# Fractional powers approach of maximal accretive operators for a class of
third order abstract Cauchy problems

Mohammed Benharrat  Ecole Nationale Polytechnique d’Oran-Maurice Audin (Ex.
ENSET d’Oran), BP 1523 Oran-El M’naouar, 31000 Oran, Algérie.
<EMAIL_ADDRESS>and  Flank D. M. Bezerra  Departamento de
Matemática, Universidade Federal da Paraíba, 58051-900 João Pessoa PB, Brazil.
<EMAIL_ADDRESS>
(Date:
The first author gratefully acknowledge the financial support from the
Laboratory of Fundamental and Applicable Mathematics of Oran (LMFAO) and the
Algerian research project: PRFU, no. C00L03ES310120220003 (D.G.R.S.D.T). The
second author gratefully acknowledge the financial support from CNPq/Brazil
grant number 303039/2021-3.)

###### Abstract.

In this paper we explore the theory of fractional powers of maximal accretive
operators to obtain results of existence, regularity and behavior asymptotic
of solutions for linear abstract evolution equations of third order in time.

Mathematics Subject Classification 2010: 34A08; 47D06; 47D03
Key words and phrases: Accretive operator; ; analytic semigroup; fractional
powers; linear evolution equation of third order; strongly continuous
semigroup.

###### Contents

  1. 1 Introduction and preliminaries
  2. 2 The linearized operator and its fractional powers
    1. 2.1 Spectral properties of the operator $\mathbb{A}$
    2. 2.2 The fractional powers of the linearized operator
  3. 3 A third order Cauchy problem associated to fractional powers of the linearized operator

###### List of Figures

  1. 1 Location of the spectrum of $\mathbb{A}$
  2. 2 Location of the spectrum of $\mathbb{A}^{\alpha^{*}}$.
  3. 3 Location of the spectrum of $\mathbb{A}^{\alpha}$ with $0<\alpha<\alpha^{*}$.

##  1\. Introduction and preliminaries

In this paper we consider the following abstract linear evolution equation of
third order,

$u^{\prime\prime\prime}(t)+Au(t)=0,\quad t>0,$ (1.1)

with initial conditions given by

$u(0)=u_{0}\in\mathcal{H}^{\frac{2}{3}},\
u^{\prime\prime}(0)=u_{1}\in\mathcal{H}^{\frac{1}{3}},\
u^{\prime\prime\prime}(0)=u_{2}\in\mathcal{H},$ (1.2)

where $\mathcal{H}$ be a separable Hilbert space and
$A:\mathcal{D}(A)\subset\mathcal{H}\to\mathcal{H}$ be a maximal
$\omega$-accretive operator, $0\leq\omega\leq\dfrac{\pi}{2}$. Here
$u^{\prime}(t)=\dfrac{du}{dt}(t).$

In this paper we explore the theory of fractional powers of maximal accretive
operators to obtain results of existence, regularity and behavior asymptotic
of solutions for (1.1)-(1.2).

To better present our results in this paper, we introduce first some notations
and terminologies. Recall that the numerical range of an operator $A$ with
domain $\mathcal{D}(A)$ in $\mathcal{H}$, is defined by

$W(A):=\\{\left\langle Au,u\right\rangle_{\mathcal{H}}:\quad
u\in\mathcal{D}(A),\quad\text{ }\left\|u\right\|_{\mathcal{H}}=1\\}.$

An operator $A$ is said to be accretive if if its numerical range is contained
in the closed right half-plane
$\overline{\mathbb{C}_{+}}:=\\{z\in\mathbb{C}:{\rm Re}(z)\geq 0\\}$, and $A$
is maximal accretive, or $m$-accretive for short, if has no proper accretive
extensions in $\mathcal{H}$. It is known (see [16]) that a closed, maximal
accretive operator is densely defined, that $A$ is closed and $m$-accretive if
and only if $A^{*}$ is, and also if and only if $-A$ is the infinitesimal
generator of a contraction semigroup $\\{\mathcal{T}(t)\\}_{t\geq 0}$, that
is, $\|\mathcal{T}(t)\|_{\mathcal{B}(\mathcal{H})}\leq 1$111Here,
$\mathcal{B}(E)$ denotes the space of bounded linear operators defined in a
Banach space $E$ into self endowed with norm
$\|S\|_{\mathcal{B}(E)}:=\sup_{x\in E,\ x\neq
0}\dfrac{\|Sx\|_{E}}{\|x\|_{E}},\ \forall S\in\mathcal{B}(E).$ ; $A$ is called
dissipative, if $-A$ is accretive. For $\omega\in[0,\pi/2)$, we denote by

$\mathcal{S}(\omega):=\left\\{z\in\mathbb{C}:|\arg(z)|<\omega\right\\},$

the open sector of $\mathbb{C}$ with semi-angle $\omega$. An operator $A$ will
be said is _$\omega$ -accretive_ if

$W(A)\subset\overline{\mathcal{S}(\omega)}:=\left\\{z\in\mathbb{C}:|\arg(z)|\leq\omega\right\\},$

or, equivalently,

$|{\rm Im}\left(\left\langle
Ax,x\right\rangle_{\mathcal{H}}\right)|\\!\leq\\!\tan(\omega)\,{\rm
Re}\left(\left\langle Ax,x\right\rangle_{\mathcal{H}}\right),\quad\text{for
all }x\in\mathcal{D}(A).$

An $\omega$-accretive operator $A$ is called m-$\omega$-accretive, if it is
$m$-accretive. We have $A$ is m-$\omega$-accretive if and only if operators
$e^{\pm i\theta}A$ are m-accretive for $\theta=\frac{\pi}{2}-\omega$,
$0<\omega\leq\pi/2$. If $\omega=\pi/2$, then we have
$\overline{\mathcal{S}(\pi/2)}=\overline{\mathbb{C}_{+}}$ and
m-$\pi/2$-accretivity means m-accretivity. Also, m-$0$-accretive is the
nonegative selfadjoint operator.

The resolvent set of an m-$\omega$-accretive operator $A$ contains the set
$\mathbb{C}\setminus\overline{\mathcal{S}(\omega)}$ and

$\|(\lambda I-A)^{-1}\|_{\mathcal{B}(\mathcal{H})}\leq\cfrac{1}{{\rm
dist}\left(\lambda,\mathcal{S}(\omega)\right)},\quad\lambda\in\mathbb{C}\setminus\overline{\mathcal{S}(\omega)}.$

Consequently, the numerical range of m-$\omega$-accretive operator,
$0\leq\omega\leq\pi/2$, has the so-called spectral inclusion property,

$\sigma(A)\subset\overline{W(A)}\subset\overline{\mathcal{S}(\omega)}.$ (1.3)

It is well known that if $A$ is m-$\omega$-accretive, $0\leq\omega<\pi/2$,
then it generates contractive $C_{0}$-semigroup $\mathcal{T}(t)=e^{-tA}$,
$t\geq 0$, and has an holomorphic continuation into the sector
$\overline{\mathcal{S}(\pi/2-\omega)}$, see Theorem IX-1.24 in [15]. In this
case, we can write $\mathcal{T}(z)=e^{-zA}$, for all
$z\in\overline{\mathcal{S}(\pi/2-\omega)}$; moreover, we have

