi}_{t}\left(\psi-\psi_{0}\right)$
(2.10)

where we have used formula (2.8) and the well-known fact ([5, Section 1.2])

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\,\Pi((t,\infty))\,dt=\frac{\Phi(\lambda)}{\lambda}.$ (2.11)

The identity $\mathcal{D}^{\Phi}_{t}\psi_{0}=\Pi((t,\infty))\psi_{0}$ follows
from the definition of $\mathcal{D}^{\Phi}_{t}$. Since $\psi$ is exponentially
bounded, the integral $\widetilde{\psi}$ is absolutely convergent for
$\lambda>\omega$. Since
$\Phi(\lambda)\widetilde{\psi}(\lambda)-\Phi(\lambda)/\lambda\,u(0)=\left(\lambda\widetilde{\psi}(\lambda)-\psi(0)\right)\Phi(\lambda)/\lambda$,
then $\mathfrak{D}^{\Phi}_{t}$ can be written as a convolution involving the
ordinary derivative $\psi^{\prime}$ and the tail $\Pi((t,\infty))$ iff
$\psi\in\mathcal{N}_{\omega}\cap C((0,\infty),\mathbb{R}_{+})$ and
$\psi^{\prime}\in\mathcal{N}_{\omega}$. In particular,

$\displaystyle\mathfrak{D}^{\Phi}_{t}\psi(t)=\int_{0}^{t}\psi^{\prime}(t-s)\,\overline{\Pi}(s)\,ds.$
(2.12)

By Young’s inequality for convolution and formula (2.11) we have that

$\displaystyle\int_{0}^{\infty}|\mathfrak{D}^{\Phi}_{t}\psi|^{p}dt\leq\left(\int_{0}^{\infty}|\psi^{\prime}|^{p}dt\right)\left(\lim_{\lambda\downarrow
0}\frac{\Phi(\lambda)}{\lambda}\right)^{p},\qquad p\in[1,\infty)$ (2.13)

where

$\displaystyle\lim_{\lambda\downarrow
0}\frac{\Phi(\lambda)}{\lambda}=\frac{d\Phi}{d\lambda}(\lambda)\bigg{|}_{\lambda=0}$
(2.14)

is finite only in some cases. The limit (2.14) will be considered again
further on and it is related with the mean value of the subordinator $H_{t}$.
Indeed, from (2.3) and the fact that $\Phi(0)=0$,

$\displaystyle\mathbf{E}_{0}[H_{t}]=-\frac{d\Phi}{d\lambda}(\lambda)\bigg{|}_{\lambda=0}.$

We notice that when $\Phi(\lambda)=\lambda$ (that is we deal with the ordinary
derivative $D_{t}$) the equality holds true (2.13) and $H_{t}=t$, $L_{t}=t$
almost surely. Some further representations of $\mathfrak{D}^{\Phi}_{t}$ in
terms of the tails of a Lévy measure $\Pi((t,\infty))$ have been given in the
recent works [11, 37] and previously in [25].

Assuming that

$\displaystyle\lim_{\lambda\downarrow 0}\frac{\Phi(\lambda)}{\lambda}<\infty,$
(2.15)

formula (2.13) says that we are looking for $\psi\in C((0,\infty))$ with
$\psi^{\prime}\in L^{1}((0,\infty))$. Thus, the minimal requirement is that
$\psi\in AC((0,\infty))$. As usual, we denote by $AC((0,\infty))$ the set of
absolutely continuous functions on $(0,\infty)$. In particular, $\psi\in
AC((0,\infty))$ if $\psi\in C((0,\infty))$ and $\psi^{\prime}=\varrho\in
L^{1}((0,\infty))$, that is we can write

$\displaystyle\psi(t)=\psi(0)+\int_{0}^{t}\varrho(s)ds.$ (2.16)

Let us denote by $C_{b}((0,\infty))$ the set of smooth and bounded functions
on $(0,\infty)$. In order to give a clear picture about the operator (2.10),
under the assumption (2.15), we now address the problem to find $\rho(t,x)$
such that $\rho\in C^{1,1}((0,\infty),(0,\infty);(0,\infty))$ and
$\forall\,x>0$, $\rho(\cdot,x)\in AC((0,\infty))$ solving

$\left\\{\begin{array}[]{ll}\displaystyle\mathfrak{D}^{\Phi}_{t}\rho(t,x)=-\frac{\partial\rho}{\partial
x}(t,x),\quad t>0,\;x>0,\\\ \displaystyle\rho(0,x)=f(x)\in
C_{b}([0,\infty)),\\\ \displaystyle\rho(t,0)=0,\quad t>0.\end{array}\right.$
(2.17)

Then, there is a (classical) solution

$\displaystyle\rho\in C^{1,1}(AC((0,\infty)),(0,\infty);(0,\infty))$ (2.18)

with probabilistic representation

$\displaystyle\rho(t,x)=\mathbf{E}_{0}[f(x-L_{t})\mathbf{1}_{(t<H_{x})}]$

where $L_{t}$ is an inverse to a subordinator $H_{t}$ with symbol $\Phi$. As
we can see $\forall\,x>0$, $\rho(\cdot,x)\notin L^{1}((0,\infty))$. We can
easily verify such results. Let us denote by
$\widehat{\rho}(t,\xi)=\int_{0}^{\infty}e^{-\xi x}\rho(t,x)\,dx$ and
$\widetilde{\rho}(\lambda,x)=\int_{0}^{\infty}e^{-\lambda t}\rho(t,x)\,dt$ the
Laplace transforms w.t. to the variables $x$ and $t$ respectively. Let
$\widehat{\widetilde{\rho}}(\lambda,\xi)$ be the double Laplace transform.
With (2.9) at hand, from the problem (2.17) we write

$\displaystyle\Phi(\lambda)\,\widehat{\widetilde{\rho}}(\lambda,\xi)-\frac{\Phi(\lambda)}{\lambda}\widehat{f}(\xi)=-\xi\widehat{\widetilde{\rho}}(\lambda,\xi)$

from which

$\displaystyle\widehat{\widetilde{\rho}}(\lambda,\xi)=\frac{\Phi(\lambda)}{\lambda}\frac{1}{\xi+\Phi(\lambda)}\widehat{f}(\xi),\quad\lambda>0,\;\xi>0.$

From Proposition 2.1 we get that

$\displaystyle\rho(t,x)=\int_{0}^{x}f(y)\,\ell(t,x-y)\,dy=\mathbf{E}_{0}[f(x-L_{t})\mathbf{1}_{(L_{t}<x)}].$

The Laplace machinery gives uniqueness. The probabilistic representation
follows by considering (2.4).

If the strong assumption (2.15) does not hold, then we have to ask for

$\displaystyle\varrho^{\prime}(t-s)\overline{\Pi}(s)\in
L^{1}((0,t)),\quad\forall\,t>0.$

Despite the minimal requirement (2.18) we notice that $\ell(\cdot,x)\in
C^{\infty}((0,\infty))$ for any $x>0$. It suffices to consider, for a given
$x>0$, the function

$\displaystyle R_{n}(\lambda)=\lambda^{n}\int_{0}^{\infty}e^{-\lambda
t}\ell(t,x)\,dt=\lambda^{n}\frac{\Phi(\lambda)}{\lambda}e^{-x\Phi(\lambda)},\quad\lambda>0,\quad
n\in\mathbb{N}_{0}.$

Since $\Phi$ is a Bernstein function with $\Phi(0)=0$, we get that

$\displaystyle\lim_{\lambda\to
0}R_{n}(\lambda)=0,\quad\lim_{\lambda\to\infty}R_{n}(\lambda)=0,\quad\forall\,n\in\mathbb{N}.$

This also prove that $\ell(\cdot,x)\notin L_{1}((0,\infty))$ for any $x>0$ as
stated before.

Furthermore, we only notice that the kernel $\ell$ can be uniquely determined
as the solution to the problem

$\left\\{\begin{array}[]{ll}\displaystyle\mathcal{D}^{\Phi}_{t}\ell(t,x)=-\frac{\partial\ell}{\partial
x}(t,x),\quad t>0,\;x>0,\\\ \displaystyle\ell(0,x)=\delta(x)\\\
\displaystyle\ell(t,0)=\Pi((t,\infty)),\end{array}\right.$ (2.19)

where $\delta$ is the Dirac function and the derivative (2.7) is considered in
place of (2.10). The Laplace technique can by applied as before by considering
the formula (2.11). The problem (2.19) has been investigated in [37]. In the
literature very often this equations are confused in the sense that, only the
first one can be written in terms of the Caputo type derivative. Sometimes the
boundary condition is omitted. Below we are interested in a kind of fractional
relaxation equation based on (2.17).

##  3\. Tempered fractional calculus

From now on we focus on the symbol

$\displaystyle\Phi(\lambda)=\sqrt{\lambda+\eta}-\sqrt{\eta},\quad\lambda\geq
0$ (3.1)

corresponding to the Lévy measure

$\displaystyle\Pi(ds)=\frac{1}{2}\frac{1}{\sqrt{\pi}}\frac{e^{-\eta
s}}{s^{\frac{1}{2}+1}}ds,\quad\eta>0.$ (3.2)

We recall that the corresponding subordinator $H_{t}$ is the tempered (also
termed relativistic) stable subordinator of order $\frac{1}{2}$. The measure
of a tempered stable processes can be obtained by multiplying the Lévy measure
of an $\alpha$-stable process by a decreasing exponential. The parameter
$\eta>0$ controls the level of tempering. The effect is to reduce the
intensity of large jumps keeping the structure of small jumps. The resulting
process has finite moments of all order and at the same time, it has an
infinite amount of (small) jumps in any finite time interval. For these
reasons these models are widely studied, see e.g. [10] for applications in
mathematical finance or or [29] and references therein for applications to
hydrology problems. Anomalous diffusion with tempered operators were
considered in [9], while a general theory for tempering stable processes was
presented in [32].

Figure 1 compares the sample paths of a stable subordinator and of a tempered
stable subordinator, showing that the presence of the tempering parameter
reduces the number of larger jumps.

\begin{overpic}[width=346.89731pt]{plot-tempered-0001.pdf} \end{overpic} (a)
sample path of $H_{t},\eta=0$

\begin{overpic}[width=346.89731pt]{plot-tempered-01.pdf} \end{overpic} (b)
sample path of $H_{t},\eta>0$

Figure 1. Comparison of a sample path of a stable subordinator ($\eta=0$) and
a tempered stable subordinator ($\eta>0$). Both pictures must be interpreted
with jumps in place of vertical lines. The paths are not continuous.

The Caputo (type) tempered derivative is given by

$\displaystyle\mathfrak{D}^{\frac{1}{2},\eta}_{t}\psi(t)=\int_{0}^{t}\psi^{\prime}(s)\overline{\Pi}(t-s)ds$
(3.3)

where $\overline{\Pi}(z)=\Pi((z,\infty))$ is the tail of the Lévy measure
$\Pi$ given in (3.2). From (2.13), we obtain that

$\displaystyle\big{\|}\mathfrak{D}^{\frac{1}{2},\eta}_{t}\psi\big{\|}_{L^{1}}\leq\frac{1}{\sqrt{2\eta}}\,\|\psi^{\prime}\|_{L^{1}}$
(3.4)

which may be of interest only if $\eta\neq 0$. It is well known that, for
$\eta=0$,

$\ell(t,x)=2e^{-\frac{x^{2}}{4t}}/\sqrt{4\pi t},\quad t>0,\,x>0.$

The symbol (3.1) for $\eta=0$ introduces the following derivatives:

  * •

the Riemann-Liouville derivative

$\displaystyle\mathcal{D}^{\frac{1}{2}}_{t}\psi(t)=\frac{1}{\sqrt{\pi}}\frac{d}{dt}\int_{0}^{t}\frac{\psi(s)}{\sqrt{t-s}}ds$

  * •

the Caputo-Djrbashian derivative

$\displaystyle\mathfrak{D}^{\frac{1}{2}}_{t}\psi(t)=\frac{1}{\sqrt{\pi}}\int_{0}^{t}\frac{\psi^{\prime}(s)}{\sqrt{t-s}}ds$

where $\psi^{\prime}=d\psi/ds$.

