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__ On a Space Fractional Stefan problem of Dirichlet type with Caputo flux

S. D. Roscani2,3, K. Ryszewska1 and L.D. Venturato2,3

1 Department of Mathematics and Information Sciences, Warsaw University of
Technology, Koszykowa 75, 00-662 Warsaw, Poland

2 Departamento de Matemática, FCE, Universidad Austral, Paraguay 1950,
S2000FZF Rosario, Argentina

3CONICET, Argentina

<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract: We study a space-fractional Stefan problem with the Dirichlet
boundary conditions. It is a model that describes superdiffusive phenomena.
Our main result is the existence of the unique classical solution to this
problem. In the proof we apply evolution operators theory and the Schauder
fixed point theorem. It appears that studying fractional Stefan problem with
Dirichlet boundary conditions requires a substantial modifications of the
approach in comparison with the existing results for problems with different
kinds of boundary conditions.

Keywords: Space fractional diffusion equation, Stefan problem, Moving boundary
problem, Caputo derivative.

MSC2010: 26A33 - 35R11 - 35R35 - 35R37.

##  1 Introduction

In recent years, there has been a significant increase in research interests
on free boundary problems governed by fractional diffusion equations. These
problems can be classified as time-fractional or space-fractional, depending
in which variable the fractional derivative is taken. Applied to the time
variable, applications are related to memory effects, while in the second case
ones speaks about nonlocality in space. These anomalous phenomena are often
associated with heterogeneity in the material or media considered.

In this paper we focus on a problem governed by a space-fractional diffusion
equation. Several models related to these kind of problems have been
extensively studied. Some examples include dendritic crystal growth [13, 14],
shoreline dynamics in ocean basins [7, 18], scaled Brownian motion [12],
anomalous transport processes [19], and infiltration of water into
heterogeneous sub-surface soils [34]. Notice that by considering different
diffusive fluxes one can obtain distinct governing equations, as proposed in
[21], where a non-local transport phenomenon is analyzed for two non-local
fluxes defined in terms of fractional derivatives.

An important difference between classical and fractional one-phase Stefan
problems in one-dimension arises in the behavior of the free boundary. In
classical Stefan problems with constant Dirichlet boundary condition on fixed
boundary, the interface grows as $s(t)\sim t^{\frac{1}{2}}$. However, for
certain fractional cases, it has been observed that the phase change boundary
moves as $s(t)\sim t^{\beta}$ where $0<\beta<\frac{1}{2}$ (sub-diffusion) or
$\frac{1}{2}<\beta<1$ (super-diffusion). This behavior appears, for example,
when heterogeneities occur across a range of length scales, with the largest
approaching the domain scale, causing standard diffusion models to break down
[19, 32]. Concerning theoretical results, the self-similar solutions for a
fractional Stefan problem with fractional Neumann-type condition were recently
obtained in [26], where the solution verifies that the advance of the free
boundary is proportional to $t^{\frac{1}{1+\alpha}}$, $\alpha\in(0,1)$ (super-
diffusion), which coincides with the result obtained in [33] for the
stationary case.

Here, we analyze a space-fractional Stefan problem with a Dirichlet boundary
conditions. More precisely, we study existence and uniqueness of classical
solutions to the following fractional Stefan problem

$\begin{array}[]{lll}(i)&u_{t}-\frac{\partial}{\partial
x}D^{\alpha}u=0,&0<x<s(t),0<t<T,\\\ (ii)&u(0,t)=0,u(t,s(t))=0,&t\in(0,T),\\\
(iii)&u(x,0)=u_{0}(x),&0<x<s(0)=b,\\\
(iv)&\dot{s}(t)=-\lim\limits_{x\rightarrow
s(t)^{-}}(D^{\alpha}u)(x,t),&t\in(0,T).\end{array}$ (1)

for $b>0$.

A formal derivation of (1) can be obtained from physical assumptions on a
model of a melting phenomena related to a phase change in an infinite slab,
due to heat transfer (see [26]).

One has to mention that in [25] and [29] the space-fractional Stefan problems
with local and nonlocal Neumann boundary conditions are considered. Although
here we only change a boundary condition, it appears that the techniques used
in the former articles have to be significantly adapted and modified in this
case. Moreover, it is well known that the different kinds of boundary
conditions must be treated very carefully when a space fractional derivative
is involved (see for example [3] for fractional diffusion or [36, Ch.1] for
general fractional ordinary differential equations) and, according to our best
knowledge, this is the first existence result for the space-fractional Stefan
problem with the Caputo derivative and Dirichlet boundary conditions.

We note that a self-similar solution for problem (1) with $b=0$ and non-
homogeneous boundary conditions, was obtained in [26]. Furthermore, the
viscosity solution to a linear initial boundary value problem
$u_{t}=\frac{\partial}{\partial x}D^{\alpha}u$ with Dirichlet boundary
conditions have been obtained in [20] by Perron’s method, and the explicit
solutions for fractional diffusion problems on bounded domains have been
established in [3]. The analytic semigroup theory for linear evolution problem
with $\frac{\partial}{\partial x}D^{\alpha}$ with different kinds of boundary
conditions was established in a PhD thesis of one of the authors ([28]).
Finally, there are several results concerning alternative problems where the
fractional Laplace operator is considered [1, 4, 6, 8].

The goal of this paper is solve the problem (1). To this end, we adapt the
results from [25], [29], and establish analogous properties and estimates in
our case. However, the case with Dirichlet boundary conditions requires
substantial change of the approach in some crucial parts of the proof. For
example, it is important to notice that discussing (1), we are not able to
control the fractional flux at the boundary $(D^{\alpha}u)(0,t)$. Thus, in
order to solve the problem via fixed point argument, we are forced to deduce a
new integral identity equivalent to the Stefan condition (1)($iv$), which does
not include $(D^{\alpha}u)(0,t)$. Furthermore, since we work with a very
specific functional spaces (see (20)) and our techniques involve interpolation
estimates, we have to provide a local estimate between our specific
interpolation spaces and fractional Sobolev spaces (Theorem 4). Our approach
relies on finding mild solutions to the moving boundary problem, increasing
their regularity and then applying the fixed point argument. Increasing the
regularity of mild solutions is very demanding and technical task, which is
illustrated in [25] and [29]. The situation is getting particularly complex
when $0<\alpha\leq 1/2$ and the iterative argument has to be used. Since the
iterative procedure has already been applied twice in [25, Lemma 3.10] and
[29, Lemma 5], in this paper we discuss only the case $\alpha>1/2$. It helps
us focus on the novelties which problem (1) delivers, without accumulating too
much of technical calculations. Notice that the assumption $\alpha>1/2$ is
only used in the regularity results (Section 3.4.2) and the rest of the proof
is valid for any $0<\alpha<1$.

The paper is organised as follows. In section 2, we recall some definitions
and properties of fractional operators and fractional Sobolev spaces. In
section 3, we firstly establish the analytic semigroup theory for the operator
$\frac{\partial}{\partial x}D^{\alpha}$ with Dirichlet boundary conditions.
Then we apply this result to solve the moving boundary problem asociated to
(1). Finally, section 4 is devoted to the proof of existence and uniqueness of
the solutions to space-fractional Stefan problem (1) for $\alpha>1/2$. The
proof follows the steps from [25] and [29], but here we need to apply
different integral identity equivalent to the Stefan condition for problem
(1).

##  2 Preliminaries

In this section, we recall some definitions and properties concerning
fractional derivatives and fractional Sobolev spaces. Let us begin with
definitions of fractional operators. We notice that the lower extreme of the
fractional operators considered in this article is always zero, thus the
classical suffix 0 will be omitted.

######  Definition 1.

Let $\alpha>0$. For $f\in L^{1}(0,1)$, we define the fractional integral
$I^{\alpha}$ as follows

$I^{\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_{0}^{x}(x-p)^{\alpha-1}f(p){\rm
d}p.$ (2)

######  Definition 2.

Let $\alpha\in(0,1)$. For $f$ regular enough, the fractional Riemann-Liouville
is defined by the formula

$\partial^{\alpha}f(x)=\frac{\partial}{\partial
x}I^{1-\alpha}f(x)=\frac{1}{\Gamma(1-\alpha)}\frac{\partial}{\partial
x}\int_{0}^{x}(x-p)^{-\alpha}f(p){\rm d}p,$ (3)

while the fractional Caputo derivative is given by

$D^{\alpha}f(x)=\frac{\partial}{\partial
x}I^{1-\alpha}[f(x)-f(0)]=\frac{1}{\Gamma(1-\alpha)}\frac{\partial}{\partial
x}\int_{0}^{x}(x-p)^{-\alpha}[f(p)-f(0)]{\rm d}p.$ (4)

######  Remark 1.

The Riemann-Liouville and Caputo fractional derivatives are well defined for
functions belonging to $AC[0,1]$. Moreover, if $f\in AC[0,1]$ then

$D^{\alpha}f(x)=I^{1-\alpha}f^{\prime}(x)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha}f^{\prime}(p){\rm
d}p.$ (5)

The absolute continuity of a function is a sufficient condition to make
$\partial^{\alpha}$ pointwisely well defined but it is not the necessary one.
In the following proposition we give an example of singular function which has
pointwise Riemann-Liouville fractional derivative and present some basic
properties of fractional operators which will be used later on.

######  Proposition 1.

[31, Chapter 2], [15, Proposition 6.5]

  1. 1.

Let $\beta>-1$. Then,

$I^{\alpha}(x^{\beta})=\frac{\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)}x^{\alpha+\beta},$

and

$\partial^{\alpha}(x^{\beta})=\frac{\Gamma(\beta+1)}{\Gamma(\beta-\alpha+1)}x^{\beta-\alpha},$

  2. 2.

Let $f\in AC[0,1]$. Then,

$(D^{\alpha}f)(x)=(\partial^{\alpha}f)(x)-\frac{x^{-\alpha}}{\Gamma(1-\alpha)}f(0),$
(6)

and

$I^{1-\alpha}f(x)=\int_{0}^{x}D^{\alpha}f(z){\rm
d}z+f(0)\frac{x^{1-\alpha}}{\Gamma(2-\alpha)}.$ (7)

In particular, if $f(0)=0$,

$I^{1-\alpha}f(x)=\int_{0}^{x}D^{\alpha}f(z){\rm d}z.$ (8)

Moreover,

$\frac{\partial}{\partial x}D^{\alpha}u=\frac{\partial}{\partial
x}I^{1-\alpha}u_{x}=\partial^{\alpha}u_{x}.$ (9)

  3. 3.

For every $f\in AC[0,1]$, $\alpha+\beta\leq 1$ there holds

$\partial^{\alpha}D^{\beta}f=D^{\alpha+\beta}f,\quad a.e.\text{ in
}(0,1),\textrm{ in particular }\partial^{1-\alpha}D^{\alpha}f=f^{\prime},\quad
a.e.\text{ in }(0,1).$ (10)

While working with fractional operators the fractional Sobolev spaces seems to
be a natural choice. Below we recall that the fractional Sobolev space
$H^{\beta}(0,1):=W^{\beta,2}(0,1)$, $\beta\in(0,1)$ is defined as

$H^{\beta}(0,1)=\left\\{f\in
L^{2}(0,1):\int_{0}^{1}\int_{0}^{1}\frac{|f(x)-f(y)|^{2}}{|x-y|^{1+2\beta}}{\rm
d}y{\rm d}x<\infty\right\\},$

equipped with the norm

$||f||_{H^{\beta}(0,1)}=\left(||f||^{2}_{L^{2}(0,1)}+\int_{0}^{1}\int_{0}^{1}\frac{|f(x)-f(y)|^{2}}{|x-y|^{1+2\beta}}{\rm
d}y{\rm d}x\right)^{\frac{1}{2}}.$

For $\beta>1$, $m=\lfloor\beta\rfloor$ and $s=\beta-\lfloor\beta\rfloor$, we
define $H^{\beta}(0,1):=W^{\beta,2}(0,1)$ as follows

$H^{\beta}(0,1)=\\{f\in H^{m}(0,1):f^{(m)}\in H^{s}(0,1)\\},$

with the norm

$||f||_{H^{\beta}(0,1)}=\left(||f||^{2}_{H^{m}(0,1)}+||f^{(m)}||^{2}_{H^{s}(0,1)}\right)^{\frac{1}{2}}.$

We will often make use of this basic property.

######  Proposition 2.

[16, Remark 12.8] For every $s\neq\frac{1}{2}$, there holds,

$\partial_{x}\in\mathcal{L}(H^{s}(0,1),H^{s-1}(0,1)).$ (11)

######  Definition 3.

Let $\gamma>0$, and

${{}_{0}}C^{\infty}[0,1]=\\{v\in C^{\infty}[0,1]:\forall
n\in{\mathbb{N}},v^{(n)}(0)=0\\},\quad{{}^{0}}C^{\infty}[0,1]=\\{v\in
C^{\infty}[0,1]:\forall n\in{\mathbb{N}},v^{(n)}(1)=0\\}.$

We define the fractional Sobolev spaces ${{}_{0}}H^{\gamma}(0,1)$ and
${{}^{0}}H^{\gamma}(0,1)$ in the following way

${{}_{0}}H^{\gamma}(0,1)=\overline{{{}_{0}}C^{\infty}[0,1]}^{H^{\gamma}(0,1)},\quad{{}^{0}}H^{\gamma}(0,1)=\overline{{{}^{0}}C^{\infty}[0,1]}^{H^{\gamma}(0,1)}.$
(12)

Let us recall the characterization for fractional Sobolev spaces.

######  Proposition 3.

[16, Theorem 11.6 and Remark 11.5], see also [36, Proposition 1] For
$\alpha\in(0,1)$ the spaces introduced in Definition 3 may be equivalently
characterized as follows

${{}_{0}}H^{\alpha}(0,1)=\begin{cases}H^{\alpha}(0,1),&\alpha\in\left(0,\frac{1}{2}\right),\\\
\\{u\in H^{\frac{1}{2}}(0,1):\int_{0}^{1}\frac{|u(x)|^{2}}{x}{\rm
d}x<\infty\\},&\alpha=\frac{1}{2},\\\ \\{u\in
H^{\alpha}(0,1):u(0)=0\\},&\alpha\in\left(\frac{1}{2},1\right),\end{cases}$
(13)

and

${{}^{0}}H^{\alpha}(0,1)=\begin{cases}H^{\alpha}(0,1),&\alpha\in\left(0,\frac{1}{2}\right),\\\
\\{u\in H^{\frac{1}{2}}(0,1):\int_{0}^{1}\frac{|u(x)|^{2}}{1-x}{\rm
d}x<\infty\\},&\alpha=\frac{1}{2},\\\ \\{u\in
H^{\alpha}(0,1):u(1)=0\\},&\alpha\in\left(\frac{1}{2},1\right),\end{cases}$
(14)

with

$||u||_{{{}_{0}}H^{\alpha}(0,1)}=||u||_{{{}^{0}}H^{\alpha}(0,1)}=||u||_{H^{\alpha}(0,1)},\quad\alpha\neq\frac{1}{2},$
$||u||_{{{}_{0}}H^{\frac{1}{2}}(0,1)}=\left(||u||^{2}_{H^{\frac{1}{2}}(0,1)}+\int_{0}^{1}\frac{|u(x)|^{2}}{x}{\rm
d}x\right)^{\frac{1}{2}},$
$||u||_{{{}^{0}}H^{\frac{1}{2}}(0,1)}=\left(||u||^{2}_{H^{\frac{1}{2}}(0,1)}+\int_{0}^{1}\frac{|u(x)|^{2}}{1-x}{\rm
d}x\right)^{\frac{1}{2}}.$

Let us give an important example.

######  Remark 2.

[36, Example 2.1] For $\beta>\alpha-\frac{1}{2}$ we have

$x^{\beta}\in{{}_{0}}H^{\alpha}(0,1).$

The following result is of fundamental importance in our further approach.

######  Proposition 4.

[37, Example 2.1], [11]. For $\alpha\in[0,1]$ the operators
$I^{\alpha}:L^{2}(0,1)\rightarrow{{}_{0}}H^{\alpha}(0,1)$ and
$\partial^{\alpha}:{{}_{0}}H^{\alpha}(0,1)\rightarrow L^{2}(0,1)$ are
isomorphism and the following inequalities hold

$c_{\alpha}^{-1}||u||_{{{}_{0}}H^{\alpha}(0,1)}\leq||\partial^{\alpha}u||_{L^{2}(0,1)}\leq
c_{\alpha}||u||_{{{}_{0}}H^{\alpha}(0,1)},\quad u\in{{}_{0}}H^{\alpha}(0,1),$
(15)
$c_{\alpha}^{-1}||I^{\alpha}f||_{{{}_{0}}H^{\alpha}(0,1)}\leq||f||_{L^{2}(0,1)}\leq
c_{\alpha}||I^{\alpha}f||_{{{}_{0}}H^{\alpha}(0,1)},\quad u\in L^{2}(0,1),$
(16)

where $c_{\alpha}$ denotes a positive constant dependent on $\alpha$.

######  Corollary 1.

[29, Corollary 1] Let $\alpha,\beta>0$. Then
$I^{\beta}:{{}_{0}}H^{\alpha}(0,1)\rightarrow{{}_{0}}H^{\alpha+\beta}(0,1)$.
Furthermore, there exists a positive constant $c$ dependent only on
$\alpha,\beta$ such that for every $f\in{}_{0}H^{\alpha}(0,1)$

$\left\|{I^{\beta}f}\right\|_{{}_{0}H^{\alpha+\beta}(0,1)}\leq
c\left\|{f}\right\|_{{}_{0}H^{\alpha}(0,1)}.$

We will also make an extensive use of the following local regularity result.

######  Lemma 1.

[29, Lemma 4] Let $f\in{}_{0}H^{\alpha}(0,1)$ for $\alpha\in(0,1)$ and
$\partial^{\alpha}f\in H^{\beta}_{loc}(0,1)$ for $\beta\in(\frac{1}{2},1]$.
Then $f\in H^{\beta+\alpha}_{loc}(0,1)$ and for every $0<\delta<\omega<1$
there exists a positive constant $c=c(\delta,\omega,\alpha,\beta)$ such that

$\left\|{f}\right\|_{H^{\beta+\alpha}(\delta,\omega)}\leq
c(\left\|{f}\right\|_{{}_{0}H^{\alpha}(0,\omega)}+\left\|{\partial^{\alpha}f}\right\|_{H^{\beta}(\frac{\delta}{2},\omega)}).$
(17)

We finish this section with some important identities for fractional
derivative.

######  Proposition 5.

[15, Proposition 6.10] If $w\in AC[0,1]$, then for any $\alpha\in(0,1)$ the
following equality holds

$\int_{0}^{1}\partial^{\alpha}w(x)\cdot w(x){\rm
d}x=\frac{\alpha}{4}\int_{0}^{1}\int_{0}^{1}\frac{\left|{w(x)-w(p)}\right|^{2}}{\left|{x-p}\right|^{1+\alpha}}{\rm
d}p{\rm d}x$
$+\frac{1}{2\Gamma(1-\alpha)}\int_{0}^{1}[(1-x)^{-\alpha}+x^{-\alpha}]\left|{w(x)}\right|^{2}{\rm
d}x.$

Hence, there exists a positive constant $c$ which depends only on $\alpha$,
such that

$\int_{0}^{1}\partial^{\alpha}w(x)\cdot w(x){\rm d}x\geq
c\left\|{w}\right\|_{H^{\frac{\alpha}{2}}(0,1)}^{2}$ (18)

######  Proposition 6.

Let $\alpha\in(0,1)$. If $f,g\in AC[0,1]$ and $g\in C^{0,\beta}([0,1])$ for
$\beta\in(\alpha,1)$, then

$\partial^{\alpha}(f\cdot
g)(x)=g(x)(\partial^{\alpha}f)(x)+\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha-1}(g(x)-g(p))f(p){\rm
d}p.$

##  3 Dirichlet boundary value problem

The strategy to solve the problem (1) follows the lines to the authors’
previous papers [25] and [29]. We firstly solve the moving boundary problem
and then we apply the Schauder’s fixed point theorem to obtain the existence
result for free boundary problem. Before we proceed with the moving boundary
problem let us firstly establish the theory concerning autonomous problem. For
this purpose, we will again relay on the semigroup theory.

