1 Introduction
In this paper we consider various forms of tempered fractional derivatives. For a function f continuous and compactly supported on the positive real line, let us consider the Marchaud type operator defined by
where
with $\eta >0,0<\alpha <1$. The operator (1.1) coincides with the classical Marchaud derivative for $\eta =0$.
(1.1)
\[ \big({\mathcal{D}^{\alpha ,\eta }}f\big)(x)={\int _{0}^{\infty }}\big(f(x)-f(x-y)\big)\hspace{0.1667em}\varPi (\mathrm{d}y)\](1.2)
\[ \varPi (\mathrm{d}y)=\frac{\alpha }{\varGamma (1-\alpha )}\frac{{e^{-\eta y}}}{{y^{\alpha +1}}}dy,\hspace{1em}y>0,\]The Laplace transform of the fractional operator (1.1) reads
Throughout the work we denote by $\tilde{f}$ the Laplace transform of f. In the Fourier analysis the factor ${(\eta +i\lambda )^{\alpha }}-{\eta ^{\alpha }}$ is the multiplier of the Fourier transform of f [7]. Tempered fractional derivatives emerge in the study of equations driving the tempered subordinators [1, 7]. In particular, the operator (1.1) is the generator of the subordinator ${H_{t}},t>0$, with Lévy measure (1.2) and density law whose Laplace transform is given by (1.3), that is,
(1.3)
\[\begin{aligned}{}{\int _{0}^{\infty }}{e^{-\lambda x}}\hspace{0.1667em}\big({\mathcal{D}^{\alpha ,\eta }}f\big)(x)\hspace{0.1667em}\mathrm{d}x& =\Bigg({\int _{0}^{\infty }}\big(1-{e^{-\lambda y}}\big)\varPi (\mathrm{d}y)\Bigg)\tilde{f}(\lambda )\\ {} & =\big({(\eta +\lambda )^{\alpha }}-{\eta ^{\alpha }}\big)\tilde{f}(\lambda ).\end{aligned}\]
\[\begin{aligned}{}\mathbb{E}{e^{-\lambda {H_{t}}}}& ={e^{-t({(\eta +\lambda )^{\alpha }}-{\eta ^{\alpha }})}}={e^{-t{\textstyle\int _{0}^{\infty }}(1-{e^{-\lambda y}})\varPi (\mathrm{d}y)}},\hspace{1em}\lambda >0.\end{aligned}\]
The process ${H_{t}}$ is called relativistic subordinator and coincides, for $\eta =0$, with a positively skewed Lévy process, that is a stable subordinator. Tempered stable subordinators can be viewed as the limits of Poisson random sums with tempered power law jumps [7].The fractional operator ${\mathcal{D}^{\alpha ,\eta }}f$ defined in (1.1) is related to the tempered upper Weyl derivatives defined by
By combining (1.4) with the lower Weyl tempered derivatives we obtain the Riesz tempered fractional derivatives $\frac{{\partial ^{\alpha ,\eta }}f}{\partial |x{|^{\alpha }}}$ from which we obtain the explicit Fourier transform in (2.5).
(1.4)
\[ \big({\hat{\mathcal{D}}_{+}^{\alpha ,\eta }}f\big)(x)=\frac{1}{\varGamma (1-\alpha )}\frac{\mathrm{d}}{\mathrm{d}x}{\int _{-\infty }^{x}}\frac{f(t)}{{(x-t)^{\alpha }}}{e^{-\eta (x-t)}}\hspace{0.1667em}\mathrm{d}t.\]We consider the Dzherbashyan–Caputo derivative of order $\frac{1}{2}$, that is,
with the Laplace transform
from which we observe that
We remark that the problems
(1.5)
\[ \big({D^{\frac{1}{2}}}f\big)(t)=\frac{1}{\sqrt{\pi }}{\int _{0}^{t}}{f^{\prime }}(s){(t-s)^{-\frac{1}{2}}}\hspace{0.1667em}\mathrm{d}s\]
\[ {\int _{0}^{\infty }}{e^{-\lambda t}}\big({D^{\frac{1}{2}}}f\big)(t)\hspace{0.1667em}\mathrm{d}t={\lambda ^{\frac{1}{2}}}\tilde{f}(\lambda )-{\lambda ^{\frac{1}{2}-1}}f(0),\hspace{1em}\lambda >0.\]
The relationship between the Riemann–Liouville and the Dzherbashyan–Caputo derivative can be given as follows,
(1.6)
\[ \big({\mathcal{D}^{\frac{1}{2}}}f\big)(t)=\big({D^{\frac{1}{2}}}f\big)(t)+\frac{{t^{\frac{1}{2}-1}}}{\varGamma (\frac{1}{2})}f(0),\](1.7)
\[ {\int _{0}^{\infty }}{e^{-\lambda t}}\big({\mathcal{D}^{\frac{1}{2}}}f\big)(t)\hspace{0.1667em}\mathrm{d}t={\lambda ^{\frac{1}{2}}}\tilde{f}(\lambda ).\]
