# Special Solutions to the Space Fractional Diffusion Problem

Tokinaga Namba<EMAIL_ADDRESS>Mathematical Science &
Technology Research Laboratory, Advanced Technology Research Laboratories,
Research & Development, Nippon Steel Corporation, 20-1 Shintomi, Futtsu, Chiba
Prefecture 293-8511, JP Piotr<EMAIL_ADDRESS>Institute of Applied
Mathematics and Mechanics
University of Warsaw
ul. Banacha 2, 02-097 Warsaw, PL Shoichi<EMAIL_ADDRESS>Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1
Komaba, Meguro-ku, Tokyo, 153-8914 JP

###### Abstract

We derive a fundamental solution ${\mathscr{E}}$ to a space-fractional
diffusion problem on the half-line. The equation involves the Caputo
derivative. We establish properties of ${\mathscr{E}}$ as well as formulas for
solutions to the Dirichlet and Neumann problems in terms of convolution of
${\mathscr{E}}$ with data. We also study integrability of derivative of
solutions given in this way. We present conditions sufficient for uniqueness.
Finally, we show the infinite speed of signal propagation.

Key words: Caputo derivative, space-fractional diffusion operator, fundamental
solution, regularity of solutions, decay of solutions, speed of propagation

2020 Mathematics Subject Classification. Primary: 35R11, Secondary: 35A08

##  1 Introduction

In [9] and [14] the authors studied the following equation

$\frac{\partial u}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x}u=0\qquad(x,t)\in\Omega\times(0,\infty),$ (1.1)

when $\Omega=(0,L)$ augmented with the boundary data and initial conditions,
here $D^{\alpha}_{x}$ is the fractional Caputo derivative with respect to the
spacial variable $x$. The interest in such problems stems from the fact that
eq. (1.1) appears in the Green–Ampt infiltration models of subsurface flows,
see [16]. In fact, (1.1) is a simplification of free boundary problem, which
was studied in [15].

In this note we derive in Theorem 3.1 a formula for ${\mathscr{E}}$, a self-
similar solution to (1.1) considered on $(x,t)\in\mathbb{R}_{+}$ (here we
write $\mathbb{R}_{+}$ for $(0,\infty)$). Studying self-similar solutions or
travelling fronts is important, when we wish to gain insight into the
structure of solution and in particular their long time behavior.

Function ${\mathscr{E}}$, which we find, is sufficiently smooth to be a
classical solution to (1.1). We show that

${\mathscr{E}}(x,t)=\frac{a_{0}}{t^{\frac{1}{1+\alpha}}}E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{x^{1+\alpha}}{(1+\alpha)t}\right),$

where $E_{\alpha,1+1/\alpha,1/\alpha}$ is the 3-parameter generalized Mittag-
Leffler function, see (2.8). We show that ${\mathscr{E}}$ is positive and the
integral of ${\mathscr{E}}(\cdot,t)$ over $\mathbb{R}_{+}$ does not depend on
$t>0$. In fact, ${\mathscr{E}}$ is a fundamental solution of (1.1), namely
solutions to the Dirichlet and Neumann problems for $\Omega=\mathbb{R}_{+}$
may be expressed by means of convolution of the initial condition with
${\mathscr{E}}$, this is the content of Theorem 3.12.

The problem of existence of a fundamental solution to various versions of
time-fractional problems has already been addressed in the literature. We name
just a few papers dealing with this issues, see [1], [8], [11], [5], [12]. The
tools used there are different from ours. However, it is not surprising that
generalized Mittag-Leffler functions play a role. We could justify positivity
of ${\mathscr{E}}$ on the grounds of the theory of Mittag-Leffler functions,
(actually we do this in the appendix). However, we would like to stress that
our proof of positivity of ${\mathscr{E}}$ is entirely based on a PDE tool,
which the maximum principle, we use this idea after [13].

Let us stress that the justification of formulas for solution based on the
convolution requires establishing a number of properties of
${\mathscr{E}}(\cdot,1)$. In particular, we show that ${\mathscr{E}}(\cdot,1)$
is monotone. For this purpose we use a PDE tool, which is the maximum
principle. We also show some sort of decay, namely $x{\mathscr{E}}(x,1)$ is
uniformly bounded, see Lemma 3.14. These properties of ${\mathscr{E}}$ seem to
be of independent interest.

Once we have a convolution formula for solutions to (1.1), we may study
properties of solutions to (1.1) when $\Omega=\mathbb{R}_{+}$. An urging
question is about uniqueness of solutions given in this way. We show that if
the initial conditions are sufficiently regular, i.e. they are absolutely
continuous with compact support, then solutions enjoy sufficient regularity
for employing the method of testing the equation with the solution itself, see
Proposition 3.16. This technique immediately yields uniqueness and decay of
solutions.

Eq. (1.1) contains a parameter $\alpha$, so does ${\mathscr{E}}$. Due to
analyticity of $E_{\alpha,1+1/\alpha,1/\alpha}$ we deduce that if $u^{\alpha}$
is a solution to (1.1) on $\mathbb{R}_{+}$, then $u^{\alpha}\to
u^{\alpha_{0}}$ in $L^{1}(\mathbb{R}_{+})$ when $\alpha$ goes to
$\alpha_{0}\in(0,1]$. We explicitly exclude $\alpha_{0}=0$, which is due to
the fact that $E_{\alpha,1+1/\alpha,1/\alpha}$ has no limit on
$\mathbb{R}_{+}$ when $\alpha\to 0$. However, this case is covered by [9,
Theorem 6.1].

With this observation we may address here the issue of the speed of the signal
propagation. This is a bit puzzling because for $\alpha=1$, eq. (1.1) becomes
the heat equation with the infinite speed of propagation, while for $\alpha=0$
problem (1.1) is the transport eq. for which the speed is finite.

Our numerical experiments presented in [10] show that an initial pulse moves
to the left with a finite speed. The same conclusions are drawn on the basis
of numerical simulations by the authors of [7] who dealt with the time-
fractional diffusion-wave equation. However, our Proposition 3.21 stated for
(1.1) with $\Omega=\mathbb{R}_{+}$ and the Neumann boundary condition shows
that actually the speed of the signal is infinite – the support of the
solution instantly becomes equal to $[0,\infty)$. This is shown with the help
of the explicit formulas employing the fundamental solution, ${\mathscr{E}}$
constructed here. This fact does not contradict numerical observation.

After presenting the content we describe the organization of the note. In
Section 2 we recall the fundamentals of the fractional calculus. Section 3 is
devoted to the derivation of a formula for a self-similar solution
${\mathscr{E}}$. Here we also study its properties collected in Theorem 3.1
and we derive the formulas for the integral representation of unique
solutions. In the Appendix we present the derivation of ${\mathscr{E}}$ which
based on the properties of the generalized Mittag-Leffler function $E_{\beta
kl}$ and not the series manipulation. We also show a short proof of positivity
of ${\mathscr{E}}$, which follows from the theory of the generalized Mittag-
Leffler function.

##  2 Preliminaries

We will simultaneously recall the definitions of the Caputo and Riemann-
Liouville fractional derivatives. For a function $f\in L^{1}(0,L)$ and
$\alpha\in(0,1)$ we introduce the fractional integration operator by setting,

$(I^{\alpha}f)(x)=\frac{1}{\Gamma(\alpha)}\int_{0}^{x}(x-z)^{\alpha-1}f(z)\,dz.$
(2.1)

For an absolutely continuous function $u\in AC[0,L]$ we define the Caputo
fractional derivative of order $\alpha\in(0,1)$ by the following formula,

$D^{\alpha}_{C}u(x)=(I^{1-\alpha}u^{\prime})(x)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{x}\frac{u^{\prime}(s)}{(x-s)^{\alpha}}\,ds,$
(2.2)

while the Riemann-Liouville fractional derivative has the form,

$D^{\alpha}_{RL}u=\frac{d}{dx}(I^{1-\alpha}u).$ (2.3)

Later, for the sake of simplicity of notation we will write $D^{\alpha}$ for
$D^{\alpha}_{C}$, occasionally with a lower index $x$ indicating the space
variable.