###  3.1 Autonomous case

The results from this subsection comes from [28, Chapter 3.2]. Let us consider
the following boundary value problem

$\left\\{\begin{array}[]{ll}u_{t}-\frac{\partial}{\partial
x}D^{\alpha}u=f&\textrm{ in }(0,1)\times(0,T),\\\ u(0,t)=0,u(1,t)=0&\textrm{
for }t\in(0,T),\\\ u(x,0)=u_{0}(x)&\textrm{ in }(0,1).\\\ \end{array}\right.$
(19)

We would like to define the operator $A=\frac{\partial}{\partial
x}D^{\alpha}:D\subseteq L^{2}(0,1)\rightarrow L^{2}(0,1)$ on the appropriate
domain $D$ in such a way that $A$ is densely defined sectorial operator and
thus generates an analytic semigroup. (For theory of semigroups and evolution
operators we refer for example to [17, 23]).

Taking into account (9), Proposition 4 and Proposition 2, one could propose

$D=\\{u\in
H^{1+\alpha}(0,1):u_{x}\in{{}_{0}}H^{1+\alpha}(0,1),u(0)=u(1)=0\\}.$

However, due to the "too many boundary conditions" in the definition of $D$,
it can be seen that $\frac{\partial}{\partial x}D^{\alpha}:D\subseteq
L^{2}(0,1)\rightarrow L^{2}(0,1)$ is not a generator of an analytic semigroup.
In order to preserve $u(0)=u(1)=0$, we omit the condition
$u_{x}\in{{}_{0}}H^{\alpha}(0,1)$ and consider

$D\left(\frac{\partial}{\partial
x}D^{\alpha}\right):=\widetilde{\mathcal{D}}_{\alpha}=\\{u=w-w(1)x^{\alpha}:w\in{{}_{0}}H^{1+\alpha}(0,1)\\},$
(20)

with the norm

$||u||_{\widetilde{\mathcal{D}}_{\alpha}}=||w||_{H^{1+\alpha}(0,1)},\quad\text{
for }\alpha\in(0,1)\setminus\left\\{\frac{1}{2}\right\\},$
$||u||_{\widetilde{\mathcal{D}}_{\alpha}}=\left(||w||_{H^{\frac{3}{2}}(0,1)}+\int_{0}^{1}\frac{|w_{x}(x)|^{2}}{x}{\rm
d}x\right)^{\frac{1}{2}},\quad\text{ for }\alpha=\frac{1}{2}.$

Now we will prove the key result of this subsection.

######  Theorem 1.

[28, Theorem 3.10] The operator $\frac{\partial}{\partial
x}D^{\alpha}:\widetilde{\mathcal{D}}_{\alpha}\subseteq L^{2}(0,1)\rightarrow
L^{2}(0,1)$ is a generator of an analytic semigroup.

###### Proof.

We begin with the analysis of the resolvent.

######  Lemma 2.

Let $\frac{\partial}{\partial
x}D^{\alpha}:\widetilde{\mathcal{D}}_{\alpha}\rightarrow L^{2}(0,1)$. Then,
for every $\lambda\in\mathbb{C}$ belonging to the sector
$\vartheta_{\alpha}:=\\{z\in\mathbb{C}\setminus\\{0\\}:\left|{\operatorname{arg}z}\right|\leq\frac{\pi(\alpha+1)}{2}\\}\cup\\{0\\}$
there holds $R(\lambda E-\frac{\partial}{\partial x}D^{\alpha})=L^{2}(0,1).$

###### Proof.

To prove the lemma we fix $g\in L^{2}(0,1)$ and $\lambda$ belonging to
$\vartheta_{\alpha}$. We must prove that there exists
$u\in\widetilde{\mathcal{D}}_{\alpha}$ such that

$\lambda u-\frac{\partial}{\partial x}D^{\alpha}u=g.$ (21)

We note that if we search for a solution in
$\widetilde{\mathcal{D}}_{\alpha}$, then it can be represented in the form
$u=w-w(1)x^{\alpha}$ and $D^{\alpha}u=D^{\alpha}w-w(1)\Gamma(\alpha+1)$. Since
$w\in{}_{0}H^{1+\alpha}(0,1)$, we have $D^{\alpha}w\in{}_{0}H^{1}(0,1)$ and
hence $(D^{\alpha}u)(0)=-w(1)\Gamma(\alpha+1)$. At first we will solve the
equation (21) with initial conditions $u(0)=0$ and $(D^{\alpha}u)(0)=a$ for
arbitrary $a\in\mathbb{C}$ and then we will choose $a$ such that $u(1)=0$. Let
us transform the equation (21) into integral form. To this end we assume that
$u$ which may be written in the form
$u=w+\frac{a}{\Gamma(1+\alpha)}x^{\alpha}$ where
$w\in{}_{0}H^{1+\alpha}(0,1)$, solves (21). Then, having integrated (21) we
obtain that

$D^{\alpha}u=(D^{\alpha}u)(0)+\lambda Iu-Ig=a+\lambda Iu-Ig.$

Applying $I^{\alpha}$ we get

$u=aI^{\alpha}1+\lambda I^{\alpha+1}u-I^{\alpha+1}g.$ (22)

We note that if we search for a solution such that there exists
$w\in{}_{0}H^{1+\alpha}(0,1)$, such that
$u=w+\frac{a}{\Gamma(1+\alpha)}x^{\alpha}$, then equation (21) is equivalent
with (22). Indeed, it follows from Proposition 4 together with Proposition 1.
We apply the operator $I^{\alpha+1}$ to (22) and we obtain

$u=aI^{\alpha}1-I^{\alpha+1}g+\lambda
aI^{2\alpha+1}1+\lambda^{2}I^{2(\alpha+1)}u-\lambda I^{2(\alpha+1)}g.$

Iterating this procedure $n$ times we arrive at

$u=a\sum_{k=0}^{n}\lambda^{k}I^{\alpha+k(\alpha+1)}1-\sum_{k=0}^{n}\lambda^{k}I^{(k+1)(\alpha+1)}g+\lambda^{n+1}I^{(n+1)(\alpha+1)}u.$
(23)

We will show, that the last expression tends to zero as $n\rightarrow\infty$.
Indeed, we may note that, since ${}_{0}H^{1+\alpha}(0,1)\subseteq
L^{\infty}(0,1)$ and due to the presence of the $\Gamma$-function in the
denominator we have

$\left|{\lambda^{n}(I^{n(\alpha+1)}u)(x)}\right|\leq\left\|{u}\right\|_{L^{\infty}(0,1)}\frac{\left|{\lambda}\right|^{n}x^{(\alpha+1)n}}{\Gamma((\alpha+1)n+1)}\leq\frac{\left\|{u}\right\|_{L^{\infty}(0,1)}\left|{\lambda}\right|^{n}}{\Gamma((\alpha+1)n+1)}\rightarrow
0\textrm{ as }n\rightarrow\infty$

for each $\lambda\in\mathbb{C}$ uniformly with respect to $x\in[0,1]$. Thus,
passing to the limit with $n$ in (23) we obtain the formula

$u=a\sum_{k=0}^{\infty}\lambda^{k}I^{\alpha+k(\alpha+1)}1-\sum_{k=0}^{\infty}\lambda^{k}I^{(k+1)(\alpha+1)}g.$
(24)

We note that

$\sum_{k=0}^{\infty}\lambda^{k}I^{\alpha+k(\alpha+1)}1=x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
x^{\alpha+1}),$

where $E_{a,b}(\cdot)$ denotes the two parameter Mittag-Leffler function.
Furthermore, one may easily show that the second series is uniformly
convergent and

$\sum_{k=0}^{\infty}\lambda^{k}I^{(k+1)(\alpha+1)}g=g*x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
x^{\alpha+1}),$

where $*$ denotes the convolution on the positive real line, i.e.
$(f*g)(x):=\int_{0}^{x}f(p)g(x-p){\rm d}p$. Together, we obtain that function
$u$ given by (24) may be equivalently written as

$u(x)=ax^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
x^{\alpha+1})-g*x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda x^{\alpha+1})$ (25)

and one may check that $u$ given by the formula (25) is a solution to (22) and
(21) with boundary conditions $u(0)=0$ and $(D^{\alpha}u)(0)=a$. It remains to
choose the value $a$ in such a way that $u(1)=0$. For this purpose, we take
$x=1$ in (25) and we obtain

$u(1)=aE_{\alpha+1,\alpha+1}(\lambda)-(g*y^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
y^{\alpha+1}))(1).$

To obtain that $u(1)=0$ we choose

$a=E_{\alpha+1,\alpha+1}(\lambda))^{-1}(g*y^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
y^{\alpha+1}))(1).$

Note that $a$ is well defined because, taking $\nu=\mu=\alpha+1$ in [24,
Theorem 4.2.1], we obtain that $E_{\alpha+1,\alpha+1}(\lambda)\neq 0$ for
$\lambda$ belonging to the sector $\vartheta_{\alpha}$. Summing up the results
we obtain that there exists a solution to (21) which belongs to
$\widetilde{\mathcal{D}}_{\alpha}$ and it is represented by the formula

$u(x)=\frac{(g*y^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
y^{\alpha+1}))(1)}{E_{\alpha+1,\alpha+1}(\lambda)}x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
x^{\alpha+1})-g*x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda x^{\alpha+1}).$

We note that here function $w$ from the definition of
$\widetilde{\mathcal{D}}_{\alpha}$ is given by

$w(x)=\frac{(g*y^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
y^{\alpha+1}))(1)}{E_{\alpha+1,\alpha+1}(\lambda)}x^{\alpha}\sum_{n=1}^{\infty}\frac{\lambda^{n}x^{(\alpha+1)n}}{\Gamma((\alpha+1)n+\alpha+1)}-g*x^{\alpha}E_{\alpha+1,\alpha+1}(\lambda
x^{\alpha+1}).$

In this way we proved the lemma. ∎

Our next aim is to prove the following.

######  Lemma 3.

For $u\in\widetilde{\mathcal{D}}_{\alpha}$ we have

$\Re(-\frac{\partial}{\partial x}D^{\alpha}u,u)\geq
c_{\alpha}\left\|{u}\right\|_{H^{\frac{1+\alpha}{2}}(0,1)}^{2}$ (26)

and

$\left|{(-\frac{\partial}{\partial x}D^{\alpha}u,u)}\right|\leq
b_{\alpha}\left\|{u}\right\|_{H^{\frac{1+\alpha}{2}}(0,1)}^{2},$ (27)

where $c_{\alpha},b_{\alpha}$ are positive constants which depends only on
$\alpha$.

###### Proof.

Let us start with (26). We fix $u\in\widetilde{\mathcal{D}}_{\alpha}$. Since
$u(0)=u(1)=0$, we may integrate by parts to obtain

$\Re\left(-\frac{\partial}{\partial
x}D^{\alpha}u,u\right)=-\Re\int_{0}^{1}(\frac{\partial}{\partial
x}D^{\alpha}u)(x)\cdot\overline{u(x)}{\rm d}x$ $=\int_{0}^{1}D^{\alpha}\Re
u(x)\cdot\frac{\partial}{\partial x}\Re u(x){\rm d}x+\int_{0}^{1}D^{\alpha}\Im
u(x)\cdot\frac{\partial}{\partial x}\Im u(x){\rm d}x.$

We note that $\widetilde{\mathcal{D}}_{\alpha}\subseteq AC[0,1]$ hence we may
apply property (10) from Proposition 1 and we get

$\Re\left(-\frac{\partial}{\partial
x}D^{\alpha}u,u\right)=\int_{0}^{1}D^{\alpha}\Re
u(x)\cdot\partial^{1-\alpha}D^{\alpha}\Re u(x){\rm
d}x+\int_{0}^{1}D^{\alpha}\Im u(x)\cdot\partial^{1-\alpha}D^{\alpha}\Im
u(x){\rm d}x.$

By the definition of $\widetilde{\mathcal{D}}_{\alpha}$ we know that
$D^{\alpha}u\in AC[0,1]$, hence we are allowed to apply inequality (18) with
$w=D^{\alpha}\Re u$ and $w=D^{\alpha}\Im u$ to obtain

$\Re\left(-\frac{\partial}{\partial x}D^{\alpha}u,u\right)\geq
c_{\alpha}\left\|{D^{\alpha}u}\right\|_{H^{\frac{1-\alpha}{2}}(0,1)}^{2}\geq
c_{\alpha}\left\|{\partial^{\frac{1-\alpha}{2}}D^{\alpha}u}\right\|_{L^{2}(0,1)}^{2}$
$=c_{\alpha}\left\|{D^{\frac{1+\alpha}{2}}\
u}\right\|_{L^{2}(0,1)}^{2}=c_{\alpha}\left\|{\partial^{\frac{1+\alpha}{2}}\
u}\right\|_{L^{2}(0,1)}^{2}\geq
c_{\alpha}\left\|{u}\right\|_{H^{\frac{1+\alpha}{2}}(0,1)}.$

Here the first equality follows from (10), while the second and third come
from Proposition 4 and the fact that $u$ vanishes at zero. It remains to show
(27). Using superposition property of fractional operators and the definition
of the Riemann-Liouville fractional derivative, similarly as in [30] we arrive
at the following sequence of identities

$\frac{\partial}{\partial
x}D^{\alpha}u=\partial^{\frac{1+\alpha}{2}}\partial^{\frac{1-\alpha}{2}}D^{\alpha}u=\frac{\partial}{\partial
x}I^{\frac{1-\alpha}{2}}D^{\frac{1+\alpha}{2}}u.$

Then, we apply integration by parts formula ( we note that the boundary terms
vanish due to $u(0)=u(1)=0$). and Fubini’s theorem to get

$\int_{0}^{1}\frac{\partial}{\partial x}D^{\alpha}u\cdot\bar{u}{\rm
d}x=-\int_{0}^{1}I^{\frac{1-\alpha}{2}}D^{\frac{1+\alpha}{2}}u\cdot\bar{u}_{x}{\rm
d}x=-\int_{0}^{1}D^{\frac{1+\alpha}{2}}u\cdot
I^{\frac{1-\alpha}{2}}_{-}\bar{u}_{x}{\rm d}x,$

where for $\beta\in(0,1)$, by $I^{\beta}_{-}$ we denote the formal adjoint to
$I^{\beta}$, which is given by the formula
$(I^{\beta}_{-}u)(x)=\frac{1}{\Gamma(\beta)}\int_{x}^{1}(p-x)^{\beta-1}u(p){\rm
d}p$. Similarly we denote
$\partial_{-}^{\beta}u=-\frac{d}{dx}I^{1-\beta}_{-}u$,
$D_{-}^{\beta}u=-\frac{d}{dx}I^{1-\beta}_{-}[u(x)-u(1)].$

Finally, we get

$\left|{\int_{0}^{1}\frac{\partial}{\partial
x}D^{\alpha}u\cdot\bar{u}dx}\right|\leq\left\|{D^{\frac{1+\alpha}{2}}u}\right\|_{L^{2}(0,1)}\left\|{D^{\frac{1+\alpha}{2}}_{-}\bar{u}}\right\|_{L^{2}(0,1)}$
$=\left\|{\partial^{\frac{1+\alpha}{2}}u}\right\|_{L^{2}(0,1)}\left\|{\partial^{\frac{1+\alpha}{2}}_{-}\bar{u}}\right\|_{L^{2}(0,1)}\leq
b_{\alpha}\left\|{u}\right\|_{H^{\frac{1+\alpha}{2}}}^{2}.$

We note that here we again used the fact that $u$ vanishes at the boundary and
we applied Proposition 4. ∎

Having proven Lemma 2 and Lemma 3 the proof of Theorem 1 follows from the
standard argument (see for example [23, Chapter 7, Theorem 2.7.]). ∎

Due to Theorem 1, and [35, Theorem 3.4 Chapter 3.2.1], we obtain the following
existence result for problem (19).

######  Corollary 2.

Let $f\in C^{o,\nu}([0,T];L^{2}(0,1))$ for $\nu\in(0,1)$. Then, there exists a
unique solution to (19) in $C([0,T];L^{2}(0,1))\cap
C((0,T];\widetilde{\mathcal{D}}_{\alpha})\cap C^{1}((0,T];L^{2}(0,1)).$

###  3.2 The moving boundary problem

Now we are able to deal with the moving boundary problem, i.e.

$\begin{array}[]{lll}(i)&u_{t}-\frac{\partial}{\partial x}D^{\alpha}u=0&\text{
in }Q_{s,T},\\\ (ii)&u(0,t)=0,u(s(t),t)=0,&t\in(0,T),\\\
(iii)&u(x,0)=u_{0}(x)\geq 0,&0<x<s(0)=b,\\\ \end{array}$ (28)

with given functions $s:[0,T]\rightarrow{\mathbb{R}}$ , and
$u_{0}\in{{}_{0}}H^{1+\alpha}(0,b)$ such that $u_{0}(0)=u_{0}(b)=0$. We assume
that

$s\in C^{0,1}[0,T],s(0)=b,\text{ and there exists }M>0\text{ such that
}0\leq\dot{s}(t)\leq M\,\text{ a.e. }t\in[0,T].$ (29)

Let us make the following basic but useful observation.

######  Remark 3.

Let $s$ be a function satisfying (29). Then,

$b\leq s(t)\leq b+MT,\quad\forall t\in[0,T].$

Similarly as in [25] and [29], we will solve (28) applying the theory of
evolution operators. To this end, we pass to a cylindrical domain applying the
standard substitution $p=\frac{x}{s(t)}$ to obtain

$\begin{array}[]{lll}(i)&v_{t}-x\frac{\dot{s}(t)}{s(t)}v_{x}-\frac{1}{s^{1+\alpha}(t)}\frac{\partial}{\partial
x}D^{\alpha}v=0,&0<x<1,0<t<T,\\\ (ii)&v(0,t)=0,v(1,t)=0,&t\in(0,T),\\\
(iii)&v(x,0)=v_{0}(x),&0<x<1,\\\ \end{array}$ (30)

where $v_{0}(x)=u_{0}(bx)$. Then, we define

$A(t):=\frac{1}{s^{1+\alpha}(t)}\frac{\partial}{\partial x}D^{\alpha},\hskip
5.69046ptA(t):\widetilde{\mathcal{D}}_{\alpha}\subseteq L^{2}(0,1)\rightarrow
L^{2}(0,1)\mbox{ for every }t\in[0,T],$ (31)

we denote $F(x,t,v):=x\frac{\dot{s}(t)}{s(t)}v_{x}$ and we may rewrite (30) as
follows:

$\begin{array}[]{lll}(i)&v_{t}=A(t)v+F(x,t,v),&0<x<1,0<t<T,\\\
(ii)&v(0,t)=0,v(1,t)=0,&t\in(0,T),\\\ (iii)&v(x,0)=v_{0}(x),&0<x<1.\\\
\end{array}$ (32)

######  Lemma 4.

For all $v\in\widetilde{\mathcal{D}}_{\alpha}$, there exists constants
$c=c(\alpha,b,M,T)$ and $C=C(\alpha,b,M,T)$ such that

$c||v||_{\widetilde{\mathcal{D}}_{\alpha}}\leq||A(t)v||_{L^{2}(0,1)}\leq
C||v||_{\widetilde{\mathcal{D}}_{\alpha}},\quad\forall t\in[0,T].$

###### Proof.

The proof is analogous to the proof of [25, Lemma 2], one just have to use
(29) and that $\frac{\partial}{\partial
x}D^{\alpha}x^{\alpha}\leavevmode\nobreak\ =\leavevmode\nobreak\ 0$. ∎

Since we deal with nonautonomous problem (30), we recall the definition of
evolution operator and the basic solvability result.