\[ \left\{\begin{array}{l}\big({D^{\frac{1}{2}}}u\big)(t)=-\displaystyle \frac{\partial u}{\partial y},\hspace{1em}t>0,y>0\\ {} u(0,y)=\delta (y)\end{array}\right.\hspace{1em}\text{and}\hspace{1em}\left\{\begin{array}{l@{\hskip10.0pt}l}\big({\mathcal{D}^{\frac{1}{2}}}u\big)(t)=-\displaystyle \frac{\partial u}{\partial y},& t>0,y>0\\ {} u(0,y)=\delta (y)\\ {} u(t,0)=\displaystyle \frac{1}{\sqrt{\pi t}},& t>0\end{array}\right.\]
have a unique solution given by the density law of an inverse to a stable subordinator, say ${L_{t}}$ (see for example [2, formulas 3.4 and 3.5]). It is well known that ${L_{t}}$ (with ${L_{0}}=0$) is identical in law to a folded Brownian motion $|{B_{t}}|$ (with ${B_{0}}=0$), that is, u is the unique solution to the problem
Thus, by considering the theory of time changes, there exist interesting connections between fractional Cauchy problems and the domains of the generators of the base processes. In our view, concerning the drifted Brownian motion, the present paper gives new results also in this direction.We denote by
the tempered Riemann–Liouville type derivative. The equality between definitions (1.8) and (1.1) can be verified by comparing the corresponding Laplace transforms. Indeed, from (1.7),
(1.8)
\[ {\mathcal{D}_{t}^{\frac{1}{2},\eta }}f:={e^{-\eta t}}{\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\eta t}}f\big)-\sqrt{\eta }f\]
\[\begin{aligned}{}{\int _{0}^{\infty }}{e^{-\lambda t}}\hspace{0.1667em}{\mathcal{D}_{t}^{\frac{1}{2},\eta }}f\hspace{0.1667em}\mathrm{d}t=& {\int _{0}^{\infty }}{e^{-(\lambda +\eta )t}}{\mathcal{D}_{t}^{\frac{1}{2}}}(g)\hspace{0.1667em}\mathrm{d}t-\sqrt{\eta }\tilde{f}(\lambda )\\ {} =& \sqrt{\lambda +\eta }{\int _{0}^{\infty }}{e^{-(\lambda +\eta )t}}g(t)\hspace{0.1667em}\mathrm{d}t-\sqrt{\eta }\tilde{f}(\lambda )\end{aligned}\]
where $g(t)={e^{\eta t}}f(t)$.Let B represent a Brownian motion starting at the origin with generator Δ. In the paper we show that the transition density $u=u(x,y,t)$ of the 1-dimensional process
satisfies the fractional equation on $(0,\infty )\times {\mathbb{R}^{2}}$
where
and
A different result concerns the reflected process
whose transition density $v=v(x,y,t)$ satisfies the equation
with initial and boundary conditions
(1.10)
\[ {\mathcal{D}_{t}^{\frac{1}{2},\eta }}v+\sqrt{\eta }\hspace{0.1667em}v=\frac{\partial v}{\partial x}+\sqrt{\eta }\tanh \big(\sqrt{\eta }(y-x)\big)v,\hspace{1em}t>0,\hspace{0.2778em}y>x>0,\]
\[\begin{aligned}{}v(x,y,0)=& \delta (y-x),\\ {} v(x,x,t)=& \frac{{e^{-\eta t}}}{\sqrt{\pi t}},\hspace{1em}t>0,\end{aligned}\]
and
The fractional equation governing the iterated Brownian motion ${B^{{\mu _{2}}}}(|{B^{{\mu _{1}}}}(t)|)$ (${B^{{\mu _{j}}}}$ being independent) has been studied in [6] and in the special case ${B^{\mu }}(|B(t)|)$ explicitly derived. For the iterated Bessel process a similar analysis is performed in [3]. A general presentation of tempered fractional calculus can be found in the paper [7].Many processes like Brownian motion, iterated Brownian motion, Cauchy process have transition functions satisfying different partial differential equations and also are solutions of fractional equations of different forms with various fractional derivatives. We here show that a similar situation arises when drifted reflecting Brownian motion is considered but in this case the corresponding fractional equations involve tempered Riemann–Liouville type derivatives.
2 A generalization of the tempered Marchaud derivative
In this section we study the tempered Weyl derivatives (upper and lower ones) and construct the Riesz tempered derivative. We are able to obtain the Fourier transform of the Riesz tempered derivatives and thus to solve some generalized fractional diffusion equation.