We notice that if $u$ is absolutely continuous and $u(0)=0,$ then we have

$\frac{d}{dx}\left(I^{1-\alpha}u\right)=I^{1-\alpha}\frac{d}{dx}u\qquad\hbox{i.e.}\qquad
D^{\alpha}_{C}u=D^{\alpha}_{{RL}}u.$ (2.4)

The following formula explains the relationship between the two types of
derivatives for a general function $u\in AC[0,L]$,

$D^{\alpha}_{x_{RL}}u=D^{\alpha}_{x_{C}}u+\frac{x^{-\alpha}}{\Gamma(1-\alpha)}u(0).$
(2.5)

The fractional integration is the inverse of the Caputo derivative up to a
constant,

$I^{\alpha}D^{\alpha}_{C}u(x)=u(x)-u(0).$ (2.6)

In our analysis we will need a more convenient representation of the operator
$(D^{\alpha}_{x}u)_{x}$. For this purpose we need to recall:

######  Lemma 2.1.

(see [9, Proposition 2.1]) Let $u:[0,l)\to\mathbb{R}$ be such that $u\in
C^{2}(0,l)\cap C[0,l)$ and $u^{\prime}\in L^{1}(0,l)$. Then,
$(D^{\alpha}_{x}u)_{x}$ exists everywhere in $(0,l)$ and

$\begin{split}(D^{\alpha}u)_{x}(x)=\frac{1}{\Gamma(1-\alpha)}&\left(\frac{\alpha(u(0)-u(x))+(\alpha+1)u^{\prime}(x)x}{x^{\alpha+1}}\right.\\\
&\quad\left.+\alpha(\alpha+1)\int_{0}^{x}[u(x-z)-u(x)+u^{\prime}(x)z]\frac{dz}{z^{\alpha+2}}\right)\end{split}$
(2.7)

for $x\in(0,l)$.

With the help of this Lemma we will compute action of $(D^{\alpha}u)_{x}$ on
scaled functions:

######  Corollary 2.2.

Let $u:[0,l)\to\mathbb{R}$ be such that $u\in C^{2}(0,l)\cap C[0,l)$ and
$u^{\prime}\in L^{1}(0,l)$. If $\lambda>0$ and we set
$v_{\lambda}(x)=u(\lambda^{\frac{1}{1+\alpha}}x)$, then
$(D^{\alpha}_{x}v_{\lambda})_{x}(x)=\lambda(D^{\alpha}_{x}u)_{y}(\lambda^{\frac{1}{1+\alpha}}x)$.

###### Proof.

We use Lemma 2.1 to calculate $(D_{x}^{\alpha}v_{\lambda})_{x}$,

$\displaystyle(D_{x}^{\alpha}v_{\lambda})_{x}(x)$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\left(\frac{\alpha(v_{\lambda}(0)-v_{\lambda}(x))+(\alpha+1)(v_{\lambda})_{x}(x)x}{x^{\alpha+1}}+\alpha(\alpha+1)\int_{0}^{x}[v_{\lambda}(x-z)-v_{\lambda}(x)+(v_{\lambda})_{x}(x)z]\frac{dz}{z^{\alpha+2}}\right)$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\left(\frac{\alpha(u(0)-u(\lambda^{\frac{1}{1+\alpha}}x))+(\alpha+1)\lambda^{\frac{1}{1+\alpha}}u_{y}(\lambda^{\frac{1}{1+\alpha}}x)x}{x^{\alpha+1}}\right.$
$\displaystyle\quad\left.+\alpha(\alpha+1)\int_{0}^{x}[u(\lambda^{\frac{1}{1+\alpha}}(x-z))-u(\lambda^{\frac{1}{1+\alpha}}x)+\lambda^{\frac{1}{1+\alpha}}u_{y}(\lambda^{\frac{1}{1+\alpha}}x)z]\frac{dz}{z^{\alpha+2}}\right)$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\left(\frac{\alpha(u(0)-u(\lambda^{\frac{1}{1+\alpha}}x))+(\alpha+1)u_{y}(\lambda^{\frac{1}{1+\alpha}}x)(\lambda^{\frac{1}{1+\alpha}}x)}{\lambda^{-1}(\lambda^{\frac{1}{1+\alpha}}x)^{\alpha+1}}\right.$
$\displaystyle\quad\left.+\alpha(\alpha+1)\int_{0}^{x}[u(\lambda^{\frac{1}{1+\alpha}}(x-z))-u(\lambda^{\frac{1}{1+\alpha}}x)+u_{y}(\lambda^{\frac{1}{1+\alpha}}x)(\lambda^{\frac{1}{1+\alpha}}z)]\frac{dz}{z^{\alpha+2}}\right).$

Changing the variable of integration by $\lambda^{\frac{1}{1+\alpha}}z=\xi$ we
see $dz/z^{\alpha+2}=\lambda d\xi/\xi^{2+\alpha}$. Hence that the last term of
the right-hand-side (RHS) is equal to

$\frac{\alpha(\alpha+1)\lambda}{\Gamma(1-\alpha)}\int_{0}^{\lambda^{\frac{1}{1+\alpha}}x}[u(\lambda^{\frac{1}{1+\alpha}}x-\xi)-u(\lambda^{\frac{1}{1+\alpha}}x)+u_{y}(\lambda^{\frac{1}{1+\alpha}}x)\xi]\frac{d\xi}{\xi^{\alpha+2}}.$

Therefore applying Lemma 2.1 again yields

$(D_{x}^{\alpha}v_{\lambda})_{x}(x)=\lambda^{1}(D_{y}^{\alpha}u)_{y}(\lambda^{\frac{1}{1+\alpha}}x).$

∎

In our construction of the self-similar solution we will use a three-parameter
generalized Mittag-Leffler function $E_{\beta,m,l}$. It is defined by the
following series for $z\in\mathbb{C}$, (see [4, formula (1.9.19)])

$E_{\beta,m,l}(z)=\sum_{n=0}^{\infty}c_{n}^{\beta ml}z^{n},\qquad\Re\beta>0,\
m>0,\ -\beta(jm+l)\not\in\mathbb{N}\setminus\\{0\\},$ (2.8)

where

$c_{n}^{\beta
ml}=\prod_{i=0}^{n-1}\frac{\Gamma(\beta(im+l)+1)}{\Gamma(\beta(im+l+1)+1)}.$

It is worth recalling that due to [2, Theorem 1], see also page 48 in [4], we
know that $E_{\beta,m,l}$ is an entire function of order $(\Re\beta)^{-1}$ and
type $m^{-1}$, i.e.

$|E_{\beta,m,l}(z)|<\exp((m^{-1}+\epsilon)|z|^{1/\beta})$

holds for $|z|\geq r_{0}(\epsilon),$ where $r_{0}(\epsilon)$ is sufficiently
large.

In several places we will use the following observation concerning the Mittag-
Leffler functions. Let us set,

$\Phi(x)=E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{x^{1+\alpha}}{1+\alpha}\right).$
(2.9)

We recall:

######  Proposition 2.3.

(see [4, Example 4.11], [3, Theorem 4]) Function $\Phi$ defined in (2.9)
satisfies the following fractional ordinary differential equation,

$D_{y_{C}}^{\alpha}v(y)+\frac{1}{1+\alpha}yv(y)=0,\quad y>0,\qquad v(0)=1.$
(2.10)

##  3 The fundamental solution and its properties

We will derive a self-similar solution, ${\mathscr{E}},$ to the following
equations,

$\frac{\partial u}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}u=0\qquad(x,t)\in(0,\infty)^{2}.$ (3.1)

This is done in the theorem below, where we also present the basic properties
of ${\mathscr{E}}$.

######  Theorem 3.1.

Function ${\mathscr{E}}:(0,\infty)^{2}\to\mathbb{R}$ given by formula

${\mathscr{E}}(x,t)=a_{0}t^{-\frac{1}{1+\alpha}}E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{x^{1+\alpha}}{(1+\alpha)t}\right).$
(3.2)

where

$\frac{1}{a_{0}}=2\int_{0}^{\infty}E_{\alpha,1+1/\alpha,1/\alpha}(-x^{1+\alpha}/(1+\alpha))\,dx$
(3.3)

is well-defined:
(1) ${\mathscr{E}}\in C^{2}((0,\infty)^{2})$ and ${\mathscr{E}}$ is a self-
similar solution to (3.1), i.e. it is invariant under the transformation
$(x,t)\mapsto(\lambda^{\frac{1}{1+\alpha}}x,\lambda t)$ for $\lambda>0$.
(2) ${\mathscr{E}}$ is positive for all $x,t>0$.
(3) For all $t>0$ function ${\mathscr{E}}(\cdot,t)$ is decreasing.
(4) For all $t>0$ function ${\mathscr{E}}(\cdot,t)$ is in $L^{1}(0,\infty)$
and the integral $\int_{0}^{\infty}{\mathscr{E}}(x,t)\,dx=\frac{1}{2}$.