######  Definition 4.

[17, Definition 6.0.1] Let $X$ be a Banach space, $T>0$. A family of linear
bounded operators $\\{G(t,\sigma):0\leq\sigma\leq t\leq T\\}$ is said to be an
evolution operator for the problem

$u^{\prime}(t)=A(t)u+f(t),0<t\leq T,u(0)=u_{0},$

where $A(\cdot)$ denotes a family of sectorial operators with common domains,
i.e. $D(A(t))\equiv D$ for every $t\in[0,T]$, if

  1. 1.

$G(t,\sigma)G(\sigma,r)=G(t,r),G(\sigma,\sigma)=E,\text{ for }0\leq
r\leq\sigma\leq t\leq T$,

  2. 2.

$G(t,\sigma)\in B(X,D)\text{ for }0\leq\sigma\leq t\leq T$,

  3. 3.

$t\mapsto G(t,\sigma)$ is differentiable in $(\sigma,T)$ with values in $B(X)$
and

$\frac{\partial}{\partial t}G(t,\sigma)=A(t)G(t,\sigma)\hskip 5.69046pt\text{
for }\hskip 5.69046pt0\leq\sigma<t\leq T.$

######  Remark 4.

We note that, if $v$ is a solution to (32)$(i)$,(32)$(iii)$, that is, $v$ is a
solution to

$v^{\prime}(\cdot,t)=A(t)v(\cdot,t)+F(\cdot,t,v),0<t\leq
T,v(\cdot,0)=v_{0}(x),$ (33)

and additionally $v(\cdot,t)\in\widetilde{\mathcal{D}}_{\alpha}$, $\forall
t\in[0,T]$, then by definition of $\widetilde{\mathcal{D}}_{\alpha}$ it
follows that $v(0,t)=v(1,t)=0$, $\forall t\in[0,T]$, thus $v$ is a solution to
(30).

######  Theorem 2.

[17, Chapter 6] Let $D$ be a Banach space continuously embedded into $X$ and
let $T>0$, $a\in(0,1)$. If for $0\leq t\leq T$ $A(t):D(A(t))\rightarrow X$
satisfies that

  1. 1.

for every $t\in[0,T]$ $A(t)$ is sectorial and $D(A(t))\equiv D$,

  2. 2.

$t\mapsto A(t)\in C^{0,a}([0,T];B(D,X))$,

then there exists a family of evolution operators for $A(t)$ given by
Definition 4.

######  Remark 5.

By Theorem 1 and (29), it follows that $A(t)$ verifies the hyphotesis of
Theorem 2 for $0\leq t\leq T$. Then,
$A(t):\widetilde{\mathcal{D}}_{\alpha}\rightarrow L^{2}(0,1)$ generates a
family of evolution operators.

######  Proposition 7.

[17, Corollary 6.1.6.(i), (iii)] Let $A(t)$ satisfies the assumptions of
Theorem 2. If $u_{0}\in X$, then $G(t,0)u_{0}\in C([0,T];X)\cap
C^{1}((0,T];X)\cap C((0,T];D)$. Furthermore, if $u_{0}\in D$, then
$G(t,0)u_{0}\in C^{1}([0,T];X)\cap C([0,T];D)$ and $\frac{\partial}{\partial
t}G(t,0)u_{0}=A(t)G(t,0)u_{0}\text{ for every }0\leq t\leq T$.

In order to develop the theory of non-homogenous problems we introduce the
notion of a mild solution.

###  3.3 Existence of a mild solution

Our next purpose is to introduce the concept of mild solution associated with
the problem (30). Recall that, if

$u^{\prime}(t)=A(t)u(t)+f(t),\hskip 5.69046pt\sigma<t\leq T,\hskip
5.69046ptu(\sigma)=u_{\sigma},$ (34)

where $A(t)$ satisfies the assumptions of Theorem 2, and
$\\{G(t,\tau):\sigma\leq\tau\leq t\leq T\\}$ the family of evolution operators
generated by $A(t)$, then for every $f\in L^{1}(\sigma,T;X)$ and
$u_{\sigma}\in X$, the function $u$ defined by the formula

$u(t)=G(t,\sigma)u_{\sigma}+\int_{\sigma}^{t}G(t,\tau)f(\tau)d\tau$ (35)

is called a mild solution to (34).

######  Proposition 8.

[17, Corollary 6.2.4.] Let $f\in C((\sigma,T];X)\cap L^{1}(\sigma,T;X)$,
$u_{\sigma}\in\overline{D}$. If problem (34) has a solution belonging to
$C^{1}((0,T];X)\cap C((0,T];D)\cap C([0,T];X)$ so that (34) is satisfied for
each $t\in(0,T]$, then $u$ is given by (35).

Returning to (30), we will use (35) to define a mild solution to this problem.

######  Definition 5.

Let us denote by $\\{G(t,\tau):\sigma\leq\tau\leq t\leq T\\}$ the family of
evolution operators generated by $A(t)$ defined in (31). We say that $v$ is a
mild solution to (30) if it verifies

$v(x,t)=G(t,0)v_{0}+\int_{0}^{t}G(t,\sigma)x\frac{\dot{s}(\sigma)}{s(\sigma)}v_{x}(x,\sigma){\rm
d}\sigma.$ (36)

Note that $v$ is involved in the right hand side of (36) too, and if we define
$f(x,t)=x\frac{\dot{s}(t)}{s(t)}v_{x}(x,t)$, then $v$ is a mild solution to
(30) if it verifies

$v(x,t)=G(t,0)v_{0}+\int_{0}^{t}G(t,\sigma)f(x,\sigma){\rm d}\sigma.$

Since $f$ depends on $v_{x}$, we will prove that there exists a mild solution
to (30), by using a fix point theorem. For this purpose, we need the following
estimates for the interpolation spaces.

######  Proposition 9.

[17, Corollary 6.1.8] Let $\\{G(t,\sigma):0\leq\sigma\leq t\leq T\\}$ be a
family of evolution operators generated by $A(t):D\rightarrow X$. Then, for
every $g\in L^{2}(0,1)$ we have

$\left\|{G(t,\sigma)g}\right\|_{X}\leq c\left\|{g}\right\|_{X}.$ (37)

If $g\in[X,D]_{\delta}$, then for any $0\leq\sigma<t\leq T$ there exists
positive constant $c=c(\theta,\delta,T)$ which is a continuous increasing
function of $T$ such that, for every $0<\delta<1$

$\left\|{G(t,\sigma)g}\right\|_{D}\leq\frac{c}{(t-\sigma)^{1-\delta}}\left\|{g}\right\|_{[X,D]_{\delta}}.$
(38)

Moreover, for any $0\leq\delta<\theta<1$, we have

$\left\|{G(t,\sigma)g}\right\|_{[X,D]_{\theta}}\leq\frac{c}{(t-\sigma)^{\theta-\delta}}\left\|{g}\right\|_{[X,D]_{\delta}}$
(39)

and for $\theta\in(0,1)$, $\delta\in(0,1]$, $\theta<\delta$

$\left\|{A(t)G(t,\sigma)g}\right\|_{[X,D]_{\theta}}\leq\frac{c}{(t-\sigma)^{1+\theta-\delta}}\left\|{g}\right\|_{[X,D]_{\delta}}.$
(40)

Furthermore, for every $0\leq\theta<\delta<1$ and $0\leq\sigma<t\leq T$

$\left\|{G(t,\sigma)g-G(r,\sigma)g}\right\|_{[X,D]_{\theta}}\leq\frac{c}{(t-r)^{\theta-\delta}}\left\|{g}\right\|_{[X,D]_{\delta}}.$
(41)

Finally, for every $0\leq\sigma<r<t\leq T$

$\left\|{A(t)G(t,\sigma)g-A(r)G(r,\sigma)g}\right\|_{X}\leq
c\left(\frac{(t-r)^{a}}{(r-\sigma)^{1-\delta}}+\frac{1}{(r-\sigma)^{1-\delta}}-\frac{1}{(t-\sigma)^{1-\delta}}\right)\left\|{g}\right\|_{[X,D]_{\delta}},$
(42)

where $a\in(0,1)$ comes from Theorem 2. The constant $c>0$ depends only on
$\alpha,\theta,\delta,T$, and the constants $b,M$ comes from (29). Moreover,
$T\mapsto c(\alpha,\theta,\delta,b,M,T)$ is an increasing function.

Now we present two substantial estimates, which will be frequently used in the
paper.

######  Lemma 5.

For every $v\in\widetilde{\mathcal{D}}_{\alpha}$, there exists a constant
$c>0$ such that

$||xv_{x}||_{{{}_{0}}H^{\alpha}(0,1)}\leq
c||v||_{\widetilde{\mathcal{D}}_{\alpha}}.$

###### Proof.

Since $v\in\widetilde{\mathcal{D}}_{\alpha}$, there exists
$w\in{{}_{0}}H^{1+\alpha}(0,1)$ such that $v(x)=w-w(1)x^{\alpha}$. Then,
$xv_{x}=xw_{x}-\alpha w(1)x^{\alpha}$, and by Corollary 1 and Sobolev
embedding, it follows that

$\begin{split}||xv_{x}||_{{{}_{0}}H^{\alpha}(0,1)}&\leq
c||w_{x}||_{{{}_{0}}H^{\alpha}(0,1)}+\alpha|w(1)|||x^{\alpha}||_{{{}_{0}}H^{\alpha}(0,1)}\leq
c(||v||_{\widetilde{\mathcal{D}}_{\alpha}}+||w||_{C[0,1]})\\\ &\leq
c(||v||_{\widetilde{\mathcal{D}}_{\alpha}}+||w||_{{{}_{0}}H^{1+\alpha}(0,1)})\leq
c||v||_{\widetilde{\mathcal{D}}_{\alpha}}.\end{split}$ (43)

and the claim holds. ∎

######  Remark 6.

For all $\theta\in(0,1)$,
$H_{0}^{(1+\alpha)\theta}(0,1)\hookrightarrow[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\theta}$
continuously. In fact,
$H_{0}^{(1+\alpha)\theta}(0,1)=[L^{2}(0,1),H_{0}^{1+\alpha}(0,1)]_{\theta}$,
and $H_{0}^{1+\alpha}(0,1)\hookrightarrow\widetilde{\mathcal{D}}_{\alpha}$
continuously. Moreover, for all $\delta<\frac{1}{2(1+\alpha)}$ we have that
$H^{(1+\alpha)\delta}(0,1)=H_{0}^{(1+\alpha)\delta}(0,1)$. Then, for every
$(1+\alpha)\delta\in\left(0,\frac{1}{2}\right)$ there holds
$H^{(1+\alpha)\delta}(0,1)\hookrightarrow[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\delta}$
and

$||u||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\delta}}\leq
c||u||_{H^{(1+\alpha)\delta}}.$ (44)

We prove now existence of a mild solution to (30).

######  Theorem 3.

Let $v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$ be. Then, there exists a unique
solution to (36) belonging to $C([0,T];\widetilde{\mathcal{D}}_{\alpha})$.

###### Proof.

The proof is analogous to that of [29, Theorem 2]. We define an operator $P$
by

$(Pv)(x,t)=G(t,0)v_{0}+\int_{0}^{t}G(t,\sigma)x\frac{\dot{s}(\sigma)}{s(\sigma)}v_{x}(x,\sigma){\rm
d}\sigma=G(t,0)v_{0}+\int_{0}^{t}G(t,\sigma)f(x,\sigma){\rm d}\sigma.$ (45)

We will apply the Banach fixed point theorem to $P$. To this end, we will show
that $P:C([0,T];\widetilde{\mathcal{D}}_{\alpha})\rightarrow
C([0,T];\widetilde{\mathcal{D}}_{\alpha})$. We note that, by [17, Corollary
6.1.6 (i), (iii)], we obtain that $G(t,0)v_{0}\in
C([0,T];\widetilde{\mathcal{D}}_{\alpha})$.

Let $v\in C([0,T];\widetilde{\mathcal{D}}_{\alpha})$. Then, for every
$0<\theta<\frac{\min\\{\alpha,1/2\\}}{1+\alpha}$, by (38), (44) and Lemma 5,
it follows that

$\begin{split}\left|\left|\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm
d}\sigma\right|\right|_{\widetilde{\mathcal{D}}_{\alpha}}&\leq\int_{0}^{t}c(t-\sigma)^{\theta-1}\left|\left|f(\cdot,\sigma)\right|\right|_{[L^{2}(0,1);\widetilde{\mathcal{D}}_{\alpha}]_{\theta}}{\rm
d}\sigma\\\
&\leq\int_{0}^{t}c(t-\sigma)^{\theta-1}\left|\left|f(\cdot,\sigma)\right|\right|_{{{}_{0}}H^{(1+\alpha)\theta}(0,1)}{\rm
d}\sigma\\\ &\leq
c\left|\left|x\frac{\dot{s}(\cdot)}{s(\cdot)}v_{x}\right|\right|_{L^{\infty}((0,T),{{}_{0}}H^{\alpha}(0,1))}t^{\theta}.\end{split}$
(46)

Then $\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm d}\sigma$ is continuous at
$t=0$ in the norm $||\cdot||_{\widetilde{\mathcal{D}}_{\alpha}}$.

On the other hand, for $0<\tau\leq t\leq T$, applying, (37), (39) and (41) it
follows that

$\begin{split}&\left|\left|\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm
d}\sigma-\int_{0}^{\tau}G(\tau,\sigma)f(\cdot,\sigma){\rm
d}\sigma\right|\right|_{L^{2}(0,1)}\\\
&\leq\int_{0}^{\tau}\left|\left|G(t,\sigma)f(\cdot,\sigma)-G(\tau,\sigma)f(\cdot,\sigma)\right|\right|_{L^{2}(0,1)}{\rm
d}\sigma+\int_{\tau}^{t}\left|\left|G(t,\sigma)f(\cdot,\sigma)\right|\right|_{L^{2}(0,1)}{\rm
d}\sigma\\\
&\leq\int_{0}^{t}c(t-\tau)^{\theta}\left|\left|f(\cdot,\sigma)\right|\right|_{[L^{2}(0,1);\widetilde{\mathcal{D}}_{\alpha}]_{\theta}}{\rm
d}\sigma+c\int_{\tau}^{t}\left|\left|f(\cdot,\sigma)\right|\right|_{L^{2}(0,1)}{\rm
d}\sigma\\\ \end{split}$ (47)

Thus, for every $0<\theta<\frac{\min\\{\alpha,1/2\\}}{1+\alpha}$, by Remark 6
and Lemma 5

$\begin{split}&\left|\left|\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm
d}\sigma-\int_{0}^{\tau}G(\tau,\sigma)f(\cdot,\sigma){\rm
d}\sigma\right|\right|_{L^{2}(0,1)}\\\ &\leq
c(t-\tau)^{\theta}\int_{0}^{\tau}\left|\left|f(\cdot,\sigma)\right|\right|_{{{}_{0}}H^{(1+\alpha)\theta}(0,1)}{\rm
d}\sigma+c\int_{\tau}^{t}\left|\left|f(\cdot,\sigma)\right|\right|_{{{}_{0}}H^{\alpha}(0,1)}{\rm
d}\sigma\\\ &\leq
c\left|\left|f\right|\right|_{L^{\infty}((0,T),{{}_{0}}H^{\alpha}(0,1))}\left(\tau(t-\tau)^{\theta}+(t-\tau)\right),\end{split}$
(48)

and the last term tends to zero when $\tau\rightarrow t$. Then
$\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm d}\sigma\in C((0,T],L^{2}(0,1))$.
By Lemma 4, we only need to prove that
$A(t)\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm d}\sigma\in
C((0,T],L^{2}(0,1))$. Let $0<\tau\leq t\leq T$ and
$0<\theta<\frac{\min\\{\alpha,1/2\\}}{1+\alpha}$. Then, applying first (40)
and (42), and then Remark 6 and Lemma 5, it follows that

$\begin{split}&\left|\left|A(t)\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm
d}\sigma-A(\tau)\int_{0}^{\tau}G(\tau,\sigma)f(\cdot,\sigma){\rm
d}\sigma\right|\right|_{L^{2}(0,1)}\\\
&\leq\int_{0}^{\tau}\left|\left|A(t)G(t,\sigma)f(\cdot,\sigma)-A(\tau)G(\tau,\sigma)f(\cdot,\sigma)\right|\right|_{L^{2}(0,1)}{\rm
d}\sigma+\int_{\tau}^{t}\left|\left|A(t)G(t,\sigma)f(\cdot,\sigma)\right|\right|_{L^{2}(0,1)}{\rm
d}\sigma\\\
&\leq\int_{0}^{t}\left(\frac{(t-\tau)^{a}}{(\tau-\sigma)^{1-\theta}}+\frac{1}{(\tau-\sigma)^{1-\theta}}-\frac{1}{(t-\sigma)^{1-\theta}}\right)\left|\left|f(\cdot,\sigma)\right|\right|_{[L^{2}(0,1);\widetilde{\mathcal{D}}_{\alpha}]_{\theta}}{\rm
d}\sigma+c\int_{\tau}^{t}(t-\sigma)^{\theta-1}\left|\left|f(\cdot,\sigma)\right|\right|_{[L^{2}(0,1);\widetilde{\mathcal{D}}_{\alpha}]_{\theta}}{\rm
d}\sigma\\\ &\leq
c\left|\left|f\right|\right|_{L^{\infty}((0,T),{{}_{0}}H^{\alpha}(0,1))}\left(\tau^{\theta}(t-\tau)^{a}+\tau^{\theta}-t^{\theta}+(t-\tau)^{\theta}\right)\\\
\end{split}$ (49)

and the last term tends to zero when $\tau\rightarrow t$. Then, we deduce
$\int_{0}^{t}G(t,\sigma)f(\cdot,\sigma){\rm d}\sigma\in
C([0,T],\widetilde{\mathcal{D}}_{\alpha})$, and we conclude $Pv\in
C([0,T],\widetilde{\mathcal{D}}_{\alpha})$.

Now, we will prove that $P$ is a contraction on
$C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})$ for $T_{1}$ small enough. Let
$v^{1},v^{2}\in C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})$. By definition
of $P$, we have

$(Pv^{1}-Pv^{2})(x,t)=\int_{0}^{t}G(t,\sigma)x\frac{\dot{s}(t)}{s(t)}(v^{1}_{x}-v^{2}_{x})(x,\sigma){\rm
d}\sigma.$ (50)

Analogously to (46),

$\begin{split}\left|\left|\int_{0}^{t}G(t,\sigma)x\frac{\dot{s}(t)}{s(t)}(v^{1}_{x}-v^{2}_{x})(x,\sigma){\rm
d}\sigma\right|\right|_{C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})}&\leq\sup_{t\in[0,T_{1}]}\int_{0}^{t}c(t-\sigma)^{\theta-1}\left|\left|x(v^{1}_{x}-v^{2}_{x})(x,\sigma)\right|\right|_{{{}_{0}}H^{\alpha}(0,1)}{\rm
d}\sigma\end{split}$ (51)

and hence, by Lemma 5,

$\begin{split}\left|\left|\int_{0}^{t}G(t,\sigma)x\frac{\dot{s}(t)}{s(t)}(v^{1}_{x}-v^{2}_{x})(x,\sigma){\rm
d}\sigma\right|\right|_{C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})}&\leq\sup_{t\in[0,T_{1}]}c\left|\left|v^{1}-v^{2}\right|\right|_{C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})}\int_{0}^{t}(t-\sigma)^{\theta-1}{\rm
d}\sigma\\\
&=\frac{cT_{1}^{\theta}}{\theta}\left|\left|v^{1}-v^{2}\right|\right|_{C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})},\end{split}$
(52)

thus, $P$ is a contraction on $C([0,T_{1}],\widetilde{\mathcal{D}}_{\alpha})$
whenever $T_{1}<\left(\frac{\theta}{c}\right)^{\frac{1}{\theta}}$. We may
extend the solution on the whole interval $[0,T]$ applying the standard
argument.