We recall the $\lambda$-potential

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\,f(L_{t})\,dt\right]=\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{\lambda}\int_{0}^{\infty}e^{-s(\sqrt{\lambda+\eta}-\sqrt{\eta})}\,f(s)\,ds$
(3.5)

(which can be obtained as special case of the formula (2.1) by considering the
Lévy measure (3.2)) and the following formula

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}^{\frac{1}{2},\eta}_{t}\psi(t)\,dt=(\sqrt{\lambda+\eta}-\sqrt{\eta})\widetilde{\psi}-\frac{1}{\lambda}(\sqrt{\lambda+\eta}-\sqrt{\eta})\psi_{0}$
(3.6)

which will be useful in the subsequent discussion. The interested reader can
consult for example [3, 15, 29] for further discussions on this operator and
tempered processes.

In the following we consider $\eta=\frac{\mu^{2}}{4}$ as a tempering
parameter. Thus in order to streamline the notation as much as possible we
write

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}\psi(t)=\mathfrak{D}^{\frac{1}{2},\eta}_{t}\psi(t),\quad\textrm{with
}\;\eta=\frac{\mu^{2}}{4},\;\mu>0$ (3.7)

######  Proposition 3.1.

Let $a,b$ be two positive constants. The unique continuous solution on the
interval $I\subseteq[0,\infty)$ of the fractional tempered relaxation equation

$\left\\{\begin{array}[]{ll}\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}r(t)+a\,r(t)=b,&t>0\\\
\displaystyle r(0)=c\in\\{0,1\\}&\end{array}\right.$ (3.8)

is given by

$\displaystyle r(t)=\frac{b}{a}\,\mathbf{P}_{0}(L_{t}\geq
T_{a})+c\,\mathbf{P}_{0}(L_{t}<T_{a})$

where $T_{a}$ is an exponential random variable (with parameter $a$)
independent from $L_{t}$ which is the inverse process with symbol (3.1).

If $c=1$, then $[0,\infty)=I\ni t\mapsto r(t)$ has the following properties:

  * i)

$b>a\;\Rightarrow\;r(t)>1$;

  * ii)

$b<a\;\Rightarrow\;r(t)<1$;

  * iii)

$b=a\;\Rightarrow\;r(t)=1$.

If $c=0$, then $r(t)$ has the following properties:

  * iv)

$b>a>0$ $\Rightarrow\,\exists\,t_{b}:r(t)\leq 1\;\textrm{ if }\;t\in
I=[0,t_{b})$;

  * v)

$b\leq a\,\Rightarrow\,r(t)\leq 1,\;t\in I=[0,\infty)$.

Moreover, $\forall\,a,b,c$ the solution $t\mapsto r(t)$ is monotone with
$r(0)=c$ and $r(t)\to b/a$ as $t\to\infty$.

###### Proof.

From (3.6), by Laplace techniques we obtain

$\displaystyle\widetilde{r}(\lambda)=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda t}r(t)\,dt,\quad\lambda\geq 0$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}\frac{b+c\sqrt{\lambda+\eta}-\sqrt{\eta}}{a+\sqrt{\lambda+\eta}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{b}{\lambda}\frac{1}{a+\sqrt{\lambda+\eta}-\sqrt{\eta}}+c\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{\lambda}\frac{1}{a+\sqrt{\lambda+\eta}-\sqrt{\eta}}.$

We recall that $\Phi(\lambda)=\sqrt{\lambda+\eta}-\sqrt{\eta}$ is a completely
monotone function for which $\Phi(0)=0$ and $\Phi(\lambda)\to\infty$ as
$\lambda\to\infty$. We immediately see that $r(t)$ is a non-negative solution.

From Proposition 2.1 with $\Phi(\lambda)=\sqrt{\lambda+\eta}-\sqrt{\eta}$ we
write

$\displaystyle r(t)=$
$\displaystyle\frac{b}{a}\,\mathbf{E}_{0}[1-e^{-aL_{t}}]+c\,\mathbf{E}_{0}[e^{-aL_{t}}]$
$\displaystyle=$ $\displaystyle\frac{b}{a}\,\mathbf{P}_{0}(L_{t}\geq
T_{a})+c\,\mathbf{P}_{0}(L_{t}<T_{a})$ $\displaystyle=$ $\displaystyle
c+\left(\frac{b}{a}-c\right)\mathbf{P}_{0}(L_{t}\geq T_{a}).$

Let us consider $c=1$. For $a=b$, if follows that
$r(t)=\mathbf{P}_{0}(T_{a}\in[0,\infty))=1$ $\forall\,t>0$. Moreover, there
exist $\varepsilon_{1}=\varepsilon_{1}(t)\geq 0$ and
$\varepsilon_{2}=\varepsilon_{2}(t)\in[0,1)$ such that, $\forall\,t>0$

$r(t)=\left\\{\begin{array}[]{ll}\displaystyle
1+\varepsilon_{1},&\textrm{if}\;b>a\\\ \displaystyle
1-\varepsilon_{2},&\textrm{if}\;b<a\end{array}\right..$ (3.9)

Since $\mathbf{P}_{0}(0=L_{0}\leq T_{a})=0$ we recover the initial condition
$r(0)=1$.

Now we focus on $c=0$.

  * ($b<a$):

Since $r(t)=(b/a)\mathbf{P}_{0}(L_{t}\geq T_{a})$, $r(t)\leq 1$ follows
immediately for $b<a$. That is,

$\displaystyle\widetilde{r}(\lambda)\leq\frac{1}{\lambda}\frac{a}{a+\sqrt{\lambda+\eta}-\sqrt{\eta}}\leq\frac{1}{\lambda}$

from which we deduce that $r(t)\leq 1$ for any $t$ with $0\leq a<b<\infty$;

  * ($b=a$):

If $b=a$ and $a\to 0$, then $r(t)\to 0$ $\forall\,t\geq 0$. If $b=a$ and
$a\to\infty$, then $r(t)\to 1$ $\forall\,t>0$. We simply have
$\widetilde{r}(\lambda)\leq 1/\lambda$ from which $r(t)\leq 1$ for any $t$
with $b=a\geq 0$;

  * ($b>a$):

Let $b<\infty$. We have that, $r(t)\to 1-\mathbf{E}_{0}[e^{-bL_{t}}]\leq 1$
uniformly in $[0,\infty)$ as $a\to b$. The crucial point is given by the fact
that $r(t)\to b\mathbf{E}_{0}[L_{t}]$ pointwise in $[0,\infty)$ as $a\to 0$
(recall that $L_{0}=0$). In this case, $r(t)\leq 1$ iff
$\mathbf{E}_{0}[L_{t}]\leq 1/b$ with $b>0$. Let us denote by $L_{t}^{0}$ the
process $L_{t}$ with $\eta=0$. Since $\lambda^{\beta}$ is $\alpha$-Hölder
continuous on $[0,\infty)$ only for $\beta=\alpha$ we obtain that
$\sqrt{\lambda+\eta}-\sqrt{\lambda}\leq\sqrt{\eta}$ which implies
$\sqrt{\lambda+\eta}-\sqrt{\eta}\leq\sqrt{\lambda}$. The comparison between
symbols and the fact that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}[L_{t}]\,dt=\frac{1}{\lambda}\frac{1}{(\sqrt{\lambda+\eta}-\sqrt{\eta})}\geq\frac{1}{\lambda}\frac{1}{\sqrt{\lambda}}=\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}[L_{t}^{0}]\,dt$

says that $\mathbf{E}_{0}[L_{t}]\geq\mathbf{E}_{0}[L_{t}^{0}]$, $t\geq 0$.
From the fact that

$\displaystyle\mathbf{E}_{0}[e^{-L_{t}^{0}}]=\sum_{k}\frac{(-1)^{k}}{k!}\mathbf{E}_{0}[(L_{t}^{0})^{k}]\quad\textrm{equals}\quad\sum_{k}\frac{(-\sqrt{t})^{k}}{\Gamma(k/2+1)}=E_{\frac{1}{2}}(-\sqrt{t})$

we get the well-known result

$\displaystyle\mathbf{E}_{0}[L_{t}^{0}]=\frac{\sqrt{t}}{\Gamma(1/2+1)}$

which implies $\mathbf{E}_{0}[L_{t}^{0}]\geq 1$ for $t\geq\pi/4$. Recall tht
$L_{0}^{0}=L_{0}=0$. Thus, $r(t)\leq 1$ only in some bounded domain
$[0,t_{b})\subset[0,\infty)$.

The monotonicity of $r(t)$ follows by considering that

$\displaystyle r(t)=c+C(a,b,c)\,\mathbf{P}_{0}(T_{a}\leq
L_{t})=c+C(a,b,c)\,\mathbf{P}_{0}(H_{T_{a}}\leq t)$

where we have used the relation (2.4). Since $\mathbf{P}_{0}(H_{T_{a}}\leq t)$
is a cumulative distribution function, the result follows. ∎

Recently fractional relaxation equations has been considered in [4]. The
authors have obtained similar results for $b=0$ and $\mu<1$ involving the
gamma random variable $\mathfrak{G}$ with density $\mathbf{P}(\mathfrak{G}\in
ds)=s^{\mu-1}/\Gamma(\mu)\,e^{-s}ds$, that is
$r(t)=\mathbf{P}_{0}(\mathfrak{G}>a^{1/\mu}\,t)$. Interesting discussions have
been made in the papers [2, 25] and the pioneering work [26]. In [2, 25] the
properties of the solutions to fractional relaxation equations in terms of
complete monotone functions have been investigated.

######  Remark 3.1.

For $c=1$ the solution $r(t)$ is monotone increasing or decreasing depending
on the ratio $b/a$, that is respectively $b/a>1$ or $b/a<1$. For $c=0$ the
solution $r(t)$ is only increasing.

######  Remark 3.2.

The initial datum $c\in\\{0,1\\}$ will be related with the fact that, for a
given Markov process and the corresponding multiplicative functional $M_{t}$
we have $M_{0}\in\\{0,1\\}$. Indeed, form the relation
$M_{t+s}=M_{t}(M_{s}\circ M_{t})$ we obtain $M_{0}=M_{0}^{2}$ which implies
that almost surely $M_{0}$ is either $0$ or $1$.

##  4\. Elastic drifted Brownian motions

We introduce and study here the elastic drifted Brownian motion, for short we
often write EDBM. We also write RBM meaning a reflecting Brownian motion.

###  4.1. EDBM with positive drift

Let us consider the process
$\widetilde{X}^{\mu}=\\{\widetilde{X}^{\mu}_{t}\\}_{t\geq 0}$ on $[0,\infty)$
with generator $(G_{\mu},D(G_{\mu}))$ where

$\displaystyle
G_{\mu}\varphi=\mu\frac{d\varphi}{dx}+\frac{d^{2}\varphi}{dx^{2}}$

and

$\displaystyle D(G_{\mu})=\left\\{\varphi,G_{\mu}\varphi\in
C_{b}((0,\infty))\,:\,\varphi^{\prime}(0^{+})=c_{\mu}\,\varphi(0^{+})\right\\}.$

The constant $c_{\mu}>0$ depending on $\mu\geq 0$ is termed elastic
coefficient.