$\left\|e^{-zA}\right\|_{\mathcal{B}(\mathcal{H})}\leq 1,\quad
z\in\overline{\mathcal{S}(\pi/2-\omega)}.$

Let $A$ is $m$-accretive and a number $\alpha\\!\in\\!(0,1)$; then there exist
several equivalent definitions of a fractional power of the operator $A$. We
give one of them: an operator $A^{\alpha}$ is the closure with
$\mathcal{D}(A)$ of the operator

$A^{\alpha}=\dfrac{\sin(\pi\alpha)}{\pi}\int_{0}^{\infty}\lambda^{\alpha-1}A(I\lambda+A)^{-1}dt,$
(1.4)

see [3]. The operators $A^{\alpha}$ are m-$(\alpha\pi/2)$-accretive and
$A^{\alpha}A^{\beta}x=A^{\alpha+\beta}x$, for all
$x\in\mathcal{D}(A^{\alpha+\beta})$, where $\beta>0$ and $\alpha+\beta<1$.
Fractional powers have the important multiplicative property

$(A^{\alpha})^{\beta}=A^{\alpha\beta}$

when $0<\alpha<1$ and $\beta>0$, [18, Theorem 4.2]. If
$\alpha\\!\in\\!(0,1/2)$, then
$\mathcal{D}(A^{\alpha})=\mathcal{D}(A^{*\alpha})$ holds in a topological
sens, where the domains of definition are equipped with corresponding graph
norms, see [16]. Such fractional powers have been defined for a more general
class of linear operators in Banach spaces by several authors (see, among
others, Amann [2], Balakrishnan [3], Hasse [13] and Martinez [18]).

Denote by $\mathcal{H}^{\alpha}=\mathcal{D}(A^{\alpha})$ for
$0\leqslant\alpha\leqslant 1$ (taking $A^{0}:=I$ on
$\mathcal{H}^{0}:=\mathcal{H}$ when $\alpha=0$). Recall that
$\mathcal{H}^{\alpha}$ is dense in $\mathcal{H}$ for all
$0\leqslant\alpha\leqslant 1$, for details see Amann [2, Theorem 4.6.5]. The
fractional power space $\mathcal{H}^{\alpha}$ endowed with the norm

$\|\cdot\|_{\mathcal{H}^{\alpha}}:=\|A^{\alpha}\cdot\|_{\mathcal{H}}$

is a Banach space. With this notation, we have
$\mathcal{H}^{-\alpha}=(\mathcal{H}^{\alpha})^{\prime}$ for all $\alpha>0$,
see Amann [2], Sobolevskiĭ [22] and Triebel [24] for the characterization of
the negative scale.

Naturally, one can convert (1.1)-(1.2) to the first order Cauchy problem.
However, the linearized operator matrix is in general not closed. Even if it
is closed, it does not generate a strongly continuous semigroup unless under
very restrictive conditions. This forces us to seek a suitable phase space to
change the awkward situation. The basic idea is to calculate the fractional
powers of order $\alpha\in(0,1)$ of the linearized operator and we are
interested in finding for which values of $\alpha\in[0,1]$ the fractional
first order system is well-posed in some sense. For such $\alpha$, we give an
explicit representation formula of the solution of the following associated
evolution equations of third order in time,

$u^{\prime\prime\prime}(t)+3\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}u^{\prime\prime}(t)+3\varUpsilon_{0}^{\alpha}A^{\frac{2\alpha}{3}}u^{\prime\prime}(t)+A^{\alpha}u(t)=0,\quad
t>0,$ (1.5)

with initial conditions given by

$u(0)=\varphi\in\mathcal{H}^{\frac{2}{3}},\
u^{\prime}(0)=\psi\in\mathcal{H}^{\frac{2-\alpha}{3}},\
u^{\prime\prime}(0)=\xi\in\mathcal{H}^{\frac{2-\alpha}{3}},$ (1.6)

where $\varUpsilon_{0}^{\alpha}=2\cos\dfrac{2\pi\alpha}{3}+1$,
$0<\alpha<\alpha^{*}$ and $\alpha^{*}=\dfrac{3\pi}{4\pi+2\omega}$.

Fractional powers approach of operators for Cauchy problems associated with
evolutionary equations has been studied in some situations, in the sense of
existence, regularity and behavior asymptotic of solutions for theses
problems, see e.g., [4, 5, 6, 7, 9, 10, 11, 25], and references therein.

We extend the work done in [8], where $A$ is assumed a self-adjoint positive
definite operator ($\omega=0$). They proved that (1.5)-(1.6) admits unique
solution, without given explicitly, for all $0<\alpha<\frac{3}{4}$. This means
that $\alpha^{*}=\frac{3}{4}$ when $\omega=0$. The analogy of this idea to the
general case, up to now is still not obvious. Also some models of continuous
mechanics are reduced to differential equation with sectorial operators, see
[1, 23] and references therein. In this cases methods, developed for self-
adjoint operators, cannot be applied. Here, we continue the analysis done in
the previously cited works.

Summaries of our main results and the structure of the paper are as follows.
In Section 2 we define the linearized operator, denoted by $\mathbb{A}$,
associated to the Cauchy problem (1.1)-(1.2) in an appropriate phase space. We
study the spectral properties of $\mathbb{A}$. We prove that $\mathbb{A}$
cannot generate a strongly continuous semigroup of any kind but it is of non-
negative type (and not necessarily accretive) operator. Moreover, under
Balakrishnan formula, see [3], we explicitly calculate the fractional powers
of the operator $\mathbb{A}$ for $0<\alpha<1$, and we study spectral
properties of the fractional power operators $\mathbb{A}^{\alpha}$. More
precisely, we establish the maximum subinterval of $(0,1)$ where $\alpha$ is
taken such that the $-\mathbb{A}^{\alpha}$ is a generator, namely
$-\mathbb{A}^{\alpha}$ generates a strongly continuous analytic semigroup if
and only if $0<\alpha<\alpha^{*}$ with $\alpha^{*}=\dfrac{3\pi}{4\pi+2\omega}$
and it generates strongly continuous semigroup if
$\alpha^{*}=\dfrac{3\pi}{4\pi+2\omega}$. Finally, Section 3 provides a result
of existence and uniqueness of the fractional partial differential equations
(1.5)-(1.6) associated with the fractional operator $\mathbb{A}^{\alpha}$. We
give an explicit representation formula of the solution of the Cauchy problem
of third order (1.5)-(1.6).