We start by giving the explicit forms of the tempered Weyl derivative
The derivative ${\hat{\mathcal{D}}_{+}^{\alpha ,\eta }}$ can be expressed in terms of ${\mathcal{D}^{\alpha ,\eta }}$ as follows:
For $0<\alpha <1$ the Riesz fractional derivative writes
In the same way we define the tempered Riesz derivative as
We now evaluate the Fourier transform of the tempered Riesz derivative
In the last step we used the following formula ([5], p. 490, formula 5)
(2.1)
\[\begin{aligned}{}& \big({\hat{\mathcal{D}}_{+}^{\alpha ,\eta }}f\big)(x)\\ {} & =\frac{1}{\varGamma (1-\alpha )}\frac{\mathrm{d}}{\mathrm{d}x}{\int _{-\infty }^{x}}\frac{f(t)}{{(x-t)^{\alpha }}}{e^{-\eta (x-t)}}\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}\frac{\mathrm{d}}{\mathrm{d}x}{\int _{0}^{\infty }}\frac{f(x-t)}{{t^{\alpha }}}{e^{-\eta t}}\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}\frac{\mathrm{d}}{\mathrm{d}x}{\int _{0}^{\infty }}f(x-t){e^{-\eta t}}{\int _{t}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\mathrm{d}w\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\mathrm{d}w{\int _{0}^{w}}{f^{\prime }}(x-t){e^{-\eta t}}\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\mathrm{d}w{\int _{x-w}^{x}}{f^{\prime }}(t){e^{-\eta (x-t)}}\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}{e^{-\eta x}}\Bigg\{f(t){e^{\eta t}}{|_{x-w}^{x}}-\eta {\int _{x-w}^{x}}f(t){e^{\eta t}}\hspace{0.1667em}\mathrm{d}t\Bigg\}\hspace{0.1667em}\mathrm{d}w\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}{e^{-\eta x}}\big[f(x){e^{\eta x}}-f(x-w){e^{\eta (x-w)}}\big]\hspace{0.1667em}\mathrm{d}w\\ {} & \hspace{1em}-\frac{\eta }{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}{\int _{x-w}^{x}}f(t){e^{-\eta (x-t)}}\hspace{0.1667em}\mathrm{d}w\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\big[f(x)-f(x-w){e^{-\eta w}}\big]\hspace{0.1667em}\mathrm{d}w\\ {} & \hspace{1em}-\frac{\eta }{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}{\int _{0}^{w}}f(x-t){e^{-\eta t}}\hspace{0.1667em}\mathrm{d}w\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\big[f(x)+f(x){e^{-\eta w}}-f(x){e^{-\eta w}}-f(x-w){e^{-\eta w}}\big]\hspace{0.1667em}\mathrm{d}w\\ {} & \hspace{1em}-\frac{\eta }{\varGamma (1-\alpha )}{\int _{0}^{\infty }}f(x-t){e^{-\eta t}}{\int _{t}^{\infty }}\alpha {w^{-\alpha -1}}\hspace{0.1667em}\mathrm{d}w\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\big(f(x)-f(x-w)\big)\hspace{0.1667em}\alpha \frac{{e^{-\eta w}}}{{w^{\alpha +1}}}\hspace{0.1667em}\mathrm{d}w\hspace{0.1667em}\mathrm{d}t\\ {} & \hspace{1em}+\frac{f(x)}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\big(1-{e^{-\eta w}}\big)\hspace{0.1667em}\mathrm{d}w-\frac{\eta }{\varGamma (1-\alpha )}{\int _{0}^{\infty }}f(x-t)\frac{{e^{-\eta t}}}{{t^{\alpha }}}\hspace{0.1667em}\mathrm{d}t\\ {} & ={\int _{0}^{\infty }}\big(f(x)-f(x-w)\big)\varPi (\mathrm{d}w)+\eta {\int _{0}^{\infty }}\big(f(x)-f(x-w)\big)\frac{{e^{-\eta w}}}{{w^{\alpha }}\varGamma (1-\alpha )}\hspace{0.1667em}\mathrm{d}w\end{aligned}\]
\[ {\hat{\mathcal{D}}_{+}^{\alpha ,\eta }}f={\mathcal{D}^{\alpha ,\eta }}f-\eta \hspace{0.1667em}{\mathcal{D}^{\alpha -1,\eta }}f.\]