Let us write, for $x\geq 0$, $y\geq 0$, $t>0$,

$\displaystyle p^{0}(t,x,y):=$ $\displaystyle
g(t,x-y)-g(t,x+y)+\int_{0}^{\infty}e^{-c_{0}\,w}\frac{(w+x+y)}{t}g(t,w+x+y)dw$
(4.1)

or the alternative representation of (4.1) given by

$\displaystyle
p^{0}(t,x,y)=g(t,x-y)+g(t,x+y)-2c_{0}\int_{0}^{\infty}e^{-c_{0}\,w}g(t,w+x+y)dw.$
(4.2)

where $g(t,z)=e^{-z^{2}/4t}/\sqrt{4\pi t}$ and $c_{0}\geq 0$. We introduce

$\displaystyle
m^{\mu}(t,x,y):=e^{-\frac{\mu^{2}}{4}t}e^{-\frac{\mu}{2}(x-y)},\quad x\geq
0,\,y\geq 0,\,t\geq 0,\quad\mu\in\mathbb{R}$

and the new density $p=p^{\mu}$, that is

$\displaystyle p(t,x,y):=m^{\mu}(t,x,y)\,p^{0}(t,x,y),\quad x\geq
0,\;y>0,\;t>0.$ (4.3)

We suppress the superscript $\mu$ in order to streamline the notation as much
as possible. Thus we write $p$ in place of $p^{\mu}$. In [23] the authors
highlight an interesting connection between the law of drifted elastic
Brownian motions (4.3) and conditional sojourn times of a Brownian motion on
the positive half-axis.

The solution to the Cauchy problem

$\displaystyle\partial_{t}u=G_{\mu}u,\quad u_{0}=f\in D(G_{\mu})$

is written as

$\displaystyle
u(t,x)=\int_{0}^{\infty}f(y)p(t,x,y)dy=\mathbf{E}_{x}[f(\widetilde{X}^{\mu}_{t})]$

and the semigroup generated by $(G_{\mu},D(G_{\mu}))$ has the probabilistic
representation

$\displaystyle
P^{\mu}_{t}f(x)=\mathbf{E}_{x}[f(X^{\mu}_{t})M_{t}^{\mu}]=\mathbf{E}_{x}[f(\widetilde{X}^{\mu}_{t})]$
(4.4)

where $\widetilde{X}^{\mu}_{t}$ is the part process of $X^{\mu}_{t}$ on
$[0,\infty)$ and $X^{\mu}_{t}$ is a drifted Brownian motion. The process
$\widetilde{X}^{\mu}_{t}$ is therefore a drifted Brownian motion on
$[0,\infty)$ reflected at $0$ and $M^{\mu}_{t}$ is the multiplicative
functional associated with the Robin boundary condition.

######  Remark 4.1.

We recall some basic facts which will be useful in the forthcoming discussion.

Let $G_{0}=\Delta$ be the infinitesimal generator for some Brownian motion on
$E$. The probabilistic representation of the solution to

$\displaystyle\frac{\partial w}{\partial t}=Gw$ $\displaystyle
w_{0}=\mathbf{1}$

with some boundary conditions can be written as
$w(t,x)=\mathbf{E}_{x}[M_{t}]=\mathbf{E}_{x}[e^{-A_{t}}]$ that is, in terms of
the multiplicative functional $M_{t}$ or equivalently in terms of the
corresponding additive functional $A_{t}$. For the Robin boundary condition
$(\partial_{\bf n}w+c\,w)|_{\partial E}=0$, we have that
$M_{t}=\mathbf{1}_{(t<\zeta)}$ where $\zeta$ is the lifetime of the process
with generator $(G,D(G))$. The additive functional to be considered is the
local time $\gamma_{t}$. In particular,

$\displaystyle
w(t,x)=\mathbf{E}_{x}[e^{-c\gamma_{t}}]=\int_{0}^{\infty}e^{-cw}\mathbf{P}_{x}(\gamma_{t}\in
dw)=1-\int_{0}^{\infty}(1-e^{-cw})\,\mathbf{P}_{x}(\gamma_{t}\in dw)$ (4.5)

or equivalently

$\displaystyle
w(t,x)=\mathbf{E}_{x}[\mathbf{1}_{(t<\zeta)}]=\int_{t}^{\infty}\mathbf{P}_{x}(\zeta\in
ds)=1-\int_{0}^{t}\mathbf{P}_{x}(\zeta\in ds).$ (4.6)

It is well known that
$\mathbf{P}_{x}(\zeta>t)=\mathbf{P}_{x}(T_{c}>\gamma_{t})$ where $T_{c}$ is an
exponential random variable with parameter $c>0$ independent from the local
time $\gamma_{t}$ on $\partial E$. The connection between (4.5) and (4.6)
immediately emerges.

It is well-known that $\gamma_{t}$ equals in law the running maximum of a
Brownian motion started at $x=0$. Moreover, such an equivalence in
distribution is maintained with the inverse to an $1/2$-stable subordinator.
Notice that, such an inverse process corresponds to $L_{t}$ with
$\eta=(\mu/2)^{2}=0$.

Our first results are concerned with the relation between the drifted Brownian
motion and the inverse to a tempered stable subordinator. In the following we
will always consider $\mu>0$.

######  Theorem 4.1.

For the positively drifted Brownian motion $X^{\mu}$ with $X^{\mu}_{0}=0$, we
have that

$\displaystyle\max_{0\leq s\leq t}X^{\mu}_{s}\stackrel{{\scriptstyle
d}}{{=}}L_{t},\quad t>0$ (4.7)

where $L$ is an inverse to a relativistic stable subordinator with symbol
(3.1), $\eta=(\mu/2)^{2}$.

###### Proof.

Formula (4.7) can be shown by a Laplace transform argument. The distribution
of the maximum of a Brownian motion with drift $\mu$ is well-known. To the
best or our knowledge, the law of the maximum has been obtained in [34, 12]
together with the joint law with its location. For our purposes we refer to
[22] (with some adaptation) and write

$\displaystyle\mathbf{P}_{x}\left(\max_{0\leq s\leq
t}X^{\mu}_{s}>\beta\right)=\int_{\beta}^{\infty}\frac{e^{-\frac{(z-x)^{2}}{4t}}}{\sqrt{4\pi
t}}e^{-\frac{\mu^{2}t}{4}-\frac{\mu}{2}x}\left[e^{\frac{\mu}{2}z}+e^{\frac{\mu}{2}(2\beta-z)}\right]\mathrm{d}z\,\,,\qquad\beta>x.$
(4.8)

A direct computation immediately shows that the Laplace transform of (4.8) is

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\mathbf{P}_{0}\left(\max_{0\leq
s\leq t}X^{\mu}_{s}>\beta\right)dt$
$\displaystyle=\int_{\beta}^{\infty}\left[e^{\frac{\mu}{2}z}+e^{\frac{\mu}{2}(2\beta-z)}\right]\int_{0}^{\infty}e^{-\left(\frac{\mu^{2}}{4}+\lambda\right)t}\frac{e^{-\frac{z^{2}}{4t}}}{\sqrt{4\pi
t}}\,\mathrm{d}t\,\mathrm{d}z$ (4.9)
$\displaystyle=\int_{\beta}^{\infty}\left[e^{\frac{\mu}{2}z}+e^{\frac{\mu}{2}(2\beta-z)}\right]\frac{e^{-z\sqrt{\lambda+\frac{\mu^{2}}{4}}}}{2\sqrt{\lambda+\frac{\mu^{2}}{4}}}\mathrm{d}z$
$\displaystyle=\frac{e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}\right)}}{2\sqrt{\lambda+\frac{\mu^{2}}{4}}}\left[\frac{1}{\sqrt{\lambda+\frac{\mu^{2}}{4}}+\frac{\mu}{2}}+\frac{1}{\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}}\right]$
$\displaystyle=\frac{1}{\lambda}e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}\right)}$

where we used the well-known formula (8.3) recalled in the Appendix.

On the other hand, by letting $\theta\to 0$ in (2.6), we immediately see that
for the inverse tempered subordinator with symbol (3.1),
$\eta=\frac{\mu^{2}}{4}$, it holds that

$\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}(L_{t}>\beta)\,\mathrm{d}t=\frac{1}{\lambda}e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}\right)}$
(4.10)

thus proving the equality in distribution (4.7).

∎

Moreover we point out a further interesting connection between the tempered
subordinator and the local time of the drifted Brownian motion. First we
introduce the process $\\{Y^{\theta,\sigma}_{t}\\}_{t\geq 0}$ as the unique
strong solution to

$\mathrm{d}Y^{\theta,\sigma}_{t}=-\theta\,\mathrm{sgn}Y^{\theta,\sigma}_{t}+\sigma\mathrm{d}B_{t}\,,\qquad
Y^{\theta,\sigma}_{0}=0$ (4.11)

where $B_{t}$ is a standard Brownian motion, $\theta\in\mathbb{R}$ and
$\sigma>0$. In the following we will restrict ourselves to the cases
$\theta=\pm\mu/2,\sigma=\sqrt{2}$ and for brevity we define
$Y^{\mu}_{t}\coloneqq Y^{\mu/2,\sqrt{2}}_{t},t\geq 0$. Denote with
$\\{\gamma_{t}(Y^{\mu})\\}_{t\geq 0}$ the local time process of
$Y^{\mu}=\\{Y^{\mu}_{t}\\}_{t\geq 0}$. Analogously we define
$Y^{-\mu}_{t}\coloneqq Y^{-\mu/2,\sqrt{2}}_{t},t\geq 0$ and
$\\{\gamma_{t}(Y^{-\mu})\\}_{t\geq 0}$ as the corresponding local time.

######  Corollary 1.

For the local time (at zero) of $Y^{\mu}$ we have that

$\displaystyle\gamma_{t}(Y^{\mu})\stackrel{{\scriptstyle d}}{{=}}L_{t},\quad
t>0.$ (4.12)

###### Proof.

In [19, Theorem 1] the authors prove the equality in distribution

$\displaystyle\left(\max_{0\leq s\leq t}X^{\mu}_{s}-X^{\mu}_{t},\max_{0\leq
s\leq t}X^{\mu}_{s}\right)\stackrel{{\scriptstyle
d}}{{=}}\big{(}|Y^{\mu}_{t}|,\gamma_{t}({Y^{\mu})}\big{)}$ (4.13)

The result follows from (4.7) and (4.13). ∎

In [19] the authors show that $|Y^{\mu}|$ constitutes a reflecting Brownian
motion with drift $-\mu$. For $\mu=0$, the relation (4.12) agrees with the
well-known equality in distribution between maximum, local time and inverse to
a $1/2$-stable subordinator as described in Remark 4.1. However, when the
presence of the drift is assumed, a fundamental difference emerges. For
$\mu>0$, that is for $\eta>0$, the inverse tempered subordinator is related to
the local time of the process $Y^{\mu}$ instead of the local time of a
Brownian motion with drift.

###  4.2. EDBM with negative drift and RBM with positive drift

We deal here with $X^{-\mu}$ with generator $(G_{-\mu},D(G_{-\mu}))$ according
with the notation introduced in the previous sections.