######  Remark 3.2.

Using $\Phi$ defined in (2.9) we can write ${\mathscr{E}}$ shortly as follows,

${\mathscr{E}}(x,t)=a_{0}t^{-\frac{1}{1+\alpha}}\Phi\left(\frac{x}{t^{\frac{1}{1+\alpha}}}\right).$

Our proof of Theorem 3.1 will be conducted in a number of steps. Here is the
first one:

######  Lemma 3.3.

Let us suppose that $u\in C^{2}((0,\infty)^{2})$. Then $u$ is a solution to
(3.1) if and only if $u_{\lambda}$ is a solution, where

$u_{\lambda}(x,t)=u(\lambda^{\frac{1}{1+\alpha}}x,\lambda t).$

###### Proof.

Now, it is straightforward to see that

$(u_{\lambda})_{t}(x,t)=\lambda^{1}u_{s}(\lambda^{\frac{1}{1+\alpha}}x,\lambda
t).$

We use Corollary 2.2 to compute $(D^{\alpha}u_{\lambda})_{x}$. We apply this
result to $v_{\lambda}$ defined as
$v_{\lambda}(x)=u(\lambda^{\frac{1}{1+\alpha}},\lambda t)$. This leads us to
the identity,

$(D^{\alpha}u_{\lambda})_{x}(x,t)=(D^{\alpha}v_{\lambda})_{x}(x)=\lambda^{1}(D^{\alpha}u)_{y}(\lambda^{\frac{1}{1+\alpha}},\lambda
t).$

Thus, we see that $u$ satisfies (3.1) if and only if $u_{\lambda}$ fulfills

$(u_{\lambda})_{t}(x,t)-(D_{x}^{\alpha}u_{\lambda})_{x}(x,t)=0$

for all $x>0,\ t>0$. ∎

The above Lemma tells us that self-similar solutions depend only on
$\xi=xt^{-\frac{1}{1+\alpha}}$. However, if we want to obtain a solution,
whose average over $\mathbb{R}_{+}$ is independent of time, then we must
consider ${\mathscr{E}}$ of the form

${\mathscr{E}}(x,t)=a_{0}t^{\gamma}v(x^{1+\alpha}/t),$ (3.4)

where $v$ is integrable. It is easy to see that for

$\gamma=-\frac{1}{1+\alpha}$

and $t_{2}>t_{1}>0$ we have

$\int_{0}^{\infty}{\mathscr{E}}(x,t_{2})\,dx=\int_{0}^{\infty}{\mathscr{E}}(x,t_{2})\,dx.$

We are now ready to derive the form of ${\mathscr{E}}$.

######  Lemma 3.4.

Let us assume that $v$ appearing in (3.4) is analytic and $v(0)=1$. Then,

$v(z)=E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{z}{1+\alpha}\right),$

where $E_{\beta,m,l}$ is the generalized Mittag-Leffler function defined in
(2.8).

###### Proof.

Let us suppose that

$v(z)=\sum_{n=0}^{\infty}c_{n}z^{n}.$

Then, inserting ${\mathscr{E}}$ defined by (3.4) into (3.1) yields

$\sum_{n=0}^{\infty}c_{n}(\gamma-n)\frac{x^{(1+\alpha)n}}{t^{n-\gamma+1}}=\sum_{n=1}^{\infty}c_{n}\frac{\Gamma((1+\alpha)n+1)[(1+\alpha)(n-1)+1]}{\Gamma((1+\alpha)(n-1)+2)}\frac{x^{(1+\alpha)(n-1)}}{t^{n-\gamma}},$

where we took into account that
$D^{\alpha}x^{\beta}=\frac{\Gamma(\beta+1)}{\Gamma(\beta+1-\alpha)}x^{\beta-\alpha}.$
Hence, we find the formula for $c_{n}$,

$c_{n}=\frac{(-1)^{n}}{(1+\alpha)^{n}}b_{n},$

where we set $c_{0}=1$ and

$b_{n}=\prod_{i=0}^{n-1}\frac{\Gamma(\alpha i+i+2)}{\Gamma(\alpha(i+1)+i+2)}.$
(3.5)

Hence,

$v(z)=\sum_{n=0}^{\infty}(-1)^{n}b_{n}\left(\frac{z}{1+\alpha}\right)^{n}.$

If we take into account the form of $b_{n}$’s, then we realize that

$v(z)=E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{z}{1+\alpha}\right),$

where $E_{\beta,m,l}$ is a generalized Mittag-Leffler function defined in
(2.8). Thus, we reache ${\mathscr{E}}$ of the form (3.2), where the
multiplicative constant $a_{0}$ has to be determined by other means. ∎

######  Remark 3.5.

The argument above is based on the series manipulation. In the Appendix, we
present the proof of this lemma, which is based only on the theory of the
generalized Mittag-Leffler function $E_{\beta,m,l}$.

An important step in our analysis is checking that ${\mathscr{E}}$ is indeed
positive. The argument we use is based on the use of the maximum principle for
(1.1). This idea was used first in the proof of a similar result in [13]. We
provide our own and extended version of the argument. In the Appendix we
present a short proof of the same result, which is of independent interest. It
is based on the theory of the three-parameter Mittag-Leffler function.

######  Lemma 3.6.

Function $\Phi$ defined in (2.9) is positive for all $x>0$.

###### Proof.

Let us set

$u(x,t):=\int_{0}^{xt^{-\frac{1}{1+\alpha}}}\Phi(z)\,dx.$

We will see that

$\left(\frac{\partial}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}\right)u<0.$ (3.6)

Indeed, $\frac{\partial u}{\partial t}(x,t)=-\frac{1}{t(1+\alpha)}y\Phi(y)$,
where $y=x{t^{-\frac{1}{1+\alpha}}}$ and by Corollary 2.2 we see that
$\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}u(x,t)=\frac{1}{t}\frac{\partial}{\partial
y}D^{\alpha}_{y_{C}}u(y,t)$. Thus, due to

$\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}u=D^{\alpha}_{x_{RL}}\frac{\partial}{\partial x}u$

and (2.5) we obtain

$\left(\frac{\partial}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}\right)u(x,t)=-\frac{1}{t}\left(\frac{y\Phi(y)}{1+\alpha}+D^{\alpha}_{y_{RL}}\Phi(y)\right)=-\frac{1}{t}\left(\frac{y\Phi(y)}{1+\alpha}+D^{\alpha}_{y_{C}}\Phi(y)\right)-\frac{y^{-\alpha}}{t\Gamma(1-\alpha)}.$

Now, we invoke Proposition 2.3 to conclude that the expression in parenthesis
vanishes. Thus, (3.6) follows.

Let us suppose our claim is not valid and the set

$A=\\{x>0:\Phi(x)<0\\}$

is not empty. Since $\Phi(0)=1$, we see that $\inf A=:x_{0}>0$. Due to the
continuity of $\Phi$ there is $x_{1}>x_{0}$ such that

$u(x,1)>0\qquad\hbox{for }x\in(0,x_{1}).$

Let us set

$\Omega=\\{(x,t)\in(0,\infty)^{2}:\
x\in(0,x_{1}t^{-1/(1+\alpha)}),t\in(1,2)\\}.$ (3.7)

By the weak maximum principle for (3.6) in non-cylindrical regions, see [15,
Lemma 8], we have

$\sup\\{u(x,t):(x,t)\in\Omega\\}=\max\\{u(x,t):(x,t)\in\partial\Omega\setminus(\\{2\\}\times\mathbb{R}_{+})\\}=:M.$

However, since $u(0,t)=0$ and
$u(x_{1}t^{-1/(1+\alpha)},t)=u(x_{1},1)<u(x_{0},1)$ for $t\in[1,2]$, then we
see that

$M=u(x_{0},1).$

If we had the strong maximum principle, then we would have reached a
contradiction. But we do not have it, thus, we have to continue our argument.
For a positive $\epsilon$ and $(x,t)\in\mathbb{R}_{+}\times(1,\infty)$ we set,

$v_{\epsilon}(x,t)=\epsilon c_{\alpha}x^{1+\alpha}+\epsilon(t-1),$

where $c_{\alpha}=\frac{1}{(1+\alpha)\Gamma(1+\alpha)}.$ Due to the choice of
$c_{\alpha}$ we see that $\frac{\partial v_{\epsilon}}{\partial
t}-\frac{\partial}{\partial x}D^{\alpha}v_{\epsilon}=0$. Thus, the sum
$u+v_{\epsilon}$ satisfies (3.6) and we may apply the weak maximum principle,
see [15, Lemma 8], which yields,

$\sup\\{(u+v_{\epsilon})(x,t):(x,t)\in\Omega\\}=\max\\{(u+v_{\epsilon})(x,t):(x,t)\in\partial\Omega\setminus(\\{2\\}\times\mathbb{R}_{+})\\}=:M_{\epsilon}.$
(3.8)

We want to select such $\epsilon$ that
$M_{\epsilon}=\max(u+v_{\epsilon})(x,1)$. We notice that
$(u+v_{\epsilon})(0,t)=\epsilon(t-1)\leq\epsilon$ and we may restrict
$\epsilon$ so that $\epsilon<u(x_{1},1)$. We take even smaller $\epsilon$ so
that

$\frac{\partial}{\partial x}(u+v_{\epsilon})(x_{1},t)<0.$

Hence, there is $x_{2}\in(x_{0},x_{1})$ such that

$\max\\{(u+v_{\epsilon})(x,1):x\in(0,x_{1})\\}=(u+v_{\epsilon})(x_{2},1)$

We want to guarantee that $M_{\epsilon}$ equals
$(u+v_{\epsilon})(x_{2},1)=u(x_{2},1)+\epsilon c_{\alpha}x_{2}^{1+\alpha}.$

We look at $u+v_{\epsilon}$ on the last part of the parabolic boundary of
$\Omega$,