In the same way we can obtain the upper Weyl derivative in the Marchaud form as
(2.2)
\[\begin{aligned}{}& \big({\hat{\mathcal{D}}_{-}^{\alpha ,\eta }}f\big)(x)\\ {} & =\frac{1}{\varGamma (1-\alpha )}\frac{\mathrm{d}}{\mathrm{d}x}{\int _{x}^{\infty }}\frac{f(t)}{{(x-t)^{\alpha }}}{e^{-\eta (x-t)}}\hspace{0.1667em}\mathrm{d}t\\ {} & ={\int _{0}^{\infty }}\frac{\alpha {w^{-\alpha -1}}}{\varGamma (1-\alpha )}\big\{{e^{-\eta w}}f(x+w)-f(x)\big\}\hspace{0.1667em}\mathrm{d}w+\eta {\int _{0}^{\infty }}\frac{{e^{\eta w}}f(x+w)}{\varGamma (1-\alpha ){w^{\alpha }}}\hspace{0.1667em}\mathrm{d}w\\ {} & =\frac{1}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\big[f(x+w)-f(x)\big]\frac{\alpha {e^{-\eta w}}}{{w^{\alpha +1}}}\\ {} & \hspace{1em}+\frac{f(x)}{\varGamma (1-\alpha )}{\int _{0}^{\infty }}\alpha {w^{-\alpha -1}}\big({e^{-\eta w}}-1\big)\hspace{0.1667em}\mathrm{d}w+\eta {\int _{0}^{\infty }}f(x+t)\frac{{e^{-\eta w}}}{\varGamma (1-\alpha ){t^{\alpha }}}\hspace{0.1667em}\mathrm{d}t\\ {} & ={\int _{0}^{\infty }}\big[f(x+w)-f(x)\big]\varPi (\mathrm{d}w)+\eta {\int _{0}^{\infty }}\big[f(x+w)-f(x)\big]\frac{{e^{-\eta w}}}{\varGamma (1-\alpha ){w^{\alpha }}}\hspace{0.1667em}\mathrm{d}w.\end{aligned}\](2.3)
\[\begin{aligned}{}\frac{{\partial ^{\alpha }}f}{\partial |x{|^{\alpha }}}& =-\frac{1}{2\cos \frac{\alpha \pi }{2}\hspace{0.1667em}\varGamma (1-\alpha )}{\int _{-\infty }^{+\infty }}\frac{f(t)}{|x-t{|^{\alpha }}}\hspace{0.1667em}\mathrm{d}t\\ {} & =-\frac{1}{2\cos \frac{\alpha \pi }{2}\hspace{0.1667em}\varGamma (1-\alpha )}\Bigg[\frac{\mathrm{d}}{\mathrm{d}x}{\int _{-\infty }^{x}}\frac{f(t)}{{(x-t)^{\alpha }}}\hspace{0.1667em}\mathrm{d}t-\frac{\mathrm{d}}{\mathrm{d}x}{\int _{x}^{\infty }}\frac{f(t)}{{(t-x)^{\alpha }}}\hspace{0.1667em}\mathrm{d}t\Bigg].\end{aligned}\]
\[ \frac{{\partial ^{\alpha ,\eta }}f}{\partial |x{|^{\alpha }}}={C_{\alpha ,\eta }}\Bigg[\frac{\mathrm{d}}{\mathrm{d}x}{\int _{-\infty }^{x}}\frac{f(t)}{{(x-t)^{\alpha }}}\frac{{e^{-\eta (x-t)}}}{\varGamma (1-\alpha )}\hspace{0.1667em}\mathrm{d}t-\frac{\mathrm{d}}{\mathrm{d}x}{\int _{x}^{\infty }}\frac{f(t)}{{(t-x)^{\alpha }}}\frac{{e^{-\eta (t-x)}}}{\varGamma (1-\alpha )}\hspace{0.1667em}\mathrm{d}t\Bigg]\]
where ${C_{\alpha ,\eta }}$ is a suitable constant which will be defined below. In view of the previous calculations we have that
(2.4)
\[\begin{aligned}{}\frac{{\partial ^{\alpha ,\eta }}f}{\partial |x{|^{\alpha }}}& ={C_{\alpha ,\eta }}\Bigg[{\int _{0}^{\infty }}\big(f(x)-f(x-w)\big)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & \hspace{1em}+\eta {\int _{0}^{\infty }}\big(f(x)-f(x-w)\big)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\\ {} & \hspace{1em}-{\int _{0}^{\infty }}\big(f(x+w)-f(x)\big)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & \hspace{1em}-\eta {\int _{0}^{\infty }}\big(f(x+w)-f(x)\big)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\Bigg]\\ {} & ={C_{\alpha ,\eta }}\Bigg[{\int _{0}^{\infty }}\big(2f(x)-f(x-w)-f(x+w)\big)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & \hspace{1em}+\eta {\int _{0}^{\infty }}\big(2f(x)-f(x-w)-f(x+w)\big)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\Bigg].\end{aligned}\](2.5)