The result of Corollary 1 relates the distribution of the inverse of a
tempered subordinator with the distribution of $Y^{\mu}$, which is in turn
related to a reflecting Brownian motion (RBM for short) with _negative_ drift.
If one starts with a RBM with positive drift, i.e. by considering the process
$Y^{-\mu}=\\{Y^{-\mu}_{t}\\}_{t\geq 0}$ and its absolute value, the symmetry
appears to break. In fact, while the equality in distribution (4.13) still
holds, relating the RBM with positive drift $|Y^{-\mu}|$ and the local time
$\gamma_{t}(Y^{-\mu})$ with a Brownian motion with negative drift $X^{-\mu}$
and its maximum, these processes are not directly related anymore to the
inverse of a tempered stable subordinator. It is instead necessary to
introduce a “truncated version” of the inverse subordinator as in the
following theorem.

######  Theorem 4.2.

For the negatively drifted Brownian motion $X^{-\mu}$ with $X^{-\mu}_{0}=0$,
we have that

$\displaystyle\max_{0\leq s\leq t}X^{-\mu}_{s}\stackrel{{\scriptstyle
d}}{{=}}L_{t}\wedge T_{\mu},\quad t>0$ (4.14)

where $T_{\mu}$ is an exponential r.v. (with parameter $\mu>0$) independent
from $L$ which is an inverse to a relativistic stable subordinator with symbol
(3.1), $\eta=(-\mu/2)^{2}$.

###### Proof.

We check that the Laplace transforms of the distribution of both sides of
(4.14) coincide. The Laplace transform of the distribution of the maximum
(4.9) in this case becomes

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\mathbf{P}_{0}\left(\max_{0\leq
s\leq t}X^{-\mu}_{s}>\beta\right)dt$
$\displaystyle=\frac{1}{\lambda}e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}+\frac{\mu}{2}\right)}.$
(4.15)

Note that (4.15) is now different from (4.10), where the tempering parameter
cannot be negative. This is why Theorem 4.1 does not apply in this case.

Now, by considering (4.15) and (4.10) we have that

$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\mathbf{P}_{0}\left(\max_{0\leq
s\leq t}X^{-\mu}_{s}>\beta\right)dt$
$\displaystyle=\frac{1}{\lambda}e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}+\frac{\mu}{2}\right)}=e^{-\mu\beta}\frac{1}{\lambda}e^{-\beta\left(\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}\right)}$
(4.16) $\displaystyle=\int_{0}^{\infty}e^{-\lambda
t}e^{-\mu\beta}\mathbf{P}_{0}(L_{t}>\beta)\,\mathrm{d}t$
$\displaystyle=\int_{0}^{\infty}e^{-\lambda
t}g^{\mu}(\beta,t)\,\mathrm{d}t\,\,,\qquad\beta>0.$

where $g^{\mu}(\beta,t)=e^{-\mu\beta}P(L_{t}>\beta)$. The quantity
$1-g^{\mu}(\beta,t)$ coincides with the distribution function of $L_{t}\wedge
T_{\mu}$, where $T_{\mu}$ is an independent exponential random variable with
parameter $\mu$ and $L_{t}$ is assumed to start from zero. In fact, by
independence,

$\displaystyle\mathbf{P}_{0}(L_{t}\wedge T_{\mu}>\beta)$
$\displaystyle=\mathbf{P}_{0}(\beta<L_{t}<T_{\mu})+\mathbf{P}_{0}(\beta<T_{\mu}\leq
L_{t})$ (4.17)
$\displaystyle=\mathbf{E}\left[\left(\int_{\beta}^{T_{\mu}}l(t,x)\,\mathrm{d}x\right)\mathbf{1}_{(T_{\mu}>\beta)}\right]+\mathbf{E}\left[\left(\int_{T_{\mu}}^{\infty}l(t,x)\,\mathrm{d}x\right)\mathbf{1}_{(T_{\mu}>\beta)}\right]$
$\displaystyle=P(L_{t}>\beta)\,\mathbf{E}\left(\mathbf{1}_{(T_{\mu}>\beta)}\right)$
(4.18)

where the last equality is justified by the independence between $L$ and
$T_{\mu}$. Thus by (4.15) the result is proved. ∎

######  Corollary 2.

For the local time (at zero) of $Y^{-\mu}$ we have that

$\gamma_{t}(Y^{-\mu})\stackrel{{\scriptstyle d}}{{=}}L_{t}\wedge
T_{\mu}\,\qquad t>0$ (4.19)

where $T_{\mu}$ is an independent exponential r.v. with parameter $\mu$.

###### Proof.

By applying the same arguments as in the proof of Corollary (1) we can show
that (4.19) holds true. We stress the fact that $Y^{-\mu}$ is the process such
that $|Y^{-\mu}|$ is a RBM with _positive drift_ $\mu$.

∎

Figure 2 shows some sample paths of the processes $Y^{\mu}$ and $Y^{-\mu}$ as
well as the corresponding reflecting processes $|Y^{\mu}|$ and $|Y^{-\mu}|$.
We see that in the case of the RBM positive drift, i.e. $|Y^{-\mu}|$, the
sample paths tend to travel further from the barrier, whereas in the case of
negative drift, i.e. $|Y^{\mu}|$, the sample path is constantly pushed towards
the barrier. This gives an intuitive explanation of the difference between the
relations (4.12) and (4.19). In the second case since the process $Y^{-\mu}$
travels away from the barrier its local time at zero tends to stop increasing.
This corresponds to the fact that the local time in this case has the same
distribution of a randomly truncated inverse subordinator.

\begin{overpic}[width=346.89731pt]{ymu-pos.pdf} \end{overpic} (a) sample path
of $Y^{-\mu}_{t}$
a

\begin{overpic}[width=346.89731pt]{abs-ymu-pos.pdf} \end{overpic} (b) sample
path of $|Y^{-\mu}_{t}|$, i.e. an RBM with positive drift.

\begin{overpic}[width=346.89731pt]{ymu-neg.pdf} \end{overpic} (c) sample path
of $Y^{\mu}_{t}$
a

\begin{overpic}[width=346.89731pt]{abs-ymu-neg.pdf} \end{overpic} (d) sample
path of $|Y^{\mu}_{t}|$, i.e. an RBM with negative drift.

Figure 2. Comparison of a sample paths of the solution to (4.11) and the
corresponding RBM with drift.

Figure 3 shows a comparison between the sample paths of an inverse stable
subordinator and the paths of the maximum of a drifted Brownian motion. In
particular 3(a) shows two sample paths of $L_{t}$ while 3(b) shows the same
sample paths randomly truncated with exponential random variables (blue
horizontal lines), i.e. realizations of $L_{t}\wedge T_{\mu}$. 3(c) shows a
sample path of a Brownian motion with _positive_ drift and its running maximum
$\max_{0\leq s\leq t}X^{\mu}_{s}$. The similarity with the sample paths in
3(a) illustrates the equality in distribution proved in Theorem 4.1. 3(d)
shows a Brownian motion with _negative_ drift and its maximum. Note that as
the sample paths travels away from zero the maximum stops increasing,
exhibiting a behavior similar to the paths in 3(b): this is the thesis of
Theorem 4.2.

\begin{overpic}[width=346.89731pt]{inv-temp.pdf} \end{overpic} (a) sample
paths of $L_{t}$

\begin{overpic}[width=346.89731pt]{inv-temp-tmu.pdf} \end{overpic} (b) sample
paths of $L_{t}\wedge T_{\mu}$

\begin{overpic}[width=346.89731pt]{maxmu-pos.pdf} \end{overpic} (c) sample
path of $X_{t}^{\mu}$ and its running maximum $\max_{0\leq s\leq
t}X^{\mu}_{s}$

\begin{overpic}[width=346.89731pt]{maxmu-neg.pdf} \end{overpic} (d) sample
path of $X_{t}^{-\mu}$ and its running maximum $\max_{0\leq s\leq
t}X^{-\mu}_{s}$

Figure 3. Comparison of a sample path of an inverse tempered stable
subordinator and the maximum of a Brownian motion with drift

##  5\. Helpful intuitive introduction to fractional boundary conditions

Here we consider a particular and instructive case which gives an helpful and
intuitive interpretation of the main result of Section 6. The proofs of the
following statements have been postponed in the Appendix.

Let us consider the generator $(G_{0},D(G_{0}))$ of the reflected Brownian
motion on $[0,\infty)$ with elastic condition at $x=0$ for which we have that

$\displaystyle
P^{0}_{t}\mathbf{1}(x)=\int_{0}^{\infty}\big{(}g(t,x-y)-g(t,x+y)\big{)}\,dy+2\int_{0}^{\infty}e^{-c_{0}\,w}g(t,w+x)\,dw.$
(5.1)

First we observe that formula (5.1) has the following representation

$\displaystyle
P^{0}_{t}\mathbf{1}(x)=1-\int_{0}^{t}\frac{x}{s}g(s,x)\,ds+2\int_{0}^{\infty}e^{-c_{0}\,w}g(t,w+x)\,dw=:F(t,x)$
(5.2)

in which the density of the lifetime $\zeta$ emerges as mentioned in Remark
4.1. Formula (5.2), in turn, can be written by considering the non-negative
and non-decreasing process $A_{t}$ as

$\displaystyle F(t,x)=$ $\displaystyle 1-\mathbf{P}_{0}(A^{-1}_{x}\leq
t)+e^{x\,c_{0}}\mathbf{E}_{0}[e^{-c_{0}\,A_{t}}\mathbf{1}_{(A_{t}>x)}],\quad
t\geq 0,\;x\geq 0$ (5.3)

for which, at the boundary point $x=0$, we get

$\displaystyle F(t,0)=\mathbf{E}_{0}[e^{-c_{0}\,A_{t}}],\quad t\geq 0.$

The process $A^{-1}_{t}=\inf\\{s\geq 0\,:\,A_{s}\geq t\\}$ is the inverse to
$A_{t}$. We have the following interesting cases at the boundary point $x=0$:

  * i)

$A_{t}=\gamma_{t}$ is the Brownian local time and the usual condition writes

$\displaystyle\frac{\partial F}{\partial x}(t,0)=c_{0}\,F(t,0).$ (5.4)

The elastic condition (5.4) introduces exponential solutions.

  * ii)

$A_{t}=L_{t}^{0}$ is the inverse to a stable subordinator (of order
$\alpha=1/2$, we use the superscript and write $L^{0}$ instead of $L$ to
underline that $\eta=0$) and

$\displaystyle D^{\frac{1}{2}}_{t}F(t,0)=-c_{0}\,F(t,0)$ (5.5)

whose solution is the Mittag-Leffler function

$F(t,0)=E_{\frac{1}{2}}(-c_{0}\,\sqrt{t}).$

The elastic condition (5.5) introduces solutions to relaxation equations.

  * iii)

The boundary condition

$\displaystyle D^{\frac{1}{2}}_{t}F(t,0)=-\frac{\partial F}{\partial x}(t,0)$
(5.6)

holds true. Despite the fact that we lose the dependence from the elastic
coefficient $c_{0}$, we get information about the additive functional. Indeed,
(5.6) is the governing equation of $L^{0}$.

Obviously $\gamma_{t}\stackrel{{\scriptstyle law}}{{=}}L^{0}_{t}$ and their
sample paths are both positive and non decreasing with
$\gamma_{0}=L^{0}_{0}=0$. Both conditions (5.4) and (5.5) give unique
characterization of the boundary behaviour of the reflected Brownian motion at
$x=0$.