$\gamma=\\{(x_{1}t^{1/(1+\alpha)},t):\ t\in(1,2)\\}.$

For $(x,t)\in\gamma$ we have

$(u+v_{\epsilon})(x_{1}t^{1/(1+\alpha)},t)=u(x_{1},1)+\epsilon(c_{\alpha}x_{1}^{1+\alpha}t+t-1)=u(x_{1},1)+\epsilon
c_{\alpha}x_{1}^{1+\alpha}+(t-1)\epsilon(c_{\alpha}x_{1}^{1+\alpha}+1).$

We see that right-hand-side above is greater than $(u+v_{\epsilon})(x_{1},1)$
for $t\in(1,2)$. At the same time we see that the left-hand-side attains its
maximum for $t=2$ and we may take $\epsilon$ so small that

$\epsilon(c_{\alpha}x_{1}^{1+\alpha}+1)+(u+v_{\epsilon})(x_{1},1)<(u+v_{\epsilon})(x_{2},1)=M_{\epsilon}.$

Now, we shall see that $\sup_{\Omega}(u+v_{\epsilon})>M_{\epsilon}$. Let us
consider points $(x,t)$ of the following form, $(x_{2}t^{1/(1+\alpha)},t)$,
$t\in(1,2).$ Now, we compute values of $(u+v_{\epsilon})$ there for $t>1$. We
obtain,

$\displaystyle(u+v_{\epsilon})(x_{2}t^{1/(1+\alpha)},t)$ $\displaystyle=$
$\displaystyle u(x_{2},1)+\epsilon
c_{\alpha}x_{2}^{1+\alpha}t+\epsilon(t-1)=u(x_{2},1)+\epsilon
c_{\alpha}x_{2}^{1+\alpha}+\epsilon(t-1)(c_{\alpha}x_{1}^{1+\alpha}+1)$
$\displaystyle>$ $\displaystyle(u+v_{\epsilon})(x_{2},1)=M_{\epsilon}.$

But this inequality violates (3.8). Thus, our claim follows. ∎

In fact, in the course of proof of Lemma 3.6, we established the following
fact:

######  Lemma 3.7.

If $U\in C^{1+\alpha}([0,\infty))$ and $u(x,t)=U(xt^{-\frac{1}{1+\alpha}})$
satisfies inequality (3.6) in $\Omega$ defined in (3.7), where $x_{1}>0$ is
now arbitrary, then $u(x,1)\equiv U(x)$ cannot attain a maximum inside
$(0,x_{1})$.

We notice that the same argument, after the necessary obvious changes, yields:

######  Lemma 3.8.

If $U\in C^{1+\alpha}([0,\infty))$ and $u(x,t)=U(xt^{-\frac{1}{1+\alpha}})$
satisfies inequality

$\left(\frac{\partial}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}\right)u>0$ (3.9)

in $\Omega$ defined in (3.7), where $x_{1}>0$ is now arbitrary, then
$u(x,1)\equiv U(x)$ cannot attain a minimum inside $(0,x_{1})$. ∎

This observation immediately implies part (3) of Theorem 3.1.

######  Lemma 3.9.

For all $t>0$ function ${\mathscr{E}}(\cdot,t)$ is decreasing.

###### Proof.

Since ${\mathscr{E}}$ is a self-similar solution, we may restrict our
attention to $t=1.$ Let us suppose our claim is false and
${\mathscr{E}}(\cdot,1)$ attains a minimum at $x_{0}$. We can find
$x_{1}>x_{0}$ such that ${\mathscr{E}}(x_{1},1)>{\mathscr{E}}(x_{0},1)$. Now,
we choose $\epsilon>0$ sufficiently small that

$v_{\epsilon}(x,t)={\mathscr{E}}(\cdot,t)-\epsilon\frac{x}{t^{\frac{1}{1+\alpha}}},$

so that $v_{\epsilon}(\cdot,1)$ attains its minimum at $x_{2}<x_{0}$. In
particular this implies that $v_{\epsilon}(x_{2},1)<v_{\epsilon}(x_{1},1).$
Moreover, for all $\epsilon>0$ inequality (3.9) is satisfied. As a result we
may apply Lemma 3.8 to $v_{\epsilon}(\cdot,1)$ to deduce that
$v_{\epsilon}(\cdot,1)$ cannot attain any minimum.

We observe that for no positive $x_{0}$ function $\Phi$ is increasing on
$(0,x_{0}).$ If such a point existed, then $D^{\alpha}\Phi\geq 0$ on
$(0,x_{0})$, but this contradicts (2.10) due to positivity of $\Phi.$ Hence,
$v_{\epsilon}$ cannot attain any maximum in $(0,\infty)$.

These observations imply that $v_{\epsilon}(\cdot,1)$ is decreasing. Indeed,
if $0<x_{1}<x_{2}$ and $v_{\epsilon}(x_{1},1)<v_{\epsilon}(x_{2},1)$. Then,
taking into account that $v_{\epsilon}(0,1)=1>v_{\epsilon}(x_{2},1)$ we deduce
that $v_{\epsilon}(\cdot,1)$ must attain a minimum in the interval $(0,x_{2})$
but this is impossible. Hence, the claim follows.

Finally, we notice that ${\mathscr{E}}(x,1)=\lim_{\epsilon\to
0^{+}}v_{\epsilon}(x,1))$, which implies monotonicity of
${\mathscr{E}}(\cdot,1).$ ∎

Now, we will show that ${\mathscr{E}}$ is integrable over the positive half-
line.

######  Lemma 3.10.

Function $\Phi$ defined in (2.9) is bounded with a bound uniform in $\alpha$
and it is integrable over $(0,\infty).$

###### Proof.

We will show first the boundedness of $\Phi$. Due to Proposition 2.3 and Lemma
3.6, we notice,

$D_{C}^{\alpha}\Phi(x)=-\frac{x}{1+\alpha}\Phi(x)<0.$

We may apply the fractional integration operator $I^{\alpha}$ to both sides of
the above inequality. Due to (2.6) we obtain,

$\Phi(x)-\Phi(0)=I^{\alpha}D_{C}^{\alpha}\Phi(x)<0.$ (3.10)

Hence,

$\Phi(x)\equiv
E_{\alpha,1+1/\alpha,1/\alpha}(-\frac{x^{1+\alpha}}{1+\alpha})<\Phi(0)=1.$
(3.11)

Let us stress that the estimate (3.11) is uniform in $\alpha.$

Now, we shall see that boundedness of $\Phi$ implies its integrability. For
this purpose we rewrite (2.10) using (2.5) as follows,

$\frac{1}{1+\alpha}\Phi(x)=\frac{x^{-1-\alpha}}{\Gamma(1-\alpha)}-\frac{x^{-1}}{\Gamma(1-\alpha)}\frac{d}{dx}\int_{0}^{x}\frac{\Phi(t)}{(x-t)^{1-\alpha}}\,dt.$

We integrate it over $[1,R]$ and we reach,

$\frac{\Gamma(1-\alpha)}{1+\alpha}\int_{1}^{R}\Phi(s)\,ds\leq\int_{1}^{R}x^{-1-\alpha}\,dx+\left|\int_{1}^{R}x^{-1}\frac{d}{dx}\int_{0}^{x}\frac{\Phi(t)}{(x-t)^{1-\alpha}}\,dtdx\right|=J_{1}+|J_{2}|.$

In the second term we integrate by parts. This yields,

$J_{2}=\int_{1}^{R}x^{-2}\int_{0}^{x}\frac{\Phi(t)}{(x-t)^{1-\alpha}}\,dtdx+\left.x^{-1}\int_{0}^{x}\frac{\Phi(t)}{(x-t)^{1-\alpha}}\,dt\right|_{x=1}^{x=R}.$

Now, we use (3.11) and positivity of $\Phi$ to see that

$J_{2}\leq\frac{1-R^{\alpha-1}}{\alpha(1-\alpha)}+\frac{R^{\alpha}}{\alpha
R}<\frac{1}{\alpha(1-\alpha)}.$

If we combine it with an easy estimate on $J_{1}$ we arrive at

$\int_{1}^{R}\Phi(s)\,ds\leq\frac{(2-\alpha)(1+\alpha)}{\alpha(1-\alpha)\Gamma(1-\alpha)}.$

Our claim follows. ∎

We notice that the estimate for the integral of $\Phi$ blows up at $\alpha=1$
and $\alpha=0$ which shows our method is not optimal, because the fundamental
solution to the heat equation is integrable due to its exponential decay at
infinity.

We are now ready to finish the proof of Theorem 3.1. The derivation is
performed in Lemmas 3.3 and 3.4. The properties of $\Phi$ were established in
Lemmas 3.6, 3.9 and 3.10. In particular, they guarantee that the integral

$\frac{1}{a_{0}}=2\int_{0}^{\infty}E_{\alpha,1+1/\alpha,1/\alpha}(-x^{1+\alpha}/(1+\alpha))\,dx$

is finite and positive. Hence, the definition of $a_{0}$ given in (3.3) is
correct and ${\mathscr{E}}$ is well-defined with the properties we stated. ∎

######  Remark 3.11.

Actually, our proof shows that for all $t>0$ we have
${\mathscr{E}}(\cdot,t)\in C^{1+\alpha}([0,\infty))$ and this regularity is
optimal.