\[\begin{aligned}{}{\int _{-\infty }^{+\infty }}{e^{i\gamma x}}\frac{{\partial ^{\alpha ,\eta }}f}{\partial |x{|^{\alpha }}}\hspace{0.1667em}\mathrm{d}x=& {C_{\alpha ,\eta }}\Bigg\{\hat{F}(\gamma ){\int _{0}^{\infty }}\big(1-{e^{i\gamma w}}\big)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & +\eta \hat{F}(\gamma ){\int _{0}^{\infty }}\big(1-{e^{i\gamma w}}\big)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\\ {} & -\hat{F}(\gamma ){\int _{0}^{\infty }}\big({e^{i\gamma w}}-1\big)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & -\eta \hat{F}(\gamma ){\int _{0}^{\infty }}\big({e^{i\gamma w}}-1\big)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\Bigg\}\\ {} =& {C_{\alpha ,\eta }}\hat{F}(\gamma )\Bigg\{2{\int _{0}^{\infty }}(1-\cos \gamma w)\frac{\alpha {e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha +1}}}\\ {} & +2\eta {\int _{0}^{\infty }}(1-\cos \gamma w)\frac{{e^{-\eta w}}\mathrm{d}w}{\varGamma (1-\alpha ){w^{\alpha }}}\Bigg\}\\ {} =& {C_{\alpha ,\eta }}\hat{F}(\gamma )\Bigg\{-2{w^{-\alpha }}(1-\cos \gamma w)\frac{{e^{-\eta w}}}{\varGamma (1-\alpha )}{|_{0}^{\infty }}\\ {} & -2\eta {\int _{0}^{\infty }}(1-\cos \gamma w)\frac{{e^{-\eta w}}\mathrm{d}w}{{w^{\alpha }}\varGamma (1-\alpha )}\\ {} & +2\gamma {\int _{0}^{\infty }}\frac{{e^{-\eta w}}\sin \gamma w}{{w^{\alpha }}\varGamma (1-\alpha )}\hspace{0.1667em}\mathrm{d}w\\ {} & +2\eta {\int _{0}^{\infty }}(1-\cos \gamma w)\frac{{e^{-\eta w}}}{{w^{\alpha }}\varGamma (1-\alpha )}\mathrm{d}w\Bigg\}\\ {} =& {C_{\alpha ,\eta }}\hat{F}(\gamma )2|\gamma |{\int _{0}^{\infty }}\frac{{e^{-\eta w}}\sin |\gamma |w}{{w^{\alpha }}\varGamma (1-\alpha )}\hspace{0.1667em}\mathrm{d}w\\ {} =& {C_{\alpha ,\eta }}\hat{F}(\gamma )\frac{2|\gamma |}{{({\eta ^{2}}+{\gamma ^{2}})^{\frac{1-\alpha }{2}}}}\sin \bigg((1-\alpha )\arctan \frac{|\gamma |}{\eta }\bigg).\end{aligned}\]
\[ {\int _{0}^{\infty }}{x^{\mu -1}}{e^{-\beta x}}\sin \delta x\hspace{0.1667em}\mathrm{d}x=\frac{\varGamma (\mu )}{{({\beta ^{2}}+{\delta ^{2}})^{\frac{\mu }{2}}}}\hspace{0.1667em}\sin \bigg(\mu \arctan \frac{\delta }{\mu }\bigg)\]
with $\mathrm{Re}\hspace{0.1667em}\mu >-1$, $\mathrm{Re}\hspace{0.1667em}\beta \ge \mathrm{Im}\hspace{0.1667em}\delta $. Remark 1.
For $\eta \to 0$ we have that
\[ \underset{\eta \to 0}{\lim }\sin \bigg((1-\alpha )\arctan \frac{|\gamma |}{\eta }\bigg)=\cos \bigg(\frac{\pi \alpha }{2}\bigg).\]
Therefore
\[ \underset{\eta \to 0}{\lim }{\int _{-\infty }^{+\infty }}{e^{i\gamma x}}\hspace{0.1667em}\frac{{\partial ^{\alpha ,\eta }}f}{\partial |x{|^{\alpha }}}\hspace{0.1667em}\mathrm{d}x=2{C_{\alpha ,0}}\hspace{0.1667em}|\gamma {|^{\alpha }}\cos \bigg(\frac{\pi \alpha }{2}\bigg)\hspace{0.1667em}\hat{F}(\gamma )\]
and thus the normalizing constant must be ${C_{\alpha ,0}}=-{(2\cos \frac{\pi \alpha }{2})^{-1}}$.This means that for $\eta \to 0$ we obtain from (2.5) the Fourier transform of the Riesz fractional derivative (2.3). This result shows that symmetric stable processes are governed by equations
see, for example, [4], where the interplay between stable laws, including subordinators and inverse subordinators, and fractional equations is considered.
Remark 2.