##  6\. Fractional boundary conditions

We discuss here the connection between the infinitesimal generators $G_{\mu}$
and $G_{-\mu}$ and the tempered derivative of order $1/2$. The order $1/2$
seems to be naturally related to the fact that $G_{\pm\mu}$ is a second order
operator (see for example [14]). The drift $\pm\mu$ is related to the
tempering parameter $\eta$ of the tempered derivative by means of the relation
$\eta=(\pm\mu/2)^{2}$.

###  6.1. The positively drifted Brownian motion

We focus on the function

$u\in C^{1,2}(AC(0,\infty)\times[0,\infty),[0,\infty))$

solving

$\begin{cases}\displaystyle\frac{\partial u}{\partial t}=\mu\frac{\partial
u}{\partial x}+\frac{\partial^{2}u}{\partial x^{2}}\\\ \displaystyle
u(0,x)=\mathbf{1}(x\geq 0)\end{cases}$ (6.1)

with the boundary condition

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+\mu\,u(t,0)+c_{\mu}\,u(t,0)=\mu,\quad
t>0.$ (6.2)

Notice that we consider here the boundary condition (6.2) in place of

$\displaystyle\frac{\partial u}{\partial x}(t,0)=c_{\mu}\,u(t,0),\quad t>0.$
(6.3)

We observe that the condition (6.2) can be rewritten as

$\displaystyle\int_{0}^{t}(G_{\mu}u)(t-s,0)\,\Pi((s,\infty))\,ds+\mu\,u(t,0)+c_{\mu}\,u(t,0)=\mu,\quad
t>0$

by following the definition (3.3). For $u\in D(G_{\mu})$, we formally have

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+\frac{\mu+c_{\mu}}{c_{\mu}}\frac{\partial
u}{\partial x}(t,0)=\mu,\quad t>0$

or equivalently

$\displaystyle\int_{0}^{t}(G_{\mu}u)(t-s,0)\,\Pi((s,\infty))\,ds+\frac{\mu+c_{\mu}}{c_{\mu}}\frac{\partial
u}{\partial x}(t,0)=\mu,\quad t>0.$

Further on we will write $\bar{\mathbf{1}}(x):=\mathbf{1}(x\geq 0)$ in order
to streamline the notation.

######  Theorem 6.1.

Let us consider $u=u(t,x)$ given in (4.4). Then, the following statements
hold:

  * i)

$u$ is the unique (classical) solution to (6.1) - (6.3);

  * ii)

$u$ is the unique (classical) solution to (6.1) - (6.2);

  * iii)

$u$ has the probabilistic representation

$\displaystyle u(t,x)=$ $\displaystyle
1-\frac{c_{\mu}}{\mu+c_{\mu}}e^{-x\mu}\mathbf{E}_{0}\left[\left(1-e^{-(\mu+c_{\mu})(L_{t}-x)}\right)\mathbf{1}_{(L_{t}\geq
x)}\right]$ (6.4)

where $L$ is an inverse to a relativistic stable subordinator with symbol
(3.1), $\eta=(\mu/2)^{2}$.

######  Remark 6.1.

The semigroup $P^{\mu}_{t}\bar{\mathbf{1}}(x)$ has the probabilistic
representation (6.4). This means that we can use the properties of $L_{t}$ in
order to obtain equivalence of functionals under $\mathbf{P}_{0}$ (in mean
$\mathbf{E}_{0}$). Since
$P^{\mu}_{t}\bar{\mathbf{1}}(x)=\mathbf{E}_{x}[M^{\mu}_{t}]$, the process
$L_{t}$ can be considered in order to obtain information about $M^{\mu}_{t}$.
We underline that, here, $L_{t}$ is independent from
$(X^{\mu}_{t},M^{\mu}_{t})$, the only advantage we may have is given by the
equivalence in law expressed by (4.4) and (6.4). A further interesting reading
will be given ahead in Theorem 6.2 and Theorem 6.3.

###### Proof of Theorem 6.1.

We obtain that (see the Appendix for the proof of (6.1))

$\displaystyle\widetilde{u}(\lambda,x)=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda t}\,u(t,x)\,dt$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-(\sqrt{\lambda+\eta}+\sqrt{\eta})\frac{1}{c_{0}+\sqrt{\lambda+\eta}}\right)e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}$
(6.5) $\displaystyle=$
$\displaystyle\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-(\Phi(\lambda)+2\sqrt{\eta})\frac{1}{c_{\mu}+2\sqrt{\eta}+\Phi(\lambda)}\right)e^{-x(\Phi(\lambda)+2\sqrt{\eta})}$

where the symbol $\Phi$ denotes

$\displaystyle\Phi(\lambda)=\sqrt{\lambda+\eta}-\sqrt{\eta}.$

By considering the fact that (see (2.1))

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}(L_{t}>x)dt=\int_{x}^{\infty}\frac{\Phi(\lambda)}{\lambda}\,e^{-s\Phi(\lambda)}\,ds$

together with Proposition 2.1 for

$\displaystyle\frac{\Phi(\lambda)}{\lambda}\frac{1}{c_{\mu}+\mu+\Phi(\lambda)}e^{-x(\Phi(\lambda)+\mu)},$

by observing that (recall that $T_{a}$ is an exponential r.v. with parameter
$a$)

$\displaystyle\frac{\mu}{\lambda}\frac{1}{c_{\mu}+\mu+\Phi(\lambda)}e^{-x(\Phi(\lambda)+\mu)}=$
$\displaystyle\frac{\mu}{\lambda}e^{-x\mu}\int_{0}^{\infty}e^{-w\,(c_{\mu}+\mu)}e^{-(w+x)\Phi(\lambda)}dw$
$\displaystyle=$ $\displaystyle\mu e^{-x\mu}\int_{0}^{\infty}e^{-\lambda
t}\left(\int_{0}^{\infty}e^{-w(c_{\mu}+\mu)}\mathbf{P}_{0}(L_{t}>x+w)\,dw\right)dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\left(\frac{\mu}{\mu+c_{\mu}}e^{-x\mu}\mathbf{P}_{0}(L_{t}-x>T_{\mu+c_{\mu}})\right)dt,$

we obtain the inverse Laplace transform of $\widetilde{u}(\lambda,x)$ given by

$\displaystyle u(t,x)=$ $\displaystyle
1-\,e^{-x\mu}\mathbf{P}_{0}(L_{t}>x)+e^{-x\mu}\mathbf{E}_{0}\left[e^{-(\mu+c_{\mu})(L_{t}-x)}\mathbf{1}_{(L_{t}\geq
x)}\right]$
$\displaystyle+\,\frac{\mu}{\mu+c_{\mu}}e^{-x\mu}\mathbf{E}_{0}\left[\left(1-e^{-(\mu+c_{\mu})(L_{t}-x)}\right)\mathbf{1}_{(L_{t}\geq
x)}\right].$

Simple manipulation leads to (6.4).

Now we show that (6.4) is the solution to (6.1) with (6.2). Taking the limit
in (6.1) we obtain

$\displaystyle\lim_{x\to
0^{+}}\widetilde{u}(\lambda,x)=\frac{\sqrt{\lambda+\eta}+\sqrt{\eta}}{\lambda}\frac{1}{c_{0}+\sqrt{\lambda+\eta}}=\widetilde{u}(\lambda,0)$

from which

$\displaystyle(c_{0}+\sqrt{\eta})\,\widetilde{u}(\lambda,0)=$
$\displaystyle\frac{\sqrt{\lambda+\eta}+\sqrt{\eta}}{\lambda}\frac{c_{0}+\sqrt{\eta}}{c_{0}+\sqrt{\lambda+\eta}}$
$\displaystyle=$
$\displaystyle\frac{\sqrt{\lambda+\eta}+\sqrt{\eta}}{\lambda}\left(1-\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{c_{0}+\sqrt{\lambda+\eta}}\right)$
$\displaystyle=$
$\displaystyle\frac{\sqrt{\lambda+\eta}+\sqrt{\eta}}{\lambda}-\frac{1}{c_{0}+\sqrt{\lambda+\eta}}.$

Let us define

$\displaystyle F_{1}(\lambda)=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)\,dt,\quad\lambda>0$

and

$\displaystyle F_{2}(\lambda)=$ $\displaystyle\lim_{x\to
0^{+}}\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,x)\,dt,\quad\lambda>0.$

From (3.6), we have that

$\displaystyle
F_{1}(\lambda)=F_{2}(\lambda)=\frac{1}{c_{0}+\sqrt{\lambda+\eta}}-\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{\lambda}.$

Since

$\displaystyle
F_{1}(\lambda)+(c_{0}+\sqrt{\eta})\,\widetilde{u}(\lambda,0)=\frac{2\sqrt{\eta}}{\lambda}$

by considering that $c_{\mu}=c_{0}-\sqrt{\eta}$, we have that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\bigg{(}\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+(c_{0}+\sqrt{\eta})u(t,0)\bigg{)}\,dt=\frac{2\sqrt{\eta}}{\lambda}$
(6.6)

which is the Laplace transform of both sides of (6.2).

The point $i)$ does not need to be proved. This concludes the proof. ∎

######  Corollary 3.

For the process $\widetilde{X}^{\mu}_{t}$, $t\geq 0$ with generator
$(G_{\mu},D(G_{\mu}))$ the multiplicative functional $\overline{M}^{\mu}_{t}$
is uniquely characterized by the boundary condition

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+\mu\,u(t,0)+c_{\mu}\,u(t,0)=\mu,\quad
t\geq 0.$

Moreover, $\overline{M}^{\mu}_{t}$ is equivalent to $M^{\mu}_{t}$.

###### Proof.

Indeed, the potential

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}\,M^{\mu}_{t}\,dt\right]$

coincides with $\widetilde{u}(\lambda,0)$ uniquely determined in (6.6) which
in turns, uniquely defines the solution to (6.1) - (6.2) . ∎

######  Remark 6.2.

Since $u(0,0)=1$, from Proposition 3.1 we know that $t\mapsto u(t,0)$ is
monotone decreasing and such that, for $c_{\mu}>0$,

$\displaystyle
u(t,0)\downarrow\frac{\mu}{\mu+c_{\mu}}\in(0,1)\quad\textrm{as}\quad
t\to\infty.$

If $c_{\mu}=0$, then $u(t,0)=1$ for any $t$.

######  Remark 6.3.

The proof of Theorem 6.1 exploits explicit probability distributions
associated to the elastic Brownian motion in order to compute the solution of
the boundary value problem (6.1)-(6.3) and then check that it also satisfies
the fractional condition (6.2). We note that it is possible to give an
alternative proof by directly solving the fractional boundary problem
(6.1)-(6.2). In fact, take Laplace transforms of (6.1) to get

$\lambda\tilde{u}(\lambda,x)-u(0,x)=\mu\partial_{x}\tilde{u}(\lambda,x)+\partial^{2}_{xx}\tilde{u}(\lambda,x).$
(6.7)

with $u(0,x)=1$. By taking Laplace transforms of (6.2) we obtain

$\Phi(\lambda)\tilde{u}(\lambda,0)-\frac{\Phi(\lambda)}{\lambda}u(0,0)+(\mu+c_{\mu})\tilde{u}=\frac{\mu}{\lambda}$
(6.8)

with $\Phi(\lambda)=\sqrt{\lambda+\frac{\mu^{2}}{4}}-\frac{\mu}{2}$ and
$u(0,0)=1$. Thus we obtain

$\tilde{u}(\lambda,0)=\frac{1}{\lambda}\frac{\sqrt{\lambda+\frac{\mu^{2}}{4}}+\frac{\mu}{2}}{\sqrt{\lambda+\frac{\mu^{2}}{4}}+c_{0}}=:c_{\lambda}$
(6.9)

Now let $v_{\lambda}(x)=\tilde{u}(\lambda,x)$. By considering (6.7) and (6.8)
we see that (6.1)-(6.2) may be rewritten as

$\begin{cases}v_{\lambda}^{\prime\prime}+\mu v_{\lambda}^{\prime}-\lambda
v_{\lambda}+1=0\\\ v_{\lambda}(0)=c_{\lambda}\\\
v_{\lambda}(x)\,\mathrm{bounded}\end{cases}$ (6.10)

which is a second order ODE that can be solved by standard techniques. It is
then immediate to check that the unique solution to (6.10) is precisely (6.1).