We constructed ${\mathscr{E}}$ on $\mathbb{R}_{+}^{2}$. In order to discuss
its properties leading to a justification of the name ’fundamental solution’
we have to extend ${\mathscr{E}}$ to $\mathbb{R}_{+}$, without changing the
notation. Actually, the function $x^{1+\alpha}$ is naturally defined for
negative argument as $|x|^{1+\alpha}$. We also set ${\mathscr{E}}$ equal to
zero on $\mathbb{R}\times(-\infty,0]$, so finally

${\mathscr{E}}(x,t)={\mathscr{E}}(|x|,t)\chi_{\mathbb{R}_{+}}(t).$

We do not want to discuss the action of $D^{\alpha}_{x_{C}}$ on
$\mathscr{D}(\mathbb{R})$ so we will not try to show that
$(\frac{\partial}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}){\mathscr{E}}=\delta_{0}$. Instead we will justify the
representation formulas for solution to (3.1) augmented with the initial and
boundary data. In fact, we will reuse the well-known formula derived with the
help of the reflection principles for solutions to the heat equation on the
half line.

######  Theorem 3.12.

Let us suppose that $g\in L^{p}(0,\infty),$ where $p\in[1,\infty)$ and $g$ has
compact support, (resp. $g\in C^{0}_{c}([0,\infty))$ and we set

${\mathscr{E}}_{t}(x)={\mathscr{E}}(x,t).$

We define functions $w_{1}$, $w_{2}$ by the following formulas,

$w_{1}(x,t)=\int_{0}^{\infty}({\mathscr{E}}(x-y,t)-{\mathscr{E}}(x+y,t))g(y)\,dy,$
(3.12)
$w_{2}(x,t)=\int_{0}^{\infty}({\mathscr{E}}(x-y,t)+{\mathscr{E}}(x+y,t))g(y)\,dy.$
(3.13)

Then,
(a) For all $t,R>0$ functions $w_{1}(\cdot,t)$ and $w_{2}(\cdot,t)$ belong to
$C^{1+\alpha}([0,R])$, they are classical solutions to

$\begin{array}[]{ll}\displaystyle{\frac{\partial w}{\partial
t}-\frac{\partial}{\partial x}D^{\alpha}w=0}&x,t>0,\\\
\nobreak\leavevmode&\nobreak\leavevmode\\\ w(x,0)=g(x)&x>0.\end{array}$ (3.14)

In addition $w_{1}$ (resp. $w_{2}$) satisfies the Dirichlet (resp. Neumann)
boundary condition,

$w(0,t)=0,\qquad(\hbox{resp. }w_{x}(0,t)=0)\qquad\hbox{for }t>0.$ (3.15)

(b) Functions $w_{1}$ and $w_{2}$ belong to
$L^{\infty}(\mathbb{R}_{+};L^{p}(\mathbb{R}_{+}))$ if $p<\infty$ (resp.
$w_{1},w_{2}\in L^{\infty}(\mathbb{R}_{+};C(\mathbb{R}_{+}))$, when $g\in
C^{0}_{c}([0,\infty))$). The initial condition is satisfied in the sense
below. However, when $g$ is continuous, we require $g(0)=0$ in case of the
Dirichlet data,

$\lim_{t\to 0}\|w_{1}(\cdot,t)-g\|_{L^{p}}=0,\qquad(\hbox{resp. }\lim_{t\to
0}\|w_{2}(\cdot,t)-g\|_{L^{\infty}}=0).$ (3.16)

###### Proof.

We will first check that $w_{1}$ and $w_{2}$ are well-defined, for this reason
we begin with the first part of (b). We will rewrite $w_{1}$ and $w_{2}$ as a
convolution of the data on $\mathbb{R}_{+}$ with ${\mathscr{E}},$

$w_{1}(x,t)=\int_{\mathbb{R}}{\mathscr{E}}(x-y,t)\tilde{g}(y)\,dy,\qquad
w_{2}(x,t)=\int_{\mathbb{R}}{\mathscr{E}}(x-y,t)\bar{g}(y)\,dy,$ (3.17)

where $\tilde{g}$ (resp. $\bar{g}$) is an odd extension, i.e.
$\tilde{g}(-y)=-g(y)$ (resp. even extension, i.e. $\bar{g}(-y)=g(y)$) for
$y>0.$ Since ${\mathscr{E}}_{t}\in L^{1}(\mathbb{R})$ and
$\int_{\mathbb{R}}{\mathscr{E}}_{t}\,dx=1$, then Young inequality for
convolutions imply that
${\mathscr{E}}_{t}*\tilde{g},{\mathscr{E}}_{t}*\bar{g}\in L^{p}(\mathbb{R})$,
when $p<\infty$ and

$\|{\mathscr{E}}_{t}*\tilde{g}\|_{L^{p}(\mathbb{R})}\leq
2\|g\|_{L^{p}(\mathbb{R}_{+})},\qquad\|{\mathscr{E}}_{t}*\bar{g}\|_{L^{p}(\mathbb{R})}\leq
2\|g\|_{L^{p}(\mathbb{R}_{+})}.$

When $g$ is bounded, then

$\|{\mathscr{E}}_{t}*\tilde{g}\|_{L^{\infty}}\leq\|g\|_{L^{\infty}},\qquad\|{\mathscr{E}}_{t}*\bar{g}\|_{L^{\infty}}\leq\|g\|_{L^{\infty}}.$

We conclude that $w_{1}$ and $w_{2}$ are well-defined.

Let us argue that $w_{1}$, $w_{2}$ are solutions to (3.14). We will provide
some details for $w_{1}$, (the proof for $w_{2}$ is the same). We first notice
that the kernel ${\mathscr{E}}_{t}$ is a composition of an analytic function
with $x\mapsto|x|^{1+\alpha}$, hence the convolution appearing in the
definition of $w_{i}$, $i=1,2$ shares this kind of smoothness. Since $w_{1}$
is a $C^{1+\alpha}$-function we may apply $D^{\alpha}_{x_{C}}$ to $v$ defined
above. We obtain,

$D^{\alpha}_{x_{C}}w_{1}(x,t)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{x}\frac{dy}{(x-y)^{\alpha}}\int_{-\infty}^{\infty}\frac{\partial}{\partial
y}{\mathscr{E}}_{t}(y+s)\tilde{g}(s)\,dy,$

where we could interchange the integral over $\mathbb{R}$ with the
differentiation due to the integrability of $\frac{\partial}{\partial
y}{\mathscr{E}}_{t}(y+s)\tilde{g}(s)$ with respect $s$ for all
$y\in\mathbb{R}$. Now, we notice that we may invoke the Fubini Theorem to
interchange the order of integrals. Thus, we conclude that the partial
integration operator $I^{1-\alpha}$ and the integration over $\mathbb{R}_{+}$
with respect to $y$ commute and we see,

$D^{\alpha}_{x_{C}}w_{1}(x,t)=\int_{-\infty}^{\infty}D^{\alpha}_{y_{C}}{\mathscr{E}}_{t}(x+y)\tilde{g}(y)\,dy.$

Now, due to regularity of ${\mathscr{E}}_{t}$ we see that

$\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}w_{1}=\int_{-\infty}^{\infty}\frac{\partial}{\partial
x}D^{\alpha}_{y_{C}}{\mathscr{E}}_{t}(x+s)\tilde{g}(s)\,dy.$

Thus, we conclude that $(\frac{\partial}{\partial t}-\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}})w_{1}=0$.

Now, we check the boundary conditions. Since $w_{1}(\cdot,t)$ is continuous up
to $x=0$ for $t>0$ we see that

$w_{1}(0,t)=\int_{0}^{\infty}({\mathscr{E}}_{t}(-y)-{\mathscr{E}}_{t}(y))g(y)\,dy=0.$

The RHS vanishes because ${\mathscr{E}}$ is even.

The argument for $w_{2}$ is similar. We use that $\frac{\partial
w_{2}}{\partial x}(\cdot,t)$ is continuous up to $\\{x=0\\}$ for $t>0$. For
positive $x$ we have

$\frac{\partial w_{2}}{\partial
x}(x,t)=-(1+\alpha)\int_{0}^{\infty}(\frac{\partial{\mathscr{E}}_{t}}{\partial
x}(x-y)\hbox{sgn}\,(x-y)|x-y|^{\alpha}+\frac{\partial{\mathscr{E}}_{t}}{\partial
x}(x+y)\hbox{sgn}\,(x-y)|x-y|^{\alpha})g(y)\,dy.$

Thus,

$\lim_{x\to 0^{+}}\frac{\partial w_{2}}{\partial
x}(x,t)=-(1+\alpha)\int_{0}^{\infty}(-\frac{\partial{\mathscr{E}}_{t}}{\partial
x}(-y)y^{\alpha}+\frac{\partial{\mathscr{E}}_{t}}{\partial
x}(y)y^{\alpha})g(y)\,dy=0.$

Now, we turn our attention to the initial condition. We recall that (3.16)
follows from the standard properties of convolution with a kernel whose
integral is one. ∎

We present here convolution formulas to solve the non-homogeneous problem.
After establishing them we may say that indeed ${\mathscr{E}}$ is a
fundamental solution, because it behaves like one.