For fractional equations of the form
the Fourier transform of the solution reads
(2.6)
\[ \left\{\begin{array}{r@{\hskip0pt}l@{\hskip10pt}r@{\hskip0pt}l}& \displaystyle \frac{\partial u}{\partial t}=\frac{{\partial ^{\alpha ,\eta }}u}{\partial |x{|^{\alpha }}},& & \displaystyle t>0,\hspace{0.2778em}x\in \mathbb{R},\\ {} & \displaystyle u(x,0)=\delta (x),& & \displaystyle x\in \mathbb{R},\end{array}\right.\]
\[\begin{aligned}{}& {\int _{-\infty }^{+\infty }}{e^{i\gamma x}}u(x,t)\hspace{0.1667em}\mathrm{d}x\\ {} & =\exp \bigg\{t\hspace{0.1667em}{C_{\alpha ,\eta }}\frac{2|\gamma |}{{({\eta ^{2}}+{\gamma ^{2}})^{\frac{1-\alpha }{2}}}}\sin \bigg((1-\alpha )\arctan \frac{|\gamma |}{\eta }\bigg)\bigg\}\\ {} & =\exp \bigg\{t\hspace{0.1667em}{C_{\alpha ,\eta }}\frac{2|\gamma |}{{({\eta ^{2}}+{\gamma ^{2}})^{1-\frac{\alpha }{2}}}}\bigg[|\gamma |\cos \bigg(\alpha \arctan \frac{|\gamma |}{\eta }\bigg)-\eta \sin \bigg(\alpha \arctan \frac{|\gamma |}{\eta }\bigg)\bigg]\bigg\}.\end{aligned}\]
3 Fractional equations governing the drifted Brownian motion
The law of the drifted Brownian motion started at x satisfies the equations
The law $u=u(x,y,t)$ of the process ${B^{\mu }}$ is given by
\[ \frac{\partial u}{\partial t}=\frac{{\partial ^{2}}u}{\partial {y^{2}}}-\mu \frac{\partial u}{\partial y},\hspace{1em}t>0,y\in \mathbb{R},\]
and
\[ \frac{\partial u}{\partial t}=\frac{{\partial ^{2}}u}{\partial {x^{2}}}+\mu \frac{\partial u}{\partial x},\hspace{1em}t>0,\hspace{0.2778em}x\in \mathbb{R}.\]
We show here that the drifted Brownian motion is related to time fractional equations with tempered derivatives. Let us consider the process
(3.1)
\[ {B^{\mu }}(t)=B(t)+\mu t+x,\hspace{1em}\mu \in \mathbb{R},\hspace{0.2778em}x\in \mathbb{R}.\]Theorem 1.
The law of ${B^{\mu }}$ solves the Cauchy problem
with
(3.3)
\[ \left\{\begin{aligned}{}& {\mathcal{D}_{t}^{\frac{1}{2},\eta }}u+\sqrt{\eta }\hspace{0.1667em}u=a(x,y)\bigg(\frac{\partial u}{\partial x}+\sqrt{\eta }\hspace{0.1667em}u\bigg),\hspace{1em}t>0,\hspace{0.2778em}x,y,\in \mathbb{R},\\ {} & {\mathcal{D}_{t}^{\frac{1}{2},\eta }}u+\sqrt{\eta }\hspace{0.1667em}u=-a(x,y)\bigg(\frac{\partial u}{\partial y}-\sqrt{\eta }\hspace{0.1667em}u\bigg),\hspace{1em}t>0,\hspace{0.2778em}x,y,\in \mathbb{R},\\ {} & u(x,y,0)=\delta (x-y)\end{aligned}\right.\]Proof.
We start by computing the Laplace–Fourier transform of the function
that is,
\[\begin{aligned}{}\hat{\tilde{g}}(y,\xi ,\lambda )& ={\int _{0}^{\infty }}{e^{-\lambda t}}{\int _{-\infty }^{+\infty }}{e^{i\xi x}}g(x,y,t)\hspace{0.1667em}\mathrm{d}x\hspace{0.1667em}\mathrm{d}t\\ {} & ={\int _{0}^{\infty }}{e^{-\lambda t}}{e^{i\xi y\hspace{0.1667em}-\hspace{0.1667em}{\xi ^{2}}t}}\hspace{0.1667em}\mathrm{d}t\\ {} & =\frac{{e^{i\xi y}}}{\lambda +{\xi ^{2}}}.\end{aligned}\]
By using the fact that
we now compute the double transform of $a(x,y)\frac{\partial g}{\partial x}$.