##  2\. The linearized operator and its fractional powers

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$. Let
consider the following phase space

$X=\mathcal{H}^{\frac{2}{3}}\times\mathcal{H}^{\frac{1}{3}}\times\mathcal{H}$

which is a Banach space equipped with the norm given by

$\|\cdot\|_{X}^{2}=\|\cdot\|^{2}_{\mathcal{H}^{\frac{2}{3}}}+\|\cdot\|^{2}_{\mathcal{H}^{\frac{1}{3}}}+\|\cdot\|^{2}_{\mathcal{H}}.$

By letting $v=\frac{du}{dt}$, $w=\frac{d^{2}u}{dt^{2}}$, we can write the
problem (1.1)-(1.2) as a Cauchy problem on $X$,

$\begin{cases}\dfrac{d{\bf u}}{dt}+\mathbb{A}{\bf u}=0,\ t>0,\\\ {\bf
u}(0)={\bf u}_{0},\end{cases}$ (2.1)

where the unbounded linear operator $\mathbb{A}:\mathcal{D}(\mathbb{A})\subset
X\to X$ is defined by

$\mathbb{A}=\begin{bmatrix}0&-I&0\\\ 0&0&-I\\\ A&0&0\end{bmatrix}$ (2.2)

and

$\mathcal{D}(\mathbb{A})=\mathcal{H}^{1}\times\mathcal{H}^{\frac{2}{3}}\times\mathcal{H}^{\frac{1}{3}}.$
(2.3)

Here

${\bf u}=\begin{bmatrix}u\\\ v\\\ w\end{bmatrix}\quad\text{ and }\quad{\bf
u}_{0}=\begin{bmatrix}u_{0}\\\ u_{1}\\\ u_{2}\end{bmatrix}.$

######  Remark 2.1.

We note that $\mathbb{A}$ is not an accretive operator on $X$. Indeed, if $u$
be a non-trivial element in $\mathcal{D}(A)$ and

$\mathbf{u}=\left[\begin{smallmatrix}u\\\ 0\\\ -u\end{smallmatrix}\right]$

then

${\rm
Re}\left<\mathbb{A}\mathbf{u},\mathbf{u}\right>_{X}=\left<\left[\begin{smallmatrix}0\\\
-u\\\ Au\end{smallmatrix}\right],\left[\begin{smallmatrix}u\\\ 0\\\
-u\end{smallmatrix}\right]\right>_{X}=-{\rm Re}\langle
Au,u\rangle_{\mathcal{H}}<0.$

Explicitly, this means that $-\mathbb{A}$ is not an infinitesimal generator of
a strongly continuous semigroup of contractions on $Y$. More precisely, we see
later, Remark 2.3, that $-\mathbb{A}$ cannot be the infinitesimal generator of
a strongly continuous semigroup of any type on $X$.

###  2.1. Spectral properties of the operator $\mathbb{A}$

######  Lemma 2.2.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$. Then
$\mathbb{A}$ is closed densely defined operator and the spectrum of
$\mathbb{A}$ is given by

$\sigma(\mathbb{A})=\\{\lambda\in\mathbb{C}:\lambda^{3}\in\sigma(A)\\}.$ (2.4)

###### Proof.

Firstly, we show that the operator $\mathbb{A}$ is closed. Indeed, if ${\bf
u}_{n}={\left[\begin{smallmatrix}u_{n}\\\ v_{n}\\\
w_{n}\end{smallmatrix}\right]}\in\mathcal{D}(\mathbb{A})$ with ${\bf
u}_{n}\to{\bf u}={\left[\begin{smallmatrix}u\\\ v\\\
w\end{smallmatrix}\right]}$ in $X$ as $n\to\infty$, and $\mathbb{A}{\bf
u}_{n}\to\varphi$ in $X$ as $n\to\infty$, where
$\varphi={\left[\begin{smallmatrix}\varphi_{1}\\\ \varphi_{2}\\\
\varphi_{3}\end{smallmatrix}\right]}$, then

$\begin{cases}v_{n}\to-\varphi_{1}\ \mbox{ in }\ \mathcal{H}^{\frac{2}{3}}\\\
w_{n}\to-\varphi_{2}\mbox{ in }\ \mathcal{H}^{\frac{1}{3}}\\\
Au_{n}\to\varphi_{3}\ \ \mbox{ in }\ \mathcal{H}\ \mbox{as}\
n\to\infty\end{cases}$

and consequently, $v=-\varphi_{1}\in\mathcal{H}^{\frac{2}{3}}$,
$w=-\varphi_{2}\in\mathcal{H}^{\frac{1}{3}}$. Finally, using the fact that $A$
is a closed operator, we have $u\in\mathcal{D}(A)$ and $Au=\varphi_{3}$; that
is, ${\bf u}\in\mathcal{D}(\mathbb{A})$ and $\mathbb{A}{\bf u}=\varphi$.

Secondly, Since the inclusions
$\mathcal{H}^{\alpha}\subset\mathcal{H}^{\beta}$ are dense for
$\alpha>\beta\geq 0$, then we conclude that $\mathcal{D}(\mathbb{A})$ is dense
in $X$.

Finally, to prove (2.4) it suffices to prove that the resolvent set of
$\mathbb{A}$ is given by

$\rho(\mathbb{A})=\\{\lambda\in\mathbb{C}:\lambda^{3}\in\rho(A)\\}.$

Suppose that $\lambda\in\mathbb{C}$ is such that $\lambda^{3}\in\rho(A)$. We
claim that $\lambda\in\rho(\mathbb{A})$. Indeed, since $-\mathbb{A}$ is closed
operator, we only need to show that

$\lambda I-\mathbb{A}:\mathcal{D}(\mathbb{A})\subset X\to X$

is bijective. We have

$\lambda I-\mathbb{A}=\begin{bmatrix}\lambda I&I&0\\\ 0&\lambda I&I\\\
-A&0&\lambda I\end{bmatrix},$ (2.5)

For injectivity consider ${\bf u}=\left[\begin{smallmatrix}u\\\ v\\\
w\end{smallmatrix}\right]\in D(\mathbb{A})$ and $(\lambda I-\mathbb{A}){\bf
u}=0$, then

$\begin{cases}\lambda u+v=0,\\\ \lambda v+w=0,\\\ -Au+\lambda w=0.\end{cases}$
(2.6)

From (2.6) we have

$(\lambda^{3}I-A)u=0.$ (2.7)

Since $\lambda^{3}\in\rho(A)$, we conclude that $u=0$ and consequently
$\textbf{u}=0$.

For surjectivity given $\varphi=\left[\begin{smallmatrix}\varphi_{1}\\\
\varphi_{2}\\\ \varphi_{3}\end{smallmatrix}\right]\in X$ we take ${\bf
u}=\left[\begin{smallmatrix}u\\\ v\\\ w\end{smallmatrix}\right]$ such that
$(\lambda I-\mathbb{A}){\bf u}=\varphi$, that is

$\begin{cases}\lambda u+v=\varphi_{1},\\\ \lambda v+w=\varphi_{2},\\\
-Au+\lambda w=\varphi_{3}.\end{cases}$

Since $\lambda^{3}\in\rho(A)$, by straightforward calculus, we obtain

$\begin{cases}u=\lambda^{2}(\lambda^{3}I-A)^{-1}\varphi_{1}-\lambda(\lambda^{3}I-A)^{-1}\varphi_{2}+(\lambda^{3}I-A)^{-1}\varphi_{3},\\\
v=-A(\lambda^{3}I-A)^{-1}\varphi_{1}+\lambda^{2}(\lambda^{3}I-A)^{-1}\varphi_{2}-\lambda(\lambda^{3}I-A)^{-1}\varphi_{3},\\\
w=\lambda
A(\lambda^{3}I-A)^{-1}\varphi_{1}-A(\lambda^{3}I-A)^{-1}\varphi_{2}+\lambda^{2}(\lambda^{3}I-A)^{-1}\varphi_{3}.\end{cases}$

Moreover, $u\in\mathcal{D}(A)$, $v\in\mathcal{D}(A^{\frac{2}{3}})$ and
$w\in\mathcal{D}(A^{\frac{1}{3}})$. Then we have ${\bf
u}\in\mathcal{D}(\mathbb{A})$ and