######  Corollary 3.13.

Let us suppose that $f\in C^{1}([0,\infty)\times\mathbb{R}_{+})\cap
C([0,\infty)^{2})\cap L^{\infty}(\mathbb{R}_{+}^{2})$ and $f$ has a compact
support. We set (in case of $w_{3}$ below we require that $f(0,t)=0$ for all
$t>0$),

$w_{3}(x,t)=\int_{0}^{t}\int_{0}^{\infty}({\mathscr{E}}(x-y,t-s)-{\mathscr{E}}(x+y,t-s))f(y,s)\,dyds,$
$w_{4}(x,t)=\int_{0}^{t}\int_{0}^{\infty}({\mathscr{E}}(x-y,t-s)+{\mathscr{E}}(x+y,t-s))f(y,s)\,dyds.$

Then, for all $R>0$ we have $w_{3},w_{4}\in C^{1+\alpha}([0,R]\times(0,R])$
and $w_{j}(x,\cdot)\in C^{\infty}(0,\infty)$ for all $x>0$, $j=3,4$. Moreover,
$w_{3},w_{4}$ are solutions to

$\begin{array}[]{ll}\displaystyle{\frac{\partial w}{\partial
t}-\frac{\partial}{\partial x}D^{\alpha}w=f}&x,t>0,\\\
\nobreak\leavevmode&\nobreak\leavevmode\\\ w(x,0)=0&x>0.\end{array}$ (3.18)

In addition, $w_{3}$ (resp. $w_{4}$) satisfies the Dirichlet (resp. Neumann)
boundary conditions.

###### Proof.

We will present an argument for $w_{3}$. The proof for $w_{4}$ goes along the
same lines.

We extend $f$ by odd reflection, $\tilde{f}(-y,t)=-f(-y,t)$, for $y>0$. Then,
$w_{3}$ takes the following form, where we use the commutativity of the
convolution,

$w_{3}(x,t)=\int_{0}^{t}\int_{\mathbb{R}}{\mathscr{E}}(x-y,t-s)\tilde{f}(y,s)\,dyds=\int_{0}^{t}\int_{\mathbb{R}}{\mathscr{E}}(y,s)\tilde{f}(x-y,t-s)\,dyds.$

Due to the compactness of the support of $\tilde{f}$, we have

$\frac{\partial w_{3}}{\partial
t}(x,t)=\int_{\mathbb{R}}{\mathscr{E}}(y,s)\tilde{f}(x-y,0)\,dy+\int_{0}^{t}\int_{\mathbb{R}}{\mathscr{E}}(y,s)\frac{\partial\tilde{f}}{\partial
t}(x-y,t-s)\,dyds.$

Then, it is easy to establish by carefully employing the integration by parts
that

$\frac{\partial w_{3}}{\partial
t}(x,t)=\int_{\mathbb{R}}{\mathscr{E}}(y,s)\tilde{f}(x-y,0)\,dy+\int_{0}^{t}\int_{\mathbb{R}}\frac{\partial}{\partial
s}{\mathscr{E}}(y,s)\tilde{f}(x-y,t-s)\,dyds.$

We have already seen in the course of proof of Theorem 3.12 that

$D^{\alpha}_{x_{C}}w_{3}(x,t)=\int_{t}\int_{\mathbb{R}}D^{\alpha}_{x_{C}}{\mathscr{E}}(x-y,t-s)\tilde{f}(y,s)\,dyds.$

Subsequently, it is easy to check that

$\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}w_{3}(x,t)=\int_{t}\int_{\mathbb{R}}\frac{\partial}{\partial
y}D^{\alpha}_{y_{C}}{\mathscr{E}}f(y,s)\tilde{(}x-y,t-s)\,dyds.$

Hence, $w_{3}$ is a solution to (3.18).

An argument used in the proof of Theorem 3.12 shows that $w_{3}$ satisfies the
Dirichlet boundary condition. Now, we shall investigate the initial condition.
Due to the boundedness of $f$ we see that

$|w_{3}(x,t)|\leq\int_{0}^{t}\int_{\mathbb{R}}{\mathscr{E}}(y,s)\|f\|_{L^{\infty}}\,dyds\leq
t\|f\|_{L^{\infty}}\,dyds.$

Our claims follow. ∎

We constructed solutions to (3.1) with the help of the convolution of the
fundamental solution with the data. Since this case is not covered neither in
[9] nor in [14] we have to show uniqueness separately. The difficulty with the
classical method of testing the equation with the solution is that
integrability of the derivatives (fractional and integer) of ${\mathscr{E}}$
is different than one might expect. Here, we present an observation which
turns out very useful.

######  Lemma 3.14.

Let us suppose that $\Phi$ is given by (2.9). Then, for all $x>0$ we have,

$0<x\,\Phi(x)\leq 2.$

Proof. We combine (2.10) and (3.10) to obtain

$\Phi(0)-\Phi(x)=\frac{1}{\Gamma(\alpha)}\int_{0}^{x}\frac{z\Phi(z)\,dz}{(x-z)^{1-\alpha}}.$

Taking into account the positivity of $\Phi$, we obtain for $x>1$ that

$1=\Phi(0)\geq\frac{1}{\Gamma(\alpha)}\int_{x-1}^{x}\frac{z\Phi(z)\,dz}{(x-z)^{1-\alpha}}.$

Since Theorem 3.1 (3) guarantees monotonicity of $\Phi$, then we obtain the
estimate

$\Gamma(\alpha)\geq(x-1)\Phi(x)\int_{x-1}^{x}\frac{dz}{(x-z)^{1-\alpha}}=\frac{1}{\alpha}(x-1)\Phi(x).$

Hence, for $x\in[0,2]$ we have $x\Phi(x)\leq 2$. For $x>2$ we obtain

$x\Phi(x)\leq\Gamma(1+\alpha)\frac{x}{x-1}\leq 2.$ ∎

We will use the above Lemma to show limited integrability of the derivatives
of solutions constructed in Theorem 3.12.

######  Lemma 3.15.

Let us suppose that $g\in W^{1,1}(0,\infty)$, $g(0)=0$ and $w_{1}$ is given by
(3.12) and $w_{2}$ is given by (3.13). We also assume that $g$ in the
definition of $w_{1}$ satisfies $g(0)=0$. Then,
(1) $\frac{\partial}{\partial t}w_{i}(\cdot,t)\in L^{1}(0,\infty)$ for all
$t>0$ and $i=1,2$;
(2) $D^{\alpha}_{x_{C}}w_{i}(\cdot,t)\in L^{\infty}(0,\infty)$ for all $t>0$
and $i=1,2$;
(3) $\frac{\partial}{\partial x}w_{i}(\cdot,t)\in L^{1}(0,\infty)$ for all
$t>0$ and $i=1,2$.

###### Proof.

We will use the representation formulas (3.17). The argument is conducted
simultaneously for $\tilde{g}$ and $\bar{g}$. For the sake of simplicity of
notation we will write here $g$ for both $\tilde{g}$ and $\bar{g}$ and also
$w$ will denote $w_{1}$ and $w_{2}$.

We will check that (1) holds. Let us compute $\frac{\partial}{\partial t}w$,
we will express it in term of $\Phi$ introduced in (2.9),

$\frac{\partial}{\partial
t}w(x,t)=-\frac{a_{0}}{(1+\alpha)t^{1+\frac{1}{1+\alpha}}}\int_{-\infty}^{\infty}\Phi\left(\frac{x-y}{t^{\frac{1}{1+\alpha}}}\right)g(y)\,dy-\frac{a_{0}}{(1+\alpha)t^{1+\frac{2}{1+\alpha}}}\int_{-\infty}^{\infty}\frac{d\Phi}{d\xi}\left(\frac{x-y}{t^{\frac{1}{1+\alpha}}}\right)g(y)\,dy.$

The RHS above is well-defined because $g$ has a compact support and $\Phi$ is
a $C^{1+\alpha}$-function. Since $g$ belongs to $W^{1,1}(\mathbb{R}_{+})$ we
may integrate the last term by parts. Here, in the case of the odd extension
of $g$ we use $g(0)=0$.

Finally, we reach

$\frac{\partial}{\partial
t}w(x,t)=-\frac{1}{(1+\alpha)t}\int_{0}^{\infty}{\mathscr{E}}_{t}(x-y)g(y)\,dy-\frac{1}{(1+\alpha)t^{1+\frac{1}{1+\alpha}}}\int_{0}^{\infty}{\mathscr{E}}_{t}(x-y)g^{\prime}(y)\,dy.$

The integrability of ${\mathscr{E}}$, $g$ and $g^{\prime}$ implies claim (1)
for $w$.