This implies, by inverting the Fourier transform, that
We recall that
thus by inverting the Laplace transform in (3.6) we obtain
and, by considering the same arguments (see (3.4)),
Returning to our initial problem, by using (3.8) and (1.8) we have that
(3.4)
\[ \tilde{g}(x,y,\lambda )=\frac{{e^{-|y-x|\sqrt{\lambda }}}}{2\sqrt{\lambda }}=\left\{\begin{array}{l}\displaystyle \frac{{e^{-(y-x)\sqrt{\lambda }}}}{2\sqrt{\lambda }},\hspace{1em}y>x,\\ {} \displaystyle \frac{{e^{-(x-y)\sqrt{\lambda }}}}{2\sqrt{\lambda }},\hspace{1em}y\le x,\end{array}\right.\](3.5)
\[\begin{aligned}{}& {\int _{-\infty }^{\infty }}{e^{i\xi x}}\big[{\mathbb{1}_{(-\infty ,y]}}(x)-{\mathbb{1}_{(y,\infty )}}(x)\big]\frac{\partial \tilde{g}}{\partial x}(x,y,\lambda )\hspace{0.1667em}\mathrm{d}x\\ {} & =\frac{1}{2}\Bigg({\int _{-\infty }^{y}}{e^{i\xi y}}{e^{-(y-x)\sqrt{\lambda }}}\hspace{0.1667em}\mathrm{d}x+{\int _{y}^{\infty }}{e^{i\xi y}}{e^{-(x-y)\sqrt{\lambda }}}\hspace{0.1667em}\mathrm{d}x\Bigg)\\ {} & =\frac{{e^{i\xi y}}}{2}\Bigg({\int _{0}^{\infty }}{e^{-i\xi x}}{e^{-x\sqrt{\lambda }}}\hspace{0.1667em}\mathrm{d}x+{\int _{0}^{\infty }}{e^{i\xi x}}{e^{-x\sqrt{\lambda }}}\hspace{0.1667em}\mathrm{d}x\Bigg)\\ {} & =\frac{{e^{i\xi y}}}{2}\bigg(\frac{1}{i\xi +\sqrt{\lambda }}+\frac{1}{-i\xi +\sqrt{\lambda }}\bigg)\\ {} & =\sqrt{\lambda }\frac{{e^{i\xi y}}}{\lambda +{\xi ^{2}}}=\sqrt{\lambda }\hat{\tilde{g}}.\end{aligned}\](3.7)
\[ {\int _{0}^{\infty }}{e^{-\lambda t}}{\mathcal{D}_{t}^{\frac{1}{2}}}g\hspace{0.1667em}\mathrm{d}t=\sqrt{\lambda }\tilde{g},\]
\[\begin{aligned}{}{\mathcal{D}_{t}^{\frac{1}{2},\frac{{\mu ^{2}}}{4}}}u& ={e^{-\frac{{\mu ^{2}}}{4}t}}{\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\frac{{\mu ^{2}}t}{4}}}u\big)-\frac{\mu }{2}u\\ {} & ={e^{+\frac{\mu }{2}(y-x)\hspace{0.1667em}-\hspace{0.1667em}\frac{{\mu ^{2}}}{4}t}}\hspace{0.1667em}{\mathcal{D}_{t}^{\frac{1}{2}}}g-\frac{\mu }{2}u\\ {} & ={e^{-\frac{{\mu ^{2}}}{4}t}}\hspace{0.1667em}{e^{\frac{\mu }{2}(y-x)}}\hspace{0.1667em}\hspace{0.1667em}a(x,y)\frac{\partial g}{\partial x}-\frac{\mu }{2}u\\ {} & =a(x,y)\bigg(\frac{\partial u}{\partial x}+\frac{\mu }{2}u\bigg)-\frac{\mu }{2}u\end{aligned}\]
and
\[\begin{aligned}{}{\mathcal{D}_{t}^{\frac{1}{2},\frac{{\mu ^{2}}}{4}}}u& ={e^{-\frac{{\mu ^{2}}}{4}t}}{\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\frac{{\mu ^{2}}t}{4}}}u\big)-\frac{\mu }{2}u\\ {} & ={e^{+\frac{\mu }{2}(y-x)\hspace{0.1667em}-\hspace{0.1667em}\frac{{\mu ^{2}}}{4}t}}\hspace{0.1667em}{\mathcal{D}_{t}^{\frac{1}{2}}}g-\frac{\mu }{2}u\\ {} & =-{e^{-\frac{{\mu ^{2}}}{4}t}}\hspace{0.1667em}{e^{\frac{\mu }{2}(y-x)}}\hspace{0.1667em}\hspace{0.1667em}a(x,y)\frac{\partial g}{\partial y}-\frac{\mu }{2}u\\ {} & =a(x,y)\bigg(-\frac{\partial u}{\partial y}+\frac{\mu }{2}u\bigg)-\frac{\mu }{2}u.\end{aligned}\]
This completes the proof. □The drifted Brownian motion has therefore a transition function satisfying a time fractional equation where the fractional derivative is a tempered Riemann–Liouville derivative with parameter η which is related to the drift by the relationship $\sqrt{\eta }=\frac{\mu }{2}$.
