# Non-local Boundary Value Problems for Brownian motions on the half line

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

We study boundary value problems involving non-local operators for the dynamic
boundary conditions. Our analysis includes a detailed description of such
operators together with their relations with random times and random
functionals. We provide some new characterizations for the boundary behaviour
of the Brownian motion based on the interplay between non-local operators and
boundary value problems.

###### Key words and phrases:

Non-local operators, subordinators, fractional boundary condition, Brownian
motion

##  1\. Introduction

We consider dynamic boundary conditions involving non-local operators and
provide the probabilistic representation of the solutions for the associated
Cauchy problems on the real line (we provide some extension for the
$d$-dimensional case). Our analysis is based on the non-local dynamic boundary
condition

$\displaystyle\eta\,\mathfrak{D}^{\Psi}_{t}u(t,0)=-\mathbf{D}^{\Phi}_{x-}u(t,0),\quad
t>0$

for the heat equation on the positive real line. The non-local operators
appearing above are given by the Caputo-Dzherbashian (type) derivative
$\mathfrak{D}^{\Psi}_{t}$ and the Marchaud (type) derivative
$\mathbf{D}^{\Phi}_{x-}$ where $\Psi$ and $\Phi$ are Bernstein symbols
characterizing the operators. The probabilistic representation of the solution
is written in terms of a Brownian motion reflected at zero, we show that the
spatial condition controls an additive part acting on the reflecting Brownian
motion whereas the time condition introduces a time change acting on the local
time (at zero) of the reflecting Brownian motion. The additive part pushes
away from zero the process and the time change forces the process to stop for
a random amount of time. Due to the right-continuity, the stop occurs
immediately after the jump. Furthermore, the process stops for an independent
amount of time and it jumps for an independent distance from the origin.

In Section 2 we introduce some basic facts and notations about subordinators
and inverses. In Section 3 we discuss the relations between the operators we
deal with and the associated processes. In particular, the left and right
Marchaud (type) derivatives $\mathbf{D}^{\Phi}_{x-}$ and
$\mathbf{D}^{\Phi}_{x+}$ can be respectively regarded as the generator and its
adjoint for a subordinator with symbol $\Phi$. From this point of view, our
analysis completely agrees with the boundary conditions introduced by Feller
in [12]. Then, we discuss the heat equation with non-local boundary conditions
in Section 4. We provide the main result (Theorem 4.1) which generalizes the
work [10] for the time operator and the work [12] for the space operator. We
also provide a discussion on the probabilistic representation of the solution
by extending the results given in [9; 10] and the probabilistic representation
given by Itô and McKean in [16; 15]. In Section 5 we discuss on the
applications and the extension of our results. In particular, the non-local
dynamic boundary condition gives new meaning to the reflection, we provide
some characterization.

##  2\. Subordinators

We recall some basic facts just to introduce some notation. Let
$H^{\Phi}=\\{H_{t}^{\Phi},\ t\geq 0\\}$ be a subordinator (see [3] for
details). Then, $H^{\Phi}$ can be characterized by the Laplace exponent
$\Phi$, that is,

$\displaystyle\mathbf{E}_{0}[\exp(-\lambda
H_{t}^{\Phi})]=\exp(-t\Phi(\lambda)),\quad\lambda\geq 0.$ (2.1)

We denote by $\mathbf{E}_{x}$ the expected value with respect to
$\mathbf{P}_{x}$ where $x$ is the starting point. Since $\Phi(\lambda)$ is the
Laplace exponent of a subordinator, it is uniquely characterized by the pair
of non-negative real numbers $(\kappa,d)$ and by the L$\acute{e}$vy measure
$\Pi^{\Phi}$ on $(0,\infty)$ such that $\int_{0}^{\infty}(1\wedge
z)\Pi^{\Phi}(dz)<\infty$. For the symbol $\Phi$, the following L$\acute{e}$vy-
Khintchine representation holds

$\displaystyle\Phi(\lambda)=\kappa+d\lambda+\int_{0}^{\infty}(1-e^{-\lambda
z})\Pi^{\Phi}(dz),\quad\lambda>0$ (2.2)

where the killing rate $\kappa$ and the drift coefficient $d$ are given by

$\displaystyle\kappa=\lim_{\lambda\to 0}\Phi(\lambda),\quad
d=\lim_{\lambda\to\infty}\frac{\Phi(\lambda)}{\lambda}.$

The symbol $\Phi$ is a Bernstein function (non-negative, non-decreasing and
continuous, see for example [29]) uniquely associated with $H^{\Phi}$. For the
reader’s convenience we also recall that

$\displaystyle\frac{\Phi(\lambda)}{\lambda}=d+\int_{0}^{\infty}e^{-\lambda
z}\overline{\Pi}^{\Phi}(z)dz,\quad\overline{\Pi}^{\Phi}(z)=k+\Pi^{\Phi}((z,\infty))$
(2.3)

where $\overline{\Pi}^{\Phi}$ is the so called tail of the L$\acute{e}$vy
measure $\Pi^{\Phi}$. In this paper we only consider symbols for which

$\displaystyle\kappa=0,\quad d=0.$

We also define the process $L^{\Phi}=\\{L_{t}^{\Phi},\ t\geq 0\\}$, with
$L_{0}^{\Phi}=0$, as the inverse of $H^{\Phi}$, that is

$\displaystyle L_{t}^{\Phi}=\inf\\{s>0\,:\,H_{s}^{\Phi}>t\\},\quad t>0.$

We focus only on strictly increasing subordinators with infinite measures
$\Pi^{\Phi}(0,\infty)=\infty$ (see [17, Theorem 21.3]). Thus, the inverse
process $L^{\Phi}$ turns out to be a continuous process. In particular,
$H^{\Phi}$ may have jumps and the inverse has non decreasing paths. Notice
that, an inverse process can be regarded as an exit time for $H^{\Phi}$. By
definition, we also have

$\displaystyle\mathbf{P}_{0}(H_{t}^{\Phi}<s)=\mathbf{P}_{0}(L_{s}^{\Phi}>t),\quad
s,t>0.$ (2.4)

Let us introduce $h,l$ for which

$\displaystyle\mathbf{P}_{0}(H_{t}^{\Phi}\in
I)=\int_{I}h(t,x)dx,\quad\mathbf{P}_{0}(L_{t}^{\Phi}\in I)=\int_{I}l(t,x)dx,$

for a given set $I\subset(0,\infty)$. From (2.1), we have that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
x}h(t,x)dx=e^{-t\Phi(\lambda)},\quad\lambda>0$

and from [23, formula (3.13)], we get

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}l(t,x)dt=\frac{\Phi(\lambda)}{\lambda}e^{-x\Phi(\lambda)},\quad\lambda>0.$
(2.5)

By using the initial value theorem for the Laplace transform, we observe that

$\displaystyle h(t,0)=\lim_{\lambda\to\infty}\lambda e^{-t\Phi(\lambda)}$

is not always zero, but depends on $\Phi(\lambda)$ and the time variable $t$
as discussed below. See [17] for the _time dependent property_. The fact that
$h(t,x)=0$ as $x\leq 0$ seems to be quite relevant in our discussion. Indeed,
this condition has non trivial consequences. The gamma subordinator is an
example.

Further on we will also introduce the processes
$H^{\Psi}=\\{H_{t}^{\Psi},t\geq 0\\}$ and $L^{\Psi}=\\{L_{t}^{\Psi},t\geq
0\\}$ with symbol

$\displaystyle\Psi(\lambda):=\int_{0}^{\infty}(1-e^{-\lambda
z})\Pi^{\Psi}(dz),\quad\lambda>0.$

##  3\. Operators

For a continuous function $u$ on $\mathbb{R}$ extended with zero on the
negative part of the real line, that is $u(x)=0$ if $x\leq 0$, we define the
right Marchaud (type) derivative

$\displaystyle\mathbf{D}_{x+}^{\Phi}u(x)=\int_{0}^{\infty}(u(x)-u(x-y))\Pi^{\Phi}(dy)$
(3.1)

and the left Marchaud (type) derivative

$\displaystyle\mathbf{D}_{x-}^{\Phi}u(x)=\int_{0}^{\infty}(u(x)-u(x+y))\Pi^{\Phi}(dy).$
(3.2)

If $\Pi^{\Phi}$ is the L$\acute{e}$vy measure associated to a stable
subordinator, formulas (3.1) and (3.2) respectively coincide with the right
and the left Marchaud derivatives, usually denoted by
$\mathbf{D}_{x+}^{\alpha}$ and $\mathbf{D}_{x-}^{\alpha}$ respectively. The
reader can consult the famous book [28, formula (5.57) and (5.58)] or the
paper [13, section 6.1] for a nice recent discussion. The operators
$\mathbf{D}_{x+}^{\Phi}$ and $\mathbf{D}_{x-}^{\Phi}$ can be defined on
different spaces depending on $\Pi^{\Phi}$. For a general definition, given
the symbol $\Phi$, we consider $u$ bounded and locally Lipschitz continuous,
then

$\displaystyle|\mathbf{D}_{x-}^{\Phi}u(x)|$
$\displaystyle\leq\int_{0}^{1}|u(x)-u(x+y)|\Pi^{\Phi}(dy)+\int_{1}^{\infty}|u(x)-u(x+y)|\Pi^{\Phi}(dy)$
$\displaystyle\leq
K\int_{0}^{1}y\Pi^{\Phi}(dy)+2||u||_{\infty}\int_{1}^{\infty}\Pi^{\Phi}(dy)$
$\displaystyle\leq(K+2||u||_{\infty})\int_{0}^{\infty}(1\wedge
y)\Pi^{\Phi}(dy)<\infty.$ (3.3)

Indeed, for the integral in $(0,1)$ we have used the Lipschitz property for a
positive constant $K>0$ whereas for the integral in $(1,\infty)$ we have used
the boundedness of $u$. Since $\int(1\wedge z)\Pi^{\Phi}(dz)<\infty$, then the
last inequality holds. The same holds for $\mathbf{D}_{x+}^{\Phi}$. Obviously,
we may take some advantage from the explicit representation of $\Pi^{\Phi}$.
For example, if $\Phi(\lambda)=\lambda^{\alpha}$ with $\alpha\in(0,1)$, then
$\Pi^{\alpha}(dy)=\frac{\alpha}{\Gamma(1-\alpha)}\frac{dy}{y^{\alpha+1}}$. The
operators (3.1) and (3.2) are therefore defined for locally
$\gamma-$H$\ddot{o}$lder continuous functions with $\gamma>\alpha$. The case
of gamma subordinator, with $\Pi^{\Phi}(dy)=a\frac{e^{-by}}{y}dy$, may be a
bit demanding. This is due to the time dependent continuity of $h(t,x)$ at
$x=0$. In [8] we have proved that (3.1) is defined for
$\gamma-$H$\ddot{o}$lder continuous functions with $\gamma>0$.