∎

###  3.4 Regularity of the mild solution

In this section, we improve the regularity of the mild solution obtained in
Theorem 3. We follow the ideas introduced in [29] that were also applied in
[25, Section 3.3]. The main novelty here, is that we deal with the
interpolation space $[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\theta}$,
which characterization is not so immediate as in previous results ([25,
Equation (3.17)]. Since we are particularly interested in the interior
regularity of the solution, we establish the continuous embedding of the space
of restrictions of $[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\theta}$ to
the interval $(\varepsilon,\omega)$ into
$H^{\frac{\theta}{1+\alpha}}(\varepsilon,\omega)$, $0<\varepsilon<\omega<1$,
which is presented in the next subsection. The regularity results are given
then in subsection 3.4.2.

####  3.4.1 A relation between the interpolation space
$[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\theta}$ and a local norm in a
fractional Sobolev space

Let us prove the following result.

######  Theorem 4.

Let $\alpha\in(0,1)$. Then, for all $0<\varepsilon<\omega<1$, for every
$0<\delta<1+\alpha$ and
$u\in[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\delta}{1+\alpha}}$,
it follows that $\tilde{u}\in H^{\delta}(\varepsilon,\omega)$ and there exists
a positive constant $c=c(\varepsilon,\omega,\alpha,\delta)$ such that

$||\tilde{u}||_{H^{\delta}(\varepsilon,\omega)}\leq
c||u||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\delta}{1+\alpha}}},$
(53)

where $\tilde{u}$ is the restriction of $u$ to the interval
$(\varepsilon,\omega)$.

We will follow the approach in [22]. Thus, let us recall some concepts and
auxiliary results for the benefit of the reader.

######  Definition 6.

Let $(X,Y)$ be an interpolation pair. By $\mathcal{F}(X,Y)$ we denote the set
of functions $f:S=\\{z\in{\mathbb{C}}:0\leq Re(z)\leq 1\\}\mapsto X+Y$ such
that

$\begin{array}[]{ll}\mathcal{F}\text{-}(i)&f\text{ is bounded and continuous
in }S,\\\ \mathcal{F}\text{-}(ii)&f\text{ is analytic in }S^{\circ},\\\
\mathcal{F}\text{-}(iii)&f(it)\in X\text{ and }f(1+it)\in Y\text{ for all
}t\in{\mathbb{R}}.\\\ \mathcal{F}\text{-}(iv)&\text{functions }t\mapsto
f(it)\text{ and t }\mapsto f(1+it)\text{ are bounded and continuous with
respect to the spaces }\\\ &X\text{ and }Y,\text{ respectively.}\\\
\end{array}$

We provide the space $\mathcal{F}(X,Y)$ with the norm

$||f||_{\mathcal{F}(X,Y)}=\max\\{\sup_{t\in{\mathbb{R}}}||f(it)||_{X},\sup_{t\in{\mathbb{R}}}||f(1+it)||_{Y}\\}.$

######  Definition 7.

Let $(X,Y)$ be an interpolation pair. We define the space

$[X,Y]_{\theta}=\\{f(\theta):f\in\mathcal{F}(X,Y)\\},$

with the norm

$||\phi||_{\theta}=\inf\\{||f||_{\mathcal{F}(X,Y)}:f(\theta)=\phi\\}.$ (54)

Before giving the proof of Theorem 4 we state the following useful result,
where we have considered $X=L^{2}(0,1)$ and
$Y=\widetilde{\mathcal{D}}_{\alpha}$.

######  Lemma 6.

Let $u\in L^{2}(0,1)$ and consider its unique decomposition in
$L^{2}(0,1)+H_{0}^{1+\alpha}(0,1)$, i.e., $u=u_{p}+u_{o}$ where $u_{p}\in
H_{0}^{1+\alpha}(0,1),u_{o}\in\left(H_{0}^{1+\alpha}(0,1)\right)^{\perp}$. Let
the norms on $L^{2}(0,1)$ given by
$||u||_{1}=||u_{o}||_{L^{2}(0,1)}+||u_{p}||_{H^{1+\alpha}(0,1)}$ and
$||u||_{2}=||u||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}.$ Then the
norms $||\cdot||_{1}$ and $||\cdot||_{2}$ are equivalent on $L^{2}(0,1)$.

###### Proof.

Let $u\in L^{2}(0,1)$. Since
$H_{0}^{1+\alpha}(0,1)\subseteq\widetilde{\mathcal{D}}_{\alpha}$, we deduce
that

$\begin{split}||u||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}&=\inf\\{||u_{1}||_{L^{2}(0,1)}+||u_{2}||_{\widetilde{\mathcal{D}}_{\alpha}}:u=u_{1}+u_{2},u_{1}\in
L^{2}(0,1),u_{2}\in\widetilde{\mathcal{D}}_{\alpha}\\}\\\
&\leq||u_{o}||_{L^{2}(0,1)}+||u_{p}||_{\widetilde{\mathcal{D}}_{\alpha}}=||u_{o}||_{L^{2}(0,1)}+||u_{p}||_{H^{1+\alpha}(0,1)}.\end{split}$
(55)

Thus, $\forall u\in L^{2}(0,1)$, $||u||_{2}\leq||u||_{1}$, and by [5,
Corollary 2.8] we conclude that there exists $c>0$ such that

$||u||_{1}\leq c||u||_{2},\quad\forall u\in L^{2}(0,1),$ (56)

and the thesis holds.∎

###### Proof of Theorem 4.

Let $g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})$. We define
a function
$G:\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})\mapsto\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))$
by $(Gg)(z)=\widetilde{g}(z)$ such that
$\widetilde{g}(z)(x)={\left.\kern-1.2ptg(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}(x)$.

We note that $H_{0}^{1+\alpha}(0,1)$ is a closed subspace of $L^{2}(0,1)$.
Then, by [27, Theorem 12.4], $\forall u\in L^{2}(0,1)$, there exists unique
$u_{p}\in
H_{0}^{1+\alpha}(0,1),u_{o}\in\left(H_{0}^{1+\alpha}(0,1)\right)^{\perp}$ such
that $u=u_{p}+u_{o}$.

We will prove first that $G$ is well defined. That is, we need to prove that
$G(g)=\tilde{g}$ verifies the conditions in Definition 6.

  * $\mathcal{F}$-(i)

Let $z,z_{0}\in S$ and consider the orthogonal representation for $g$ to
obtain,

$\widetilde{g}(z)-\widetilde{g}(z_{0})=[{\left.\kern-1.2ptg_{o}(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptg_{o}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}]+[{\left.\kern-1.2ptg_{p}(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptg_{p}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}].$
(57)

Then,

$\begin{split}||\widetilde{g}(z)-\widetilde{g}(z_{0})||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}&\leq||g_{o}(z)-g_{o}(z_{0})||_{L^{2}(0,1)}+||g_{p}(z)-g_{p}(z_{0})||_{H^{1+\alpha}(0,1)}.\end{split}$
(58)

By Lemma 6, there exists $c>0$, which does not depend on $z$ or $z_{0}$, such
that

$||g_{o}(z)-g_{o}(z_{0})||_{L^{2}(0,1)}+||g_{p}(z)-g_{p}(z_{0})||_{H^{1+\alpha}(0,1)}\leq
c||g(z)-g(z_{0})||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}.$ (59)

And from (58) and (59), we deduce that

$||\widetilde{g}(z)-\widetilde{g}(z_{0})||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}\leq
c||g(z)-g(z_{0})||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}.$

Since $g$ is continuous in $S$, we conclude that $\widetilde{g}$ is continuous
in $S$.

Anagolously, we can deduce that

$||\widetilde{g}(z)||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}\leq||g(z)||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}$
(60)

which implies that $\widetilde{g}$ is bounded in $S$, since $g$ is bounded in
$S$.

  * $\mathcal{F}$-(ii)

Let $z_{0}\in S^{\circ}$. We know that $g$ is analytic in $S^{\circ}$. Then,
there exists $g^{\prime}(z_{0})$ such that

$\left|\left|\frac{g(z_{0}+h)-g(z_{0})}{h}-g^{\prime}(z_{0})\right|\right|_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}\rightarrow
0,\quad\text{ when }z\rightarrow z_{0}.$

A candidate for $\widetilde{g}^{\prime}(z_{0})$ is

$j(z_{0})={\left.\kern-1.2ptg^{\prime}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}.$

In fact, by Lemma 6 there exists $c>0$ such that

$\begin{split}&\left|\left|\frac{\widetilde{g}(z_{0}+h)-\widetilde{g}(z_{0})}{h}-j(z_{0})\right|\right|_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}\\\
&\leq\left|\left|\frac{{\left.\kern-1.2ptg_{o}(z_{0}+h)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptg_{o}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}}{h}-{\left.\kern-1.2ptg_{o}^{\prime}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}\right|\right|_{L^{2}(\varepsilon,\omega)}+\left|\left|\frac{{\left.\kern-1.2ptg_{p}(z_{0}+h)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptg_{p}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}}{h}-{\left.\kern-1.2ptg_{p}^{\prime}(z_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}\right|\right|_{H^{1+\alpha}(\varepsilon,\omega)}\\\
&\leq
c\left|\left|\frac{g(z_{0}+h)-g(z_{0})}{h}-g^{\prime}(z_{0})\right|\right|_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}\end{split}$
(61)

and the last expression tends to zero as $z\rightarrow z_{0}$. Thus,
$\widetilde{g}$ is analytic in $S^{\circ}$.

  * $\mathcal{F}$-(iii)

For all $g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})$ and
$t\in{\mathbb{R}}$, we have

$\widetilde{g}(it)(x)={\left.\kern-1.2ptg(it)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}(x)\in
L^{2}(\varepsilon,\omega)$

and

$\widetilde{g}(1+it)(x)={\left.\kern-1.2ptg(1+it)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}(x).$

Since $g(1+it)\in\widetilde{\mathcal{D}}_{\alpha}$, there exists a unique
$w_{g,t}\in{{}_{0}}H^{1+\alpha}(0,1)$ such that

$g(1+it)=w_{g,t}-w_{g,t}(1)x^{\alpha}.$

Note that, $x^{\alpha}\in H^{1+\alpha}(\varepsilon,\omega)$ and thus
$g(1+it)\in H^{1+\alpha}(\varepsilon,\omega)$.

  * $\mathcal{F}$-(iv)

We will only prove that $t\mapsto\widetilde{g}(1+it)$ is bounded and
continuous with respect to $H^{1+\alpha}(\varepsilon,\omega)$, because this
part of the statement is more involved. Let us consider $\widetilde{g}(1+it)$.
By hypothesis $g(1+it)\in\widetilde{\mathcal{D}}_{\alpha}$, thus
$g(1+it)=w_{g,t}-w_{g,t}(1)x^{\alpha}$, for
$w_{g,t}\in{{}_{0}}H^{1+\alpha}(0,1)$. Hence,

$\begin{split}||\widetilde{g}(1+it)-\widetilde{g}(1+it_{0})||_{H^{1+\alpha}(\varepsilon,\omega)}&=||{\left.\kern-1.2ptg(1+it)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptg(1+it_{0})\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}||_{H^{1+\alpha}(\varepsilon,\omega)}\\\
&\leq||{\left.\kern-1.2ptw_{g,t}\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}-{\left.\kern-1.2ptw_{g,t_{0}}\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}||_{H^{1+\alpha}(\varepsilon,\omega)}+||(w_{g,t_{0}}(1)-w_{g,t}(1))x^{\alpha}||_{H^{1+\alpha}(\varepsilon,\omega)}\\\
&\leq||w_{g,t}-w_{g,t_{0}}||_{H^{1+\alpha}(0,1)}+c|w_{g,t_{0}}(1)-w_{g,t}(1)|\\\
&=||g(1+it)-g(1+it_{0})||_{\widetilde{\mathcal{D}}_{\alpha}}+c|w_{g,t_{0}}(1)-w_{g,t}(1)|.\\\
\end{split}$ (62)

Then, since $t\mapsto g(1+it)$ is continuous with respect to
$\widetilde{\mathcal{D}}_{\alpha}$, we deduce that

$||g(1+it)-g(1+it_{0})||_{\widetilde{\mathcal{D}}_{\alpha}}\rightarrow
0,\quad\text{ as }t\rightarrow t_{0},$

or equivalently

$||w_{g,t}-w_{g,t_{0}}||_{H^{1+\alpha}(0,1)}\rightarrow 0,\quad\text{ as
}t\rightarrow t_{0}.$

Since $H^{1+\alpha}(0,1)\hookrightarrow C([0,1])$, there exists $c>0$ such
that

$|w_{g,t}(1)-w_{g,t_{0}}(1)|\leq\sup_{x\in[0,1]}|w_{g,t}(x)-w_{g,t_{0}}(x)|\leq
c||w_{g,t}-w_{g,t_{0}}||_{H^{1+\alpha}(0,1)}\rightarrow 0,\quad\text{ as
}t\rightarrow t_{0}.$

Thus, we conclude that $t\mapsto\widetilde{g}(1+it)$ is continuous with
respect to $H^{1+\alpha}(\varepsilon,\omega)$. On the other hand,

$||\widetilde{g}(1+it)||_{H^{1+\alpha}(\varepsilon,\omega)}\leq
c||\widetilde{g}(1+it)||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}\leq
c||g(1+it)||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}\leq
c||g(1+it)||_{\widetilde{\mathcal{D}}_{\alpha}}.$

Then, since $t\mapsto g(1+it)$ is bounded with respect to
$\widetilde{\mathcal{D}}_{\alpha}$, we deduce that
$t\mapsto\widetilde{g}(1+it)$ is bounded with respect to
$H^{1+\alpha}(\varepsilon,\omega)$.

We can conclude now that
$\widetilde{g}\in\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))$
and hence, $G$ is well defined.

We will prove now that $G$ is surjective. Let
$h\in\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))$.
We define $g:S\mapsto L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}$ by
$g(z)=\eta{\left.\kern-1.2pt\rho
h(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(0,1)}}$, where $\eta$ is a
fixed function such that $\eta\in C^{\infty}_{0}(0,1)$, $\eta(x)=1$, $\forall
x\in(\varepsilon,\omega)$, and $\rho$ is the extension operator defined in
[16, Theorem 8.1]. Note that by [16, Theorem 9.1],
$\rho\in\mathcal{L}(H^{1+\alpha}(\varepsilon,\omega),H^{1+\alpha}({\mathbb{R}}))$.
For simplicity, we denote $\eta{\left.\kern-1.2pt\rho
h(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(0,1)}}=\eta\rho h(z)$.

By definition of $g$, we get immediately that
$(Gg)(z)={\left.\kern-1.2ptg(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}={\left.\kern-1.2pt\eta\rho
h(z)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}=h(z),\quad\forall
z\in S$. Thus, we need to prove that
$g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})$.

  * $\mathcal{F}$-(i)

Let $z\in S$. Note that, for every decomposition $h(z)=u_{1,z}+u_{2,z}$ with
$u_{1,z}\in L^{2}(\varepsilon,\omega),u_{2,z}\in
H^{1+\alpha}(\varepsilon,\omega)$, the following decomposition in
$L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}$ for $g(z)$ holds for all $z\in
S$

$g(z)=\eta\rho u_{1,z}(z)+\eta\rho u_{2,z}(z).$

Additionally, since
$\rho\in\mathcal{L}(L^{2}(\varepsilon,\omega),L^{2}({\mathbb{R}}))$

$||\eta\rho u_{1,z}||_{L^{2}(0,1)}\leq
c||u_{1,z}||_{L^{2}(\varepsilon,\omega)},\quad\text{ and }\quad||\eta\rho
u_{2,z}||_{\widetilde{\mathcal{D}}_{\alpha}}\leq
c||u_{2,z}||_{H^{1+\alpha}(\varepsilon,\omega)}.$ (63)

Thus, by (63), there exists $c>0$ such that

$||\eta\rho u_{1,z}||_{L^{2}(0,1)}+||\eta\rho
u_{2,z}||_{\widetilde{\mathcal{D}}_{\alpha}}\leq
c(||u_{1,z}||_{L^{2}(\varepsilon,\omega)}+||u_{2,z}||_{H^{1+\alpha}(\varepsilon,\omega)})$
(64)

for every decomposition $h(z)=u_{1,z}+u_{2,z}$. Then

$\begin{split}||g(z)||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}&=\inf\\{||w_{1,z}||_{L^{2}(0,1)}+||w_{2,z}||_{\widetilde{\mathcal{D}}_{\alpha}}:g(z)=w_{1,z}+w_{2,z},\,w_{1,z}\in
L^{2}(0,1),w_{2,z}\in\widetilde{\mathcal{D}}_{\alpha}\\}\\\
&\leq\inf\\{||\eta\rho u_{1,z}||_{L^{2}(0,1)}+||\eta\rho
u_{2,z}||_{\widetilde{\mathcal{D}}_{\alpha}}:h(z)=u_{1,z}+u_{2,z},\,u_{1,z}\in
L^{2}(\varepsilon,\omega),u_{2,z}\in H^{1+\alpha}(\varepsilon,\omega)\\}\\\
&\leq
c\inf\\{||u_{1,z}||_{L^{2}(\varepsilon,\omega)}+||u_{2,z}||_{H^{1+\alpha}(\varepsilon,\omega)}:h(z)=u_{1,z}+u_{2,z},\,u_{1,z}\in
L^{2}(\varepsilon,\omega),u_{2,z}\in H^{1+\alpha}(\varepsilon,\omega)\\}\\\
&=c||h(z)||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)},\end{split}$
(65)

and using the fact that
$h\in\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))$
we conclude that $g$ is bounded in $S$.

Similarly, for $z,z_{0}\in S$, we deduce that

$||g(z)-g(z_{0})||_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}\leq
c||h(z)-h(z_{0})||_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)},$
(66)

where the last expresion tends to $0$ when $z\rightarrow z_{0}$. Hence, $g$ is
continuous in $S$.

  * $\mathcal{F}$-(ii)

A candidate for $g^{\prime}(z_{0})$ is $j(z_{0})=\eta\rho(h^{\prime}(z_{0}))$.
In fact, from (64),

$\left|\left|\frac{g(z_{0}+\xi)-g(z_{0})}{\xi}-j(z_{0})\right|\right|_{L^{2}(0,1)+\widetilde{\mathcal{D}}_{\alpha}}\leq
c\left|\left|\frac{h(z_{0}+\xi)-h(z_{0})}{\xi}-h^{\prime}(z_{0})\right|\right|_{L^{2}(\varepsilon,\omega)+H^{1+\alpha}(\varepsilon,\omega)}$
(67)

and the last expresion tends to $0$ when $\xi\rightarrow 0$. Hence, $g$ is
analytic in $S^{\circ}$.

  * $\mathcal{F}$-(iii)

It is clear that $g(it)\in L^{2}(0,1)$ and
$g(1+it)\in\widetilde{\mathcal{D}}_{\alpha}$.

  * $\mathcal{F}$-(iv)

It is easy to prove that $t\mapsto g(it)$ and $t\mapsto g(1+it)$ are
continuous and bounded with respect to $L^{2}(0,1)$ and
$\widetilde{\mathcal{D}}_{\alpha}$ respectively, following the proof for
$\tilde{g}$.

We conclude that
$g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})$ and then, $G$
is surjective.

As before, for all $t\in{\mathbb{R}}$ we deduce that

$||(Gg)(it)||_{L^{2}(\varepsilon,\omega)}\leq||g(it)||_{L^{2}(0,1)}$

and

$||(Gg)(1+it)||_{H^{1+\alpha}(\varepsilon,\omega)}\leq
c||g(1+it)||_{\widetilde{\mathcal{D}}_{\alpha}}.$

Thus, there exists $c>0$ such that

$\max\\{\sup_{t\in{\mathbb{R}}}||Gg(it)||_{L^{2}(\varepsilon,\omega)},\sup_{t\in{\mathbb{R}}}||Gg(1+it)||_{H^{1+\alpha}(\varepsilon,\omega)}\\}\leq
c\max\\{\sup_{t\in{\mathbb{R}}}||g(it)||_{L^{2}(0,1)},\sup_{t\in{\mathbb{R}}}||g(1+it)||_{\widetilde{\mathcal{D}}_{\alpha}}\\},$

or equivalently

$||Gg||_{\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))}\leq
c||g||_{\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})}.$ (68)

We pass now to the interpolation spaces. By definition,

$H^{\delta}(\varepsilon,\omega)=[L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega)]_{\frac{\delta}{1+\alpha}}=\left\\{f\left(\frac{\delta}{1+\alpha}\right):f\in\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))\right\\},$
$[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\delta}{1+\alpha}}=\left\\{g\left(\frac{\delta}{1+\alpha}\right):g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})\right\\}.$

Let
$u\in[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\delta}{1+\alpha}}$.
Then, there exists
$g\in\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})$ such that
$g\left(\frac{\delta}{1+\alpha}\right)=u$.