4 Fractional equation governing the folded drifted Brownian motion
We here consider the process
This process has distribution
for $y>x$ and $t>0$. We now prove the following theorem.
\[\begin{aligned}{}P\big(|B(t)+\mu t|+x<y\big)=& P\big(x-y-\mu t<B(t)<y-x-\mu t\big)\\ {} =& {\int _{x-y-\mu t}^{y-x-\mu t}}\frac{{e^{-\frac{{w^{2}}}{4t}}}}{\sqrt{4\pi t}}\hspace{0.1667em}\mathrm{d}w\end{aligned}\]
and therefore its transition function is
(4.2)
\[\begin{aligned}{}P\big(|{B^{\mu }}(t)|+x\in \mathrm{d}y\big)/\mathrm{d}y& =\frac{{e^{-\frac{{(y-x-\mu t)^{2}}}{4t}}}}{\sqrt{4\pi t}}+\frac{{e^{-\frac{{(y-x+\mu t)^{2}}}{4t}}}}{\sqrt{4\pi t}}\\ {} & =\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}{e^{-{\mu ^{2}}\frac{t}{4}}}\big[{e^{-\frac{\mu }{2}(y-x)}}+{e^{\frac{\mu }{2}(y-x)}}\big]\\ {} & =v(x,y,t)\end{aligned}\]Theorem 2.
The law v of $|{B^{\mu }}(t)|+x$ satisfies the fractional equation
with initial and boundary conditions
and
(4.3)
\[ {\mathcal{D}_{t}^{\frac{1}{2},\eta }}v=-\frac{\partial v}{\partial y}+v\hspace{0.1667em}\sqrt{\eta }\tanh \big(\sqrt{\eta }(y-x)\big)-\frac{\mu }{2}v,\hspace{1em}y>x>0,\]Proof.
From (4.2) and (1.8) we have that
\[\begin{aligned}{}{\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\frac{{\mu ^{2}}t}{4}}}v\big)& =2\cosh \bigg(\frac{\mu }{2}(y-x)\bigg)\hspace{0.1667em}\hspace{0.1667em}{\mathcal{D}_{t}^{\frac{1}{2}}}\bigg(\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\bigg)\end{aligned}\]
Let ${E_{\frac{1}{2}}}$ be the Mittag-Leffler function of order $1/2$ and g be the function
Since
\[\begin{aligned}{}{\int _{0}^{\infty }}{e^{-\lambda t}}{\int _{x}^{\infty }}{e^{-\xi y}}g(x,y,t)\hspace{0.1667em}\mathrm{d}y\hspace{0.1667em}\mathrm{d}t=& {e^{-\xi x}}{\int _{0}^{\infty }}{e^{-\lambda t}}{E_{\frac{1}{2}}}\big(-\xi {t^{\frac{1}{2}}}\big)\hspace{0.1667em}\mathrm{d}t\\ {} =& {e^{-\xi x}}\frac{{\lambda ^{\frac{1}{2}-1}}}{\xi +{\lambda ^{\frac{1}{2}}}}\end{aligned}\]
we obtain that
Then, for $y>x$,
\[\begin{aligned}{}& {\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\frac{{\mu ^{2}}t}{4}}}v\big)\\ {} & =2\cosh \bigg(\frac{\mu }{2}(y-x)\bigg)\hspace{0.1667em}\hspace{0.1667em}\bigg(-\frac{\partial }{\partial y}\bigg(\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\bigg)\bigg)\\ {} & =-\frac{\partial }{\partial y}\bigg(2\cosh \bigg(\frac{\mu }{2}(y-x)\bigg)\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\bigg)+\frac{\mu }{2}\cdot 2\sinh \bigg(\frac{\mu }{2}(y-x)\bigg)\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\\ {} & =-\frac{\partial }{\partial y}\big({e^{\frac{{\mu ^{2}}}{4}t}}v(x,y,t)\big)+\mu \hspace{0.1667em}\sinh \bigg(\frac{\mu }{2}(y-x)\bigg)\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\end{aligned}\]
with boundary condition
In view of (1.8) we obtain that
\[\begin{aligned}{}{\mathcal{D}_{t}^{\frac{1}{2},\eta }}v+\frac{\mu }{2}v& ={e^{-\frac{{\mu ^{2}}}{4}t}}{\mathcal{D}_{t}^{\frac{1}{2}}}\big({e^{\frac{{\mu ^{2}}}{4}t}}v\big)\\ {} & =-\frac{\partial v}{\partial y}-\mu {e^{-\frac{{\mu ^{2}}}{4}t}}\sinh \bigg(\frac{\mu }{2}(y-x)\bigg)\frac{{e^{-\frac{{(y-x)^{2}}}{4t}}}}{\sqrt{4\pi t}}\\ {} & =-\frac{\partial v}{\partial y}-\frac{\mu }{2}\hspace{0.1667em}\tanh \bigg(\frac{\mu }{2}(y-x)\bigg)v.\end{aligned}\]
□This result shows that the structure of the governing equation of the process $|B(t)+\mu t|+x$ is substantially different from that of $B(t)+\mu t+x$. The difference between (3.3) and (4.3) consists in the non-constant coefficient $\tanh \frac{\mu }{2}(y-x)$ which converges to one as the difference $|y-x|$ tends to infinite. Thus the two equations emerging in this analysis coincide for $|x-y|\to \infty $.