Let us introduce the spaces

$\displaystyle W^{1,p}(0,\infty)=\\{u\in L^{p}(0,\infty):\;u^{\prime}\in
L^{p}(0,\infty)\\}$

and

$\displaystyle W^{1,p}_{0}(0,\infty)=\\{u\in W^{1,p}(0,\infty):\,u(0)=0\\}$

with $p\in[1,\infty]$. Observe that $AC(I)$ coincides with $W^{1,1}(I)$ only
if $I\subset(0,\infty)$ is bounded. We recall that $AC$ denotes the set of
absolutely continuous functions. For the interval $I$, $W^{1,\infty}(I)$
coincides with the space of Lipschitz continuous functions. If $u\in
W^{1,\infty}(0,\infty)$, then the Marchaud (type) derivatives (3.1) and (3.2)
are well defined almost everywhere. Indeed, with the first inequality of (3)
at hand, we use the fact that $u$ is essentially bounded and locally Lipschitz
almost everywhere (its derivative is bounded).

The operator $\mathbf{D}_{x+}^{\Phi}$ can be obtained by the Pillips’
representation ([25]) in the set of functions extended with zero on the
negative part of the real line. Indeed, for the shift semigroup
$\mathcal{S}_{y}u(x)=u(x-y)$ we have the representation

$\displaystyle\mathbf{D}_{x+}^{\Phi}u(x)=\int_{0}^{\infty}(u(x)-\mathcal{S}_{y}u(x))\Pi^{\Phi}(dy)$

for which

$\displaystyle\int_{0}^{\infty}e^{-\lambda
x}\mathbf{D}_{x+}^{\Phi}u(x)dx=\Phi(\lambda)\int_{0}^{\infty}e^{-\lambda
x}u(x)\,dx$ (3.4)

where we used (2.2) with $\kappa=0$ and $d=0$. We will investigate in a deep
detail the role of the Marchaud (type) derivatives
$\mathbf{D}_{x+}^{\Phi}u,\mathbf{D}_{x-}^{\Phi}u$ in the governing equations
of $H$. In order to define the Riemann-Liouville (type) derivatives on the
positive real line, we first consider a closed interval
$\bar{I}\subset(0,\infty)$ and $u\in AC(\bar{I})$. Then we extend the result
on $\mathbb{R}^{+}$ as in [18, page 79]. We now introduce the Riemann-
Liouville (type) derivatives

$\displaystyle\mathcal{D}_{(x,\infty)}^{\Phi}u(x):=-\frac{d}{dx}\int_{x}^{\infty}u(y)\overline{\Pi}^{\Phi}(y-x)dy$
(3.5)

and

$\displaystyle\mathcal{D}_{(0,x)}^{\Phi}u(x):=\frac{d}{dx}\int_{0}^{x}u(y)\overline{\Pi}^{\Phi}(x-y)dy$
(3.6)

respectively defined for function $u$ such that

$\displaystyle u(y)\overline{\Pi}^{\Phi}(y-x)\in
L^{1}(x,\infty),\quad\text{and}\quad u(y)\overline{\Pi}^{\Phi}(x-y)\in
L^{1}(0,x)\quad\forall x.$

Let us focus on (3.5). We observe that

$\displaystyle\mathcal{D}_{(x,\infty)}^{\Phi}u(x)$
$\displaystyle=-\frac{d}{dx}\int_{x}^{\infty}u(y)\overline{\Pi}^{\Phi}(y-x)dy$
$\displaystyle=\lim_{b\to\infty}-\frac{d}{dx}\int_{x}^{b}u(y)\overline{\Pi}^{\Phi}(y-x)dy$
$\displaystyle=\lim_{b\to\infty}-\frac{d}{dx}\int_{0}^{b-x}u(z+x)\overline{\Pi}^{\Phi}(z)dz$
$\displaystyle=\lim_{b\to\infty}u(b)\overline{\Pi}^{\Phi}(b-x)-\int_{0}^{b-x}u^{\prime}(z+x)\overline{\Pi}^{\Phi}(z)dz$
$\displaystyle=\lim_{b\to\infty}u(b)\overline{\Pi}^{\Phi}(b-x)-\int_{x}^{\infty}u^{\prime}(y)\overline{\Pi}^{\Phi}(y-x)dy$

where, from the final value theorem for the Laplace transform in (2.3), we
have

$\displaystyle 0=\Phi(0)=\lim_{b\to\infty}\overline{\Pi}^{\Phi}(b).$

Assuming that the growth of $u$ is asymptotically bounded, then

$\displaystyle\mathcal{D}_{(x,\infty)}^{\Phi}u(x)=-\int_{x}^{\infty}u^{\prime}(y)\overline{\Pi}^{\Phi}(y-x)dy.$
(3.7)

The right-hand side in (3.7) can be regarded as a Caputo-Dzherbashian (type)
derivative. A further relevant fact for (3.5) is the equivalence with (3.2).
If (3.7) holds, then

$\displaystyle\mathbf{D}_{x-}^{\Phi}u(x)=\mathcal{D}_{(x,\infty)}^{\Phi}u(x).$
(3.8)

Indeed, using first (3.7) and then the second formula in (2.3), we have

$\displaystyle\mathcal{D}_{(x,\infty)}^{\Phi}u(x)$
$\displaystyle=-\int_{x}^{\infty}u^{\prime}(y)\overline{\Pi}^{\Phi}(y-x)dy$
$\displaystyle=-\int_{0}^{\infty}\frac{d}{dy}u(x+y)\overline{\Pi}^{\Phi}(y)dy$
$\displaystyle=-\int_{0}^{\infty}\frac{d}{dy}u(x+y)\Pi^{\Phi}((y,\infty))dy$
$\displaystyle{=}-\int_{0}^{\infty}\int_{0}^{z}\frac{d}{dy}u(x+y)dy\Pi^{\Phi}(dz)$
$\displaystyle=-\int_{0}^{\infty}(u(x+z)-u(x))\Pi^{\Phi}(dz)$
$\displaystyle=\mathbf{D}_{x-}^{\Phi}u(x).$

Concerning (3.6), the analogous result for $\mathbf{D}_{x+}^{\Phi}$ and
$\mathcal{D}_{(0,x)}^{\Phi}$ can be proved.

For the operators (3.2) and (3.1) we introduce the following integration by
parts formula.

######  Theorem 3.1.

If $u,v\in W_{0}^{1,1}(0,\infty)$ and
$\mathbf{D}_{x-}^{\Phi}v(x),\mathbf{D}_{x+}^{\Phi}u(x)\in L^{1}(0,\infty)$,
then

$\displaystyle\int_{0}^{\infty}u(x)\left(\mathbf{D}_{x-}^{\Phi}v(x)\right)dx=\int_{0}^{\infty}\left(\mathbf{D}_{x+}^{\Phi}u(x)\right)v(x)dx.$
(3.9)

###### Proof.

First of all, from H$\ddot{o}$lder’s inequality, we observe that

$\displaystyle||u\mathbf{D}_{x-}^{\Phi}v||_{1}\leq||u||_{\infty}||\mathbf{D}_{x-}^{\Phi}v||_{1}<\infty$

because $\mathbf{D}_{x-}^{\Phi}v\in L^{1}(0,\infty)$ and
$W_{0}^{1,1}(0,\infty)$ embeds into $L^{\infty}(0,\infty)$ (see [22, section
11.2]). Similarly for $\mathbf{D}_{x+}$. By Fubini’s theorem, from the
definition (3.2),

$\displaystyle\int_{0}^{\infty}u(x)\left(\mathbf{D}_{x-}^{\Phi}v(x)\right)dx$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}u(x)[v(x)-v(x+y)]\Pi^{\Phi}(dy)dx$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}[u(x)-u(x-y)+u(x-y)][v(x)-v(x+y)]\Pi^{\Phi}(dy)dx.$

By using (3.1),

$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}[u(x)-u(x-y)]v(x)\Pi^{\Phi}(dy)dx=\int_{0}^{\infty}\left(\mathbf{D}_{x+}^{\Phi}u(x)\right)v(x)dx,$

and

$\displaystyle\int_{0}^{\infty}u(x)\left(\mathbf{D}_{x-}^{\Phi}v(x)\right)dx=\int_{0}^{\infty}\left(\mathbf{D}_{x+}^{\Phi}u(x)\right)v(x)dx\
+F(u,v)$

where

$\displaystyle F(u,v)$
$\displaystyle:=\int_{0}^{\infty}\int_{0}^{\infty}\left(u(x-y)[v(x)-v(x+y)]-[u(x)-u(x-y)]v(x+y)\right)\Pi^{\Phi}(dy)dx$
$\displaystyle=\int_{0}^{\infty}\int_{y}^{\infty}u(x-y)v(x)dx\Pi^{\Phi}(dy)-\int_{0}^{\infty}\int_{0}^{\infty}u(x)v(x+y)\Pi^{\Phi}(dy)dx$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}u(x)v(x+y)dx\Pi^{\Phi}(dy)-\int_{0}^{\infty}\int_{0}^{\infty}u(x)v(x+y)\Pi^{\Phi}(dy)dx$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}u(x)v(x+y)\Pi^{\Phi}(dy)dx-\int_{0}^{\infty}\int_{0}^{\infty}u(x)v(x+y)\Pi^{\Phi}(dy)dx=0,$

hence (3.9) holds. ∎

######  Remark 3.1.

The same result for left and right Marchaud derivatives
$\mathbf{D}_{x-}^{\alpha}$ and $\mathbf{D}_{x+}^{\alpha}$ when
$x\in\mathbb{R}$ is presented in [28, (6.27)] and in [20, Exercise 1.8.2].

We are now ready to discuss the connection between operators and
subordinators.

######  Theorem 3.2.

For $0<x<y$ and $t>0$, the density $h(t,y-x)$ of $x+H_{t}^{\Phi}$ solves

$\displaystyle\partial_{t}h=-\mathbf{D}_{y+}^{\Phi}h=-\mathbf{D}_{x-}^{\Phi}h$
(3.10)

with $h(0,y-x)=\delta(y-x)$, the Dirac delta function.