We now recall the condition

$\displaystyle\mathcal{D}^{\frac{1}{2},\mu}_{t}u(t,0)+(\mu+c_{\mu})\,u(t,0)=\mu,\quad
t>0$

and the standard condition

$\displaystyle\frac{\partial u}{\partial x}(t,0)=c_{\mu}\,u(t,0),\quad t>0$

which is associated to (6.1). Then, we conclude with the following two
results.

######  Theorem 6.2.

The solution to (6.1) - (6.2) has the probabilistic representation

$\displaystyle\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-e^{-c_{\mu}\,(L_{t}\wedge
T_{\mu}-x)}\right)\mathbf{1}_{(L_{t}\wedge T_{\mu}>x)}\right]$ (6.11)

where $T_{\mu}$ is an exponential r.v. (with parameter $\mu>0$) independent
from $L$ which is an inverse to a relativistic stable subordinator with symbol
(3.1), $\eta=(\mu/2)^{2}$

###### Proof.

Let us write (6.1) as follows

$\displaystyle\widetilde{u}(t,x)=\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-(\sqrt{\lambda+\mu^{2}/4}+\mu/2)\frac{1}{c_{0}+\sqrt{\lambda+\mu^{2}/4}}\right)e^{-x(\sqrt{\lambda+\mu^{2}/4}+\mu/2)}.$
(6.12)

From Theorem 4.2, we have that

$\displaystyle\frac{1}{\lambda}e^{-x(\sqrt{\lambda+\mu^{2}/4}+\mu/2)}=\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}(L_{t}\wedge T_{\mu}>x)\,dt$

and

$\displaystyle\frac{\sqrt{\lambda+\mu^{2}/4}+\mu/2}{\lambda}e^{-x(\sqrt{\lambda+\mu^{2}/4}+\mu/2)}dx=\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}(L_{t}\wedge T_{\mu}\in dx)\,dt.$

Thus, the integral

$\displaystyle\frac{\sqrt{\lambda+\mu^{2}/4}+\mu/2}{\lambda}\int_{0}^{\infty}e^{-w(c_{0}-\mu/2+(\sqrt{\lambda+(\mu/2)^{2}}+\mu/2)}e^{-x(\sqrt{\lambda+\mu^{2}/4}+\mu/2)}dw$

takes the form

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-(c_{0}-\mu/2)(L_{t}\wedge
T_{\mu}-x)}\mathbf{1}_{(L_{t}\wedge T_{\mu}>x)}\right]dt$

where, by definition $c_{0}-\mu/2=c_{\mu}$. By collecting all the previous
parts, we get that

$\displaystyle u(t,x)=1-\mathbf{E}_{0}[\mathbf{1}_{(L_{t}\wedge
T_{\mu}>x)}]+\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-c_{\mu}\,(L_{t}\wedge
T_{\mu}-x)}\mathbf{1}_{(L_{t}\wedge T_{\mu}>x)}\right]$

which is the claimed result. ∎

######  Remark 6.4.

Formula (6.11) can be succinctly represented as

$u(t,x)=\mathbf{P}_{0}(L_{t}\wedge T_{\mu}-x<T_{c_{\mu}})$ (6.13)

where $T_{c_{\mu}}$ is an exponential random variable of parameter $c_{\mu}$
independent from $L_{t}$ and $T_{\mu}$, provided that $c_{\mu}>0$. This can be
justified as follows

$\displaystyle\mathbf{P}_{0}(L_{t}\wedge
T_{\mu}-x<T_{c_{\mu}}\cap((L_{t}\wedge T_{\mu}-x>0)\cup(L_{t}\wedge
T_{\mu}-x<0)))$ $\displaystyle=\mathbf{P}_{0}(L_{t}\wedge
T_{\mu}-x<0)+\mathbf{E}_{0}\big{[}\,\mathbf{E}[\mathbf{1}_{(T_{c_{\mu}}>L_{t}\wedge
T_{\mu}-x)}|L_{t}\wedge T_{\mu}]\,\mathbf{1}_{(L_{t}\wedge
T_{\mu}-x>0)}\big{]}$ $\displaystyle=\mathbf{P}_{0}(L_{t}\wedge
T_{\mu}<x)+\mathbf{E}_{0}\left[e^{-c_{\mu}(L_{t}\wedge
T_{\mu}-x)}\mathbf{1}_{(L_{t}\wedge T_{\mu}>x)}\right].$

We now present the last result for the positively drifted Brownian motion.

######  Theorem 6.3.

The solution to (6.1) - (6.2) has the probabilistic representation

$\displaystyle\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-e^{-c_{\mu}\,(S^{-\mu}_{t}-x)}\right)\mathbf{1}_{(S^{-\mu}_{t}>x)}\right]$
(6.14)

where

$\displaystyle S^{-\mu}_{t}=\max_{0\leq s\leq t}X^{-\mu}_{s},\quad
t>0,\;\mu>0.$

###### Proof.

The proof follows immediately from Theorem 4.2. ∎

######  Remark 6.5.

(About the reading of (6.2)) Assume that the boundary conditions

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+(c_{0}+\frac{\mu}{2})\,u(t,0)=\mu$

and

$\displaystyle\frac{\partial u}{\partial x}(t,0)=(c_{0}-\frac{\mu}{2})u(t,0)$

are equivalent. Are we able to obtain information about $M^{\mu}_{t}$ from the
previous conditions? It seems that the first equation gives immediately the
answer we are looking for. If we consider the boundary condition

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+c_{1}\,u(t,0)=c_{2}$

we are able to characterize $M^{\mu}_{t}$ in terms of the coefficients
$c_{1},c_{2}$ as in the discussion below.

We analyse some different cases concerned with (6.4) and in particular with
the lifetime $\zeta^{\mu}$ of the process $\widetilde{X}^{\mu}$. Recall that

$\displaystyle P^{\mu}_{t}\bar{\mathbf{1}}(x)=\mathbf{P}_{x}(\zeta^{\mu}>t).$

  * -

(Drift coefficient). Let us consider

$c_{\mu}\geq\mu=0.$

Then, $\forall\,c_{0}\geq 0$,

$\displaystyle
P_{t}^{0}\bar{\mathbf{1}}(x)=\mathbf{P}_{0}(L_{t}<x)+\mathbf{E}_{0}\left[e^{-c_{0}(L_{t}-x)}\,\mathbf{1}_{(L_{t}\geq
x)}\right],\quad t\geq 0$ (6.15)

solves the equation (6.1) with

$\displaystyle\mathfrak{D}^{\frac{1}{2},0}_{t}u(t,0)+c_{0}\,u(t,0)=0,\quad
t\geq 0$ (6.16)

where the tempered derivative becomes the Caputo derivative
$D^{\frac{1}{2}}_{t}$ (see formula (3.7) with $\eta=0$). Notice that the
corresponding multiplicative functional is associated with the elastic
Brownian motion with no drift. In particular,
$P^{0}_{t}\bar{\mathbf{1}}(x)=\mathbf{E}_{x}[M^{0}_{t}]$ where
$M^{0}_{t}=\exp(-c_{0}\gamma_{t})$ emerges in case of Robin boundary
condition. Let us consider $x=0$ for the sake of simplicity. We immediately
see that

$\displaystyle M^{0}_{t}\stackrel{{\scriptstyle law}}{{=}}e^{-c_{0}\,L_{t}}$

where the right-hand side comes out from (6.15). The equation (6.16) can be
associated with $r(t)$ in Proposition 3.1 with $c_{0}=a>b=0$. In particular,
$u(t,0)$ coincides with

$\displaystyle
r(t)=E_{\frac{1}{2}}(-c_{0}\sqrt{t})=\mathbf{E}_{0}[e^{-c_{0}\,L_{t}}]$

which is the Mittag-Leffler function introduced in (5.5).

We also notice that $\mathbf{P}_{0}(L_{0}<0)=1-\mathbf{P}_{0}(L_{0}=0)=0$ and
$\mathbf{1}_{(L_{0}\geq x)}\to 1$ as $x\to 0^{+}$. On the other hand, for
$t>0$, as $x\to 0^{+}$, $\mathbf{P}_{0}(L_{t}<x)\to 0$ and
$\mathbf{1}_{(L_{t}\geq x)}\to 1$.

  * -

(Elastic coefficient). For

$\mu\geq c_{\mu}=0$

we obtain that

$\displaystyle P_{t}^{\mu}\bar{\mathbf{1}}(x)=1,\quad\forall\,x.$

The boundary behaviour is characterized by

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+\mu\,u(t,0)=\mu,\quad
t>0$ (6.17)

for which the function $u(t,0)$ can be associated with $r(t)$ in Proposition
3.1 with $a=b=\mu$. It follows that

$\displaystyle r(t)=1\quad\forall\,t\geq 0$

coincides with $u(t,0)$, $t\geq 0$.

The lifetime $\zeta^{\mu}$ of the process is infinite almost surely. Indeed,
$\forall\,x\in[0,\infty)$, $\mathbf{P}_{x}(\zeta^{\mu}>t)=1$ for any $t\geq
0$. The multiplicative functional $M^{\mu}_{t}=\mathbf{1}_{(t<\infty)}$ is
associated with reflection at $x=0$ of the drifted Brownian motion (see
formula (8.1)).

  * -

Let us consider

$c_{\mu}\in(0,\infty).$

For $0<c_{\mu}<\mu$, we obtain $\mu+c_{\mu}=a>b=\mu$. For $c_{\mu}>\mu$, we
still have $\mu+c_{\mu}=a>b=\mu$. Then, from Proposition 3.1, $a-b=c_{\mu}\geq
0$ implies that $0<r(t)<1$ $\forall\,t>0$. In particular, $u(t,0)$ coincides
with

$\displaystyle r(t)=\frac{\mu}{\mu+c_{\mu}}\mathbf{P}_{0}(L_{t}\geq
T_{\mu+c_{\mu}})+\mathbf{P}_{0}(L_{t}<T_{\mu+c_{\mu}}).$ (6.18)

Equivalently, by using representation (6.13) we have that

$r(t)=\mathbf{P}_{0}(L_{t}\wedge T_{\mu}<T_{c_{\mu}})$

  * -

Let us consider

$c_{\mu}\to\infty\quad\textrm{with}\quad\mu\geq 0.$

Obviously, this special case does not completely agree with the inital datum.
Formulas (6.4) and (6.11) take the form

$\displaystyle u(t,x)=1-e^{-x\mu}\,\mathbf{P}_{0}(L_{t}\geq x)=1-P(L_{t}\wedge
T_{\mu}\geq x)$

for which $u(t,0)=1-\mathbf{P}_{0}(L_{t}\geq 0)=0$. The formal limit in (6.2)
gives the Dirichlet boundary condition. This corresponds to the fact that, by
exploiting the representation (6.14), we obtain

$u(t,x)=1-\mathbf{P}_{0}\left(\max_{0\leq s\leq t}X^{-\mu}_{s}\geq
x\right)=\mathbf{P}_{x}\left(\min_{0\leq s\leq
t}X^{\mu}_{s}>0\right)=\mathbf{P}_{x}(\zeta^{\mu}>t)$

where $\zeta^{\mu}$ now represents the lifetime of a drifted Brownian motion
with an absorbing barrier at zero.