We define
$\widetilde{u}={\left.\kern-1.2ptu\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}$.
Then
$\widetilde{u}={\left.\kern-1.2ptu\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}={\left.\kern-1.2ptg\left(\frac{\delta}{1+\alpha}\right)\mathchoice{\vphantom{\big{|}}}{}{}{}\right|_{(\varepsilon,\omega)}}=(Gg)\left(\frac{\delta}{1+\alpha}\right)=:\widetilde{g}\left(\frac{\delta}{1+\alpha}\right)$,
and by definition of $G$ it holds that
$\widetilde{g}\in\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))$.
Then,
$\widetilde{u}=\widetilde{g}\left(\frac{\delta}{1+\alpha}\right)\in[L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega)]_{\frac{\delta}{1+\alpha}}$,
and using (68) it holds that,

$\begin{split}||u||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\delta}{1+\alpha}}}&=\inf\left\\{||g||_{\mathcal{F}(L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha})}:g\left(\frac{\delta}{1+\alpha}\right)=u\right\\}\\\
&\geq
c\inf\left\\{||Gg||_{\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))}:(Gg)\left(\frac{\delta}{1+\alpha}\right)=\widetilde{u}\right\\}\\\
&=c\inf\left\\{||\widetilde{g}||_{\mathcal{F}(L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega))}:\widetilde{g}\left(\frac{\delta}{1+\alpha}\right)=\widetilde{u}\right\\}\\\
&=c||\widetilde{u}||_{[L^{2}(\varepsilon,\omega),H^{1+\alpha}(\varepsilon,\omega)]_{\frac{\delta}{1+\alpha}}}\\\
&=c||\widetilde{u}||_{H^{\delta}(\varepsilon,\omega)}.\end{split}$ (69)

In this way we complete the proof of Theorem 4. ∎

####  3.4.2 Regularity results

The statements of the following lemmas are analogous to [29, Lemma 1 and Lemma
2], but the adaptation of the proof for the latter needs to be done carefully,
taking into account some details related to estimations when working with the
interpolation spaces $[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\theta}$.

######  Lemma 7.

The mild solution $v$ obtained in Theorem 3 verifies that $v_{t}\in
L^{\infty}_{loc}(0,T;L^{2}(0,1))$ and

$v_{t}=\frac{1}{s^{1+\alpha}(t)}\frac{\partial}{\partial
x}D^{\alpha}v+x\frac{\dot{s}(t)}{s(t)}v_{x},\quad\text{ in }L^{2}(0,1)\text{
for all }t\in(0,T).$

We ommit the proof of Lemma 7 because it is analogous to the given in [29,
Lemma 1].

Below, we will focus on the case $\alpha>1/2$, since for smaller values of
$\alpha$, the proofs of the following results are more technical (see [29,
Lemma 3] and [25, Lemma 3.10]).

######  Lemma 8.

Let us assume that $\alpha\in(\frac{1}{2},1)$ and
$u_{\sigma}\in{}_{0}H^{\alpha}(0,1)$. We denote by $u$ the solution to the
equation

$\left\\{\begin{array}[]{ll}u_{t}=A(t)u&\textrm{ for
}0<x<1,0\leq\sigma<t<T,\\\ u(x,\sigma)=u_{\sigma}(x)&\textrm{ for }0<x<1,\\\
\end{array}\right.$ (70)

given by the evolution operator generated by the family $A(t)$. Then, for
every $0<\gamma<\alpha$, for every $0<\varepsilon<\omega<1$ there exists a
positive constant $c=c(\alpha,b,M,T,\varepsilon,\omega,\gamma)$, where $b,M$
comes from (29), such that for every $t\in(\sigma,T]$ there holds

$\left\|{A(t)u(\cdot,t)}\right\|_{H^{\gamma}(\varepsilon,\omega)}\leq
c(t-\sigma)^{-\frac{1+\gamma}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{}_{0}H^{\alpha}(0,1)}.$
(71)

###### Proof.

The proof follows the ideas of the one in [29, Lemma 2] but since the
characterization in [25, Eq. (3.17)] is not applicable to the interpolation
spaces under consideration, we will apply Theorem 4 several times.

Note that since $\alpha>\frac{1}{2}$ and $u_{\sigma}\in{}_{0}H^{\alpha}(0,1)$,
it follows that $u_{\sigma}\in H^{(1+\alpha)\delta}(0,1)$, for all
$\delta<\frac{1}{2(1+\alpha)}$.

On the other hand, by Proposition 8 to (70), it follows that there exists a
unique solution $u\in C([\sigma,T];L^{2}(0,1))\cap
C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})\cap
C^{1}((\sigma,T];L^{2}(0,1))$ such that $u(\cdot,t)=G(t,\sigma)u_{\sigma}$,
and from (38) it follows that

$||u(\cdot,t)||_{\widetilde{\mathcal{D}}_{\alpha}}\leq
c(t-\sigma)^{\delta-1}||u_{\sigma}||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\delta}}.$

We recall that the interpolation constant $c$ depends on the parameters of
interpolation as well as on $\alpha,T$ and $b,M$, that is
$c=c(\alpha,b,M,T,\delta)$. However, as before we neglect this dependency in
notation and leave it just in the final results. The same comment applies on
the parameters $\varepsilon$ and $\omega$ and we note that the constant $c>0$
may change from line to line.

We fix $0<\varepsilon<\omega<1$ and we set $\omega_{*}=\frac{1+\omega}{2}$.
Let us discuss the approximate problem. We choose a sequence
$\\{\varphi^{k}\\}\subseteq C_{0}^{\infty}(0,1)$ such that

$\\{\varphi^{k}\\}\subseteq\widetilde{\mathcal{D}}_{\alpha},\hskip
5.69046pt\varphi^{k}\rightarrow u_{\sigma}\mbox{ in
}{}_{0}H^{\alpha}(0,\omega_{*})\mbox{ and }\varphi^{k}\rightarrow
u_{\sigma}\mbox{ in }H^{\bar{\gamma}}(0,1)\mbox{ for every
}\bar{\gamma}<\frac{1}{2}.$ (72)

See [28, Lemma 4.4] for the existence proof.

Consider now the following approximate problem

$\left\\{\begin{array}[]{ll}u^{k}_{t}=A(t)u^{k}&\textrm{ for }\hskip
5.69046pt0<x<1,\hskip 5.69046pt0\leq\sigma<t<T,\\\
u^{k}(x,\sigma)=\varphi^{k}(x)&\textrm{ for }\hskip 5.69046pt0<x<1,\\\
\end{array}\right.$ (73)

Then, by [17, Corollary 6.2.4] there exists a unique solution $u^{k}\in
C([\sigma,T];L^{2}(0,1))\cap
C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})\cap
C^{1}((\sigma,T];L^{2}(0,1))$, and $u^{k}(\cdot,t)=G(t,\sigma)\varphi^{k}$.
Hence, by Proposition 7, it follows that $u^{k}\in
C([\sigma,T];\widetilde{\mathcal{D}}_{\alpha})\cap
C^{1}([\sigma,T];L^{2}(0,1))$

Let $u,u^{k}$ be the solutions of (70) and (73) respectively. Thus, by (40)
and (44) it follows that, for all
$0<\bar{\gamma}<\bar{\gamma_{1}}<\frac{1}{2}$,

$||A(t)(u-u^{k})||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\overline{\gamma}}{1+\alpha}}}\leq
c(t-\sigma)^{-1-\frac{\overline{\gamma}-\overline{\gamma}_{1}}{1+\alpha}}||u_{\sigma}-\varphi^{k}||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\overline{\gamma_{1}}}{1+\alpha}}}\leq
c(t-\sigma)^{-1-\frac{\overline{\gamma}-\overline{\gamma}_{1}}{1+\alpha}}||u_{\sigma}-\varphi^{k}||_{H^{\overline{\gamma_{1}}}(0,1)}$
(74)

and the last term tends to $0$ when $k\rightarrow\infty$. Hence, for almost
all $t\in(\sigma,T]$

$A(t)u^{k}(\cdot,t)\rightarrow A(t)u(\cdot,t),\quad\text{ in
}[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\overline{\gamma}}{1+\alpha}}\quad\forall\overline{\gamma}<\frac{1}{2}.$

Then, taking into account (29), for every $0\leq\bar{\gamma}<\frac{1}{2}$ and
almost all $t\in(\sigma,T]$

$\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)\rightarrow\frac{\partial}{\partial
x}D^{\alpha}u(\cdot,t)\hskip 5.69046pt\mbox{ in }\hskip
5.69046pt[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\overline{\gamma}}{1+\alpha}}.$
(75)

Applying (40) first, and subsequently (44) and (72), for $k$ large enough and
every $0\leq\bar{\gamma}<\bar{\gamma_{1}}<\frac{1}{2}$ we have

$||A(t)u^{k}(\cdot,t)||_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\overline{\gamma}}{1+\alpha}}}\leq
c(t-\sigma)^{-1-\frac{\overline{\gamma}-\overline{\gamma}_{1}}{1+\alpha}}||\varphi^{k}||_{H^{\overline{\gamma}_{1}}(0,1)}\leq
c(t-\sigma)^{-1-\frac{\overline{\gamma}-\overline{\gamma}_{1}}{1+\alpha}}||u_{\sigma}||_{H^{\overline{\gamma}_{1}}(0,1)}.$
(76)

In order to obtain $k$ \- independent local estimate of $A(t)u^{k}(\cdot,t)$
in more regular space we introduce a smooth non-negative cutoff function
$\eta$ such that $\eta\equiv 0$ on
$[0,\frac{\varepsilon}{2}]\cup[\omega_{*},1]$, and $\eta\equiv 1$ on
$[\varepsilon,\omega]$. At first, we apply $\partial^{\alpha}$ to (73) and
multiply the resulting identity by $\eta$.

Applying Proposition 6 we arrive at

$\eta\partial^{\alpha}\frac{\partial}{\partial
x}D^{\alpha}u^{k}=-\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha-1}(\eta(x)-\eta(p))\frac{\partial}{\partial
x}D^{\alpha}u^{k}(p){\rm d}p+\partial^{\alpha}(\frac{\partial}{\partial
x}D^{\alpha}u^{k}\cdot\eta)$
$=-\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha-1}(\eta(x)-\eta(p))\frac{\partial}{\partial
x}D^{\alpha}u^{k}(p){\rm d}p+\partial^{\alpha}\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})-\partial^{\alpha}(\eta^{\prime}D^{\alpha}u^{k}).$

Since $u^{k}(0,t)=0$ for every $t\in(0,T]$, from (6) we have

$\partial^{\alpha}\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(x,t)=\partial^{\alpha}\frac{\partial}{\partial
x}(\eta\partial^{\alpha}u^{k}(x,t)).$

In view of identity (9) we obtain that $\\{u^{k}\\}$ satisfy the system of
equations

$\left\\{\begin{array}[]{ll}(\partial^{\alpha}u^{k}\cdot\eta)_{t}-A(t)(\partial^{\alpha}u^{k}\cdot\eta)=F^{k}&\textrm{
for }0<x<1,\hskip 5.69046pt0\leq\sigma<t<T,\\\
(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,\sigma)=\partial^{\alpha}\varphi^{k}\cdot\eta&\textrm{
for }0<x<1,\\\ \end{array}\right.$ (77)

where

$F^{k}:=\frac{1}{s^{1+\alpha}(t)}\left[\frac{-\alpha}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha-1}(\eta(x)-\eta(p))\frac{\partial}{\partial
x}D^{\alpha}u^{k}(p,t)dp-\partial^{\alpha}(\eta^{\prime}D^{\alpha}u^{k})\right].$

Our aim now is to apply Proposition 8 to problem (77). We need to prove that:

  1. (i)

$F^{k}\in C((\sigma,T];L^{2}(0,1))\cap L^{1}(\sigma,T;L^{2}(0,1))$,

  2. (ii)

$\partial^{\alpha}\varphi^{k}\cdot\eta\in\widetilde{\mathcal{D}}_{\alpha}$,

  3. (iii)

$\partial^{\alpha}u^{k}\cdot\eta\in C([\sigma,T];L^{2}(0,1))\cap
C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})\cap
C^{1}((\sigma,T];L^{2}(0,1))$.

At first, we note that for every $x,p\in[0,1]$, $x\neq p$ we have

$\left|{\frac{\eta(x)-\eta(p)}{x-p}}\right|\leq\left\|{\eta}\right\|_{W^{1,\infty}(0,1)},$
(78)

hence,

$\frac{1}{\Gamma(1-\alpha)}\left|{\int_{0}^{x}(x-p)^{-\alpha-1}(\eta(x)-\eta(p))A(t)u^{k}(p,t){\rm
d}p}\right|\leq\left\|{\eta}\right\|_{W^{1,\infty}(0,1)}I^{1-\alpha}\left|{A(t)u^{k}(x,t)}\right|.$

Since $I^{1-\alpha}$ is bounded on $L^{2}(0,1)$ we obtain

$\left\|{\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{x}(x-p)^{-\alpha-1}(\eta(x)-\eta(p))A(t)u^{k}(p,t){\rm
d}p}\right\|_{L^{2}(0,1)}\leq
c\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}.$ (79)

By Proposition 4 and (29), we may write

$\begin{split}&\frac{1}{s^{1+\alpha}(t)}\left\|{\partial^{\alpha}(\eta^{\prime}D^{\alpha}u^{k})}\right\|_{L^{2}(0,1)}\leq\frac{c}{s^{1+\alpha}(t)}\left\|{(\eta^{\prime}D^{\alpha}u^{k})}\right\|_{{}_{0}H^{\alpha}(0,1)}\leq\frac{c}{s^{1+\alpha}(t)}\left\|{(\eta^{\prime}D^{\alpha}u^{k})}\right\|_{{}_{0}H^{1}(0,1)}\\\
&\leq\frac{c}{s^{1+\alpha}(t)}\left\|{(\eta^{\prime\prime}D^{\alpha}u^{k})}\right\|_{L^{2}(0,1)}+\frac{c}{s^{1+\alpha}(t)}\left\|{(\eta^{\prime}\frac{\partial}{\partial
x}D^{\alpha}u^{k})}\right\|_{L^{2}(0,1)}\\\ &\leq
c\left\|{D^{\alpha}u^{k}}\right\|_{L^{2}(0,1)}+c\left\|{A(t)u^{k}}\right\|_{L^{2}(0,1)}\\\
\end{split}$ (80)

Additionally, taking into account that $u^{k}=w^{k}-w^{k}(1,t)x^{\alpha}$,
with $w^{k}(\cdot,t)\in{{}_{0}}H^{1+\alpha}(0,1)$, applying Proposition 4 and
[16, Theorem 9.8], we have

$\begin{split}\left\|{D^{\alpha}u^{k}}\right\|_{L^{2}(0,1)}&=\left\|{\partial^{\alpha}u^{k}}\right\|_{L^{2}(0,1)}\leq\left\|{\partial^{\alpha}w^{k}}\right\|_{L^{2}(0,1)}+|w^{k}(1,t)|\left\|{\partial^{\alpha}x^{\alpha}}\right\|_{L^{2}(0,1)}\\\
&\leq
c\left\|{w^{k}}\right\|_{{{}_{0}}H^{\alpha}(0,1)}+c\left\|{w^{k}}\right\|_{C([0,1])}\leq
c\left\|{w^{k}}\right\|_{{{}_{0}}H^{1}(0,1)}+c\left\|{w^{k}}\right\|_{C([0,1])}\\\
&=c\left\|{w^{k}}\right\|_{{{}_{0}}H^{1+\alpha}(0,1)}+c\left\|{w^{k}}\right\|_{C([0,1])}\leq
c||u^{k}||_{\widetilde{\mathcal{D}}_{\alpha}}.\end{split}$ (81)

Hence, by (80), (81) and Lemma 4 we deduce that

$\frac{1}{s^{1+\alpha}(t)}\left\|{\partial^{\alpha}(\eta^{\prime}D^{\alpha}u^{k})}\right\|_{L^{2}(0,1)}\leq
c\left\|{A(t)u^{k}}\right\|_{L^{2}(0,1)}.$ (82)

Combining (79), (82), and (76) with $\overline{\gamma}=0$, we obtain that for
every $0<\gamma<\frac{1}{2}$

$\left\|{F^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}\leq
c\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}\leq
c(t-\sigma)^{\frac{\gamma}{(1+\alpha)}-1}\left\|{u_{\sigma}}\right\|_{H^{\gamma}(0,1)}$
(83)

and hence $F^{k}\in L^{1}(\sigma,T;L^{2}(0,1))$.

Recalling that $u^{k}\in C([\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$, from
definition of $F^{k}$, it follows that $F^{k}\in C((\sigma,T];L^{2}(0,1))$. In
fact, for $0<\tau<t\leq T$, we have that

$\begin{split}\left\|{F^{k}(\cdot,t)-F^{k}(\cdot,\tau)}\right\|_{L^{2}(0,1)}&\leq\alpha\left\|{\eta}\right\|_{W^{1,\infty}(0,1)}\left\|{I^{1-\alpha}[A(t)u^{k}(\cdot,t)-A(\tau)u^{k}(\cdot,\tau)]}\right\|_{L^{2}(0,1)}\\\
&\quad+\left\|{\partial^{\alpha}\left(\frac{1}{s^{1+\alpha}(\tau)}(\eta^{\prime}D^{\alpha}u^{k})(\cdot,\tau)-\frac{1}{s^{1+\alpha}(t)}(\eta^{\prime}D^{\alpha}u^{k})(\cdot,t)\right)}\right\|_{L^{2}(0,1)}\end{split}$
(84)

The first term tends to zero because fractional integral is bounded in $L^{2}$
and $A(\cdot)u^{k}\in C([\sigma,T];L^{2}(0,1))$. Furthermore, since $u^{k}\in
C([\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$, we have
$\eta^{\prime}D^{\alpha}u^{k}\in C([\sigma,T];{}_{0}H^{1}(0,1))$. In
particular, taking into account (29),
$\frac{1}{s^{1+\alpha}(t)}\eta^{\prime}D^{\alpha}u^{k}\in
C([\sigma,T];L^{2}(0,1))$. Then, by Lemma 4 and (78), we deduce that $F^{k}\in
C((\sigma,T];L^{2}(0,1))$, and (i) holds.

To verify (ii) we proceed as follows. Since $\varphi^{k}\in
C_{0}^{\infty}(0,1)\subseteq{{}_{0}}H^{1+2\alpha}(0,1)$, it follows that
$\partial^{\alpha}\varphi^{k}=\frac{\partial}{\partial
x}I^{1-\alpha}\varphi^{k}\in H^{1+\alpha}(0,1)$ and then,
$\partial^{\alpha}\varphi^{k}\cdot\eta\in
H_{0}^{1+\alpha}(0,1)\subseteq\widetilde{\mathcal{D}}_{\alpha}$.

Finally, we will prove (iii). From (6) we have
$\partial^{\alpha}u^{k}=D^{\alpha}u^{k}$. Thus, it is enough to show that
$D^{\alpha}u^{k}\cdot\eta\in C([\sigma,T];L^{2}(0,1))\cap
C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$. Recall that
$D^{\alpha}u^{k}\cdot\eta\in C([\sigma,T];L^{2}(0,1))$. Let us show that
$D^{\alpha}u^{k}\cdot\eta\in C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$.