###### Proof.

First we focus on $\partial_{t}h=-\mathbf{D}_{x-}^{\Phi}h$. For $\lambda>0$,
consider the Laplace transform $\widetilde{h}(t,x,\lambda)$ of the density
$h(t,x,y)=h(t,y-x)$ given by

$\displaystyle\widetilde{h}(t,x,\lambda)=\mathbf{E}_{x}[e^{-\lambda
H_{t}^{\Phi}}]=\mathbf{E}_{0}[e^{-\lambda x-\lambda H_{t}^{\Phi}}]=e^{-\lambda
x-t\Phi(\lambda)},\quad x>0,\;t>0.$

From (3.8) and (3.7), we have

$\displaystyle-\mathbf{D}_{x-}^{\Phi}\widetilde{h}(t,x,\lambda)=$
$\displaystyle-\mathcal{D}_{(x,\infty)}^{\Phi}\widetilde{h}(t,x,\lambda)$
$\displaystyle=$ $\displaystyle\int_{x}^{\infty}\left(-\lambda e^{-\lambda
y-t\Phi(\lambda)}\right)\overline{\Pi}^{\Phi}(y-x)\,dy$ $\displaystyle=$
$\displaystyle-\lambda e^{-t\Phi(\lambda)}\int_{x}^{\infty}e^{-\lambda
y}\overline{\Pi}^{\Phi}(y-x)\,dy$ $\displaystyle=$ $\displaystyle-\lambda
e^{-\lambda x-t\Phi(\lambda)}\int_{0}^{\infty}e^{-\lambda
z}\overline{\Pi}^{\Phi}(z)\,dz$ $\displaystyle=$
$\displaystyle-\Phi(\lambda)\widetilde{h}(t,x,\lambda)$

where in the last step we have used (2.3). Thus, for $\lambda>0$, the function
$\widetilde{h}$ solves the equation

$\displaystyle\frac{\partial\widetilde{h}}{\partial
t}=-\mathbf{D}_{x-}^{\Phi}\widetilde{h},\quad\widetilde{h}(0,x)=e^{-\lambda
x},\quad x>0.$ (3.11)

We now focus on $\partial_{t}h=-\mathbf{D}_{y+}^{\Phi}h$ for $y>x>0$, $t>0$
for which we have that

$\displaystyle\mathbf{D}_{y+}^{\Phi}h(t,y-x)\mathbf{1}_{(y>x)}=-\int_{0}^{\infty}\left(h(t,y-s-x)\mathbf{1}_{(x,\infty)}(y-s)-h(t,y-x)\mathbf{1}_{(x,\infty)}(y)\right)\Pi^{\Phi}(ds)$

and therefore, the Laplace transform is given by

$\displaystyle\int_{0}^{\infty}e^{-\lambda
y}\mathbf{D}_{y+}^{\Phi}h(t,y-x)dy=$
$\displaystyle-\int_{0}^{\infty}\left(e^{-\lambda(s+x)-t\Phi(\lambda)}-e^{-\lambda
x-t\Phi(\lambda)}\right)\Pi^{\Phi}(ds)$ $\displaystyle=$
$\displaystyle-e^{-\lambda x-t\Phi(\lambda)}\int_{0}^{\infty}\left(e^{-\lambda
s}-1\right)\Pi^{\Phi}(ds)$ $\displaystyle=$
$\displaystyle\Phi(\lambda)e^{-\lambda x-t\Phi(\lambda)}.$

Thus, we get that

$\displaystyle\frac{\partial\widetilde{h}}{\partial
t}=-\Phi(\lambda)\widetilde{h}=\int_{0}^{\infty}e^{-\lambda
y}\left(-\mathbf{D}_{y+}^{\Phi}h\mathbf{1}_{(y>x)}\right)dy$

which is the claimed result. ∎

Let $M>0$ and $w\geq 0$. Let $\mathcal{M}_{w}$ be the set of (piecewise)
continuous function on $[0,\infty)$ of exponential order $w$ such that
$|u(t)|\leq Me^{wt}$. Let $u\in\mathcal{M}_{w}\cap C([0,\infty)$ with
$u^{\prime}\in\mathcal{M}_{w}$. Then we define the Caputo-Dzherbashian (type)
derivative as

$\displaystyle\mathfrak{D}_{x}^{\Phi}u(x):=\int_{0}^{x}u^{\prime}(y)\overline{\Pi}^{\Phi}(x-y)dy$
(3.12)

which is a convolution type operator. Indeed, the Laplace transform writes

$\displaystyle\int_{0}^{\infty}e^{-\lambda
x}\mathfrak{D}_{x}^{\Phi}u(x)dx=\Phi(\lambda)\tilde{u}(\lambda)-\frac{\Phi(\lambda)}{\lambda}u(0),\quad\lambda>w$
(3.13)

where, as above, $\widetilde{u}$ is the Laplace transform of $u$. For explicit
representations of the operator $\mathfrak{D}^{\Phi}_{x}$ see also the recent
works [19; 30; 7].

We remark that, from the Young’s inequality, we have

$\displaystyle||\mathfrak{D}_{x}^{\Phi}u||_{p}\leq||u^{\prime}||_{p}\left(\lim_{\lambda\to
0}\frac{\Phi(\lambda)}{\lambda}\right)$ (3.14)

where $\lim_{\lambda\to 0}\Phi(\lambda)/\lambda$ is finite only in some cases
(see [5]). If $\lim_{\lambda\to 0}\Phi(\lambda)/\lambda$ is finite, then
(3.14) gives a clearly information about the cases of measurable or
essentially bounded functions. In particular, we have that for $u^{\prime}\in
L^{1}(0,\infty)$, if

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

then (3.9) holds. Indeed, by considering the relation

$\displaystyle\mathfrak{D}_{x}^{\Phi}u(x)=\mathcal{D}_{(0,x)}^{\Phi}\left(u(x)-u(0)\right)$

we have equivalence between derivatives as $u(0)=0$. We can easily check that,
in this case, (3.13) coincides with (3.4).

For our purposes the operator (3.12) is used as time derivative for the
equation governing the inverse process $L^{\Phi}$, while (3.2) and (3.1) are
used as space derivatives for equations governing the process $H^{\Phi}$. Such
operators will be considered in the next section in the boundary conditions
associated with equations governing Brownian motions.

The governing equations of $H^{\Phi}$ and $L^{\Phi}$ have been studied in the
literature (see for example [19; 30]) with special attention only about the
fundamental solutions. For the reader’s convenience we conclude this section
by giving a clear statement for such equations based on the previous
discussion. Excluding densities $h$ with _time dependent property_ (as the
case of gamma subordinator), we have that

$\displaystyle C^{1,1}((0,\infty),W^{1,1}_{0}(0,\infty))\ni
h_{f}(t,x)=\int_{0}^{x}f(x-y)h(t,y)dy=\mathbf{E}_{0}[f(x-H^{\Phi}_{t})\mathbf{1}_{(t<L^{\Phi}_{t})}]$

is the unique solution to

$\begin{cases}\displaystyle\frac{\partial}{\partial
t}h_{f}(t,x)=-\mathbf{D}^{\Phi}_{x+}h_{f}(t,x),\quad(t,x)\in(0,\infty)\times(0,\infty)\\\
\displaystyle h_{f}(t,0)=0,\quad t>0\\\ \displaystyle h_{f}(0,x)=f(x)\in
W^{1,1}_{0}(0,\infty)\end{cases}$ (3.15)

and

$\displaystyle C^{1,1}(W^{1,\infty}(0,\infty),(0,\infty)\ni
l_{f}(t,x)=\int_{0}^{x}f(x-y)l(t,y)dy=\mathbf{E}_{0}[f(x-L^{\Phi}_{t})\mathbf{1}_{(t<H^{\Phi}_{t})}]$

is the unique solution to

$\begin{cases}\displaystyle\mathfrak{D}^{\Phi}_{t}l_{f}(t,x)=-\frac{\partial}{\partial
x}l_{f}(t,x),\quad(t,x)\in(0,\infty)\times(0,\infty)\\\ \displaystyle
l_{f}(t,0)=0,\quad t>0\\\ \displaystyle l_{f}(0,x)=f(x)\in
L^{p}(0,\infty)\end{cases}$ (3.16)

Notice that $l_{f}(t,\cdot)\in L^{p}(0,\infty),\forall t>0$.

Let us write

$\bar{h}_{f}(x)=\int_{0}^{\infty}h_{f}(t,x)dt\quad\textrm{and}\quad\bar{l}_{f}(x)=\int_{0}^{\infty}l_{f}(t,x)dt.$
(3.17)

We can immediately check that the Abel (type) equation
$f(x)=\mathbf{D}^{\Phi}_{x+}\bar{h}_{f}(x)$ gives the elliptic problem
associated with (3.15). On the other hand, the elliptic problem associated
with (3.16) exists only if $\lim_{\lambda\downarrow
0}\Phi(\lambda)/\lambda<\infty$, this gives a clear meaning to (3.14). Such a
result is not surprising, indeed by considering $f=\mathbf{1}$,

$\displaystyle\bar{l}_{\mathbf{1}}(x)=\int_{0}^{\infty}\mathbf{E}_{0}[\mathbf{1}_{(t<H^{\Phi}_{x})}]dt=\mathbf{E}_{0}[H^{\Phi}_{x}]=x\lim_{\lambda\downarrow
0}\frac{\Phi(\lambda)}{\lambda}$

In case the elliptic problem exists, it takes the form

$\displaystyle f(x)=\left(\lim_{\lambda\downarrow
0}\frac{\lambda}{\Phi(\lambda)}\right)\frac{\partial}{\partial
x}\bar{l}_{f}(x)$

Moving to the elliptic problems associated with (3.10) we notice that the
solution to

$\displaystyle-\mathbf{D}_{x-}^{\Phi}w_{-}(x)=-f(x)+\lambda w_{-}(x),\quad
x>0,\lambda>0$ (3.18)

is given by

$\displaystyle w_{-}(x)=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(x+H_{t}^{\Phi})dt\right].$ (3.19)

Indeed, from (3.19),

$\displaystyle w_{-}(x)=$
$\displaystyle\int_{x}^{\infty}f(y)\int_{0}^{\infty}e^{-\lambda
t}h(t,y-x)\,dt\,dy$

and by using (3.10),

$\displaystyle-\mathbf{D}_{x-}^{\Phi}w_{-}(x)=$
$\displaystyle\int_{x}^{\infty}f(y)\int_{0}^{\infty}e^{-\lambda
t}(-\mathbf{D}_{x-}^{\Phi}h(t,y-x))\,dt\,dy$ $\displaystyle=$
$\displaystyle\int_{x}^{\infty}f(y)\int_{0}^{\infty}e^{-\lambda
t}(\partial_{t}h(t,y-x))\,dt\,dy$ $\displaystyle=$
$\displaystyle\int_{x}^{\infty}f(y)\left(-\delta(y-x)\right)dy+\lambda\int_{x}^{\infty}f(y)\int_{0}^{\infty}e^{-\lambda
t}h(t,y-x)\,dt\,dy$ $\displaystyle=$ $\displaystyle-f(x)+\lambda w_{-}(x).$