  * -

Assume now that

$c_{0}=0\leq\mu.$

Thus $c_{\mu}=-\frac{\mu}{2}$. Then, the condition (6.2) becomes

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+\frac{\mu}{2}\,u(t,0)=\mu,\quad
t\geq 0.$

From Proposition 3.1 with $\mu/2=a<b=\mu$ we see that $r(t)>1$ $\forall\,t>0$.

###  6.2. The negatively drifted Brownian motion

We focus on the function

$u\in C^{1,2}(AC(0,\infty)\times[0,\infty),[0,\infty))$

solving

$\begin{cases}\displaystyle\frac{\partial u}{\partial t}=-\mu\frac{\partial
u}{\partial x}+\frac{\partial^{2}u}{\partial x^{2}}\\\ \displaystyle
u(0,x)=\mathbf{1}(x\geq 0)\end{cases}$ (6.19)

with the boundary condition

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+c_{-\mu}\,u(t,0)=0,\quad
t>0$ (6.20)

where

$\displaystyle c_{-\mu}:=c_{0}+\frac{\mu}{2}$

plays now the role of the elastic coefficient in the condition

$\frac{\partial u}{\partial x}(t,0)=c_{-\mu}\,u(t,0),\quad t>0.$ (6.21)

Formally, the relation between the conditions (6.20) and (6.21) is justified
by the equation

$\displaystyle\mathfrak{D}^{\frac{1}{2},\mu}_{t}\ell(t,x)=-\frac{\partial}{\partial
x}\ell(t,x)\quad t>0,\,x>0.$

corresponding to (2.17) in the tempered stable case.

For the negatively drifted Brownian motion
$\\{\widetilde{X}^{-\mu}_{t}\\}_{t\geq 0}$ with generator
$(G_{-\mu},D(G_{-\mu}))$ we write

$\displaystyle P_{t}^{-\mu}f(x)=\mathbf{E}_{x}[f(X^{-\mu}_{t})\,M^{-\mu}_{t}]$
(6.22)

which is the analogue of (4.4).

######  Theorem 6.4.

Let us consider $u=u(t,x)$ given in (6.22). Then, the following statements
hold:

  * i)

$u$ is the unique (classical) solution to (6.19) - (6.20);

  * ii)

$u$ is the unique (classical) solution to (6.19) - (6.21);

  * iii)

$u$ has the probabilistic representation

$\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-e^{-c_{-\mu}\,(L_{t}-x)}\right)\mathbf{1}_{(L_{t}>x)}\right]$

where $L$ is an inverse to a relativistic stable subordinator with symbol
(3.1), $\eta=(-\mu/2)^{2}$.

###### Proof.

Point iii). Let us consider (6.1) in the following form (6.12) and write
$-\mu$ in place of $\mu$. We get

$\displaystyle\widetilde{u}(\lambda,x)=\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-(\sqrt{\lambda+\mu^{2}/4}-\mu/2)\frac{1}{c_{0}+\sqrt{\lambda+\mu^{2}/4}}\right)e^{-x(\sqrt{\lambda+\mu^{2}/4}-\mu/2)}$

that is

$\displaystyle\widetilde{u}(\lambda,x)=\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-\Phi(\lambda)\frac{1}{c_{0}+\mu/2+\Phi(\lambda)}\right)e^{-x\Phi(\lambda)}$
(6.23)

where, as usual, $\Phi(\lambda)=\sqrt{\lambda+\mu^{2}/4}-\mu/2$. Since

$\displaystyle\frac{e^{-x\Phi(\lambda)}}{\lambda}=\int_{0}^{\infty}e^{-\lambda
t}\mathbf{P}_{0}(L_{t}>x)\,dt$

and

$\displaystyle\frac{\Phi(\lambda)}{\lambda}\int_{0}^{\infty}e^{-w\,(c_{0}+\mu/2)}e^{-(x+w)\Phi(\lambda)}dw=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\left(\int_{0}^{\infty}e^{-w\,(c_{0}+\mu/2)}\ell(t,x+w)\,dw\right)dt$
$\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathbf{E}_{0}\left[e^{-(c_{0}+\mu/2)(L_{t}-x)}\,\mathbf{1}_{(L_{t}>x)}\right]dt$

the solution $u$ can be written as

$\displaystyle
u(t,x)=1-\mathbf{E}_{0}[\mathbf{1}_{(L_{t}>x)}]+\mathbf{E}_{0}\left[e^{-(c_{0}+\mu/2)(L_{t}-x)}\,\mathbf{1}_{(L_{t}>x)}\right]$

that is

$\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-e^{-(c_{0}+\mu/2)(L_{t}-x)}\right)\mathbf{1}_{(L_{t}>x)}\right].$

By observing that $c_{0}+\mu/2=c_{-\mu}$ we get the result claimed in iii).

In order to prove point ii) we check that $u(t,0)$ satisfies (6.21). In fact
from (6.23) we get

$\displaystyle\tilde{u}(\lambda,0)=\frac{\Phi(\lambda)}{\lambda}\frac{1}{c_{0}+\mu/2+\Phi(\lambda)}$
(6.24)

and then from (3.6) we obtain

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\bigg{(}\mathfrak{D}^{\frac{1}{2},\mu}_{t}u(t,0)+(c_{0}+\mu/2)u(t,0)\bigg{)}\,dt$
(6.25)
$\displaystyle=\frac{1}{\lambda}\frac{(\sqrt{\lambda+\eta}-\sqrt{\eta})^{2}}{c_{0}+\sqrt{\lambda+\eta}}-\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{\lambda}+\frac{(c_{0}+\sqrt{\eta})}{\lambda}\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{c_{0}+\sqrt{\lambda+\eta}}$
$\displaystyle=\frac{1}{\lambda}\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{c_{0}+\sqrt{\lambda+\eta}}\left(\sqrt{\lambda+\eta}-\sqrt{\eta}+c_{0}+\sqrt{\eta}\right)-\frac{\sqrt{\lambda+\eta}-\sqrt{\eta}}{\lambda}$
$\displaystyle=0$

Point i) does not need to be proved. This concludes the proof. ∎

As in Corollary 3, also in this case we are able to show that
$\overline{M}^{-\mu}_{t}$ is equivalent to $M^{-\mu}_{t}$ associated with
$X^{-\mu}$ and therefore, with the generator $(G_{-\mu},D(G_{-\mu}))$. In
particular, $\overline{M}^{-\mu}_{t}$ is uniquely characterized by the
boundary condition (6.20).

Below we present the last result for the negatively drifted Brownian motion.

######  Theorem 6.5.

The solution to (6.19) - (6.20) has the probabilistic representation

$\displaystyle\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-e^{-c_{-\mu}\,(S^{\mu}_{t}-x)}\right)\mathbf{1}_{(S^{\mu}_{t}>x)}\right]$

where

$\displaystyle S^{\mu}_{t}=\max_{0\leq s\leq t}X^{\mu}_{s},\quad t>0,\;\mu>0.$

###### Proof.

The proof follows immediately from Theorem 4.1. ∎

##  7\. Conclusion

We observe that Theorem 6.5 is the analogue to Theorem 6.3. Such results give
clear representations of the solutions, in both cases, in which we have
positive or negative drift,

$\displaystyle
u(t,x)=1-\mathbf{E}_{0}\left[\left(1-E^{-c_{\mp}(S^{{}^{\pm}\mu}_{t}-x)}\right),S^{{}^{\pm}\mu}_{t}>x\right],\quad
t>0,\;x>0.$

Moreover, in our view, Theorem 6.2 and Theorem 6.4 seems to be quite
interesting with regards to the applications. Indeed, they are written in
terms of very simple processes, that is, non-decreasing processes on
$(0,\infty)$. Formula (4.13) suggests also a representation in terms of the
local time which is usually sneaky. Some fruitful applications of such
representations may arise in numerical solutions, optimization, inverse
problems and so forth. Indeed, in these contexts, it is important to obtain
fast and accurate simulations. On the other hand, the proposed algorithms may
result in high demanding computational tasks, as for the Monte Carlo
approximations for instance. For a description of simulation algorithms for a
Brownian motion on the half-line with boundary conditions the interested
reader can consult [7] and references therein. Such algorithms are based on
spatial discretizations for the generator of the process. Clearly our results
provide a simpler and immediate alternative which only requires the simulation
of the increments of a tempered subordinator, which is a straightforward task
(see e.g. [29]).

##  8\. Appendix

###  8.1. Some elementary facts

Let us recall some elementary aspects concerning (4.1) and (4.3). We have that
$\partial_{t}p^{0}=\partial_{xx}p^{0}=\partial_{yy}p^{0}$ with Robin boundary
conditions

$\displaystyle\partial_{x}p^{0}(t,0,y)=$
$\displaystyle\frac{y}{t}g(t,y)+\int_{0}^{\infty}e^{-c_{0}\,w}\frac{1}{t}g(t,w+y)dw+\int_{0}^{\infty}e^{-c_{0}\,w}\frac{w+y}{t}\frac{d}{dw}g(t,w+y)\,dw$
$\displaystyle=$ $\displaystyle
c_{0}\,\int_{0}^{\infty}e^{-c_{0}\,w}\frac{w+y}{t}g(t,w+y)\,dw$
$\displaystyle=$ $\displaystyle c_{0}\,\,p^{0}(t,0,y)$

and

$\displaystyle\partial_{y}p^{0}(t,x,0)=$
$\displaystyle\frac{x}{t}g(t,x)+\int_{0}^{\infty}e^{-c_{0}\,w}\frac{1}{t}g(t,w+x)dw+\int_{0}^{\infty}e^{-c_{0}\,w}\frac{w+x}{t}\frac{d}{dw}g(t,w+x)\,dw$
$\displaystyle=$ $\displaystyle
c_{0}\int_{0}^{\infty}e^{-c_{0}\,w}\frac{w+x}{t}g(t,w+x)\,dw$ $\displaystyle=$
$\displaystyle c_{0}\,p^{0}(t,x,0)$

with $c_{0}>0$. Since $p(t,x,y)=m^{\mu}(t,x,y)\,p^{0}(t,x,y)$ we immediately
see that $\partial_{t}p=G_{\mu}p$ and

$\displaystyle\partial_{x}p(t,0,y)=$
$\displaystyle-\frac{\mu}{2}m^{\mu}(t,0,y)p^{0}(t,0,y)+m^{\mu}(t,0,y)\partial_{x}p^{0}(t,0,y)$
$\displaystyle=$
$\displaystyle-\frac{\mu}{2}m^{\mu}(t,0,y)p^{0}(t,0,y)+c_{0}\,m^{\mu}(t,0,y)p^{0}(t,0,y)$
$\displaystyle=$ $\displaystyle\left(c_{0}-\frac{\mu}{2}\right)p(t,0,y).$
(8.1)

For $c_{0}=0$, formula (4.3) takes the form $p=m^{\mu}\,p^{0}$ where
$p^{0}(t,x,y)=g(t,x-y)+g(t,x+y)$ is the Neumann kernel. The corresponding
process is a drifted Brownian motion reflected at $x=0$.