$(\lambda I-\mathbb{A}){\bf u}=\varphi.$

Now suppose that $\lambda\in\rho(\mathbb{A})$. If $u\in D(A)$ is such that
$(\lambda^{3}I-A)u=0$, taking ${\bf u}=\left[\begin{smallmatrix}u\\\ v\\\
w\end{smallmatrix}\right]\in\mathcal{D}(\mathbb{A})$ we have

$(\lambda I-\mathbb{A}){\bf u}=0.$ (2.8)

Since $\lambda\in\rho(\mathbb{A})$, it follows that ${\bf u}=0$ and
consequently $u=0$, which proves the injectivity of $\lambda^{3}I-A$. Given
$f\in\mathcal{H}$, consider $\varphi=\left[\begin{smallmatrix}0\\\ 0\\\
f\end{smallmatrix}\right]\in X$. By the surjectivity of $\lambda I-\mathbb{A}$
there exists ${\bf u}=\left[\begin{smallmatrix}u\\\ v\\\
w\end{smallmatrix}\right]\in\mathcal{D}(\mathbb{A})$ such that

$(\lambda I-\mathbb{A}){\bf u}=\varphi$ (2.9)

which gives

$(\lambda^{3}I-A)u=f$

and the proof is complete. ∎

######  Remark 2.3.

Let $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$. In the
following some interesting consequences of Lemma 2.2 and its proof,

  1. (1)

The resolvent set of $\mathbb{A}$ is given by

$\rho(\mathbb{A})=\\{\lambda\in\mathbb{C}:\lambda^{3}\in\rho(A)\\},$ (2.10)

and

$(\lambda
I-\mathbb{A})^{-1}=\begin{bmatrix}\lambda^{2}(\lambda^{3}I-A)^{-1}&-\lambda(\lambda^{3}I-A)^{-1}&(\lambda^{3}I-A)^{-1}\\\
-A(\lambda^{3}I-A)^{-1}&\lambda^{2}(\lambda^{3}I-A)^{-1}&-\lambda(\lambda^{3}I-A)^{-1}\\\
\lambda
A(\lambda^{3}I-A)^{-1}&-A(\lambda^{3}I-A)^{-1}&\lambda^{2}(\lambda^{3}I-A)^{-1}\end{bmatrix}$

for all $\lambda\in\rho(\mathbb{A})$.

  2. (2)

By (2.10), we have

$(-\infty,0)\subset\rho(\mathbb{A}).$

  3. (3)

We have, for every $r\geq 0$,

$re^{i\theta}\in\sigma(A)\Longleftrightarrow
r^{1/3}e^{i\frac{\theta}{3}},r^{1/3}e^{i\frac{\theta+2\pi}{3}},r^{1/3}e^{i\frac{\theta-2\pi}{3}}\in\sigma(\mathbb{A}).$

Thus, we can see that the spectrum of $\mathbb{A}$ is a reunion of possibly
three disjointed closed parts $\varLambda_{1}$, $\varLambda_{2}$ and
$\varLambda_{3}$, with

$\varLambda_{1}\subset\overline{\mathcal{S}(\omega/3)},\qquad\varLambda_{2}\subset
e^{i\frac{2\pi}{3}}\overline{\mathcal{S}(\omega/3)},\qquad\text{ and
}\qquad\varLambda_{3}\subset
e^{-i\frac{2\pi}{3}}\overline{\mathcal{S}(\omega/3)}.$

Consequently,

$\mathcal{S}(\pi-\omega/3)\cup\left\\{\lambda\in\mathbb{C}:\frac{\pi}{2}-\omega/3<\arg(\lambda)<\frac{\pi}{2}\right\\}\cup\left\\{\lambda\in\mathbb{C}:-\frac{\pi}{2}<\arg(\lambda)<-\frac{\pi}{2}-\omega/3\right\\}\subset\rho(\mathbb{A}).$

See the figure below.

  4. (4)

Clearly,

$0\in\rho(A)\Longleftrightarrow 0\in\rho(\mathbb{A}),$

in this case, $\varLambda_{1}\cap\varLambda_{2}\cap\varLambda_{3}=\emptyset$
and $\sigma(\mathbb{A})=\varLambda_{1}\cup\varLambda_{2}\cup\varLambda_{3}.$

  5. (5)

If $A$ a self adjoint operator with a compact resolvent, then $-\mathbb{A}$
cannot be the infinitesimal generator of a strongly continuous semigroup on
$X$. Indeed, if $-\mathbb{A}$ generates a strongly continuous semigroup, it
follows from Pazy [21, Theorem 1.2.2] that there exist constants $\omega\geq
0$ such that

$\\{\lambda\in\mathbb{C}:{\rm Re}\lambda>\omega\\}\subset\rho(-\mathbb{A}).$
(2.11)

But, we have

$\sigma_{p}(-\mathbb{A})=\\{\lambda\in\mathbb{C}:\lambda^{3}\in\sigma_{p}(-A)\\},$

where $\sigma_{p}(-\mathbb{A})$ and $\sigma_{p}(-A)$ denote the point spectrum
set of $-\mathbb{A}$ and $-A$, respectively. Since
$\sigma_{p}(-A)=\\{-\mu_{j}:j\in\mathbb{N}\\}$ with $\mu_{j}\in\sigma_{p}(A)$
for each $j\in\mathbb{N}$ and $\mu_{j}\to\infty$ as $j\to\infty$, we conclude
that

$\sigma_{p}(-\mathbb{A})\cap\\{\lambda\in\mathbb{C}:{\rm
Re}\lambda>\omega\\}\neq\emptyset$

See figure 1. This contradicts the equation (2.11) and therefore $-\mathbb{A}$
can not be the infinitesimal generator of a strongly continuous semigroup on
$X$.

$\scriptstyle{\rm Re}(\lambda)$$\scriptstyle{\rm Im}(\lambda)$Sector
$\overline{\mathcal{S}(\omega/3)}$ containing $\varLambda_{1}$Sector
$e^{i\frac{2\pi}{3}}\overline{\mathcal{S}(\omega/3)}$ containing
$\varLambda_{2}$Sector $e^{-i\frac{2\pi}{3}}\overline{\mathcal{S}(\omega/3)}$
containing $\varLambda_{3}$$\omega$ Figure 1. Location of the spectrum of
$\mathbb{A}$

###  2.2. The fractional powers of the linearized operator

Let $\mathbb{A}:\mathcal{D}(\mathbb{A})\subset X\to X$ be the unbounded linear
operator defined in (2.2)-(2.3). In this section we prove that $\mathbb{A}$ is
an operator of non-negative type in the sens of [19, Definition 1.1.1]. These
are the operators that one can define the fractional power, for more details
see Pazy [21, Section 2.2.6] and Amann [2, Section 3.4.6].

######  Lemma 2.4.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$.
Then, $\mathbb{A}$ is closed, densely defined,
$\left(-\infty,0\right)\subset\rho(\mathbb{A})$ and there exists a positive
constant $M$, such that

$\|(\lambda I+\mathbb{A})^{-1}\|_{\mathcal{B}(X)}\leqslant\frac{M}{\lambda},\
\mbox{for all}\ \lambda>0.$ (2.12)

###### Proof.

By Lemma 2.2, the operator $\mathbb{A}$ is closed and densely defined and
$\left(-\infty,0\right)\subset\rho(\mathbb{A})$.