Thus, the solution to (3.18) is given by (3.19). Similarly, we observe that
the solution to

$\displaystyle-\mathbf{D}_{y+}^{\Phi}w_{+}(y)=-f(y)+\lambda w_{+}(y),\quad
y>0,\lambda>0$ (3.20)

is given by

$\displaystyle w_{+}(y)=\int_{0}^{y}\int_{0}^{\infty}e^{-\lambda
t}h(t,y-x)dtf(x)dx.$

Indeed, by using (3.10), we have

$\displaystyle-\mathbf{D}_{y+}^{\Phi}w_{+}(y)=$
$\displaystyle\int_{0}^{y}f(x)\int_{0}^{\infty}e^{-\lambda
t}(-\mathbf{D}_{y+}^{\Phi}h(t,y-x))\,dt\,dx$ $\displaystyle=$
$\displaystyle\int_{0}^{y}f(x)\int_{0}^{\infty}e^{-\lambda
t}(\partial_{t}h(t,y-x))\,dt\,dx$ $\displaystyle=$
$\displaystyle\int_{0}^{y}f(x)\left(-\delta(y-x)\right)dx+\lambda\int_{0}^{y}f(x)\int_{0}^{\infty}e^{-\lambda
t}h(t,y-x)\,dt\,dx$ $\displaystyle=$ $\displaystyle-f(y)+\lambda w_{+}(y).$

######  Remark 3.2.

If $\lambda\to 0$ in (3.20), then the solution can be written as

$\displaystyle w_{+}(y)=(\mathcal{I}^{\Phi}f)(y)$

where $\mathcal{I}^{\Phi}$ is a non-local integral associated to $\Phi$.
Indeed, the convolution between $f$ and $\int_{0}^{\infty}h(t,x)dt$ can be
represented as $\mathcal{I}^{\Phi}$. For example, for the $\alpha-$stable
subordinator we have that

$\displaystyle\int_{0}^{\infty}\mathbf{E}_{0}[e^{-\xi
H_{t}}]dt=\frac{1}{\xi^{\alpha}}\quad\text{where}\quad\frac{1}{\xi^{\alpha}}=\int_{0}^{\infty}e^{-\xi
x}\frac{x^{\alpha-1}}{\Gamma(\alpha)}dx.$

Then, we deduce that

$\displaystyle\int_{0}^{\infty}h(t,x)dt=\frac{x^{\alpha-1}}{\Gamma(\alpha)}$

must be considered in order to write the fractional integral.

##  4\. Non-local boundary conditions

Let us consider the heat equation on the positive half line with a Neumann
boundary condition at $x=0$, that is

$\displaystyle\begin{cases}\frac{\partial}{\partial t}u(t,x)=\Delta
u(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\ \frac{\partial}{\partial
x}u(t,x)\big{|}_{x=0}=0\quad\quad\quad\quad\quad&t>0\end{cases}.$

The probabilistic representation of the solution can be written in terms of
the reflecting Brownian motion $B^{+}=\\{B_{t}^{+},t\geq 0\\}$. We denote by
$\gamma_{t}=\gamma_{t}(B^{+})$ the local time at zero of the reflecting
Brownian motion $B^{+}$ written as

$\displaystyle\gamma_{t}=\int_{0}^{t}\mathbf{1}_{(B_{s}^{+}=0)}ds.$

Recall that $H^{\Phi}$ is the subordinator associated to $\Pi^{\Phi}$ through
(2.2) and $L^{\Phi}$ is the inverse of $H^{\Phi}$. We introduce the main
object we deal with, that is the process $B^{\bullet}=\\{B_{t}^{\bullet},t\geq
0\\}$ written as

$\displaystyle
B_{t}^{\bullet}=H^{\Phi}L^{\Phi}\gamma_{t}-\gamma_{t}+B_{t}^{+},\quad t>0.$
(4.1)

We use the following notation

$\displaystyle H^{\Phi}L^{\Phi}\gamma_{t}=H^{\Phi}\circ
L^{\Phi}\circ\gamma_{t}$

for the composition of $H^{\Phi},L^{\Phi},\gamma$. The process
$L^{\Phi}\gamma_{t}$ can be regarded as the local time of $B^{\bullet}$ at
zero [15, section 14]. We also observe that $B_{t}=B_{t}^{+}-\gamma_{t}(B)$ in
law, from the Tanaka’s formula we may consider an alternative representation
for the process (4.1) given by
$B_{t}^{\bullet}=H^{\Phi}L^{\Phi}\gamma_{t}+B_{t}$, where $B=\\{B_{t},t\geq
0\\}$ is the one-dimensional Brownian motion with local time at zero
$\gamma_{t}(B)$.

The probabilistic reading of (4.1) says that $B^{\bullet}$ is a reflecting
Brownian motion jumping away from the origin, with a length of the jump given
by the subordinator $H^{\Phi}$ running with the clock $L^{\Phi}\gamma_{t}$.
Since $\Pi^{\Phi}(0,\infty)=\infty$, then in each time interval the number of
jumps of $H^{\Phi}$ are infinite. A detailed analysis of the sample paths has
be given in [15, section 12]. We provide a discussion at the end of the paper.

We remark that $L^{\Phi}H^{\Phi}=t$ almost surely, whereas the composition
$H^{\Phi}L^{\Phi}$ is a bit demanding. In particular, for the subordinator
$H^{\Phi}$ and the inverse $L^{\Phi}$ we have that
$\mathbf{P}_{0}(H^{\Phi}{L^{\Phi}_{t}}-<t=H^{\Phi}{L^{\Phi}_{t}})=0$ ([2,
Proposition 2, section III.2]). Moreover, if $d=0$ in (2.2), then
$\mathbf{P}_{0}(H^{\Phi}{L^{\Phi}_{t}}>t)=1$ for every $t>0$ ([2, Theorem 4,
section III.2]). Thus, for a zero drift subordinator, the passage above any
given level is almost surely realized by a jump.

Now we focus on the problem with non-local dynamic boundary conditions and
provide the probabilistic representation of the solution in terms of Brownian
motions. We deal with positive functions $u$ on unbounded domain for which
$|u(t,x)|\leq Me^{cx^{2}}$, for $M,c>0$. Thus, standard arguments guarantees
uniqueness of the solution in terms of positivity (Widder’s theorem [31]) and
exponential growth ([11, section 2.3.3]) .

Let us introduce $D_{0}=D_{1}\cup D_{2}$ where

$\displaystyle D_{1}=\left\\{\varphi:\forall\,x,\;\varphi(\cdot,x)\in
W^{1,\infty}(0,\infty)\;\;\textrm{and}\;\lim_{x\downarrow
0}\mathfrak{D}^{\Psi}_{t}\varphi(t,x)\;\textrm{exists}\right\\},$

$\displaystyle D_{2}=\left\\{\varphi:\forall\,t,\;\varphi(t,\cdot)\in
W^{1,\infty}(0,\infty)\;\textrm{and}\;\;\lim_{x\downarrow
0}\mathbf{D}^{\Phi}_{x-}\varphi(t,x)\;\textrm{exists}\right\\}.$

######  Theorem 4.1.

The solution $\upsilon\in C^{1,2}((0,\infty),(0,\infty))\cap D_{0}$ to the
problem

$\displaystyle\begin{cases}\frac{\partial}{\partial
t}\upsilon(t,x)=\Delta\upsilon(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\eta\mathfrak{D}_{t}^{\Psi}\upsilon(t,x)\big{|}_{x=0}=-\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0}\quad&t>0\\\
\upsilon(0,x)=f(x)\quad&x\geq 0\end{cases}$

with $f\in C[0,\infty)\cap L^{\infty}(0,\infty)$ and $\eta\geq 0$ is unique.
Moreover, the solution $\upsilon$ has the probabilistic representation

$\displaystyle\upsilon(t,x)=\mathbf{E}_{x}\left[f(B^{\bullet}_{S_{t}^{-1}})\right],$

where $B^{\bullet}_{t}$ is defined in (4.1) and $S_{t}^{-1}$ is the right-
inverse of $S_{t}=t+H^{\Psi}\eta L^{\Phi}_{\gamma_{t}}$.

###### Proof.

The solution $\upsilon$ can be written as

$\displaystyle\upsilon(t,x)=Q_{t}^{D}f(x)+\int_{0}^{t}\frac{x}{\tau}g(\tau,x)\upsilon(t-\tau,0)d\tau$
(4.2)

where

$\displaystyle Q_{t}^{D}f(x)=\int_{0}^{\infty}(g(t,x-y)-g(t,x+y))f(y)dy$

is the Dirichlet semigroup and $g(t,z)=e^{-z^{2}/4t}/\sqrt{4\pi t}$ for which
we recall the well-known transforms

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}g(t,x)dt=\frac{1}{2}\frac{e^{-x\sqrt{\lambda}}}{\sqrt{\lambda}},\quad\lambda>0$
(4.3)

and

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\frac{x}{t}g(t,x)dt=e^{-x\sqrt{\lambda}},\quad\lambda>0.$ (4.4)

Assume that

$\displaystyle\upsilon(t,x)=\mathbf{E}_{x}\left[f(B^{\bullet}\circ
S_{t}^{-1})\right],$ (4.5)

and write the $\lambda-$potential

$\displaystyle
R_{\lambda}f(x)=\mathbf{E}_{x}\left[\int_{0}^{\infty}e^{-\lambda
t}f(B^{\bullet}\circ S_{t}^{-1})dt\right],\quad\lambda>0.$ (4.6)

From the representation (4.2) we must have

$\displaystyle R_{\lambda}f(x)=R^{D}_{\lambda}f(x)+\bar{R}_{\lambda}f(x)$

where, from (4.3),

$\displaystyle R^{D}_{\lambda}f(x)$
$\displaystyle=\int_{0}^{\infty}e^{-\lambda t}Q_{t}^{D}f(x)dt$
$\displaystyle=\frac{1}{2}\int_{0}^{\infty}\left(\frac{e^{-|x-y|\sqrt{\lambda}}}{\sqrt{\lambda}}-\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}\right)f(y)dy$
(4.7)
$\displaystyle=\frac{1}{2}\int_{0}^{\infty}\frac{e^{(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}f(y)dy-\frac{1}{2}\int_{0}^{\infty}\frac{e^{-(x+y)\sqrt{\lambda}}}{\sqrt{\lambda}}f(y)dy\
+$
$\displaystyle-\frac{1}{2}\int_{0}^{x}\left(\frac{e^{(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}-\frac{e^{-(x-y)\sqrt{\lambda}}}{\sqrt{\lambda}}\right)f(y)dy,\quad\lambda>0$

and, from (4.4),

$\displaystyle\bar{R}_{\lambda}f(x)=e^{-x\sqrt{\lambda}}R_{\lambda}f(0),\quad\lambda>0$

with

$\displaystyle R_{\lambda}f(x)=\int_{0}^{\infty}e^{-\lambda
t}\upsilon(t,x)dt,\quad\lambda>0$