We note that for any $0<\beta<1+\alpha$

$\left\|{\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,t)}\right\|_{H^{\beta}(0,1)}\leq\left\|{\eta^{\prime}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}(0,1)}+\left\|{\eta\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}(0,1)}.$

Applying Theorem 4 and estimate (40) we obtain that

$\begin{split}\left\|{\eta\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}(0,1)}&\leq\left\|{\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq\left\|{\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\beta}{1+\alpha}}}\leq
c(t-\sigma)^{-\frac{\beta}{1+\alpha}}\left\|{\varphi^{k}}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}.\end{split}$
(85)

Moreover,

$\left\|{\eta^{\prime}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}(0,1)}\leq
c\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}\left(\frac{\varepsilon}{2},\omega^{*}\right)}.$

Besides, if $\beta\in(0,1]$,

$\begin{split}\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}\left(\frac{\varepsilon}{2},\omega^{*}\right)}&\leq
c\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{1}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{1}(0,1)}\\\ &\leq
c\left(\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}+\left\|{\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}\right).\\\ \end{split}$ (86)

Furthermore,

$\begin{split}\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}&\leq\left\|{\partial^{\alpha}w^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}+c\left\|{w^{k}}\right\|_{C([0,1])}\\\
&\leq
c\left\|{w^{k}(\cdot,t)}\right\|_{{{}_{0}}H^{1+\alpha}(0,1)}=c\left\|{u^{k}(\cdot,t)}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}\leq
c\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}.\\\ \end{split}$ (87)

Hence,

$\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)},\quad 0<\beta\leq 1.$ (88)

Consider now $\beta\in(1,1+\alpha)$. Since $\beta-1\in(0,1)$, by (85) and
(88), it follows that

$\begin{split}\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta}\left(\frac{\varepsilon}{2},\omega^{*}\right)}&=\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{L^{2}\left(\frac{\varepsilon}{2},\omega^{*}\right)}+\left\|{\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta-1}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\\\
&\leq\left\|{D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta-1}\left(\frac{\varepsilon}{2},\omega^{*}\right)}+c\left\|{\eta\frac{\partial}{\partial
x}D^{\alpha}u^{k}(\cdot,t)}\right\|_{H^{\beta-1}(0,1)}\\\ &\leq
c\left(\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}+(t-\sigma)^{-\frac{\beta-1}{1+\alpha}}\left\|{\varphi^{k}}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}\right),\end{split}$
(89)

Thus, by Lemma (4), (85), (88) and (89), we deduce

$\left\|{\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,t)}\right\|_{H^{\beta}(0,1)}\leq
c\left(\left\|{A(t)u^{k}(\cdot,t)}\right\|_{L^{2}(0,1)}+(t-\sigma)^{-\frac{\beta}{1+\alpha}}\left\|{\varphi^{k}}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}\right),$
(90)

which implies

$\frac{\partial}{\partial x}(\eta D^{\alpha}u^{k})\in
L^{\infty}_{loc}((\sigma,T];H^{\beta}(0,1))\mbox{ for every
}0<\beta<1+\alpha.$

Taking into account the definition of $\eta$, and the fact that $||\eta
D^{\alpha}u^{k}||_{\widetilde{\mathcal{D}}_{\alpha}}=||\eta
D^{\alpha}u^{k}||_{{{}_{0}}H^{1+\alpha}(0,1)}$ we deduce that

$\left\|{\eta D^{\alpha}u^{k}(\cdot,t)-\eta
D^{\alpha}u^{k}(\cdot,\tau)}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}^{2}=\left\|{\eta
D^{\alpha}u^{k}(\cdot,t)-\eta
D^{\alpha}u^{k}(\cdot,\tau)}\right\|_{{}_{0}H^{1+\alpha}(0,1)}^{2}$
$=\left\|{\eta D^{\alpha}u^{k}(\cdot,t)-\eta
D^{\alpha}u^{k}(\cdot,\tau)}\right\|_{L^{2}(0,1)}^{2}+\left\|{\frac{\partial}{\partial
x}(\eta D^{\alpha}u^{k})(\cdot,t)-\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,\tau)}\right\|_{{}_{0}H^{\alpha}(0,1)}^{2}.$ (91)

Since $D^{\alpha}u^{k}\cdot\eta\in C([\sigma,T];L^{2}(0,1))$, it follows that
the first norm in (91) tends to zero as $\tau\rightarrow t$.

For the last term in (91), we use the characterization [25, (3.17)] and [17,
Corollary 1.2.7.] to deduce that

$\left\|{\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,t)-\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,\tau)}\right\|_{{}_{0}H^{\alpha}(0,1)}^{2}$ $\leq
c\left\|{\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,t)-\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,\tau)}\right\|_{{}_{0}H^{\beta}(0,1)}^{\frac{\alpha}{\beta}}\left\|{\frac{\partial}{\partial
x}(\eta D^{\alpha}u^{k})(\cdot,t)-\frac{\partial}{\partial x}(\eta
D^{\alpha}u^{k})(\cdot,\tau)}\right\|_{L^{2}(0,1)}^{1-\frac{\alpha}{\beta}}$

for every $\alpha<\beta<1+\alpha$. We note that, by (90), for every
$\sigma<\tau,t\leq T$ the first norm is bounded while the second tends to zero
as $\tau\rightarrow t$ because $u^{k}\in
C([\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$. Hence,
$D^{\alpha}u^{k}\cdot\eta\in C((\sigma,T];\widetilde{\mathcal{D}}_{\alpha})$.

Furthermore, $D^{\alpha}u^{k}\cdot\eta\in C^{1}((\sigma,T];L^{2}(0,1))$.
Indeed, $A(t)u^{k}\in C((\sigma,T];L^{2}(0,1))$ and by (85) $A(t)u^{k}\in
L^{\infty}_{loc}((\sigma,T];H^{\beta}(0,1))$ for every $\beta\in(0,1+\alpha)$.
Thus, applying again the interpolation estimate, in particular we obtain that
$A(t)u^{k}\in C((\sigma,T];H^{1}(0,1))$ and hence $u^{k}_{t}\in
C((\sigma,T];H^{1}(0,1))$ which implies
$\partial^{\alpha}u^{k}_{t}\cdot\eta=D^{\alpha}u^{k}_{t}\cdot\eta\in
C((\sigma,T];L^{2}(0,1))$, and condition (iii) is verified.

We can apply now Proposition 8 to problem (77) to obtain that
$\partial^{\alpha}u^{k}\cdot\eta$ satisfies the integral identity

$(\partial^{\alpha}u^{k}\cdot\eta)(x,t)=G(t,\sigma)(\partial^{\alpha}\varphi^{k}\cdot\eta)(x)+\int_{\sigma}^{t}G(t,\tau)F^{k}(x,\tau)d\tau.$
(92)

We fix $\gamma_{0}\in(0,1+\alpha)$. Since $\partial^{\alpha}u^{k}\cdot\eta\in
L^{2}(0,1)$ and $F^{k}\in L^{1}(\sigma,T;L^{2}(0,1))$, by (39) and Theorem 4
it follows that

$\begin{split}\left\|{(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma_{0}}(0,1)}&=\left\|{(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma_{0}}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c\left\|{(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\gamma_{0}}{1+\alpha}}}\\\
&\leq
c\left\|{G(t,\sigma)(\partial^{\alpha}\varphi^{k}\cdot\eta)}\right\|_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\gamma_{0}}{1+\alpha}}}+c\int_{\sigma}^{t}\left\|{G(t,\tau)F^{k}(\cdot,\tau)}\right\|_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\gamma_{0}}{1+\alpha}}}d\tau\\\
&\leq
c(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{\partial^{\alpha}\varphi^{k}\cdot\eta}\right\|_{L^{2}(0,1)}+c\int_{\sigma}^{t}(t-\tau)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{F^{k}(\cdot,\tau)}\right\|_{L^{2}(0,1)}d\tau.\end{split}$
(93)

Using the estimate (83) we get that for every $0<\bar{\gamma}<\frac{1}{2}$
there holds

$\begin{split}\left\|{(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma_{0}}(0,1)}&\leq
c(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{\partial^{\alpha}\varphi^{k}}\right\|_{L^{2}(0,\omega_{*})}+c\int_{\sigma}^{t}(t-\tau)^{-\frac{\gamma_{0}}{1+\alpha}}(\tau-\sigma)^{\frac{\bar{\gamma}}{1+\alpha}-1}d\tau\left\|{u_{\sigma}}\right\|_{H^{\bar{\gamma}}(0,1)}\\\
&=c(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{\partial^{\alpha}\varphi^{k}}\right\|_{L^{2}(0,\omega_{*})}+c(t-\sigma)^{\frac{\bar{\gamma}-\gamma_{0}}{1+\alpha}}B\left(\frac{\bar{\gamma}}{1+\alpha},1-\frac{\gamma}{1+\alpha}\right)\left\|{u_{\sigma}}\right\|_{H^{\bar{\gamma}}(0,1)},\end{split}$
(94)

where $B$ is the Beta function.

By Proposition 4 and (72) we have
$\left\|{\partial^{\alpha}\varphi^{k}}\right\|_{L^{2}(0,\omega_{*})}\leq
c\left\|{\varphi^{k}}\right\|_{{{}_{0}}H^{\alpha}(0,\omega_{*})}\leq
c\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,\omega_{*})}$ for $k$ big
enough.

Hence,

$\begin{split}\left\|{(\partial^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma_{0}}(0,1)}\leq
c\left[(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,\omega_{*})}+(t-\sigma)^{\frac{\bar{\gamma}-\gamma_{0}}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{H^{\bar{\gamma}}(0,1)}\right]\end{split}$
(95)

Applying (6) and by [9, Proposition 2.1] we deduce that

$\left\|{(D^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma_{0}}(0,1)}\leq
c(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,1)}.$
(96)

Since $\eta$ is smooth, we may apply (95) with $\eta=\eta^{\prime}$ to obtain

$\left\|{(D^{\alpha}u^{k}\cdot\eta^{\prime})(\cdot,t)}\right\|_{H^{\gamma_{0}}(0,1)}\leq
c(t-\sigma)^{-\frac{\gamma_{0}}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,1)}.$
(97)

Then, combining (96) and (97), we deduce that for all $0<\gamma<\alpha$

$\begin{split}\left\|{\left(\frac{\partial}{\partial
x}D^{\alpha}u^{k}\cdot\eta\right)(\cdot,t)}\right\|_{H^{\gamma}(0,1)}&=\left\|{\frac{\partial}{\partial
x}(D^{\alpha}u^{k}\cdot\eta)(\cdot,t)}\right\|_{H^{\gamma}(0,1)}+\left\|{(D^{\alpha}u^{k}\cdot\eta^{\prime})(\cdot,t)}\right\|_{H^{\gamma}(0,1)}\\\
&\leq
c(t-\sigma)^{-\frac{1+\gamma}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,1)}.\end{split}$
(98)

Hence, by (29),

$\left\|{A(t)u^{k}(\cdot,t)}\right\|_{H^{\gamma}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c(t-\sigma)^{-\frac{1+\gamma}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{H^{\alpha}(0,1)}.$
(99)

Finally, from the weak lower semicontinuity of the norm, we deduce that

$\left\|{A(t)u(\cdot,t)}\right\|_{H^{\gamma}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c\left\|{A(t)u^{k}(\cdot,t)}\right\|_{H^{\gamma}\left(\frac{\varepsilon}{2},\omega^{*}\right)}\leq
c(t-\sigma)^{-\frac{1+\gamma}{1+\alpha}}\left\|{u_{\sigma}}\right\|_{{{}_{0}}H^{\alpha}(0,1)}$
(100)

where $c=c(\alpha,b,M,\varepsilon,\omega,T,\gamma)$. Since $\varepsilon$ and
$\omega$ are arbitrary numbers, (71) holds.

∎

Now we are able to increase the regularity of solutions to (30) in the
interior.

######  Lemma 9.

Let $\alpha\in\left(\frac{1}{2},1\right)$,
$v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$, and $v$ be the mild solution to
(30) obtained in Theorem 3. Then the following results hold:

  1. 1.

There exists $\beta>\frac{1}{2}$ such that

$v\in L^{\infty}_{loc}(0,T;H^{\beta+1+\alpha}_{loc}(0,1))\mbox{ and
}\partial^{\alpha}v_{x}\in L^{\infty}_{loc}(0,T;H^{\beta}_{loc}(0,1)).$ (101)

  2. 2.

Moreover, there exists $\beta>\frac{1}{2}$ such that

$v\in C((0,T];H^{\beta+1+\alpha}_{loc}(0,1))\mbox{ and
}\partial^{\alpha}v_{x}\in C((0,T];H^{\beta}_{loc}(0,1)).$ (102)

###### Proof.

1\. Let $\beta\in\left(\frac{1}{2},\alpha\right)$. By Definition 5

$v(x,t)=G(t,0)v_{0}(x)+\int_{0}^{t}G(t,\sigma)f(x,\sigma){\rm d}\sigma.$ (103)

In order to use Theorem 4, we introduce the parameters
$0<\varepsilon<\omega<1$. We apply the operator $A(t)$ to (103) and estimate
its norm

$\left\|{A(t)v}\right\|_{H^{\beta}(\varepsilon,\omega)}\leq\left\|{A(t)G(t,0)v_{0}}\right\|_{H^{\beta}(\varepsilon,\omega)}+\int_{0}^{t}\left\|{A(t)G(t,\sigma)f(\cdot,\sigma)}\right\|_{H^{\beta}(\varepsilon,\omega)}{\rm
d}\sigma.$ (104)

Let $\sigma\in[0,T]$ and denote $w(x,t):=G(t,\sigma)f(x,\sigma)$. Then,

$w_{t}(x,t)=\frac{\partial}{\partial
t}G(t,\sigma)f(x,\sigma)=A(t)G(t,\sigma)f(x,\sigma)=A(t)w(x,t)\quad\text{ and
}$ $w(x,\sigma)=G(\sigma,\sigma)f(x,\sigma)=f(x,\sigma).$

Hence, by Theorem 4, Lemma 8 and (40), it follows that

$\begin{split}\left\|{A(t)v}\right\|_{H^{\beta}(\varepsilon,\omega)}&\leq\left\|{A(t)G(t,0)v_{0}}\right\|_{H^{\beta}(\varepsilon,\omega)}+\int_{0}^{t}\left\|{A(t)G(t,\sigma)f(\cdot,\sigma)}\right\|_{H^{\beta}(\varepsilon,\omega)}{\rm
d}\sigma\\\ &\leq
c\left\|{A(t)G(t,0)v_{0}}\right\|_{[L^{2}(0,1),\widetilde{\mathcal{D}}_{\alpha}]_{\frac{\beta}{1+\alpha}}}+c\int_{0}^{t}(t-\sigma)^{-\frac{1+\beta}{1+\alpha}}\left\|{f(\cdot,\sigma)}\right\|_{{{}_{0}}H^{\alpha}(0,1)}{\rm
d}\sigma\\\ &\leq
ct^{-\frac{\beta}{1+\alpha}}\left\|{v_{0}}\right\|_{\widetilde{\mathcal{D}}_{\alpha}}+c\frac{1+\alpha}{\alpha-\beta}t^{\frac{\alpha-\beta}{1+\alpha}}\left\|{f}\right\|_{L^{\infty}(0,T;{{}_{0}}H^{\alpha}(0,1))}.\end{split}$
(105)

Hence, by (29) it follows that $\partial^{\alpha}v_{x}\in
L^{\infty}_{loc}(0,T;H^{\beta}_{loc}(0,1))$ for all
$\frac{1}{2}<\beta<\alpha$.

On the other hand, since $v(\cdot,t)\in\widetilde{\mathcal{D}}_{\alpha}$ for
every $t$,

$v(x,t)=w(x,t)-w(1,t)x^{\alpha},\quad w(\cdot,t)\in{{}_{0}}H^{1+\alpha}(0,1),$

and then, $w_{x}(\cdot,t)\in{{}_{0}}H^{\alpha}(0,1)$. Additionally, since
$\frac{\partial}{\partial x}D^{\alpha}x^{\alpha}=0$, it follows that
$A(t)v=A(t)w$, thus
$\partial^{\alpha}w_{x}(\cdot,t)=\partial^{\alpha}v_{x}(\cdot,t)\in
H^{\beta}_{loc}(0,1)$ for all $\frac{1}{2}<\beta<\alpha$. Hence, applying
Lemma 1, we deduce that $w_{x}\in
L^{\infty}_{loc}(0,T;H^{\beta+\alpha}_{loc}(0,1))$. Finally

$\begin{split}\left\|{v_{x}(\cdot,t)}\right\|_{H^{\beta+\alpha}(\varepsilon,\omega)}&\leq\left\|{w_{x}(\cdot,t)}\right\|_{H^{\beta+\alpha}(\varepsilon,\omega)}+\left\|{\alpha
w(1,t)x^{\alpha-1}}\right\|_{H^{\beta+\alpha}(\varepsilon,\omega)}\\\
\end{split}$ (106)

and taking into account that $x^{\alpha-1}\in
H^{\beta+\alpha}(\varepsilon,\omega)$, we deduce that $v_{x}\in
L^{\infty}_{loc}(0,T;H^{\beta+\alpha}_{loc}(0,1))$ and (101) holds.

2\. Theorem 3 states that $v\in C([0,T];\widetilde{\mathcal{D}}_{\alpha})$.
Since for arbitrary $0<\varepsilon<\omega<1$ and for every
$0<\overline{\beta}<\beta$ there holds

$H^{\overline{\beta}+\alpha+1}(\varepsilon,\omega)=[H^{1+\alpha}(\varepsilon,\omega),H^{\beta+\alpha+1}(\varepsilon,\omega)]_{\frac{\overline{\beta}}{\beta}},$

we may estimate by the interpolation theorem ([17, Corollary 1.2.7])

$\left\|{v(\cdot,t)-v(\cdot,\tau)}\right\|_{H^{\overline{\beta}+\alpha+1}(\varepsilon,\omega)}\leq
c\left\|{v(\cdot,t)-v(\cdot,\tau)}\right\|_{H^{1+\alpha}(\varepsilon,\omega)}^{1-\frac{\overline{\beta}}{\beta}}\left\|{v(\cdot,t)-v(\cdot,\tau)}\right\|_{H^{\beta+\alpha+1}(\varepsilon,\omega)}^{\frac{\overline{\beta}}{\beta}},$

where $c=c(\beta,\overline{\beta},\varepsilon)$. By Lemma 9-1, there exists
$\beta>\frac{1}{2}$ such that the second norm on the right hand side above is
bounded on every compact interval contained in $(0,T]$, and the first term
tends to zero because $v\in C([0,T];\widetilde{\mathcal{D}}_{\alpha})$. We
conclude that $v\in C((0,T];H^{\bar{\beta}+1+\alpha}_{loc}(0,1))$.

In order to obtain the claim for $\partial^{\alpha}v_{x}$ we note that,
applying again the interpolation theorem we obtain that for every
$0<\varepsilon<\omega<1$, $0<\tau<t\leq T$ and every
$0<\overline{\beta}<\beta$,

$\begin{split}\left\|{\partial^{\alpha}v_{x}(\cdot,t)-\partial^{\alpha}v_{x}(\cdot,\tau)}\right\|_{H^{\overline{\beta}}(\varepsilon,\omega)}\leq
c\left\|{\partial^{\alpha}v_{x}(\cdot,t)-\partial^{\alpha}v_{x}(\cdot,\tau)}\right\|_{L^{2}(0,1)}^{1-\frac{\overline{\beta}}{\beta}}\left\|{\partial^{\alpha}v_{x}(\cdot,t)-\partial^{\alpha}v_{x}(\cdot,\tau)}\right\|_{H^{\beta}(\varepsilon,\omega)}^{\frac{\overline{\beta}}{\beta}}.\end{split}$
(107)

Then, the first norm on the right hand side tends to zero as $\tau\rightarrow
t$, while the second one is bounded on every compact interval contained in
$(0,T]$ due to Lemma 9-1. Then, we may choose $\bar{\beta}>\frac{1}{2}$ such
that $\partial^{\alpha}v_{x}\in C((0,T];H^{\bar{\beta}}_{loc}(0,1))$. ∎

######  Corollary 3.

Let us assume that $v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$. Then, the
solution to (30) obtained in Theorem 3 satisfies that

$v\in C([0,1]\times[0,T]).$ (108)

###### Proof.

It is a direct consequence of the embedding
$\widetilde{\mathcal{D}}_{\alpha}\hookrightarrow C([0,1])$. ∎

######  Corollary 4.

Under the assumptions of Lemma 9 for every
$\alpha\in\left(\frac{1}{2},1\right)$ there exists $\beta>\frac{1}{2}$ such
that for every $0<\varepsilon<\omega<1$ there holds

$v_{x}\in C((0,T];H^{\alpha+\beta}(\varepsilon,\omega)).$

######  Remark 7.

The regularity in Corollary 4 is a fundamental tool for the next section,
where the maximum principle for the fractional diffusion equation (1)-$(i)$ is
used to claim the existence and uniqueness of the solution to the fractional
Stefan problem (1).