We are going to establish part (2). We have already seen that
$D^{\alpha}_{x_{C}}$ commute with integration, so we have,

$D^{\alpha}_{x_{C}}w=\int_{-\infty}^{\infty}D^{\alpha}_{x_{C}}{\mathscr{E}}_{t}(x-y)g(y)\,dy.$

Since
${\mathscr{E}}_{t}(x)=\frac{a_{0}}{t^{\frac{1}{1+\alpha}}}\Phi(xt^{-\frac{1}{1+\alpha}})$,
we have to calculate the Caputo derivative of a scaled function. Let us
suppose that $f$ is absolutely continuous on $[0,\infty$). We set
$f_{\lambda}(x)=f(\lambda x)$. We compute $D^{\alpha}_{x_{C}}f_{\lambda}$,

$D^{\alpha}_{x_{C}}f_{\lambda}(x)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{x}\frac{df_{\lambda}}{ds}(s)(x-s)^{-\alpha}\,ds=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{x}\lambda\frac{f^{\prime}(\lambda
s)}{(x-s)^{\alpha}}\,ds.$

After changing the variables $\lambda s=z$ we obtain,

$D^{\alpha}_{x_{C}}f_{\lambda}(x)=\frac{\lambda^{\alpha}}{\Gamma(1-\alpha)}\int_{0}^{\lambda
x}\frac{f^{\prime}(z)}{(\lambda
x-z)^{\alpha}}\,dz=\lambda^{\alpha}D^{\alpha}_{y_{C}}f(\lambda x).$ (3.19)

Taking into account(3.19) yields,

$D^{\alpha}_{x_{C}}w=\frac{a_{0}}{t^{\frac{2}{1+\alpha}}}\int_{-\infty}^{\infty}D^{\alpha}_{\xi_{C}}\Phi((x-y)t^{-\frac{1}{1+\alpha}})g(y)\,dy.$

Now, due to (2.10) and Lemma 3.14 we conclude that

$|D^{\alpha}_{x_{C}}v_{2}|=\left|a_{0}t^{-\frac{1}{1+\alpha}}\int_{-\infty}^{\infty}\frac{(x-y)}{t^{\frac{1}{1+\alpha}}}\Phi\left(\frac{(x-y)}{t^{\frac{1}{1+\alpha}}}\right)g(y)\,dy\right|\leq
a_{0}2t^{-\frac{1}{1+\alpha}}.$

Part (2) follows.

Part (3) is established along the lines of the proof of (1). Let us compute
the derivative of $w$, then we see

$\frac{\partial}{\partial
x}w(x,t)=\int_{-\infty}^{\infty}\frac{\partial}{\partial
x}{\mathscr{E}}_{t}(x+y)g(y)\,dy=-\int_{-\infty}^{\infty}{\mathscr{E}}_{t}(x+y)g^{\prime}(y)\,dy,$

where we used the boundedness of the support of $g$. ∎

We would like to state our uniqueness result. For this purpose we define a
class of functions, which we find suitable,

$\displaystyle X=$ $\displaystyle\,C([0,\infty)^{2})\cap
C^{1+\alpha}([0,\infty)\times(0,\infty))\cap$ $\displaystyle\\{u\in
L^{\infty}(\mathbb{R}_{+};L^{2}(\mathbb{R}_{+})):\ \forall t>0\
u_{t}(\cdot,t),u_{x}(\cdot,t)\in L^{1}(\mathbb{R}_{+}),\
D^{\alpha}_{x_{c}}u(\cdot,t)\in L^{\infty}(\mathbb{R}_{+})\\}.$

######  Proposition 3.16.

If $w$ is a solution to (3.14) with either Dirichlet or Neumann boundary data
(3.15) and $w\in X$, then $w$ is unique.

###### Proof.

We take the difference $w$ of two solutions $u_{1}$ and $u_{2}$ from $X$, we
multiply them by $w$ and integrate over $\mathbb{R}_{+}$. The definition of
the class $X$ permits us to write

$\frac{1}{2}\frac{d}{dt}\|w(t)\|^{2}_{L^{2}}=\langle
w_{t},w\rangle\equiv\int_{0}^{\infty}\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}w(x,t)w(x,t)\,dx.$

Since $w(\cdot,t)$ is continuous on $\mathbb{R}_{+}$ as well as it is in
$L^{2}$, then there exists a sequence $R_{n}$ converging to infinity, such
that $w(R_{n},t)D^{\alpha}w(R_{n},t)$ goes to zero, when $n\to\infty.$ Thus,
after integration by parts over $[0,R_{n}]$ the RHS above takes the following
form,

$\int_{0}^{\infty}\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}w(x,t)w(x,t)\,dx=\lim_{n\to\infty}\int_{0}^{R_{n}}\frac{\partial}{\partial
x}D^{\alpha}_{x_{C}}w(x,t)w(x,t)\,dx=-\lim_{n\to\infty}\int_{0}^{R_{n}}D^{\alpha}_{x_{C}}w(x,t)\frac{\partial}{\partial
x}w(x,t)\,dx,$

where we also take into account the zero Dirichlet data at $x=0$ or vanishing
$D^{\alpha}w(0)$. In order to estimate the RHS we recall that [6, Proposition
6.10] implies that

$\int_{0}^{R_{n}}D^{\alpha}_{x_{C}}w(x,t)\frac{\partial}{\partial
x}w(x,t)\,dx\geq 0.$

See formula (5) in the proof of [14, Theorem 1] for more details. As a result,
we conclude that

$\frac{d}{dt}\|w(t)\|^{2}_{L^{2}}\leq 0.$

Hence, for all $t>0$ we have $\|w(t)\|_{L^{2}}^{2}\leq\|w(0)\|_{L^{2}}^{2}=0$.
∎

It is interesting to check when a solution belongs to class $X.$ We do not
offer a full answer, however, Lemma 3.15 gives us a hint. We note :

######  Corollary 3.17.

Let us suppose that $f$ is in $C_{c}(\mathbb{R}_{+}^{2})$. Then,
(1) If $g$ is in $W^{1,1}(\mathbb{R}_{+})$ with bounded support and $g(0)=0$,
then $w_{1}+w_{3}$ is a unique solution to (3.18) with initial condition
(3.142) and boundary condition (3.151).
(2) If $g$ is in $W^{1,1}(\mathbb{R}_{+})$ with bounded support, then
$w_{2}+w_{4}$ is a unique solution to (3.18) with initial condition (3.142)
and boundary condition (3.152).

###### Proof.

Lemma 3.15 shows that indeed $w_{1}$ and $w_{2}$ are in class $X$. The
calculations we performed in the course of proof of Theorem 3.12 show that
$w_{3}$, $w_{4}$ also belong to $X$. Then, we use Proposition 3.16 to finish
the proof. ∎

Having established an integral representation of solution we may draw
conclusions about their asymptotic behavior. Here we note a decay property.

######  Proposition 3.18.

Let us suppose that the assumptions of the uniqueness theorem, Proposition
3.16, hold and $g\in W^{1,1}(\mathbb{R}_{+})$ has bounded support. If $u$ is a
unique solution to (3.14) corresponding solution to $g$, then

$\sup_{x\in\mathbb{R}_{+}}|u(x,t)|\leq Ct^{-1/(1+\alpha)}\|g\|_{L^{1}}$

###### Proof.

We use the representation formula and the boundedness of ${\mathscr{E}}$. ∎

We proved in [9] that viscosity solutions depend continuously upon
$\alpha\in[0,1].$ We can use the representation formula to establish the same
result. Indeed, we can show:

######  Proposition 3.19.

If $w^{\alpha}_{i}$ , $i=1,2$ then

$\lim_{\alpha\to 1^{-}}w^{\alpha}_{i}=w^{0}_{i},\qquad i=1,2,$

where $w^{0}_{i}$ is the solution to the heat equation.

###### Proof.

We use here the uniform convergence of the Mittag-Leffler functions
$E_{\alpha,1+1/\alpha,1/\alpha}$ to $E_{1,2,1}$. This follows from the uniform
boundedness of the family $E_{\alpha,1+1/\alpha,1/\alpha}$ and the Montel
Theorem. ∎

######  Remark 3.20.

This proof breaks down if we try to pass to the limit as $\alpha$ goes to 0.
In this case $E_{\alpha,1+1/\alpha,1/\alpha}$ converges to $\frac{1}{1-z}$ for
$|z|<1.$

The observation made in the Proposition above has a bit surprising
consequences. It suggests that for small $\alpha$ we should see phenomena
typical for the hyperbolic problems.