We can easily check that

$\displaystyle\Delta R_{\lambda}f(x)=\lambda
R_{\lambda}f(x)-f(x)=\int_{0}^{\infty}e^{-\lambda t}\frac{\partial}{\partial
t}\upsilon(t,x)dt,\quad\lambda>0,x>0.$ (4.8)

We have to characterize $R_{\lambda}f(0)$ and therefore $\upsilon(t,0)$ in
order to obtain a solution of the form given in (4.2). Thus, we focus on the
boundary condition for which, first we observe that

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}_{t}^{\Psi}\upsilon(t,x)\big{|}_{x=0}dt=\int_{0}^{\infty}e^{-\lambda
t}\mathfrak{D}_{t}^{\Psi}\upsilon(t,x)dt\bigg{|}_{x=0},\quad\lambda>0$

and, thanks to (3.13), we have

$\displaystyle\int_{0}^{\infty}e^{-\lambda
t}\eta\mathfrak{D}_{t}^{\Psi}\upsilon(t,x)dt\bigg{|}_{x=0}=\eta\Psi(\lambda)R_{\lambda}f(x)-\eta\frac{\Psi(\lambda)}{\lambda}f(0),\quad\lambda>0.$
(4.9)

A key point in our proof is that

$\displaystyle\mathbf{D}_{x-}^{\Phi}R_{\lambda}f(x)\big{|}_{x=0}=\big{(}\mathbf{D}_{x-}^{\Phi}R_{\lambda}^{D}f(x)+\mathbf{D}_{x-}^{\Phi}\bar{R}_{\lambda}f(x)\big{)}\big{|}_{x=0}$

can be written by considering the definition (3.2) and by taking into account
the properties of $R_{\lambda}f$. In particular, we use the fact that

$\displaystyle\lim_{x\downarrow
0}\int_{0}^{\infty}\left(R^{D}_{\lambda}f(x)-R^{D}_{\lambda}f(x+y)\right)\Pi^{\Phi}(dy)$
$\displaystyle=\int_{0}^{\infty}\left(R^{D}_{\lambda}f(0)-R^{D}_{\lambda}f(y)\right)\Pi^{\Phi}(dy)$
$\displaystyle=-\int_{0}^{\infty}R^{D}_{\lambda}f(y)\,\Pi^{\Phi}(dy)$

and

$\displaystyle\lim_{x\downarrow
0}\int_{0}^{\infty}\left(\bar{R}_{\lambda}f(x)-\bar{R}_{\lambda}f(x+y)\right)\Pi^{\Phi}(dy)$
$\displaystyle=\lim_{x\downarrow
0}\int_{0}^{\infty}\left(e^{-x\sqrt{\lambda}}-e^{-(x+y)\sqrt{\lambda}}\right)\Pi^{\Phi}(dy)\,R_{\lambda}f(0)$
$\displaystyle=\Phi(\sqrt{\lambda})\,R_{\lambda}f(0).$

Notice that, for $|f(x)|\leq K$,

$\displaystyle\int_{0}^{\infty}R^{D}_{\lambda}f(y)\,\Pi^{\Phi}(dy)\leq\frac{K}{\lambda}\int_{0}^{\infty}(1-e^{-\sqrt{\lambda}y})\,\Pi^{\Phi}(dy)=\frac{K}{\lambda}\Phi(\sqrt{\lambda}),\quad\lambda>0.$

We therefore obtain

$\displaystyle-\mathbf{D}_{x-}^{\Phi}R_{\lambda}f(x)\big{|}_{x=0}=\int_{0}^{\infty}R^{D}_{\lambda}f(y)\,\Pi^{\Phi}(dy)-\Phi(\sqrt{\lambda})\,R_{\lambda}f(0).$

This together with (4.9) leads to

$\displaystyle
R_{\lambda}f(0)=\frac{\frac{\eta}{\lambda}\Psi(\lambda)f(0)+\int_{0}^{\infty}R_{\lambda}^{D}f(l)\
\Pi^{\Phi}(dy)}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})},\quad\lambda>0.$
(4.10)

Now we observe that

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}f(B^{\bullet}\circ
S_{t}^{-1})dt\right]=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
S_{t}}f(B^{\bullet}_{t})dS_{t}\right]=I_{1}^{\Psi}+I_{2}^{\Psi}$

where, due to the fact that $S_{t}=t+H^{\Psi}\eta L^{\Phi}_{\gamma_{t}}$,

$\displaystyle I_{1}^{\Psi}=\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}e^{-\lambda H^{\Psi}\eta
L^{\Phi}_{\gamma_{t}}}f(B^{\bullet}_{t})dt\right],\quad\lambda>0$

and

$\displaystyle
I_{2}^{\Psi}=f(0)\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}e^{-\lambda H^{\Psi}\eta L^{\Phi}_{\gamma_{t}}}dH^{\Psi}\eta
L^{\Phi}_{\gamma_{t}}\right],\quad\lambda>0$

can be obtained as follows. The computation of $I_{1}^{\Psi}$ follows from the
computation of $e_{1}$ in [15, page 215], recalling that

$\displaystyle\mathbf{E}_{0}[e^{-\lambda H^{\Psi}_{\eta t}}]=e^{-\eta
t\Psi(\lambda)},\quad\lambda>0.$

Thus, we have that

$\displaystyle I_{1}^{\Psi}=\frac{\int_{0}^{\infty}R_{\lambda}^{D}f(l)\
\Pi^{\Phi}(dl)}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})},\quad\lambda>0.$

For $I_{2}^{\Psi}$, we get that

$\displaystyle I_{2}^{\Psi}$
$\displaystyle=f(0)\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}e^{-\lambda H^{\Psi}\eta L^{\Phi}_{\gamma_{t}}}d(H^{\Psi}\eta
L^{\Phi}_{\gamma_{t}})\right]$
$\displaystyle=f(0)\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\gamma^{-1}H^{\Phi}_{t}}e^{-\lambda
H^{\Psi}_{\eta t}}dH^{\Psi}_{\eta t}\right]$
$\displaystyle=\frac{f(0)}{\lambda}\mathbf{E}_{0}\left[-1-\int_{0}^{\infty}e^{-\lambda
H^{\Psi}_{\eta t}}d\left(e^{-\gamma^{-1}H^{\Phi}_{t}}\right)\right]$
$\displaystyle=\frac{f(0)}{\lambda}\mathbf{E}_{0}\left[-1-\int_{0}^{\infty}e^{-\eta
t\Psi(\lambda)}d\left(e^{-\gamma^{-1}H^{\Phi}_{t}}\right)\right]$
$\displaystyle=\frac{f(0)}{\lambda}\mathbf{E}_{0}\left[-1-\left(-1-\eta\Psi(\lambda)\int_{0}^{\infty}e^{-\eta
t\Psi(\lambda)}e^{-\lambda\gamma^{-1}{H^{\Phi}_{t}}}dt\right)\right]$
$\displaystyle=\frac{\eta}{\lambda}\Psi(\lambda)f(0)\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\eta
t\Psi(\lambda)}e^{-\lambda\gamma^{-1}H^{\Phi}_{t}}dt\right]$
$\displaystyle=\frac{\eta}{\lambda}\Psi(\lambda)f(0)\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\eta\Psi(\lambda)L^{\Phi}_{\gamma_{t}}}e^{-\lambda
t}dL^{\Phi}_{\gamma_{t}}\right],\quad\lambda>0.$

Now set $A_{t}:=L^{\Phi}_{\gamma_{t}}$. Then

$\displaystyle\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\eta\Psi(\lambda)A_{t}}e^{-\lambda
t}dA_{t}\right]$
$\displaystyle=-\frac{1}{\eta\Psi(\lambda)}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}de^{-\eta\Psi(\lambda)A_{t}}\right]$
$\displaystyle=-\frac{1}{\eta\Psi(\lambda)}\mathbf{E}_{0}\left[-1+\lambda\int_{0}^{\infty}e^{-\lambda
t}e^{-\eta\Psi(\lambda)A_{t}}dt\right]$
$\displaystyle=\frac{1}{\eta\Psi(\lambda)}-\frac{\lambda}{\eta\Psi(\lambda)}\mathbf{E}_{0}\left[\int_{0}^{\infty}e^{-\lambda
t}e^{-\eta\Psi(\lambda)L^{\Phi}_{\gamma_{t}}}dt\right]$
$\displaystyle=\frac{1}{\eta\Psi(\lambda)}-\frac{\lambda}{\eta\Psi(\lambda)}\mathbf{E}_{0}\left[\int_{0}^{\infty}\frac{\sqrt{\lambda}}{\lambda}e^{-\sqrt{\lambda}w}e^{-\eta\Psi(\lambda)L^{\Phi}_{w}}dw\right]$
$\displaystyle=\frac{1}{\eta\Psi(\lambda)}-\frac{\lambda}{\eta\Psi(\lambda)}\int_{0}^{\infty}\frac{\sqrt{\lambda}}{\lambda}e^{-\eta\Psi(\lambda)z}\frac{\Phi(\sqrt{\lambda})}{\sqrt{\lambda}}e^{-z\Phi(\sqrt{\lambda})}dz$
$\displaystyle=\frac{1}{\eta\Psi(\lambda)}-\frac{\Phi(\sqrt{\lambda})}{\eta\Psi(\lambda)}\frac{1}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})}$
$\displaystyle=\frac{1}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})},$

where we have used (2.5). We arrive at

$\displaystyle
I_{2}^{\Psi}=\frac{\eta}{\lambda}\Psi(\lambda)f(0)\frac{1}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})},\quad\lambda>0.$

By collecting all pieces together, we obtain

$\displaystyle I_{1}^{\Psi}+I_{2}^{\Psi}$
$\displaystyle=\frac{\int_{0}^{\infty}R_{\lambda}^{D}f(l)\
\Pi^{\Phi}(dl)}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})}+\frac{\eta}{\lambda}\Psi(\lambda)f(0)\frac{1}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})}$
$\displaystyle=\frac{\frac{\eta}{\lambda}\Psi(\lambda)f(0)+\int_{0}^{\infty}R_{\lambda}^{D}f(l)\
\Pi^{\Phi}(dl)}{\eta\Psi(\lambda)+\Phi(\sqrt{\lambda})}.$

which coincides with (4.10). Thus, we conclude that (4.6) and (4.5) hold true.
In particular, $\upsilon$ is a continuous function with Laplace transform
$R_{\lambda}f$, then $\upsilon$ is unique. ∎

######  Remark 4.1.

The lifetime of the process $B_{S^{-}1}^{\bullet}$ (which basically moves on a
given path of a reflecting Brownian motion) is infinite, that is
$R_{0}\mathbf{1}=\infty$ and $\forall\,x$, $\upsilon(\cdot,x)\notin
L^{1}(0,\infty)$ for a bounded initial datum $f$. However, there is no hope to
find $\upsilon(\cdot,x)\in L^{1}(0,\infty)$ by also asking for $f\in
L^{1}(0,\infty)$ for which we only have

$\displaystyle R_{0}^{D}f(x)=2\int_{0}^{\infty}(y\wedge x)f(y)dy<\infty$

in the potential $R_{\lambda}f=R_{\lambda}^{D}f+\bar{R}_{\lambda}f$. The last
formula can be obtained from (4). Then we ask for $f\in L^{\infty}$ in order
to obtain $\upsilon(\cdot,x)\in L^{\infty}(0,\infty)$, $\forall x$.