###  3.5 Existence of a classical solution to the moving boundary problem

We are able to prove existence and regularity of solutions to the moving
boundary problem (30).

######  Theorem 5.

Let $b>0$ and $\alpha\in\left(\frac{1}{2},1\right)$. Suppose that $s$
satisfies (29), and $v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$, where
$v_{0}(x)=u_{0}(bx)$. Then, there exists a unique solution $u$ to (28) such
that $u\in C(\overline{Q_{s,T}})$, $u_{t},\frac{\partial}{\partial
x}D^{\alpha}u\in C(Q_{s,T})$, and for all $t\in(0,T]$, $u(\cdot,t)\in
AC[0,s(t)]$, $u_{t}(\cdot,t),\frac{\partial}{\partial
x}D^{\alpha}u(\cdot,t)\in L^{2}(0,s(t))$. Furthermore, there exists
$\beta>\frac{1}{2}$ such that for all $t\in(0,T]$ and
$0<\varepsilon<\omega<s(t)$ we have $u_{x}(\cdot,t)\in
H^{\alpha+\beta}(\varepsilon,\omega)$.

###### Proof.

Let $v$ be the mild solution to (30) given by Theorem 3. By hypothesis
$v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$. Then, by Theorem 3, there exists a
unique solution $v\in C([0,T];\widetilde{\mathcal{D}}_{\alpha})$ to (36).
Since $v(\cdot,t)\in\widetilde{\mathcal{D}}_{\alpha}$ for all $t\in[0,T]$, $v$
satisfies (30)-($ii$). Moreover, by Lemma 7, $v$ verifies (30)-($i$). Hence,
$v$ is a solution to (30).

On the other hand, $v$ has the regularity given by Lemma 9 and Corolary 3. In
particular, $v\in C([0,1]\times[0,T])$. Moreover, by Corolary 4 there exists
$\beta>\frac{1}{2}$ such that, for $0<\varepsilon<\omega<1$, $v_{x}\in
C((0,T];H^{\alpha+\beta}(\varepsilon,\omega))$.

Define $u(x,t)=v\left(\frac{x}{s(t)},t\right)$. Then, $u$ is a solution to
(28). Taking into account that $v\in C([0,1]\times[0,T])$ and (29), it follows
that $u\in C(\overline{Q_{s,T}})$. On the other hand, since $v_{x}\in
C((0,T];H^{\alpha+\beta}(\varepsilon,\omega))$ for every
$0<\varepsilon<\omega<1$, we have $u_{x}(\cdot,t)\in
H^{\alpha+\beta}(\varepsilon,\omega)$ for all $t\in(0,T]$ and for every
$0<\varepsilon<\omega<s(t)$. Also, we know that $v\in
C([0,T],\widetilde{\mathcal{D}}_{\alpha})$. Then, we deduce that
$u(\cdot,t)\in AC[0,s(t)]$ for all $t\in[0,t]$.

We note that, by Proposition 4 we have that $\frac{\partial}{\partial
x}D^{\alpha}v(\cdot,t)=\partial^{\alpha}v_{x}(\cdot,t)\in L^{2}(0,1)$, for all
$t\in(0,T]$. Then, if we define $p=\frac{x}{s(t)}$,

$\frac{\partial}{\partial
x}D^{\alpha}u(x,t)=\frac{1}{s^{1+\alpha}(t)}\left(\frac{\partial}{\partial
p}D^{\alpha}v\right)(p,t),$

and by (29) we deduce that $u_{t}(\cdot,t),\frac{\partial}{\partial
x}D^{\alpha}u(\cdot,t)\in L^{2}(0,s(t))$, for all $t\in(0,T]$.

In addition, by Lemma 9-4 we have that $\partial^{\alpha}v_{x}\in
C((0,T];H^{\beta}_{loc}(0,1))$, from where, by the Sobolev embedding [16,
Chapter 9, Theorem 9.8] it follows that $\partial^{\alpha}v_{x}\in
C((0,1)\times(0,T])$. Hence, $\partial^{\alpha}u_{x}\in C(Q_{s,T})$ and we
conclude that $\frac{\partial}{\partial x}D^{\alpha}u,u_{t}\in C(Q_{s,T})$.

We prove now the uniqueness of solution. Let $u_{1},u_{2}$ be solutions to
(28) with the regularity given in the theorem. We define $u=u_{1}-u_{2}$.
Then, $u$ is a solution to (28) for $u_{0}\equiv 0$. Thus, multiplying by $u$,
integrating over $Q_{s,t_{0}}$ for $0<t_{0}<T$ the equation (28)-(i), and
applying Fubini’s Theorem, we obtain

$\begin{split}\int_{0}^{t_{0}}\int_{0}^{s(\tau)}u_{t}(x,\tau)\cdot
u(x,\tau){\rm d}x{\rm
d}\tau-\int_{0}^{t_{0}}\int_{0}^{s(\tau)}\frac{\partial}{\partial
x}D^{\alpha}u(x,\tau)\cdot u(x,\tau){\rm d}x{\rm d}\tau&=0\\\
\frac{1}{2}\int_{0}^{s(t_{0})}\int_{s^{-1}(x)}^{t_{0}}\frac{d}{dt}(u(x,\tau)^{2}){\rm
d}\tau{\rm d}x+\int_{0}^{t_{0}}\int_{0}^{s(\tau)}-\frac{\partial}{\partial
x}D^{\alpha}u(x,\tau)\cdot u(x,\tau){\rm d}x{\rm d}\tau&=0\\\
\frac{1}{2}\int_{0}^{s(t_{0})}|u(x,t_{0})|^{2}{\rm
d}x+\int_{0}^{t_{0}}\left(-\frac{\partial}{\partial
x}D^{\alpha}u(\cdot,t),u(\cdot,t)\right){\rm d}\tau&=0.\end{split}$ (109)

From (26) we infer that

$\left(-\frac{\partial}{\partial x}D^{\alpha}u(\cdot,t),u(\cdot,t)\right)\geq
c_{\alpha}||u(\cdot,t)||^{2}_{H^{\frac{1+\alpha}{2}}(0,s(t))},$ (110)

where $c_{\alpha}$ is a positive constants that depend on $\alpha$. Then,

$\begin{split}0&=\frac{1}{2}\int_{0}^{s(t_{0})}|u(x,t_{0})|^{2}{\rm
d}x+\int_{0}^{t_{0}}\left(-\frac{\partial}{\partial
x}D^{\alpha}u(\cdot,t),u(\cdot,t)\right){\rm d}\tau\geq 0\end{split}$ (111)

from where $\displaystyle\frac{1}{2}\int_{0}^{s(t_{0})}|u(x,t_{0})|^{2}{\rm
d}x=0$. Hence, since $u\in C(\overline{Q_{s,T}})$, it follows that
$u(\cdot,t_{0})=0$. Being $0<t_{0}<T$ an arbitrary number, it results that
$u\equiv 0$, and then $u_{1}\equiv u_{2}$.

∎

##  4 Existence of a classical solution to the Stefan Problem with Dirichlet
boundary condition

Let us return to the problem

$\begin{array}[]{lll}(i)&u_{t}-\frac{\partial}{\partial
x}D^{\alpha}u=0,&\text{ in }Q_{s,T}\\\
(ii)&u(0,t)=0,u(t,s(t))=0,&t\in(0,T),\\\ (iii)&u(x,0)=u_{0}(x),&0<x<s(0)=b,\\\
(iv)&\dot{s}(t)=-\lim\limits_{x\rightarrow
s(t)^{-}}(D^{\alpha}u)(x,t),&t\in(0,T).\end{array}$ (112)

######  Remark 8.

Note that, from [25, Proposition 2.9], if $u$ is the solution to (112) given
by Theorem 5 for a fixed function $s$, we deduce that $D^{\alpha}u(\cdot,t)\in
C(0,s(t)]$, for all $t\in(0,T]$, and then $\lim\limits_{x\rightarrow
s(t)^{-}}(D^{\alpha}u)(x,t)$ is well defined for all $t\in(0,T)$.

######  Theorem 6.

Let $u$ be a solution to the equation

${u}_{t}-\frac{\partial}{\partial x}D_{x}^{\alpha}u=f\quad\text{in }\,Q_{s,T}$

such that $u\in C(\overline{Q_{s,T}})$, $u_{t},\frac{\partial}{\partial
x}\,D_{x}^{\alpha}u\in C(Q_{s,T})$ and $u(\cdot,t)\in AC[0,s(t)]$ $\forall
t\in(0,t)$. Also suppose that $s$ verifies (29) and $u$ verifies the local
regularity condition

$\exists\,\beta>\frac{1}{2}\,\text{ such that, for every }t\in(0,T)\mbox{ and
every }0<\varepsilon<\omega<s(t),\hskip 5.69046ptu_{x}(\cdot,t)\in
H^{\alpha+\beta}(\varepsilon,\omega).$ (113)

Then $u$ attains its maximum on $\partial\Gamma_{s,T}$ if $f\leq 0$, and $u$
attains its minimum on $\partial\Gamma_{s,T}$ if $f\geq 0$.

The following results are analogous to [28, Lemma 4.14] and [28, Lemma 4.16]
but we recall them for the benefit of the reader.

######  Lemma 10.

Let $u$ be the solution to (28) given by Theorem 5, for a fixed function $s$.
Then $(D^{\alpha}u)(s(t),t)\leq 0$ for all $t\in(0,T]$. Furthermore, if
$u_{0}\not\equiv 0$, then for all $t\in(0,T]$, $(D^{\alpha}u)(s(t),t)<0$.

######  Lemma 11.

Let $u$ be the solution to (28) given by Theorem 5. Suppose that $s$ satisfies
(29), and $u_{0}$ verifies

$0\leq u_{0}(x)\leq\frac{M}{2\Gamma(1+\alpha)}(b^{\alpha}-x^{\alpha}),\quad
x\in[0,b].$ (114)

where $M$ is the constant in (29). Then,

$(D^{\alpha}u)(s(t),t)\geq-\frac{M}{2},\quad t\in(0,T),$ (115)

and

$0\leq
u(x,t)\leq\frac{M}{2\Gamma(1+\alpha)}(s^{\alpha}(t)-x^{\alpha}),\quad(x,t)\in
Q_{s,T}.$ (116)

We establish now an integral condition equivalent to (112)$-(iv)$.

######  Proposition 10.

Let $(u,s)$ be a solution to (112) having the regularity in Theorem 6. Then,
the condition (112)$-(iv)$ is equivalent to the following integral condition

$s^{2}(t)=b^{2}+2\int_{0}^{b}xu_{0}(x){\rm
d}x-2\int_{0}^{t}I^{1-\alpha}u(s(\tau),\tau){\rm
d}\tau-2\int_{0}^{s(t)}xu(x,t){\rm d}x.$ (117)

###### Proof.

Suppose that $u$ satisfies (28) and let us define $h(x)=\begin{cases}0&0<x\leq
b,\\\ s^{-1}(x)&x>b.\end{cases}$. Then, multiplying equation (28)$-(i)$ by $x$
and integrating in time from $h(x)$ to $t$ we get

$xu(x,t)-xu(x,h(x))=\int_{h(x)}^{t}x\frac{\partial}{\partial
x}D^{\alpha}u(x,\tau){\rm d}\tau$

Now we integrate in space, apply Fubini’s theorem and integrate by parts to
obtain

$\int_{0}^{s(t)}xu(x,t){\rm d}x-\int_{0}^{s(t)}xu(x,h(x)){\rm
d}x=\int_{0}^{s(t)}\int_{h(x)}^{t}x\frac{\partial}{\partial
x}D^{\alpha}u(x,\tau){\rm d}\tau{\rm d}x$ $\int_{0}^{s(t)}xu(x,t){\rm
d}x-\int_{0}^{b}xu(x,0){\rm
d}x=\int_{0}^{t}s(\tau)D^{\alpha}u(s(\tau),\tau){\rm
d}\tau-\int_{0}^{t}\int_{0}^{s(t)}D^{\alpha}u(x,\tau){\rm d}x{\rm d}\tau$

Finally, from (8) we deduce

$\int_{0}^{s(t)}xu(x,t){\rm d}x-\int_{0}^{b}xu(x,0){\rm
d}x=\int_{0}^{t}s(\tau)D^{\alpha}u(s(\tau),\tau){\rm
d}\tau-\int_{0}^{t}I^{1-\alpha}u(s(\tau),\tau){\rm d}\tau$ (118)

Now, if $(u,s)$ is a solution to the Stefan problem (112) then (117) easily
follows by placing (112)$-(iv)$ in (118). For the converse, suppose that the
pair $(u,s)$ verifies $\eqref{MBP}$ and $\eqref{CondInt2general}$ Then by
differentiating both sides in (117) and using (28)$-(ii)$ we get

$\begin{split}2s(t)\dot{s}(t)&=-2I^{1-\alpha}u(s(t),t)-2\int_{0}^{s(t)}xu_{t}(x,t){\rm
d}x=-2I^{1-\alpha}u(s(t),t)-2\int_{0}^{s(t)}x\frac{\partial}{\partial
x}D^{\alpha}u(x,t){\rm d}x\end{split}$ (119)

and the claim follows by integration by parts.

∎

Now we present the monotone dependence upon data result. It is analogous to
[25, Theorem 3], but we need to modify the estimates in the proof, due to the
new integral condition given in Proposition 10. In the proof we will use the
following generalized Gronwall’s Lemma.

######  Lemma 12.

[10, Theorem 2.1] Let $p>q\geq 0$, $x(t)$ and $h(t)$ be real-valued non-
negative continuous functions defined on $I=[0,\infty)$, and $n(t)$ be a
positive non-decreasing continuous function defined on $I$. If the following
inequality holds

$x^{p}(t)\leq n^{p}(t)+\int_{0}^{t}h(s)x^{q}(s)ds,\quad\forall t\in I,$

then

$x(t)\leq
n(t)\left[1+\frac{p-q}{p}\int_{0}^{t}h(s)n^{-(p-q)}(s)ds\right],\quad\forall
t\in I.$

######  Theorem 7.

Let $s_{i}$, $i=1,2$ verify (29), and $u^{i}$ be the solutions to (28) given
by Theorem 5, corresponding to $s_{i}$, $b_{i}$ and $u_{0}^{i}$ for $i=1,2$
respectively. If $b_{1}\leq b_{2}$ and $u_{0}^{1}\leq u_{0}^{2}$, then for
every $t\in[0,T]$ we have $s_{1}(t)\leq s_{2}(t)$.

###### Proof.

We consider two cases.

Case 1: $b_{1}<b_{2}$, $u_{0}^{1}\leq u_{0}^{2}$ and $u_{0}^{1}\not\equiv
u_{0}^{2}$ on $[0,b_{1}]$.

It is identical to the one in [25, Theorem 3].

Case 2: $b_{1}\leq b_{2}$, $u_{0}^{1}\leq u_{0}^{2}$.

Let $\delta>0$, and $u_{0}^{\delta}$ be a smooth funtion on $[0,b_{2}+\delta]$
such that

$u_{0}^{\delta}\equiv 0,\text{ on
}\left[b_{2}+\frac{\delta}{2},b_{2}+\delta\right],\quad u_{0}^{\delta}\geq
u_{0}^{2},\text{ on }[0,b_{2}],$
$\max_{x\in[0,b_{2}]}(u_{0}^{\delta}(x)-u_{0}^{2}(x))=\delta,\quad\max_{x\in\left[b_{2},b_{2}+\frac{\delta}{2}\right]}u_{0}^{\delta}(x)=\delta.$

We denote by $(u^{\delta},s_{\delta})$ a solution to (112) corresponding to
the data $b_{2}+\delta$ and $u_{0}^{\delta}$.

By the preceding case, we have that $s_{1}\leq s_{\delta}$ and $s_{2}\leq
s_{\delta}$. Moreover, by Proposition 10 we deduce that

$\begin{split}s_{\delta}^{2}(t)-s_{2}^{2}(t)&=2\delta
b_{2}+\delta^{2}+2\int_{b_{2}}^{b_{2}+\frac{\delta}{2}}xu_{0}^{\delta}(x){\rm
d}x+2\int_{0}^{b_{2}}x(u_{0}^{\delta}-u_{0}^{2})(x){\rm d}x\\\
&\quad+2\int_{0}^{t}[I^{1-\alpha}u^{2}(s_{2}(\tau),\tau)-I^{1-\alpha}u^{\delta}(s_{\delta}(\tau),\tau)]{\rm
d}\tau\\\ &\quad-2\int_{0}^{s_{2}(t)}x(u^{\delta}-u^{2})(x,t){\rm
d}x-2\int_{s_{2}(t)}^{s_{\delta}(t)}xu^{\delta}(x,t){\rm d}x\end{split}$ (120)

On the other hand, by Theorem 6, we have $u^{\delta}\geq 0$,
$u^{\delta}-u^{2}\geq 0$. Additionally,
$||u_{0}^{\delta}-u_{0}^{2}||_{L^{\infty}(0,b_{2})}=\delta$. Therefore,

$\begin{split}s_{\delta}^{2}(t)-s_{2}^{2}(t)&\leq 2\delta
b_{2}+\delta^{2}+\delta\left(b_{2}\delta+\frac{\delta^{2}}{4}\right)+b_{2}^{2}\delta+2\int_{0}^{t}[I^{1-\alpha}u^{2}(s_{2}(\tau),\tau)-I^{1-\alpha}u^{\delta}(s_{\delta}(\tau),\tau)]{\rm
d}\tau.\\\ \end{split}$ (121)

We note that,

$\begin{split}B(\tau)&:=I^{1-\alpha}u^{2}(s_{2}(\tau),\tau)-I^{1-\alpha}u^{\delta}(s_{\delta}(\tau),\tau)\\\
&\leq\frac{1}{\Gamma(1-\alpha)}\left[\int_{0}^{s_{2}(\tau)}\frac{u^{2}(x,\tau)}{(s_{2}(\tau)-x)^{\alpha}}-\frac{u^{\delta}(x,\tau)}{(s_{2}(\tau)-x)^{\alpha}}{\rm
d}x\right.\\\
&\quad\left.+\int_{0}^{s_{2}(\tau)}\frac{u^{\delta}(x,\tau)}{(s_{2}(\tau)-x)^{\alpha}}-\frac{u^{\delta}(x,\tau)}{(s_{\delta}(\tau)-x)^{\alpha}}{\rm
d}x\right],\\\ \end{split}$ (122)

and by (116) and Remark 3,

$0\leq
u^{\delta}(x,\tau)\leq\frac{M}{2\Gamma(1+\alpha)}(s_{\delta}^{\alpha}(\tau)-x^{\alpha})\leq\frac{M(b_{2}+\delta+MT)^{\alpha}}{2\Gamma(1+\alpha)}.$

Moreover, we know that $s_{2}\leq s_{\delta}$ and $u^{2}\leq u^{\delta}$, thus

$\begin{split}B(\tau)&\leq\frac{M(b_{2}+\delta+MT)^{\alpha}}{2\Gamma(1+\alpha)\Gamma(2-\alpha)}(s_{\delta}(\tau)-s_{2}(\tau))^{1-\alpha}\end{split}$
(123)

Additionally,

$2b_{2}(s_{\delta}(t)-s_{2}(t))\leq(s_{\delta}(t)+s_{2}(t))(s_{\delta}(t)-s_{2}(t))=s_{\delta}^{2}(t)-s_{2}^{2}(t).$
(124)

Combining (121), (123) and (124), we obtain

$s_{\delta}(t)-s_{2}(t)\leq\delta\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)+\frac{M(b_{2}+\delta+MT)^{\alpha}}{b_{2}\Gamma(1+\alpha)\Gamma(2-\alpha)}\int_{0}^{t}(s_{\delta}(\tau)-s_{2}(\tau))^{1-\alpha}{\rm
d}\tau.$ (125)

We want to apply a generalized Grönwall’s inequality to (125). More precisely,
we apply Theorem 12 with $x(t)=s_{\delta}(t)-s_{2}(t)$,
$n(t)=\delta\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)$,
$h(t)=\frac{M(b_{2}+\delta+MT)^{\alpha}}{b_{2}\Gamma(1+\alpha)\Gamma(2-\alpha)}$,
$p=1$ and $q=1-\alpha$. Then,

$\begin{split}s_{\delta}(t)-s_{2}(t)&\leq\delta\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)+\alpha\delta^{1-\alpha}\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)^{1-\alpha}\frac{M(b_{2}+\delta+MT)^{\alpha}}{b_{2}\Gamma(1+\alpha)\Gamma(2-\alpha)}T,\end{split}$
(126)

which leads to

$s_{1}(t)\leq s_{\delta}(t)\leq
s_{2}(t)+\delta\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)+\alpha\delta^{1-\alpha}\left(1+\frac{b_{2}}{2}+\frac{\delta}{2b_{2}}+\frac{\delta}{2}+\frac{\delta^{2}}{8b_{2}}\right)^{1-\alpha}\frac{M(b_{2}+\delta+MT)^{\alpha}}{b_{2}\Gamma(1+\alpha)\Gamma(2-\alpha)}T,$
(127)

Finally, taking $\delta\rightarrow 0$, it follows that

$s_{1}(t)\leq s_{2}(t),\quad\forall t\in[0,T].$

∎

By Theorem 5 and Theorem 7 the uniqueness result follows easily.