For $\varphi=\left[\begin{smallmatrix}\varphi_{1}\\\ \varphi_{2}\\\
\varphi_{3}\end{smallmatrix}\right]$ , ${\bf u}=\left[\begin{smallmatrix}u\\\
v\\\ w\end{smallmatrix}\right]$ in $X$ and any $\lambda>0$ we have, ${\bf
u}=(\lambda I+\mathbb{A})^{-1}\varphi$ if and only if

$\begin{cases}u=\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{1}+\lambda(\lambda^{3}I+A)^{-1}\varphi_{2}+(\lambda^{3}I+A)^{-1}\varphi_{3},\\\
v=-A(\lambda^{3}I+A)^{-1}\varphi_{1}+\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{2}+\lambda(\lambda^{3}I+A)^{-1}\varphi_{3},\\\
w=-\lambda
A(\lambda^{3}I+A)^{-1}\varphi_{1}-A(\lambda^{3}I+A)^{-1}\varphi_{2}+\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{3}.\end{cases}$

In order to verify the equation (2.12) for $\mathbb{A}$, it is sufficient to
show that for $\|\varphi\|_{X}\leq 1$ there exists a constant
$M_{\mathbb{A}}>0$ such that

$\|u\|_{\mathcal{H}^{\frac{2}{3}}}+\|v\|_{\mathcal{H}^{\frac{1}{3}}}+\|w\|_{\mathcal{H}}\leqslant\frac{M_{\mathbb{A}}}{\lambda}.$
(2.13)

We first observe that

$A(\lambda^{3}I+A)^{-1}=I-\lambda^{3}(\lambda^{3}I+A)^{-1}.$

This and the fact that $A$ is m-accretive and $\lambda^{3}\in\rho(A)$,

$\|A(\lambda^{3}I+A)^{-1}\|_{\mathcal{B}(\mathcal{H})}\leqslant
1+\|\lambda^{3}(\lambda^{3}I+A)^{-1}\|_{\mathcal{B}(\mathcal{H})}\leqslant
2.\\\ $

Now, for $x\in\mathcal{H}$, from the inequality of [21, Theorem 6.10] we have

$\begin{split}\|A^{\alpha}(\lambda^{3}I+A)^{-1}x\|_{\mathcal{H}}&\leqslant
C\|(\lambda^{3}I+A)^{-1}x\|_{\mathcal{H}}^{1-\alpha}\|A(\lambda^{3}I+A)^{-1}x\|_{\mathcal{H}}^{\alpha}\\\
&\leqslant\dfrac{2^{\alpha}C}{\lambda^{3(1-\alpha)}}\|x\|_{\mathcal{H}}\end{split}$

for some $C>0$. Thus,

$\displaystyle\|u\|_{\mathcal{H}^{\frac{2}{3}}}$
$\displaystyle\leq\left\|\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{1}\right\|_{\mathcal{H}^{\frac{2}{3}}}+\left\|\lambda(\lambda^{3}I+A)^{-1}\varphi_{2}\right\|_{\mathcal{H}^{\frac{2}{3}}}+\left\|(\lambda^{3}I+A)^{-1}\varphi_{3}\right\|_{\mathcal{H}^{\frac{2}{3}}}$
$\displaystyle\leq\dfrac{1}{\lambda}\left\|\varphi_{1}\right\|_{\mathcal{H}^{\frac{2}{3}}}+\dfrac{2^{1/3}C}{\lambda}\left\|\varphi_{2}\right\|_{\mathcal{H}^{\frac{1}{3}}}+\dfrac{2^{2/3}C}{\lambda}\left\|\varphi_{3}\right\|_{\mathcal{H}}$
$\displaystyle\leq\max\left\\{1,2^{1/3}C,2^{2/3}C\right\\}\dfrac{1}{\lambda},$
$\displaystyle\|v\|_{\mathcal{H}^{\frac{1}{3}}}$
$\displaystyle\leq\left\|A(\lambda^{3}I+A)^{-1}\varphi_{1}\right\|_{\mathcal{H}^{\frac{1}{3}}}+\left\|\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{2}\right\|_{\mathcal{H}^{\frac{1}{3}}}+\lambda\left\|(\lambda^{3}I+A)^{-1}\varphi_{3}\right\|_{\mathcal{H}^{\frac{1}{3}}}$
$\displaystyle\leq\dfrac{2^{2/3}C}{\lambda}\left\|\varphi_{1}\right\|_{\mathcal{H}^{\frac{2}{3}}}+\dfrac{1}{\lambda}\left\|\varphi_{2}\right\|_{\mathcal{H}^{\frac{1}{3}}}+\dfrac{2^{1/3}C}{\lambda}\left\|\varphi_{3}\right\|_{\mathcal{H}}$
$\displaystyle\leq\max\left\\{1,2^{1/3}C,2^{2/3}C\right\\}\dfrac{1}{\lambda}$

and

$\displaystyle\|w\|_{\mathcal{H}}$ $\displaystyle\leq\left\|\lambda
A(\lambda^{3}I+A)^{-1}\varphi_{1}\right\|_{\mathcal{H}}+\left\|A(\lambda^{3}I+A)^{-1}\varphi_{2}\right\|_{\mathcal{H}}+\left\|\lambda^{2}(\lambda^{3}I+A)^{-1}\varphi_{3}\right\|_{\mathcal{H}}$
$\displaystyle\leq\dfrac{2^{1/3}C}{\lambda}\left\|\varphi_{1}\right\|_{\mathcal{H}^{\frac{2}{3}}}+\dfrac{2^{2/3}C}{\lambda}\left\|\varphi_{2}\right\|_{\mathcal{H}^{\frac{1}{3}}}+\dfrac{1}{\lambda}\left\|\varphi_{3}\right\|_{\mathcal{H}}$
$\displaystyle\leq\max\left\\{1,2^{1/3}C,2^{2/3}C\right\\}\dfrac{1}{\lambda}$

Hence,

$\|u\|_{\mathcal{H}^{\frac{2}{3}}}+\|v\|_{\mathcal{H}^{\frac{1}{3}}}+\|w\|_{\mathcal{H}}\leqslant\frac{M_{\mathbb{A}}}{\lambda},$

with $M_{\mathbb{A}}=3\max\left\\{1,2^{1/3}C,2^{2/3}C\right\\}$. ∎

According to Lemma 2.4 and the terminology of [19, Definition 1.1.1], the
operator $\mathbb{A}$ is a non-negative type, see also [20, P. 686]. As a
particular case of this definition we see that an operator is m-accretive if
and only it is non-negative type and its non-negative constant is equal to
$1$, see [19, Theorem 1.3.3].