We could relate this observation to the behavior the discretization scheme. We
presented in [10] the $\frac{1}{2}$-shifted Grünwald approximation. We write
out the scheme, for the Dirichlet data, in terms of the Grünwald weights. The
approximation of $u(i\Delta x,k\Delta t)$ takes the following form,

$\displaystyle u_{0}^{k+1}=u_{0}^{k},\qquad u_{n}^{k+1}=u_{n}^{k},$ (3.20)
$\displaystyle
u_{i}^{k+1}=\sum_{j=0}^{i-1}{\beta(g_{i+1-j}-g_{i-j})u_{j}^{k}}+\left(1+\beta(g_{1}-g_{0})\right)u_{i}^{k}+\beta
g_{0}u_{i+1}^{k},$

where and the Grünwald weights are given by

$\beta=\frac{\Delta t}{(\Delta x)^{1+\alpha}},\quad g_{0}=1,\quad
g_{i}=\frac{i-1-\alpha}{i}g_{i-1},\quad i=1,2,\ldots.$ (3.21)

We see that when $\alpha\to 0$, then (3.20)–(3.21) converge to an explicit
finite difference scheme for the transport equation. This suggests a finite
speed of propagation. This is supported by our simulations, presented there
which showed an initial pulse tending toward the left.

There are two sides of the same coin. For $\alpha=1$, eq. (3.1) becomes a heat
equation, where the speed of propagation is infinite. It means that if the
initial perturbation is non-negative and it has a compact support, then
solutions to (3.1) will be positive everywhere for $t>0$. Interestingly, when
$\alpha\to 1$, then (3.20)–(3.21) converges to an explicit finite difference
scheme for the heat equation. We also noticed the smearing out effect, see
Fig. 2 in [10].

Actually, we can show:

######  Proposition 3.21.

For all $\alpha\in(0,1]$ the speed of signal propagation for solutions to
(3.1) with initial condition $u(x,0)=f(x)$ and the zero Neumann data is
infinite, i.e. if $f\geq 0$, $f\neq 0$ and $\hbox{supp}\,{f}\subset(0,R)$,
where $R>0$, then for all $x,t>0$ we have

$w_{2}(x,t)>0.$

###### Proof.

This is an immediate consequence of the definition of $w_{2}$ and positivity
of ${\mathscr{E}}$ and $f$. ∎

We have chosen the Neumann data, because of the simplicity of the formula for
a solution.

## Appendix

Here we present a different and more elegant approach to the derivation of
${\mathscr{E}}$ and its positivity. It depends on the properties of the three-
parameter generalized Mittag-Leffler function and the arguments are shorter.

We first derive the form of ${\mathscr{E}}$ assuming that $u$ defined as

$u(x,t)=t^{-\frac{1}{1+\alpha}}v(xt^{-\frac{1}{1+\alpha}})$

is a solution to (3.1). Then, Corollary (2.2) yields,

$(D^{\alpha}_{x}u)_{x}(x,t)=t^{-1-\frac{1}{1+\alpha}}(D^{\alpha}_{y}v)_{y}(y),$

where $y=xt^{-\frac{1}{1+\alpha}}.$ Moreover, it is easy to check directly
that

$u_{t}(x,t)=-\frac{1}{1+\alpha}t^{-1-\frac{1}{1+\alpha}}(v(y)+yv^{\prime}(y)).$

Combining these observations with (3.1) yields,

$-\frac{1}{1+\alpha}t^{-1-\frac{1}{1+\alpha}}(v(y)+yv^{\prime}(y))=(D^{\alpha}_{y}v)_{y}(y).$

This can be rewritten as

$\frac{d}{dy}\left(D_{y}^{\alpha}v(y)+\frac{1}{1+\alpha}yv(y)\right)=0.$

Since we are interested in smooth solutions then, due to [10, Proposition 3.1]
we see that

$D_{y}^{\alpha}v(y)+\frac{1}{1+\alpha}yv(y)=0,\qquad y>0.$ $None$

According to Proposition 2.3 a solution to this equation is given by the
formula

$v(y)=v(0)E_{\alpha,1+1/\alpha,1/\alpha}\left(-\frac{y^{1+\alpha}}{1+\alpha}\right).$

Here $E_{\alpha,1+1/\alpha,1/\alpha}$ is a generalized Mittag-Leffler function
defined in (2.8).

We can also offer much shorter and easier proof of the positivity of
${\mathscr{E}}$. It is based again on the theory of the Mittag-Leffler
function and ODEs.

Lemma A. Function $(0,\infty)\ni x\mapsto v(x)\equiv
E_{\alpha,1+\frac{1}{\alpha},\frac{1}{\alpha}}(-\frac{x^{1+\alpha}}{1+\alpha})$
is positive for all $x>0$.

###### Proof.

We consider eq. (*) for $v$ with the initial condition $v(0)=1$ Then,
$v(x)\equiv
E_{\alpha,1+\frac{1}{\alpha},\frac{1}{\alpha}}(-\frac{x^{1+\alpha}}{1+\alpha})$
is a solution to (*). Let us suppose our claim is not valid and the set

$A:=\\{x>0:v(x)\leq 0\\}$

is not empty. Due to the continuity of $v$, there is $x_{0}$ such that

$v(x_{0})=0\qquad\hbox{and}\qquad v(x)>0\quad\mbox{for}\,\,x\in(0,x_{0}).$

However, we have

$\displaystyle\displaystyle v(x_{0})$
$\displaystyle=-\frac{1+\alpha}{x_{0}}D_{x}^{\alpha}v(x_{0})$
$\displaystyle=-\frac{1+\alpha}{x_{0}\Gamma(1-\alpha)}\left(\frac{v(x_{0})-v(0)}{x_{0}^{\alpha}}+\alpha\int_{0}^{x_{0}}\frac{v(x_{0})-v(x_{0}-z)}{z^{\alpha+1}}dz\right)$
$\displaystyle=\frac{1+\alpha}{x_{0}\Gamma(1-\alpha)}\left(\frac{1}{x_{0}^{\alpha}}+\alpha\int_{0}^{x_{0}}\frac{v(x_{0}-z)}{z^{\alpha+1}}dz\right)>0.$

This is a contradiction. ∎

### Acknowledgments

PR was in part supported by the National Science Centre, Poland, through the
grant number 2017/26/M/ST1/00700. This research was initiated during a PR
visit to the University of Tokyo in 2019, whose hospitality is greatly
appreciated.

## References

  * [1] S.Dipierro, B.Pellacci, E.Valdinoci, G.Verzini, Time-fractional equations with reaction terms: fundamental solutions and asymptotics, arXiv:1903.11939.
  * [2] R.Gorenflo, A.A.Kilbas, S.V. Rogosin, On the generalized Mittag-Leffler type functions. Integral Transform. Spec. Funct. 7 (1998), no. 3-4, 215-224.
  * [3] A.A. Kilbas, M.Saigo, On mittag-leffler type function, fractional calculus operators and solutions of integral equations, Integral Transform. Spec. Funct., 4 (1996) no 4, 355–370.
  * [4] A.A.Kilbas, H.M.Srivastava, J.J.Trujillo, Theory and applications of fractional differential equations. North-Holland Mathematics Studies, 204. Elsevier Science B.V., Amsterdam, 2006.
  * [5] K.-H.Kim, D.Park, J.Ryu, An $L^{q}(L^{p})$-theory for diffusion equations with space-time nonlocal operators, J. Differential Equations, 287 (2021), 376-427.
  * [6] A.Kubica, M. Yamamoto, Initial-Boundary Value Problems For Fractional Diffusion Equations With Time-Dependent Coefficients, Fract. Calc. Appl. Anal., 21 (2018), 276–311.
  * [7] Y.Luchko, F.Mainardi, Y.Povstenko, Propagation speed of the maximum of the fundamental solution to the fractional diffusion-wave equation, Comput. Math. Appl., 66 (2013), no. 5, 774–784.
  * [8] F.Mainardi, Y.Luchko, G.Pagnini, The fundamental solution of the space-time fractional diffusion equation, Fract. Calc. Appl. Anal., 4 (2001), no. 2, 153–192.
  * [9] T.Namba, P.Rybka, On viscosity solutions of space-fractional diffusion equations of Caputo type, SIAM J. Math. Anal 52, (2020), no 1, 653–681.
  * [10] T.Namba, P.Rybka, V.Voller, Some comments on using fractional derivative operators in modeling non-local diffusion processes, J. Comput. Appl. Math. 381 (2021), 113040.
  * [11] Y.Povstenko, J.Klekot, Fractional heat conduction with heat absorption in a sphere under Dirichlet boundary condition, Comput. Appl. Math. 37 (2018), no. 4, 4475–4483.
  * [12] A.V.Pskhu, The fundamental solution of a diffusion-wave equation of fractional order, Izv. Math. 73 (2009) 141–182.
  * [13] S. D. Roscani, D. A. Tarzia and L. Venturato, The similarity method and explicit solutions for the fractional space one-phase Stefan problems, arXiv:2009.12977
  * [14] K.Ryszewska, An analytic semigroup generated by a fractional differential operator, J. Math. Anal. Appl. 483 (2020), no. 2, 123654, 17 pp.
  * [15] K.Ryszewska, A space-fractional Stefan problem, Nonlinear Anal. 199 (2020), 112027, 30 pp.
  * [16] V.R. Voller, On a fractional derivative form of the Green-Ampt infiltration model, Adv. Water Res., 34 (2010), 257–262.