######  Corollary 5.

The solution to (112) is unique in a class of functions satisfying the
regularity assumptions from Theorem 5.

We are ready to prove existence and uniqueness of solution to (112). The proof
is similar to the one given in [25, Theorem 3.14], where an operator $P$ was
defined in such a way that its fixed point satisfies an equivalent integral
condition for a Stefan Problem similar to (112) where a fractional Neumann
condition was considered. However, we cannot use the same operator $P$, since
(117) is not equivalent to the integral condition obtained in [25, Theorem
2.13].

######  Theorem 8.

Let $b,T>0$, $\alpha\in(1/2,1)$. Furthermore, let $u_{0}$ be a nonnegative
function defined on $[0,b]$ such that $v_{0}(x):=u_{0}(bx)$ verifies
$v_{0}\in\widetilde{\mathcal{D}}_{\alpha}$. Moreover, suppose that there
exists a positive constant $M$ such that,

$u_{0}(x)\leq\frac{M}{2\Gamma(1+\alpha)}(b^{\alpha}-x^{\alpha}),\quad
x\in[0,b].$ (128)

Then, there exists a unique solution $(u,s)$ to (112), such that $s\in
C^{0,1}([0,T])$, $0<\dot{s}(t)\leq M$ for all $t\in(0,T]$, $u\in
C(\overline{Q_{s,T}})$, $u_{t},\frac{\partial}{\partial x}D^{\alpha}u\in
C(Q_{s,T})$, and for all $t\in(0,T]$, $u(\cdot,t)\in AC[0,s(t)]$,
$u_{t}(\cdot,t),\frac{\partial}{\partial x}D^{\alpha}u(\cdot,t)\in
L^{2}(0,s(t))$. Furthermore, there exists $\beta>\frac{1}{2}$ such that for
all $t\in(0,T]$ and $0<\varepsilon<\omega<s(t)$ we have $u_{x}(\cdot,t)\in
H^{\alpha+\beta}(\varepsilon,\omega)$.

###### Proof.

We follow the ideas introduced in [2], that was used also in [25] and [29],
but proposing a new operator $P$, destined to be related to the the new
integral condition given in Proposition 10.

We define the set

$\Sigma:=\\{s\in C^{0,1}[0,T]:0\leq\dot{s}\leq M,s(0)=b\\}.$

We will apply the Schauder fixed point Theorem. We note that $\Sigma$ is a
compact and convex subset of the Banach space $C([0,T])$ with the maximum
norm.

For every $s\in\Sigma$ we define

$(Ps)(t)=\left(b^{2}-2\int_{0}^{t}D^{\alpha}u(s(\tau),\tau)s(\tau){\rm
d}\tau\right)^{\frac{1}{2}},$ (129)

where $u$ is the unique solution to the MBP (28) which exists and is unique
due to Theorem 5.

Let’s prove that $P:\Sigma\to\Sigma$. Define

$H(t)=b^{2}-2\int_{0}^{t}D^{\alpha}u(s(\tau),\tau)s(\tau){\rm d}\tau.$

Note that $H(0)=b^{2}$ and from Lemma 10 $H^{\prime}(t)\geq 0$ for every
$t>0$. Then $Ps$ is well defined and from the regularity of $u$ we deduce that
$Ps$ is a continuous function.

From the other side, from Lemma 11 we have

$0\leq(Ps)^{\prime}(t)=\frac{-2D^{\alpha}u(s(t),t)s(t)}{2H^{1/2}(t)}\leq\frac{M}{2b}(b+MT)\leq
M\quad\text{ if }T\leq\frac{b}{M}$ (130)

and then $Ps\in\Sigma$.

Now, in order to prove that the operator P is continuous, we use the relation
(118) to state that $P$ may be written as follows

$(Ps)(t)=\left(b^{2}+2j(t)\right)^{1/2}$ (131)

where

$j(t)=\int_{0}^{b}xu_{0}(x){\rm
d}x-\int_{0}^{t}I^{1-\alpha}u(s(\tau),\tau){\rm
d}\tau-\int_{0}^{s(t)}xu(x,t){\rm d}x>0.$

Let $s_{1},s_{2}\in\Sigma$ and $s_{min}(t):=\min\\{s_{1}(t),s_{2}(t)\\}$,
$s_{max}(t):=\max\\{s_{1}(t),s_{2}(t)\\}$, $i:[0,T]\rightarrow\\{1,2\\}$ with
$i(t)=1$ if $s_{max}(t)=s_{1}(t)$ and $i(t)=2$ otherwise.

Let $u_{1},u_{2}$ be solutions to (28) given by Theorem 5, corresponding to
$s_{1}$ and $s_{2}$ respectively. We define $v(x,t)=u_{1}(x,t)-u_{2}(x,t)$.
Then, $v$ satisfies

$\begin{array}[]{lll}(i)&v_{t}-\frac{\partial}{\partial
x}D^{\alpha}v=0,&\text{ in }Q_{s_{min},T}\\\ (ii)&v(0,t)=0,&t\in(0,T),\\\
(iii)&v(s_{min}(t),t)=(-1)^{i(t)+1}u_{i(t)}(s_{min}(t),t)&t\in(0,T),\\\
(iv)&v(x,0)=0,&0<x<b.\end{array}$ (132)

By Theorem 6, $v$ attains its maximum and minimum on the parabolic boundary.
Hence,

$\max_{Q_{s_{min},T}}|v|\leq\max_{t\in[0,T]}|u_{i(t)}(s_{min}(t),t)|,$

and applying (116), we obtain that

$\max_{Q_{s_{min},T}}|v|\leq\frac{M}{2\Gamma(1+\alpha)}\max_{t\in[0,T]}|s^{\alpha}_{1}(t)-s^{\alpha}_{2}(t)|.$
(133)

Using the real estimate

$b^{\alpha}-a^{\alpha}\leq\frac{\alpha}{a^{1-\alpha}}(b-a),\text{ for }0<a<b,$
(134)

it follows that

$\begin{split}|(Ps_{2})(t)-(Ps_{1})(t)|&=\left|\left(b^{2}+2j_{2}(t)\right)^{1/2}-\left(b^{2}+2j_{1}(t)\right)^{1/2}\right|\leq\frac{2\left|j_{2}(t)-j_{1}(t)\right|}{2b}\\\
&\leq\frac{1}{b}\left(\int_{0}^{t}|I^{1-\alpha}u_{1}(s_{1}(\tau),\tau)-I^{1-\alpha}u_{2}(s_{2}(\tau),\tau)|{\rm
d}\tau+\int_{0}^{s_{min}(t)}x|v(x,t)|{\rm
d}x+\int_{s_{min}(t)}^{s_{max}(t)}xu_{i(t)}(x,t){\rm d}x\right)\end{split}$
(135)

Now,

$\begin{split}&\Gamma(1-\alpha)|I^{1-\alpha}u_{2}(s_{2}(\tau),\tau)-I^{1-\alpha}u_{1}(s_{1}(\tau),\tau)|\\\
&\leq\int_{0}^{s_{min}(\tau)}\frac{M}{2\Gamma(1+\alpha)}(s^{\alpha}_{max}(\tau)-x^{\alpha})\left(\frac{1}{(s_{min}(\tau)-x)^{\alpha}}-\frac{1}{(s_{max}(\tau)-x)^{\alpha}}\right){\rm
d}x\\\
&\quad+\frac{M}{2\Gamma(1+\alpha)}\int_{s_{min}(\tau)}^{s_{max}(\tau)}\frac{s^{\alpha}_{max}(\tau)-x^{\alpha}}{(s_{max}(\tau)-x)^{\alpha}}{\rm
d}x+\int_{0}^{s_{min}(\tau)}\frac{|v(x,\tau)|}{(s_{min}(\tau)-x)^{\alpha}}{\rm
d}x\\\
&\leq\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{s_{min}(\tau)}(s_{max}^{\alpha}(\tau)-x^{\alpha})\left(\frac{1}{(s_{min}(\tau)-x)^{\alpha}}-\frac{1}{(s_{max}(\tau)-x)^{\alpha}}\right){\rm
d}x\\\
&\quad+\frac{M}{2\Gamma(1+\alpha)}\int_{s_{min}(\tau)}^{s_{max}(\tau)}\frac{s_{max}^{\alpha}(\tau)-x^{\alpha}}{(s_{max}(\tau)-x)^{\alpha}}{\rm
d}x+\frac{M}{2\Gamma(1+\alpha)}\max_{\tau\in[0,T]}|s^{\alpha}_{1}(\tau)-s^{\alpha}_{2}(\tau)|\int_{0}^{s_{min}(\tau)}(s_{min}(\tau)-x)^{-\alpha}{\rm
d}x\\\ &=:F_{1}+F_{2}+F_{3}.\end{split}$ (136)

Applying the sustitution $w=\frac{x}{s_{min}(\tau)}$ in $F_{1}$, and using
again (134), it follows that

$\begin{split}F_{1}&=\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{1}(s_{max}(\tau)^{\alpha}-w^{\alpha}s_{min}(\tau)^{\alpha})\left(\frac{1}{(s_{min}(\tau)-ws_{min}(\tau))^{\alpha}}-\frac{1}{(s_{max}(\tau)-ws_{min}(\tau))^{\alpha}}\right)s_{min}(\tau){\rm
d}w\\\
&=\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{1}\left(\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}\right)^{\alpha}-w^{\alpha}\right)\left((1-w)^{-\alpha}-\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)^{-\alpha}\right)s_{min}(\tau){\rm
d}w\\\ &\leq\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{1}\alpha
w^{\alpha-1}\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)\left((1-w)^{-\alpha}-\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)^{-\alpha}\right)s_{min}(\tau){\rm
d}w\\\ &=\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{1}\alpha
w^{\alpha-1}(1-w)^{-\alpha}\left(\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)-(1-w)^{\alpha}\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)^{1-\alpha}\right)s_{min}(\tau){\rm
d}w\\\ &\leq\frac{M}{2\Gamma(1+\alpha)}\int_{0}^{1}\alpha
w^{\alpha-1}(1-w)^{-\alpha}\left(\left(\frac{s_{max}(\tau)}{s_{min}(\tau)}-w\right)-(1-w)^{\alpha}(1-w)^{1-\alpha}\right)s_{min}(\tau){\rm
d}w\\\ &=\frac{\alpha
M}{2\Gamma(1+\alpha)}(s_{max}(\tau)-s_{min}(\tau))\int_{0}^{1}w^{\alpha-1}(1-w)^{-\alpha}{\rm
d}w\\\
&\leq\frac{M\Gamma(1-\alpha)}{2}\max_{t\in[0,T]}|s_{1}(t)-s_{2}(t)|.\end{split}$
(137)

On the other hand, by (29) and (134), it is easy to see that $F_{2}$ and
$F_{3}$ in (136) are bounded by $\max_{t\in[0,T]}|s_{1}(t)-s_{2}(t)|$ as
follows,

$F_{2}\leq\frac{M(b+MT)^{1-\alpha}}{b^{1-\alpha}\Gamma(\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|,$
$F_{3}\leq\frac{M(b+MT)^{1-\alpha}}{b^{1-\alpha}(1-\alpha)\Gamma(\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|.$

Hence, if
$c=\max\left\\{\frac{M\Gamma(1-\alpha)}{2},\frac{M(b+MT)^{1-\alpha}}{b^{1-\alpha}(1-\alpha)\Gamma(\alpha)}\right\\}$,
we have

$\Gamma(1-\alpha)|I^{1-\alpha}u_{2}(s_{2}(\tau),\tau)-I^{1-\alpha}u_{1}(s_{1}(\tau),\tau)|\leq
c\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|,$ (138)

Applying (116), (133) and (138) to (135), we obtain that

$\begin{split}&|(Ps_{2})(t)-(Ps_{1})(t)|\\\
&\leq\frac{1}{b}\left(\frac{cT}{\Gamma(1-\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|\right.\\\
&\quad\left.+\frac{\alpha
Mb^{\alpha-1}}{2\Gamma(1+\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|\int_{0}^{s_{min}(t)}x{\rm
d}x+\frac{M}{2\Gamma(1+\alpha)}\int_{s_{min}(t)}^{s_{max}(t)}x(s_{max}^{\alpha}(t)-x^{\alpha}){\rm
d}x\right)\\\
&\leq\frac{1}{b}\left(\frac{cT}{\Gamma(1-\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|\right.\\\
&\quad\left.+\frac{\alpha
Mb^{\alpha-1}(b+MT)^{2}}{4\Gamma(1+\alpha)}\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|+\frac{Ms_{max}^{\alpha}(t)}{2\Gamma(1+\alpha)}\frac{s_{max}(t)^{2}-s_{min}(t)^{2}}{2}\right)\\\
&\leq\frac{1}{b}\left(\frac{cT}{\Gamma(1-\alpha)}+\frac{\alpha
Mb^{\alpha-1}(b+MT)^{2}}{4\Gamma(1+\alpha)}+\frac{Ms_{max}^{\alpha}(t)}{2\Gamma(1+\alpha)}\frac{s_{max}(t)+s_{min}(t)}{2}\right)\max_{t\in[0,T]}|s_{max}(t)-s_{min}(t)|\\\
\end{split}$ (139)

We conclude that, for
$C=\frac{1}{b}\left(\frac{cT}{\Gamma(1-\alpha)}+\frac{\alpha
Mb^{\alpha-1}(b+MT)^{2}}{4\Gamma(1+\alpha)}+\frac{M(b+MT)^{1+\alpha}}{2\Gamma(1+\alpha)}\right)$

$|(Ps_{1})(t)-(Ps_{2})(t)|\leq C\max_{t\in[0,T]}|s_{1}(t)-s_{2}(t)|.$

Hence, $P$ is continuous in the maximum norm. From the Schauder fixed point
Theorem, we obtain that for $T\leq\frac{b}{M}$, there exists a unique
$s\in\Sigma$ such that $Ps=s$ and the pair $(u,s)$ verifies (28) and (117).
Applying Proposition 10 we deduce that $(u,s)$ is a solution to (112) and the
uniqueness follows from Corollary 5. Thus, we proved Theorem (8) for
$T\leq\frac{b}{M}$.

We can extend the result for larger T by a standard argument, by considering a
time step of length $\frac{b}{2M}$ and taking as the initial condition value
of solution obtained in a previous step in point $t_{i}=\frac{b}{4M}$. Due to
Theorem 5, in each step the initial condition is regular enough to repeat the
proof of the first step. In fact, let $(u^{0},s^{0})$ be the solution to (112)
for $T=\frac{b}{2M}$, and consider a new problem given by (112) with
$u^{1}_{0}(x)=u^{0}(x,\frac{b}{4M})$,
$s^{1}(0)=b^{1}:=s^{0}\left(\frac{b}{4M}\right)\geq b$. Then $u^{1}_{0}$
satisfy the assumption (128) due to Lemma 11, and we can repeat the previous
reasoning obtaining a solution $(u^{1},s^{1})$ to (112) in $[0,\frac{b}{2M}]$,
but we have started from $t_{1}=\frac{b}{4M}$. Note that from Proposition 10
$s^{0}$ and $s^{1}$ verify both the fractional Stefan condition (112)$-(iv)$
and from Corollary 5 we have that the solutions $(u^{0},s^{0})$ and
$(u^{1},s^{1})$ agree in the region $\left\\{0\leq x\leq
s^{0}(t)=s^{1}(t-\frac{b}{4M}),\frac{b}{4M}\leq t<\frac{b}{2M}\right\\}$.
Then, we can state that the pair $(u,s)$ given by
$s(t)=\begin{cases}s^{0}(t),&0\leq t<\frac{b}{4M}\\\
s^{1}(t-\frac{b}{4M})&\frac{b}{4M}\leq t<\frac{b}{2M}\end{cases}$ and
$u(x,t)=\begin{cases}u^{0}(x,t)&0\leq x\leq s(t),0\leq t<\frac{b}{4M}\\\
u^{1}(x,t-\frac{b}{4M})&0\leq x\leq s(t),\frac{b}{4M}\leq
t<\frac{b}{2M}\end{cases}$ is a solution to (112) in $Q_{s,\frac{b}{2M}}$ and
has the regularity requested in the claim.

Finally, by repeating this argument $k$ times for $k$ such that
$k\frac{b}{4M}>T$, we can obtain a solution $(u,s)$ to (112), defining
$s:[0,T]\mapsto{\mathbb{R}}$ and $u:Q_{s,T}\mapsto{\mathbb{R}}$ as follows

$s(t)=s^{i}\left(t-i\frac{b}{4M}\right),\text{ for
}t\in\left(i\frac{b}{4M},(i+1)\frac{b}{4M}\right],\quad i=0,...,k$
$u(x,t)=u^{i}\left(x,t-i\frac{b}{4M}\right),\text{ for }x\in[0,s^{i}(t)],\quad
t\in\left(i\frac{b}{4M},(i+1)\frac{b}{4M}\right],\quad i=0,...,k.$

∎

##  5 Conclusions

We have studied a space-fractional Stefan problem with the Dirichlet
homogeneus boundary conditions, and have proven the existence and uniqueness
of a regular solution. In the proof, we utilized several established facts for
similar fractional Stefan problems, and developed additional results,
including an estimate between fractional Sobolev spaces and interpolation
spaces, and an integral equation equivalent to the Stefan condition. We
applied the latter to prove the monotone dependence on data. Aditionally, we
adapted regularity results given for a similar problem with a fractional
Neumann condition.

It remains an open problem to study the existence and uniqueness of the
solution to problem (1) with $b=0$.

##  6 Acknowledgements

The first and third author were partly supported by the Project INV-006-00030
from Universidad Austral and PICT-2021-I-INVI-00317. The second author was
partly supported by the National Sciences Center, Poland, through grant
2017/26/M/ST1/00700.

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