Now, if we assume fuhrer, $0\in\rho(A)$, then there exists $0<\varepsilon\leq
1$ such that

${\rm Re}\left\langle
Ax,x\right\rangle_{\mathcal{H}}\geq\varepsilon\left\|x\right\|^{2}_{\mathcal{H}}\quad\text{
for all }x\in\mathcal{D}(A).$

Moreover, for every $\mu\geq 0$, we have

$(\mu+\varepsilon)\left\|x\right\|^{2}_{\mathcal{H}}\leq{\rm
Re}\left\langle(\mu+A)x,x\right\rangle_{\mathcal{H}}\geq\left\|(\mu+A)x\right\|_{\mathcal{H}}\left\|x\right\|_{\mathcal{H}}\quad\text{
for all }x\in\mathcal{D}(A).$

implies,

$\left\|(\mu+A)^{-1}\right\|_{\mathcal{B}(\mathcal{H})}\leq\dfrac{1}{\mu+\varepsilon}.$

Thus,

$\left\|(\mu+A)^{-1}\right\|_{\mathcal{B}(\mathcal{H})}\leq\dfrac{1}{\mu+1},\text{
for all }\mu\geq 0.$

Using this last inequality, Lemma 2.4, reads as follows

######  Corollary 2.5.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$ and
$0\in\rho(A)$. Then, $\mathbb{A}$ is closed, densely defined,
$\left(-\infty,0\right]\subset\rho(\mathbb{A})$ and there exists a positive
constant $K$, such that

$\|(\lambda
I+\mathbb{A})^{-1}\|_{\mathcal{B}(X)}\leqslant\frac{K}{\lambda+1},\ \mbox{for
all}\ \lambda\geq 0.$ (2.14)

The operator $\mathbb{A}$ is called of positive type; [19, Definition 1.1.2]
and [19, Proposition 1.1.1]. Also a boundedly invertible m-accretive operator
is of positive type.

It is well known that, the generator of a strongly continuous semigroup which
decays exponentially is of positive type, see Amann [2, Page 156]. However, it
possible that an operator of positive type do not generate a semigroup of any
kind, see Remark 2.3-(5).

######  Theorem 2.6.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$ and
$0\in\rho(A)$. Then the fractional powers $\mathbb{A}^{\alpha}$ of
$\mathbb{A}$ can be defined for $0\leqslant\alpha\leqslant 1$ by
$\mathbb{A}^{\alpha}:D(\mathbb{A}^{\alpha})\subset X\to X$,

$\mathcal{D}(\mathbb{A}^{\alpha})=\mathcal{H}^{\frac{\alpha+2}{3}}\times\mathcal{H}^{\frac{\alpha+1}{3}}\times\mathcal{H}^{\frac{\alpha}{3}},$

and

$\mathbb{A}^{\alpha}=\dfrac{1}{3}\begin{bmatrix}\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}&-\varUpsilon_{2}^{\alpha}A^{\frac{\alpha-1}{3}}&\varUpsilon_{1}^{\alpha}A^{\frac{\alpha-2}{3}}\\\
&&\\\
-\varUpsilon_{1}^{\alpha}A^{\frac{\alpha+1}{3}}&\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}&-\varUpsilon_{2}^{\alpha}A^{\frac{\alpha-1}{3}}\\\
&&\\\
\varUpsilon_{2}^{\alpha}A^{\frac{\alpha+2}{3}}&-\varUpsilon_{1}^{\alpha}A^{\frac{\alpha+1}{3}}&\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}\end{bmatrix}$
(2.15)

where,

$\varUpsilon_{j}^{\alpha}=2\cos\dfrac{2\pi(\alpha+j)}{3}+1,\quad\text{ for
}j=0,1,2.$ (2.16)

###### Proof.

By Lemma 2.4, the operator $\mathbb{A}$ is a non-negative type, hence we can
define the fractional powers $\mathbb{A}^{\alpha}$ for $0<\alpha<1$ by
$\mathbb{A}^{\alpha}:D(\mathbb{A}^{\alpha})\subset X\to X$, and

$\mathbb{A}^{\alpha}=\dfrac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\lambda^{\alpha-1}\mathbb{A}(\lambda
I+\mathbb{A})^{-1}d\lambda.$

On the other hand,

$(\lambda
I+\mathbb{A})^{-1}=\begin{bmatrix}\lambda^{2}(\lambda^{3}I+A)^{-1}&\lambda(\lambda^{3}I+A)^{-1}&(\lambda^{3}I+A)^{-1}\\\
-A(\lambda^{3}I-A)^{-1}&\lambda^{2}(\lambda^{3}I+A)^{-1}&-\lambda(\lambda^{3}I+A)^{-1}\\\
-\lambda
A(\lambda^{3}I+A)^{-1}&-A(\lambda^{3}I+A)^{-1}&\lambda^{2}(\lambda^{3}I+A)^{-1}\end{bmatrix}$

for all $\lambda>0$. Consequently,

$\mathbb{A}(\lambda
I+\mathbb{A})^{-1}=\begin{bmatrix}A(\lambda^{3}I+A)^{-1}&-\lambda^{2}(\lambda^{3}I+A)^{-1}&-\lambda(\lambda^{3}I+A)^{-1}\\\
\lambda
A(\lambda^{3}I-A)^{-1}&A(\lambda^{3}I+A)^{-1}&-\lambda^{2}(\lambda^{3}I+A)^{-1}\\\
\lambda^{2}A(\lambda^{3}I+A)^{-1}&\lambda
A(\lambda^{3}I+A)^{-1}&A(\lambda^{3}I+A)^{-1}\end{bmatrix}$

for all $\lambda>0$. Now we apply in each entry $[\mathbb{A}(\lambda
I+\mathbb{A})^{-1}]_{ij}$ the fractional formula for $A$ given by (1.4), and
after the change of variable $\mu=\lambda^{3}$ ($\lambda=\mu^{\frac{1}{3}}$
and $d\lambda=\frac{1}{3}\mu^{\frac{-2}{3}}d\mu$) for $i\geqslant j$, we
obtain the following:

$\begin{split}[\mathbb{A}(\lambda
I+\mathbb{A})^{-1}]_{ij}&=\dfrac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\lambda^{i-j}\lambda^{\alpha-1}A(\lambda^{3}I+A))^{-1}d\lambda\\\
&=\dfrac{1}{3}\dfrac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\mu^{\frac{i-j}{3}}\mu^{\frac{\alpha}{3}-1}A(\mu
I+A)^{-1}d\mu\\\
&=\dfrac{(-1)^{i-j}}{3}\dfrac{\sin\Big{(}(\frac{\alpha+i-j}{3})3\pi\Big{)}}{\pi}\int_{0}^{\infty}\mu^{\frac{\alpha+i-j}{3}-1}A(\mu
I+A)^{-1}d\mu,\end{split}$ (2.17)

Now, for any $\theta$, we have,

$\begin{split}\sin(3\theta\pi)&=\sin(\theta\pi+2\theta\pi)=\sin(\theta\pi)\cos(2\theta\pi)+\sin(2\theta\pi)\cos(\theta\pi)\\\
&=\sin(\theta\pi)(\cos(2\theta\pi)+2\cos^{2}(\theta\pi))\\\
&=\sin(\theta\pi)(2\cos(2\theta\pi)+1)\end{split}$ (2.18)

Thus, for $\theta=\frac{\alpha+i-j}{3}$, we obtain

$\sin\Big{(}(\frac{\alpha+i-j}{3})3\pi\Big{)}=\Big{(}2\cos\Big{(}(\frac{\alpha+i-j}{3})\pi\Big{)}+1\Big{)}\sin\Big{(}(\frac{\alpha+i-j}{3})\pi\Big{)}$
(2.19)

and by (2.17), (2.19) and $0<\alpha+i-j<3$, we have

$[\mathbb{A}(\lambda
I+\mathbb{A})^{-1}]_{ij}=\dfrac{(-1)^{i-j}}{3}\Big{(}2\cos\Big{(}(\frac{\alpha+i-j}{3})\pi\Big{)}+1\Big{)}A^{\frac{\alpha+i-j}{3}}.$
(2.20)