We now briefly discuss the composition $H^{\Psi}L^{\Phi}$ of a subordinator
with an inverse to a subordinator. The process can be seen as a non-Markovian
time-changed process, being the random time $L^{\Phi}$ non-Markovian. We use
the following notation, for a set $I\subset(0,\infty)$,

$\displaystyle\mathbf{P}_{0}(H_{t}^{\Psi}\in
I)=\int_{I}h^{\Psi}(t,x)dx\quad\mathbf{P}_{0}(L_{t}^{\Phi}\in
I)=\int_{I}l^{\Phi}(t,x)dx,$

and

$\displaystyle\mathbf{P}_{0}(H^{\Psi}L^{\Phi}_{t}\in I)=\int_{I}w(t,x)dx.$

We have that

$\displaystyle\mathfrak{D}_{t}^{\Phi}w(t,x)=-\mathbf{D}_{x+}^{\Psi}w(t,x),\quad
t>0,x>0$ (4.11)

where $\mathfrak{D}_{t}^{\Phi}$ is defined in (3.12) and
$\mathbf{D}_{x+}^{\Psi}$ is defined in (3.1). For the reader’s convenience we
give the following sketch of proof. By applying the $\lambda-$time Laplace
transform in both sides of (4.11), from (3.13), we obtain

$\displaystyle\Phi(\lambda)\widetilde{w}(\lambda,x)-\frac{\Phi(\lambda)}{\lambda}\delta(x)=-\mathbf{D}_{x+}^{\Psi}\widetilde{w}(\lambda,x).,\quad\lambda>0,t>0.$

From (3.4), by taking into account the $\xi-$space Laplace transform, we have

$\displaystyle\Phi(\lambda)\widetilde{w}(\lambda,\xi)-\frac{\Phi(\lambda)}{\lambda}=-\Psi(\xi)\widetilde{w}(\lambda,\xi),\quad\lambda>0,\xi>0$

which leads to

$\displaystyle\widetilde{w}(\lambda,\xi)=\frac{\Phi(\lambda)}{\lambda}\frac{1}{\Phi(\lambda)+\Psi(\xi)},\quad\lambda>0,\xi>0.$

The inverse of $\widetilde{w}(\lambda,\xi)$ gives the density function of
$H^{\Psi}L^{\Phi}_{t}$ which is written as

$\displaystyle w(t,x)=\int_{0}^{\infty}h^{\Psi}(s,x)l^{\Phi}(t,s)ds,\quad
t>0,x>0$

as the independence between $H^{\Psi}$ and $L^{\Phi}_{t}$ entails.

Concerning the special case of dependent processes, that is $L^{\Phi}$ the
inverse of $H^{\Phi}$, we focus on the case of stable subordinators.

######  Proposition 4.1.

Let $H^{\alpha}$ be an $\alpha-$stable subordinator and $L^{\alpha}$ its
inverse. We have:

  * (i)

For $z>t>0$

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{t}}\in
dz)=\frac{\sin(\pi\alpha)}{\pi}\frac{1}{z}\left(\frac{t}{z-t}\right)^{\alpha}dz.$

  * (ii)

For $0<y<t$

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{t}-}\in
dy)=\frac{\sin(\pi\alpha)}{\pi}\frac{1}{y}\left(\frac{y}{(t-y)}\right)^{\alpha}dy.$

###### Proof.

(i) From [2, Proposition 2, section III.2] we have that, for each $t\geq 0$
and $0\leq y\leq t<z$,

$\displaystyle\mathbf{P}_{0}(H^{\alpha}L^{\alpha}_{t}-\in
dy,H^{\alpha}L^{\alpha}_{t}\in dz)=U(dy)\Pi^{\alpha}(dz-y),$ (4.12)

where $U(dy)$ is the potential measure for which

$\displaystyle\int_{0}^{\infty}e^{-\lambda
y}U(dy)=\frac{1}{\lambda^{\alpha}},\quad\lambda>0.$

It turns out that $\Pi^{\alpha}$ and $U$ are written as

$\displaystyle\Pi^{\alpha}(dz)=\frac{\alpha}{\Gamma(1-\alpha)}\frac{dz}{z^{\alpha+1}}\quad\text{and}\quad
U(dy)=\frac{1}{\Gamma(\alpha)}\frac{dy}{y^{-\alpha+1}}.$

Hence, by integrating (4.12),

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{t}}\in dz)$
$\displaystyle=\int_{0}^{t}\mathbf{P}_{0}(H^{\alpha}L^{\alpha}_{t^{-}}\in
dy,H^{\alpha}L^{\alpha}_{t}\in dz)$
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha){\Gamma(\alpha)}}\int_{0}^{t}\frac{y^{\alpha-1}}{(z-y)^{\alpha+1}}dy\,dz$
$\displaystyle=\frac{1}{\Gamma(\alpha)\Gamma(1-\alpha)}\frac{1}{z}\left(\frac{t}{z-t}\right)^{\alpha}\,dz.$

(ii) From [3, Lemma 1.10], we have

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{t}-}\in
dy)=\overline{\Pi}^{\alpha}(t-y)U(dy).$

We obtain

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{t}-}\in
dy)=\frac{1}{\Gamma(1-\alpha)}\frac{1}{(t-y)^{\alpha}}\frac{y^{\alpha-1}}{\Gamma(\alpha)}dy.$

which is the claim. ∎

We observe that, for $t=1$, $H^{\alpha}{L^{\alpha}_{1}-}$ has the so-called
generalized arcsine law

$\displaystyle\mathbf{P}_{0}(H^{\alpha}{L^{\alpha}_{1}-}\in
dy)=\frac{\sin\alpha\pi}{\pi}y^{\alpha-1}(1-y)^{-\alpha}dy,$

and for $\alpha=\frac{1}{2}$ we get the well known arcsine law for the
standard Brownian motion.

###  4.1. The case $\eta=0$: Non-local space conditions

For $\eta=0$ in Theorem 4.1, since $H^{\Psi}_{0}=0$, the probabilistic
representation of the solution $\upsilon$ can be written only in terms of
$B^{\bullet}_{t}$. We observe that

$\displaystyle\int_{0}^{\infty}(\upsilon(t,l)-\upsilon(t,0))\Pi^{\Phi}(dl)$
$\displaystyle=\lim_{x\to
0}\int_{0}^{\infty}(\upsilon(t,x+l)-\upsilon(t,x))\Pi^{\Phi}(dl)$
$\displaystyle=-\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0},$ (4.13)

then our Marchaud (type) derivative coincides with the integral operator
defined in [12]. As a confirmation of this, K. Itô and P. McKean in [15,
section 12] introduced the process $B^{\bullet}$ to write the probabilistic
representation of the solution of

$\displaystyle\begin{cases}\frac{\partial}{\partial
t}\upsilon(t,x)=\Delta\upsilon(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\int_{0}^{\infty}(\upsilon(t,l)-\upsilon(t,0))\Pi^{\Phi}(dl)=0\quad&t>0.\end{cases}$

The identification of the integral operator with a Marchaud-type derivative
allows us to use some techniques from fractional calculus. Thus we write

$\displaystyle\begin{cases}\frac{\partial}{\partial
t}\upsilon(t,x)=\Delta\upsilon(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0}=0\quad&t>0.\end{cases}$
(4.14)

We now discuss the special case of Marchaud derivatives and $\alpha-$stable
subordinators $H^{\alpha}$, characterized by

Figure 1. A simulation of the $\alpha-$stable subordinator if $\alpha=0.5$,
based on [24, section 5.2].

$\displaystyle\Phi(\lambda)=\lambda^{\alpha}=\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{\infty}(1-e^{-\lambda
x})\frac{dx}{x^{\alpha+1}},\quad\alpha\in(0,1).$

Then, we write $\mathbf{D}_{x-}^{\alpha}$ in place of $\mathbf{D}_{x-}^{\Phi}$
which coincides with the well known Marchaud left derivative. We now use the
fact that (see [18, (2.2.5)] for example)

$\displaystyle\mathbf{D}_{x+}^{\alpha}u(x)$ $\displaystyle\to u(x)\quad\ \
\alpha\downarrow 0,$ (4.15) $\displaystyle\mathbf{D}_{x+}^{\alpha}u(x)$
$\displaystyle\to u^{\prime}(x)\quad\ \alpha\uparrow 1,$ (4.16)

in order to arrive at the analogue

$\displaystyle\mathbf{D}_{x-}^{\alpha}u(x)$ $\displaystyle\to u(x)\quad\ \
\alpha\downarrow 0,$ (4.17) $\displaystyle\mathbf{D}_{x-}^{\alpha}u(x)$
$\displaystyle\to-u^{\prime}(x)\quad\alpha\uparrow 1.$ (4.18)

Indeed, formula (3.9) which still holds for $\mathbf{D}_{x\pm}^{\Phi}u\in
L^{\infty}(0,\infty)$, together with (4.15) and (4.16) say that (4.17) and
(4.18) hold in some sense. Thus, roughly speaking, for $\alpha\downarrow 0$
the problem (4.14) takes the form

$\displaystyle\begin{cases}\frac{\partial}{\partial
t}\upsilon(t,x)=\Delta\upsilon(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\upsilon(t,0)=0\quad&t>0.\end{cases}$ (4.19)

and for $\alpha\uparrow 1$, (4.14) becomes

$\displaystyle\begin{cases}\frac{\partial}{\partial
t}\upsilon(t,x)=\Delta\upsilon(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\frac{\partial}{\partial x}\upsilon(t,x)\big{|}_{x=0}=0\quad&t>0.\end{cases}$
(4.20)

As reported in [3, Section 3.1.1], for $\alpha=0$ the subordinator dies
immediately, and for $\alpha=1$, it is the elementary subordinator $t$. Then,
the process (4.1) for $\alpha\downarrow 0$ becomes a reflected Brownian motion
killed at $x=0$, that is a killed Brownian motion as expected for the solution
of (4.19). For $\alpha\uparrow 1$, the subordinator becomes $H_{t}^{\alpha}=t$
and the process (4.1) is a reflected Brownian motion that does not jump at the
boundary point. It reflects and never dies. Indeed, it coincides with a
reflected Brownian motion on $[0,\infty)$ as expected for the solution of
(4.20).