###  8.2. Proof of the statements in Section 5

###### Proof of the formula (5.2).

Since

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\,g(t,x-y)\,dt=\frac{1}{2}\frac{e^{-|x-y|\sqrt{\lambda}}}{\sqrt{\lambda}}\quad\textrm{and}\quad\int_{0}^{\infty}e^{-\lambda
t}\,g(t,x+y)\,dt=\frac{1}{2}\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}$

we have that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\int_{0}\big{(}g(t,x-y)-g(t,x+y)\big{)}dy\,dt=\frac{1}{\lambda}-\frac{1}{\lambda}e^{-x\sqrt{\lambda}}$
(8.2)

Now we note that

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

for which we have

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\,g(s,x)\,ds=\frac{1}{2}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}$ (8.3)

and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
s}\,\frac{x}{s}g(s,x)\,dx=-\,\frac{\partial}{\partial
x}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}=e^{-x\sqrt{\lambda}}.$

Then,

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\left(1-\int_{0}^{t}\frac{x}{s}g(s,x)\,ds\right)dt=\frac{1}{\lambda}-\frac{1}{\lambda}e^{-x\sqrt{\lambda}}$
(8.4)

By comparing (8.2) with (8.4) we prove that, $\forall\,x$,

$\displaystyle P^{0}_{t}\mathbf{1}(x)=F(t,x).$

Moreover, for our convenience, we observe that

$\displaystyle 2\int_{0}^{\infty}e^{-\lambda
t}\int_{0}^{\infty}e^{-c_{0}\,w}g(t,w+x)\,dw\,dt=$
$\displaystyle\int_{0}^{\infty}e^{-c_{0}w}\frac{e^{-(w+x)\sqrt{\lambda}}}{\sqrt{\lambda}}\,dw$
$\displaystyle=$
$\displaystyle\frac{e^{-x\sqrt{\lambda}}}{\lambda}\frac{\sqrt{\lambda}}{c_{0}+\sqrt{\lambda}}$

and therefore

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\,P^{0}_{t}\mathbf{1}(x)\,dt=\frac{1}{\lambda}-\frac{e^{-x\sqrt{\lambda}}}{\lambda}\frac{c_{0}}{c_{0}+\sqrt{\lambda}}.$
(8.5)

∎

###### Proof of the formula (5.3).

Let us consider the non-negative and non-decreasing process $A_{t}$, $t\geq 0$
with probability

$\displaystyle\mathbf{P}_{0}(A_{t}>x)=2\int_{x}^{\infty}g(t,s)\,ds.$

Consider the inverse $A^{-1}_{t}$, $t\geq 0$ which is, by construction, a non-
negative and non-decreasing process. By definition we have that

$\displaystyle\mathbf{P}_{0}(A^{-1}_{x}<t)=\mathbf{P}_{0}(A_{t}>x)$

with

$\displaystyle\frac{\partial}{\partial
t}\mathbf{P}_{0}(A^{-1}_{x}<t)=2\int_{x}^{\infty}\frac{\partial^{2}}{\partial
s^{2}}g(t,s)\,ds=\frac{x}{t}g(t,x).$

Since $\mathbf{P}_{0}(A^{-1}_{0}<t)=\mathbf{P}_{0}(A_{t}>0)=1$ and $A_{t}$ is
continuous we obtain $F(t,0)=\mathbf{E}_{0}[e^{-c_{0}\,A_{t}}]$. ∎

###### Proof of the formula (5.4).

It holds that

$\displaystyle\frac{\partial F}{\partial
x}(t,x)\bigg{|}_{x=0}=c_{0}\,F(t,x)\bigg{|}_{x=0}.$

Indeed, $F$ is the density law of the elastic Brownian motion on $[0,\infty)$.
∎

###### Proof of the formula (5.5).

Since $L^{0}_{t}$ is the inverse to a stable subordinator, the density
$\ell(t,x)=2\,g(t,x)$ is such that

$\displaystyle D^{\frac{1}{2}}_{t}\ell=-\frac{\partial\ell}{\partial x}$

and the potential

$\displaystyle\hat{\ell}(t,c_{0})=\int_{0}^{\infty}e^{-c_{0}\,w}\ell(t,w)\,dw=E_{\frac{1}{2}}(-c_{0}\,\sqrt{t})$

is the Mittag-Leffler function. It is well-known that the Mittag-Leffler is an
eigenfunction for the Caputo derivative. That is,

$\displaystyle D^{\frac{1}{2}}_{t}\hat{\ell}=-c_{0}\,\hat{\ell}.$

By observing that $\hat{\ell}(t,c_{0})=F(t,0)$ we get formula (5.5). The
problem to check that

$\displaystyle
D^{\frac{1}{2}}_{t}F(t,x)\bigg{|}_{x=0}=-c_{0}\,F(t,x)\bigg{|}_{x=0}$

is part of the results presented in this work. ∎

###  8.3. Proof of Theorem 6.1: some minor results

###### Proof of equation 6.1.

We write

$\displaystyle u(t,x)=e^{-\eta
t-\sqrt{\eta}x}\big{(}G(t,x)-2c_{0}\,V(t,x)\big{)}$

where

$\displaystyle
G(t,x)=\int_{0}^{\infty}e^{\sqrt{\eta}y}\big{(}g(t,x+y)+g(t,x-y)\big{)}\,dy$
(8.6)

and

$\displaystyle
V(t,x)=\int_{0}^{\infty}e^{\sqrt{\eta}y}\int_{0}^{\infty}e^{-c_{0}\,w}g(t,x+y+w)\,dw\,dy$
(8.7)

We obtain that

$\displaystyle\widetilde{G}(\lambda,x)=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}G(t,x)\,dt=\widetilde{G}_{1}(\lambda,x)+\widetilde{G}_{2}(\lambda,x)$

where

$\displaystyle\widetilde{G}_{1}(\lambda,x)=$
$\displaystyle\int_{0}^{\infty}e^{\sqrt{\eta}y}\frac{1}{2}\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}\,dy=\frac{1}{2}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}-\sqrt{\eta}},\quad\lambda>\eta$
(8.8)

and

$\displaystyle\widetilde{G}_{2}(\lambda,x)=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}e^{\sqrt{\eta}y}\frac{e^{-|x-y|\sqrt{\lambda}}}{\sqrt{\lambda}}\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{x}e^{\sqrt{\eta}y}\frac{e^{-(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}dy+\frac{1}{2}\int_{x}^{\infty}e^{\sqrt{\eta}y}\frac{e^{-(y-x)\sqrt{\lambda}}}{\sqrt{\lambda}}\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{0}^{x}e^{y(\sqrt{\lambda}+\sqrt{\eta})}\,dy+\frac{1}{2}\frac{e^{\sqrt{\lambda}x}}{\sqrt{\lambda}}\int_{x}^{\infty}e^{-y(\sqrt{\lambda}-\sqrt{\eta})}\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\left(\frac{e^{x(\sqrt{\lambda}+\sqrt{\eta})}-1}{\sqrt{\lambda}+\sqrt{\eta}}\right)+\frac{1}{2}\frac{e^{\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{e^{-x(\sqrt{\lambda}-\sqrt{\eta})}}{\sqrt{\lambda}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{e^{x(\sqrt{\lambda}+\sqrt{\eta})}}{\sqrt{\lambda}+\sqrt{\eta}}-\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}+\sqrt{\eta}}+\frac{1}{2}\frac{e^{\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{e^{-x(\sqrt{\lambda}-\sqrt{\eta})}}{\sqrt{\lambda}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{e^{x\sqrt{\eta}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}+\sqrt{\eta}}-\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}+\sqrt{\eta}}+\frac{1}{2}\frac{e^{x\sqrt{\eta}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{e^{x\sqrt{\eta}}}{\lambda-\eta}-\frac{1}{2}\frac{e^{-\sqrt{\lambda}x}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}+\sqrt{\eta}},\quad\lambda>\eta$
(8.9)

Thus,

$\displaystyle\widetilde{G}(\lambda,x)=$
$\displaystyle\frac{1}{2}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}-\sqrt{\eta}}+\frac{e^{x\sqrt{\eta}}}{\lambda-\eta}-\frac{1}{2}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}+\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{e^{x\sqrt{\eta}}}{\lambda-\eta}+\frac{\sqrt{\eta}}{\sqrt{\lambda}}\frac{e^{-x\sqrt{\lambda}}}{\lambda-\eta},\quad\lambda>\eta$

From (8.7) we obtain the Laplace transform

$\displaystyle\widetilde{V}(\lambda,x)=$
$\displaystyle\int_{0}^{\infty}e^{-\lambda t}V(t,x)dt$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}e^{\sqrt{\eta}y}\int_{0}^{\infty}e^{-c_{0}\,w}\frac{e^{-(x+y+w)\sqrt{\lambda}}}{\sqrt{\lambda}}\,dw\,dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\frac{1}{c_{0}+\sqrt{\lambda}}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}}\frac{1}{\sqrt{\lambda}-\sqrt{\eta}},\quad\lambda>\eta.$
(8.10)

We get that

$\displaystyle\widetilde{u}(\lambda,x)=e^{-\sqrt{\eta}x}\left(\widetilde{G}(\lambda+\eta,x)-2c_{0}\,\widetilde{V}(\lambda+\eta,x)\right)$

from which, we explicitly obtain

$\displaystyle\widetilde{u}(\lambda,x)=$
$\displaystyle\frac{1}{\lambda}+\frac{1}{\lambda}e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}\frac{\sqrt{\eta}}{\sqrt{\lambda+\eta}}-\frac{c_{0}}{c_{0}+\sqrt{\lambda+\eta}}\frac{e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}}{\sqrt{\lambda+\eta}}\frac{1}{\sqrt{\lambda+\eta}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}+\frac{1}{\lambda}e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}\frac{\sqrt{\eta}}{\sqrt{\lambda+\eta}}-\frac{e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}}{\sqrt{\lambda+\eta}}\frac{1}{\sqrt{\lambda+\eta}-\sqrt{\eta}}$
$\displaystyle+\frac{1}{c_{0}+\sqrt{\lambda+\eta}}e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}\frac{1}{\sqrt{\lambda+\eta}-\sqrt{\eta}}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}+\left(\frac{1}{\lambda}\frac{\sqrt{\eta}}{\sqrt{\lambda+\eta}}+\frac{1}{\sqrt{\lambda+\eta}-\sqrt{\eta}}\left(\frac{1}{c_{0}+\sqrt{\lambda+\eta}}-\frac{1}{\sqrt{\lambda+\eta}}\right)\right)e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}+\left(\frac{1}{\lambda}\frac{\sqrt{\eta}}{\sqrt{\lambda+\eta}}+\frac{\sqrt{\lambda+\eta}+\sqrt{\eta}}{\lambda}\left(\frac{1}{c_{0}+\sqrt{\lambda+\eta}}-\frac{1}{\sqrt{\lambda+\eta}}\right)\right)e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}$
$\displaystyle=$
$\displaystyle\frac{1}{\lambda}-\frac{1}{\lambda}\left(1-(\sqrt{\lambda+\eta}+\sqrt{\eta})\frac{1}{c_{0}+\sqrt{\lambda+\eta}}\right)e^{-x(\sqrt{\lambda+\eta}+\sqrt{\eta})}$

∎

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