Similarly, for $i<j$, we obtain the following:

$\begin{split}[\mathbb{A}(\lambda
I+\mathbb{A})^{-1}]_{ij}=&-\dfrac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\lambda^{\alpha+3+i-j-1}(\lambda^{3}I+A))^{-1}d\lambda\\\
&=-\dfrac{1}{3}\dfrac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\mu^{\frac{\alpha+i-j}{3}}(\mu
I+A)^{-1}d\mu\\\
&=\dfrac{(-1)^{i-j+4}}{3}\dfrac{\sin\Big{(}(\frac{\alpha+i-j+3}{3})3\pi\Big{)}}{\pi}\int_{0}^{\infty}\mu^{\frac{\alpha+i-j+3}{3}-1}(\mu
I+A)^{-1}d\mu,\end{split}$ (2.21)

As above from (2.21), (2.18) and $0<\alpha+i-j+3<3$, we get

$[\mathbb{A}(\lambda
I+\mathbb{A})^{-1}]_{ij}=\dfrac{(-1)^{i-j}}{3}\Big{(}2\cos\Big{(}(\frac{\alpha+i-j}{3})\pi\Big{)}+1\Big{)}A^{\frac{\alpha+i-j}{n}}.$
(2.22)

and thanks to (2.20) and (2.22) we get (2.15) for $0<\alpha<1$.

Finally, it is easy to see that the matrix representation of
$\mathbb{A}^{\alpha}$ holds for any $0\leqslant\alpha\leqslant 1$. ∎

######  Remark 2.7.

  1. (1)

If we take $\omega=0$, we obtain [8, Theorem 2.7].

  2. (2)

The assumption that $0\in\rho(A)$ and therefore a whole neighborhood of zero
is in $\rho(A)$ was made mainly for convenience. The Theorem 2.6 remains true
even if $0\notin\rho(A)$.

  3. (3)

We can see that the matrix representation of $\mathbb{A}^{\alpha}$ is a
Toeplitz matrix for any $0\leqslant\alpha\leqslant 1$.

The next result is an immediate consequence of [20, Theorem 3.6], see also
[20, Theorem 3.2] and [20, Theorem 3.4].

######  Corollary 2.8.

$\sigma(\mathbb{A}^{\alpha})=\\{\lambda^{\alpha}:\lambda\in\sigma(\mathbb{A})\\}=\\{\lambda^{\alpha}:\lambda^{3}\in\sigma(A)\\}$
(2.23)

for any $0<\alpha<1$.

We have, for every $r\geq 0$,

$re^{i\theta}\in\sigma(A)\Longleftrightarrow
r^{\alpha/3}e^{i\frac{\alpha\theta}{3}},r^{\alpha/3}e^{\alpha
i\frac{\theta+2\pi}{3}},r^{\alpha/3}e^{i\alpha\frac{\theta-2\pi}{3}}\in\sigma(\mathbb{A}^{\alpha}).$

Thus, we can see that the spectrum of $\mathbb{A}^{\alpha}$ is a reunion of
possibly three disjointed closed parts $\varGamma_{1}$, $\varGamma_{2}$ and
$\varGamma_{3}$, with

$\varGamma_{1}\subset\overline{\mathcal{S}(\alpha\omega/3)},\qquad\varGamma_{2}\subset
e^{i\alpha\frac{2\pi}{3}}\overline{\mathcal{S}(\alpha\omega/3)},\qquad\text{
and }\qquad\varGamma_{3}\subset
e^{-i\alpha\frac{2\pi}{3}}\overline{\mathcal{S}(\alpha\omega/3)}.$ (2.24)

Consequently,

$\mathcal{S}(\pi-\frac{\alpha\omega}{3})\cup\left\\{\lambda\in\mathbb{C}:\frac{\pi}{2}-\frac{\alpha\omega}{3}<\arg(\lambda)<\frac{\pi}{2}\right\\}\cup\left\\{\lambda\in\mathbb{C}:-\frac{\pi}{2}<\arg(\lambda)<-\frac{\pi}{2}-\frac{\alpha\omega}{3}\right\\}\subset\rho(\mathbb{A}^{\alpha}).$

From [2, Remark 4.6.12], we can deduce that $-\mathbb{A}^{\alpha}$ generates a
strongly continuous analytic semigroup on $X$ for $0<\alpha\leq\frac{1}{2}$.
In the following proposition we prove that $\alpha$ is given in term of the
angle $\omega$ of the sector for which $A$ is m-$\omega$ accretive. This is a
particular interest because $\alpha$ maybe strictly greater than to $1/2$. For
example, if $A$ is a self adjoint operator, i.e $\omega=0$, then
$-\mathbb{A}^{\alpha}$ generates a strongly continuous analytic semigroup on
$X$ for $0<\alpha<\frac{3}{4}$. The last case is proved in [8, Theorem 2.8].
Our main result cover this case and we give more direct proof.

######  Proposition 2.9.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$ and
$0\in\rho(A)$. Set $\alpha^{*}=\dfrac{3\pi}{4\pi+2\omega}$. Then
$-\mathbb{A}^{\alpha}$ generates a strongly continuous semigroup on $X$ for
$\alpha=\alpha^{*}$ and generates a strongly continuous analytic semigroup on
$X$ for $0<\alpha<\alpha^{*}$.

###### Proof.

Firstly, $\mathbb{A}^{\alpha}$ is closed densely defined and there exists a
positive constant $K$, such that

$\|(\lambda
I+\mathbb{A}^{\alpha})^{-1}\|_{\mathcal{B}(X)}\leqslant\frac{K}{\lambda+1},\
\mbox{for all}\ \lambda\geq 0.$

Form inclusions in (2.24), we can see that the two boundary rays of the sector
$e^{i\alpha\frac{2\pi}{3}}\overline{\mathcal{S}(\alpha\omega/3)}$ are the
complex numbers of argument $\alpha\frac{2\pi}{3}+\frac{\alpha\omega}{3}$ and
$\alpha\frac{2\pi}{3}-\frac{\alpha\omega}{3}$, respectively. To forces this
sector to be in the closed right complex half plane it suffices that
$\alpha\frac{2\pi}{3}+\frac{\alpha\omega}{3}\leq\frac{\pi}{2}$. That is
$\alpha\leq\dfrac{3\pi}{2(2\pi+\omega)}$. The same argument for the sector
$e^{-i\alpha\frac{2\pi}{3}}\overline{\mathcal{S}(\alpha\omega/3)}$ in the
opposite direction. Consequently,

$\sigma(\mathbb{A}^{\alpha})\subset\overline{\mathcal{S}(\pi/2)}\qquad\text{
for }\alpha=\dfrac{3\pi}{4\pi+2\omega}$

and

$\sigma(\mathbb{A}^{\alpha})\subset\overline{\mathcal{S}(\varpi)}\qquad\text{
with }\qquad\varpi<\pi/2\qquad\text{ for }\alpha<\dfrac{3\pi}{4\pi+2\omega}.$

Therefore,

$\mathcal{S}(\pi/2)\subset\rho(-\mathbb{A}^{\alpha})\qquad\text{ for
}\alpha=\dfrac{3\pi}{4\pi+2\varpi}$

and

$\mathcal{S}(\frac{\pi}{2}+\varpi)\subset\rho(-\mathbb{A}^{\alpha})\qquad\text{
with }\qquad\varpi<\pi/2\qquad\text{ for }\alpha<\dfrac{3\pi}{4\pi+2\omega}.$

∎

######  Remark 2.10.

We include below figures 2 and 3 which reflects, in particular, the loss of a
“good” sectoriality property as $\alpha\nearrow\alpha^{*}$. It is remarkable
to see that when $\alpha$ decreases from one there is a rotation of the two
sectors located in the left half-plane given in figure 1 to be in the right
half-plane and at the same time the angle of sectoriality decreases. Moreover,
e.g. for $\alpha^{*}<\alpha\leqslant 1$ the figures 1 and 2 reflects, in
particular, the loss of well-posedness of the Cauchy problem (2.1) and hence
those of (1.1)-(1.2).