Figure 2. To the left a simulation of the $\alpha-$stable subordinator if
$\alpha=0.01$, near zero. To the right a simulation of the $\alpha-$stable
subordinator if $\alpha=0.99$, near one.

###  4.2. Non-local dynamic conditions. The case $\Phi=Id$

If $\Phi=Id$, from (4.1), we have $B^{\bullet}=B^{+}$. Then, Theorem 4.1
generalizes [10, Theorem 3.1] and the solution has the following probabilistic
representation

$\displaystyle\upsilon(t,x)=\mathbf{E}_{x}\left[f(B^{+}_{\bar{V}_{t}^{-1}})\right]$

where $\bar{V}_{t}=t+H^{\Psi}\eta{\gamma_{t}}$.

###  4.3. Dynamic condition and non-local space conditions. The case $\Psi=Id$

We notice that the condition can be written as
$\Delta\upsilon(t,x)\big{|}_{x=0}=-\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0}$
by means of which we can underline the sticky nature of the motion. Indeed,
$\mathfrak{D}_{t}^{\Psi}\upsilon=\partial_{t}\upsilon$ in Theorem 4.1 and
$\upsilon$ satisfies the heat equation on $[0,\infty)$. The solution has the
probabilistic representation

$\displaystyle\upsilon(t,x)=\mathbf{E}_{x}\left[f(B^{\bullet}_{V_{t}^{-1}})\right]$
(4.21)

where $B^{\bullet}_{t}$ is defined in (4.1) and $V_{t}=t+\eta
L^{\Phi}_{\gamma_{t}}$.

###  4.4. The reflected Brownian motion in an orthant

Our results can be extended to the $d$-dimensional case by considering the
reflected Brownian motion on a positive orthant. In particular, we consider
the semimartingale reflecting Brownian motion (SRBM in short), that is a
continuous process, say $Z=\\{Z_{t}\\}_{t>0}$, with representation $Z=B+RY$
where

  * -

$B=\\{B_{t}\\}_{t>0}$ is a $d$-dimensional Brownian motion;

  * -

$Y=\\{Y_{t}\\}_{t>0}$ is a positive, continuous and non-decreasing
$d$-dimensional process with $Y_{0}=0$ and components
$\\{Y^{i}_{t}\\}_{i=1,\ldots,d}$ for which

$\displaystyle\int_{0}^{t}\mathbf{1}_{R^{d}_{+}\setminus
F_{k}}(Z_{s})dY^{k}_{s}=0$

with $F_{k}=\\{x\in\mathbb{R}^{d}_{+}\,:\,x_{k}=0\\}$ for
$k\in(1,2,\ldots,d)$;

  * -

$R$ is a (non-negative) reflection matrix. For example, if $R$ is (positive)
diagonal, then the direction of reflection at each point of the boundary is a
positive inward normal.

Under such conditions, the process $Z$ is a weak solution to
$dZ_{t}=dB_{t}+RdY_{t}$. The process behaves like a Brownian motion inside the
orthant and it reflects instantaneously on the boundary. The process $Y$ plays
the role of local time. A discussion about SRBM can be found in [14; 27; 32].

The vector $G\circ Y$ with components $G^{i}\circ Y^{i}:=G^{i}_{Y^{i}}$ where
$G^{i}_{t}=HL_{t}-t$, $i=1,2,\ldots,d$ can be considered in order to obtain
the $d$-dimensional process $B^{\bullet}$. Indeed, for $R=diag\\{1\\}$,

$\displaystyle B^{\bullet}_{t}=B_{t}+G\circ Y_{t}+Y_{t}$ (4.22)

is a Brownian motion pushed inside the orthant by the components
$HLY^{i}_{t}-Y^{i}_{t}$, $i=1,2,\dots,d$. We can also consider
$G_{t}^{(i)}=H^{i}L^{i}_{t}-t$ where $H^{i}\perp L^{i}$ introduces the
independent vectors $\\{H^{i}_{t}\\}$ and $\\{L^{i}_{t}\\}$. The processes
$G^{i}$ and $G^{(i)}$ give rise to different problems.

Figure 3. The path of $B^{\bullet}$ in two dimensions. The starting point is
A, the process hits the boundary at B, then it jumps to the point C.

Figure 4. The path of $B^{\bullet}_{S^{-1}}$ in two dimensions. The process
can be obtained by that path of Figure 3. The new motion is trapped on the
vertical line after the jump at the point C. Then, it moves on the vertical
line as a 1-dimensional Brownian motion (see Figure 5 ) for a random amount of
time and it starts afresh from the point D as 2-dimensional process
$B^{\bullet}$

.

Figure 5. The path of
$B^{\bullet}_{S^{-1}}\big{|}({{}_{1}B^{\bullet}_{S^{-1}}=const})$ where
$B^{\bullet}_{t}$ can be written as the vector
$({{}_{1}B^{\bullet}_{t}},{{}_{2}B^{\bullet}_{t}})$.
Indeed,${{}_{1}B^{\bullet}_{t}}$ hits the zero point and jumps according with
$G^{1}_{t}$, then it stops for a random amount of time according with
$S^{-1}$. This path is represented by the vertical line of Figure 4.

###  4.5. Reflected Brownian motion in a bounded domain

Here we have no results. We only underline the main problem in dealing with
$B^{\bullet}_{S^{-1}}$ in a bounded domain. Since we have jumps as the process
approaches the boundary, then we have to control the jumping measure so that
the process can not jump outside. As far as we know, there are no results in
this direction. The authors are working on this case. Concerning $\Phi=Id$, we
have a clear picture of the problem given in [10] for a smooth domain
$\Omega\in\mathbb{R}^{d}$.

##  5\. Discussion

###  5.1. Sample paths description

In this section we recap briefly the dynamics of the processes introduced in
Section 4. The process $B^{\bullet}$, defined in (4.1), is a reflected
Brownian motion that jumps at $0$ inside $(0,\infty)$ in a random point given
by the last jump of $H^{\Phi}$. This process turns out to be right-continuous
since $H^{\Phi}L^{\Phi}$ is the composition of the right-continuous
subordinator $H^{\Phi}$ with its inverse (and continuous process) $L^{\Phi}$.

Figure 6. A possible path for $B^{\bullet}$. The case $\eta=0$.

We recall that the reflected sticky Brownian motion, say $B^{s}$, can be
represented as a time-changed reflected Brownian motion. The solution of the
problem

$\displaystyle\begin{cases}\frac{\partial}{\partial t}v_{s}(t,x)=\Delta
v_{s}(t,x)\quad&(t,x)\in(0,\infty)\times(0,\infty)\\\
\eta\frac{\partial}{\partial t}v_{s}(t,0)=\frac{\partial}{\partial
x}v_{s}(t,0)\quad&t>0\\\ v_{s}(0,x)=f(x)\quad&x\geq 0\end{cases}$

for $\eta\geq 0$ can be written as

$\displaystyle
v(t,x)=\mathbf{E}_{x}\left[f(B^{s}_{t})\right]=\mathbf{E}_{x}\left[f(B^{+}_{T_{t}^{-1}})\right],$

where $T_{t}=t+\eta\gamma_{t}$. As $B^{+}$ hits the origin, then $\gamma_{t}$
increases in a such a way that $T^{-1}$ runs slowly and slows down $B^{s}$. In
the last section we have seen that the solution of the heat equation with
boundary condition
$\eta\partial_{t}\upsilon(t,0)=-\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0}$
is given by (4.21). The process $L^{\Phi}_{\gamma_{t}}$ can be regarded as the
local time at $0$ of the process $B^{\bullet}$. Due to right-continuity of
$H^{\Phi}$, determining the jump, the time change $V^{-1}$ slows down
$B^{\bullet}$ immediately after the jump. The same arguments apply for the
boundary condition
$\eta\mathfrak{D}_{t}^{\Psi}\upsilon(t,x)\big{|}_{x=0}=-\mathbf{D}_{x-}^{\Phi}\upsilon(t,x)\big{|}_{x=0}\quad$
for which we only underline that the waiting time after a given jump is now
independent from the process. This is due to the fact that the plateaus after
the jumps are given by $H^{\Psi}$ independently from $H^{\Phi}$.

Figure 7. A possible path for $B^{\bullet}_{S^{-1}}$. The case $\eta>0$. Since
$\Psi\neq Id$ the random plateaus are given by $H^{\Psi}$ which is independent
from $B^{\bullet}$.

###  5.2. Delayed and rushed reflection

Let us consider the construction given in [10]. We respectively denote by
$\\{e_{i},\,i\in\mathbb{N}\\}$ and $\\{H^{\Psi}_{e_{i}},\,i\in\mathbb{N}\\}$
the sequences of holding times (at the boundary) for the reflecting Brownian
motion $X^{+}$ and the partially delayed reflecting Brownian motion $\bar{X}$.
The latter agrees with a delayed reflection as a boundary behavior whereas,
the process $X^{+}$ is instantaneously reflected. We say that $\bar{X}$ is
partially delayed in the sense that a delay effect only occurs on the
boundary. Thus, $B^{\bullet}_{S^{-1}_{t}}$ equals in law $\bar{X}$ in case
$\Phi=Id$. Moreover, the process $\bar{X}$ is obtained as the composition
$X^{+}_{\bar{V}^{-1}_{t}}$ written in terms of the inverse to
$\bar{V}_{t}=S_{t}:=t+H^{\Psi}_{\tau_{t}}$ with
$\tau_{t}:=\eta\,\gamma_{t}(X)$. In particular, it turns out that
$H^{\Psi}_{e_{i}}$, $i\in\mathbb{N}$ are i.i.d. random variables for which

$\displaystyle\mathbf{E}_{x}[H^{\Psi}_{e_{i}}|X^{+}]=\mathbf{E}_{x}[e_{1}|X^{+}]\mathbf{E}_{0}[H^{\Psi}_{1}]=\mathbf{E}_{x}[e_{1}|X^{+}]\,\lim_{\lambda\downarrow
0}\frac{\Psi(\lambda)}{\lambda}$ (5.1)

gives the mean holding time on the boundary for the process $\bar{X}$. In the
same spirit of the results in [6] we are able to give a classification for the
reflection near the boundary. Write

$\displaystyle c:=\lim_{\lambda\downarrow 0}\frac{\Psi(\lambda)}{\lambda}$

and assume that (5.1) holds $\forall\,x$. The process $\bar{X}$ may have (in
terms of mean holding time on the boundary):

  * B1)

delayed boundary behavior if $c<1$;

  * B2)

rushed boundary behavior if $c>1$;

  * B3)

base (reflected) boundary behavior if $c=1$.

A special case of subordinator including either delayed or rushed effect if
the gamma subordinator for which, for $a,b>0$,

$\displaystyle\Psi(\lambda)=a\ln(1+\lambda/b)\quad\textrm{and}\quad\lim_{\lambda\downarrow
0}\frac{\Psi(\lambda)}{\lambda}=\frac{a}{b}$

and $c<1$, $c>1$ or $c=1$ depending on the ratio $a/b$. On the other hand, by
considering $\Psi=Id$, we have that $B^{\bullet}$ jumps away from the boundary
according with the last jump of $H^{\Phi}$. Due to nature of the subordinator,
the process $B^{\bullet}$ never hits the boundary, it reflects instantaneously
before hitting the boundary. In this case, the time the process spend on the
boundary can be only given (eventually) by the starting point as a point of
the boundary, case in which the process is pushed immediately away once again
for the nature of the subordinator. We recall that our analysis is based only
on strictly increasing subordinators.

Our analysis introduces the following further characterization:

  * R1)

delayed/rushed reflection;

  * R2)

instantaneous jumping reflection;

  * R3)

reflection.

The delayed reflection can be related with a reflection on the boundary (like
an insulating boundary for example) whereas the rushed reflection seems to be
related with a reflection near the boundary (like an inflating boundary for
example). These characterizations are therefore given according with the
comparison between the total (mean) amounts of time the processes $B^{+}$ and
$\bar{X}$ would spend on the boundary. By base boundary behavior we mean the
behavior of the reflecting process $B^{+}$. We observe that instantaneous
reflection means that the time the process spends on the boundary has zero
Lebesgue measure. The reflection R1 is obtained via time change (by
considering $S^{-1}_{t}$) and depends on the symbol $\Psi$ of the boundary
condition. R1 gives rise to the behaviors B1 and B2, notice that in this case
we may consider a boundary diffusion (in the sense of Dirichlet-Neumann
generator) which can be delayed or rushed with respect to the base boundary
diffusion (according with the definition given in [6]). R2 turns out to be
associated with the symbol $\Phi$ of the boundary condition. Concerning R3 we
easily make the connection with B3, that is the case of Neumann boundary
condition.

The behaviors B1-B2-B3 can be associated with a macroscopic description of
motions on irregular domains. Indeed, irregular domains in a microscopic point
of view need a geometric characterization in order to be completely described.
By irregular domain here we mean domains with irregular boundaries, for
example the Koch domain still is regular. The definition given in [4] of trap
domain says that if a reflecting process has an infinite (mean) lifetime in a
bounded domain including a Dirichlet boundary, then that process is trapped in
the reflecting part of the boundary. That is to say, the boundary and
therefore the domain is trap for that process. In a deep detail, given a
$d$-dimensional reflecting Brownian motion $B^{+}$ on a domain
$\Omega\setminus\mathcal{B}$ where $\mathcal{B}$ is a ball with killing
boundary $\partial\mathcal{B}$ and $\Omega$ has reflecting boundary
$\partial\Omega$, then $\Omega$ is trap for the Brownian motion if

$\displaystyle\sup_{x\in\Omega\setminus\mathcal{B}}\mathbf{E}_{x}[\inf\\{t\,:\,B^{+}_{t}\notin\Omega\setminus\mathcal{B}\\}]=\infty.$
(5.2)

From the analytical point of view we ask for

$\displaystyle\sup_{x\in\Omega\setminus\mathcal{B}}\int_{\Omega\setminus\mathcal{B}}G^{+}(x,y)dy=\infty\quad\textrm{where}\quad
G^{+}(x,y)dy=\int_{0}^{\infty}\mathbf{P}_{x}(B^{+}_{t}\in dy)\,dt.$

Concerning the process $B^{\bullet}_{S^{-1}}$ on
$\Omega^{*}\subset\mathbb{R}^{d}$ with $\Phi=Id$ we provide the following
conjecture as a possible link between macroscopic and microscopic viewpoints.
We have $B^{\bullet}_{S^{-1}}\stackrel{{\scriptstyle
law}}{{=}}X^{+}_{\bar{V}^{-1}}$ as discussed above in terms of $\bar{X}$ on
$\Omega^{*}$. Our statement is the following:

The behavior of $B^{+}$ on the irregular domain $\Omega$ can be captured by
$\bar{X}$ on the smooth domain $\Omega^{*}$.

This statement is justified by the fact that the time change $\bar{V}^{-1}$
for the reflecting Brownian motion $X^{+}$ introduces the behaviors B1-B2-B3
and the (mean) total amount of time the process spend on the boundary if given
in terms of (5.1). This well accords with (5.2). We argue that different
symbols may be considered (in the macroscopic analysis) in order to describe
different boundaries (in the microscopic analysis).

We conclude this section by underlying an interesting connection with Brownian
excursions. An excursion of $B^{+}$ can be associated with no crossing trough
a boundary for $B^{+}$ and therefore, with no Poison events associated with
the crossing. For the holding time,

$\displaystyle\mathbf{P}_{x}(H^{\Psi}_{e_{i}}>t|X^{+})=\mathbf{P}_{x}(e_{i}>L^{\Psi}_{t}|X^{+})=\mathbf{E}_{0}[e^{-L^{\Psi}_{t}}]$

assuming that the exponential random variables $e_{i}$ are of parameter $1$.
The excursions of $B^{\bullet}_{S^{-1}}$ introduces the non-local Poisson
process as we can immediately see for $\Psi(\lambda)=\lambda^{\alpha}$,
$\alpha\in(0,1)$. In this case,

$\displaystyle\mathbf{P}_{x}(H^{\Psi}_{e_{i}}>t|X^{+})=E_{\alpha}(-t^{\alpha})=:\mathbf{P}_{0}(N_{L_{t}}=0)$

where $L$ is an inverse to a stable subordinator and $N$ is a Poisson process
with $\mathbf{P}_{0}(N_{t}=k)=(t^{k}/k!)e^{-t}$. The fractional Poisson
process has been introduced in [21], the literature is really vast, we refer
to [1, formula (3.19)] and [26].

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$E_{1}$ is simple, the first claim clearly hold for $E$, so we only need to
verify the second claim. That holds because, by the form of $E$, every node
$v$ is in $\llbracket E\rrbracket^{G^{\prime}}(v)$, but not in $\llbracket
r\rrbracket^{G^{\prime}}(v)$, as $G$ does not have any self-loops.

Finally, assume $E$ is of the form $E_{1}/E_{2}$. Note that if $E_{1}$ or
$E_{2}$ is simple, clearly claim one holds because $\llbracket
r\rrbracket^{G^{\prime}}\subseteq\llbracket E\rrbracket^{G^{\prime}}$. The
argument that follows will therefore also apply when $E_{1}$ or $E_{2}$ is
simple. We will be careful not to apply the induction hypothesis for the
second statement to $E_{1}$ and $E_{2}$.

We distinguish two cases.

  * •

If $\llbracket r\rrbracket^{G^{\prime}}\subseteq\llbracket
E_{2}\rrbracket^{G^{\prime}}$, then we show that $\llbracket
r\rrbracket^{G^{\prime}}\subseteq\llbracket E\rrbracket^{G^{\prime}}$. Let
$v\in X_{i}$. We verify the following two inclusions:

    * –

$\llbracket E\rrbracket^{G}(v)\supseteq X_{i}$. Let $u\in X_{i}$. If $u\neq
v$, choose a third node $w\in X_{i}$. Since $X_{i}$ is a clique,
$(v,w)\in\llbracket E_{1}\rrbracket^{G}$ because the first claim holds for
$E_{1}$. By $\llbracket r\rrbracket^{G^{\prime}}\subseteq\llbracket
E_{2}\rrbracket^{G^{\prime}}$, we also have $(w,u)\in\llbracket
E_{2}\rrbracket^{G^{\prime}}$, whence $u\in\llbracket
E\rrbracket^{G^{\prime}}(v)$ as desired. If $u=v$, we similarly have
$(v,w)\in\llbracket E_{1}\rrbracket^{G^{\prime}}$ and $(w,u)\in\llbracket
E_{2}\rrbracket^{G^{\prime}}$ as desired.

    * –

$\llbracket E\rrbracket^{G}(v)\supseteq X_{\mathit{next}(i)}$. Let $u\in
X_{\mathit{next}(i)}$ and choose $w\neq v\in X_{i}$. Because the first claim
holds for $E_{1}$, we have $(v,w)\in\llbracket E_{1}\rrbracket^{G}$. By
$\llbracket r\rrbracket^{G^{\prime}}\subseteq\llbracket
E_{2}\rrbracket^{G^{\prime}}$, we also have $(w,u)\in\llbracket
E_{2}\rrbracket^{G^{\prime}}$, whence $u\in\llbracket
E\rrbracket^{G^{\prime}}(v)$ as desired.

We conclude that $\llbracket E\rrbracket^{G^{\prime}}(v)\supseteq X_{i}\cup
X_{\mathit{next}(i)}\supseteq\llbracket r\rrbracket^{G^{\prime}}$ as desired.

  * •

If $\llbracket r^{-}\rrbracket^{G^{\prime}}\subseteq\llbracket
E_{2}\rrbracket^{G^{\prime}}$, then we show that $\llbracket
r^{-}\rrbracket^{G^{\prime}}\subseteq\llbracket E\rrbracket^{G^{\prime}}$.
This is analogous to the previous case, now verifying that $\llbracket
E\rrbracket^{G}(v)\supseteq X_{i}\cup X_{\mathit{prev}(i)}$.

In both cases, the second statement now follows for every node $v$. Indeed,
$v\in X_{i}\subseteq\llbracket E\rrbracket^{G^{\prime}}(v)$ but
$v\notin\llbracket r\rrbracket^{G^{\prime}}(v)$.