$\scriptstyle{\rm Re}(\lambda)$$\scriptstyle{\rm Im}(\lambda)$Sectors
containing the spectrum of $\mathbb{A}^{\alpha^{*}}$ Figure 2. Location of the
spectrum of $\mathbb{A}^{\alpha^{*}}$. $\scriptstyle{\rm
Re}(\lambda)$$\scriptstyle{\rm Im}(\lambda)$Sectors containing the spectrum of
$\mathbb{A}^{\alpha}$ Figure 3. Location of the spectrum of
$\mathbb{A}^{\alpha}$ with $0<\alpha<\alpha^{*}$.

##  3\. A third order Cauchy problem associated to fractional powers of the
linearized operator

From Theorem 2.6, we showed that it is possible to calculate explicitly the
fractional power $\mathbb{A}^{\alpha}$ of the operator $\mathbb{A}$ for
$0<\alpha<1$. Furthermore, $-\mathbb{A}^{\alpha}$ generates a strongly
continuous analytic semigroup on $X$ for $0<\alpha<\alpha^{*}$, by Proposition
2.9. This enables us to consider the following fist order Cauchy problem
associated to the fractional power of the linearized operator,

$\begin{cases}\dfrac{d{\bf u}}{dt}+\mathbb{A}^{\alpha}{\bf u}=0,\ t>0,\quad
0<\alpha<\alpha^{*},\\\ {\bf u}(0)={\bf u}_{0}.\end{cases}$ (3.1)

Here, $\mathbb{A}^{\alpha}:\mathcal{D}(\mathbb{A}^{\alpha})\subset X\to X$
denotes the fractional power operator of $\mathbb{A}$.

By Proposition 2.9 and [21, Corollary 1.5, Chap 4.], we can assert that for
any ${\bf u}_{0}\in X$ the problem (3.1) admits unique solution given by ${\bf
u}(t)=e^{-t\mathbb{A}^{\alpha}}{\bf u}_{0}$ for all $0<\alpha<\alpha^{*}$.
Also, by Proposition 2.9 and [21, Theorem 1.3, Chap 4.], the initial value
problem (3.1) has a unique solution ${\bf u}(t)=\mathcal{T}(t){\bf u}_{0}$,
for every initial value ${\bf u}_{0}\in\mathcal{D}(\mathbb{A}^{\alpha^{*}})$,
where $\mathcal{T}(t)$ is the strongly continuous semigroup generated by
$\mathbb{A}^{\alpha^{*}}$. This allows us to define the associated third order
Cauchy problem for which (3.1) will be it associated linearized problem.

$u^{\prime\prime\prime}(t)+3\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}u^{\prime\prime}(t)+3\varUpsilon_{0}^{\alpha}A^{\frac{2\alpha}{3}}u^{\prime\prime}(t)+A^{\alpha}u(t)=0,\quad
t>0,$ (3.2)

with initial conditions given by

$u(0)=\varphi,\ u^{\prime}(0)=\psi,\ u^{\prime\prime}(0)=\xi,$ (3.3)

where $\varUpsilon_{0}^{\alpha}=2\cos\dfrac{2\pi\alpha}{3}+1$ and
$0<\alpha<\alpha^{*}=\dfrac{3\pi}{4\pi+2\omega}$.

######  Theorem 3.1.

Assume that $A$ be m-$\omega$-accretive operator, $0\leq\omega\leq\pi/2$ and
$0\in\rho(A)$. Let $0<\alpha<\alpha^{*}$. Then for every
$\varphi\in\mathcal{H}^{\frac{2}{3}}$,
$\psi\in\mathcal{H}^{\frac{2-\alpha}{3}}$ and
$\xi\in\mathcal{H}^{\frac{2-2\alpha}{3}}$ the Cauchy problem (3.2)-(3.3) has a
unique solution

$u\in C^{3}((0,+\infty);\mathcal{H})\cap
C^{2}((0,+\infty);\mathcal{H}^{\frac{\alpha}{3}})\cap
C^{1}((0,+\infty);\mathcal{H}^{\frac{2\alpha}{3}})\cap
C((0,+\infty);\mathcal{H}^{\alpha}),$

and given by

$\displaystyle
u(t)=e^{taA^{\frac{\alpha}{3}}}\varphi-\dfrac{a}{b-a}\Big{[}e^{tbA^{\frac{\alpha}{3}}}-e^{taA^{\frac{\alpha}{3}}}\Big{]}\varphi$
(3.4)
$\displaystyle+\dfrac{ab}{1+b}\Big{[}\dfrac{1}{b-a}e^{tbA^{\frac{\alpha}{3}}}-\dfrac{1+b}{(1+a)(b-a)}e^{taA^{\frac{\alpha}{3}}}+\dfrac{1}{1+a}e^{-tA^{\frac{\alpha}{3}}}\Big{]}\varphi$
$\displaystyle+\dfrac{1}{b-a}A^{-\frac{\alpha}{3}}\Big{[}e^{tbA^{\frac{\alpha}{3}}}-e^{taA^{\frac{\alpha}{3}}}\Big{]}\psi$
$\displaystyle-\dfrac{a+b}{1+b}A^{-\frac{\alpha}{3}}\Big{[}\dfrac{1}{b-a}e^{tbA^{\frac{\alpha}{3}}}-\dfrac{1+b}{(1+a)(b-a)}e^{taA^{\frac{\alpha}{3}}}+\dfrac{1}{1+a}e^{-tA^{\frac{\alpha}{3}}}\Big{]}\psi$
$\displaystyle+\dfrac{1}{1+b}A^{-\frac{2\alpha}{3}}\Big{[}\dfrac{1}{b-a}e^{tbA^{\frac{\alpha}{3}}}-\dfrac{1+b}{(1+a)(b-a)}e^{taA^{\frac{\alpha}{3}}}+\dfrac{1}{1+a}e^{-tA^{\frac{\alpha}{3}}}\Big{]}\xi,$

with $a=e^{i\pi{\frac{3-2\alpha}{3}}}$ and $b=e^{-i\pi{\frac{3+2\alpha}{3}}}$.

###### Proof.

First, the Cauchy problem (3.2)-(3.3) is equivalent to the first order problem

$\dfrac{d}{dt}\left[\begin{smallmatrix}u\\\ v\\\
w\end{smallmatrix}\right]+\mathbb{A}^{\alpha}\left[\begin{smallmatrix}u\\\
v\\\ w\end{smallmatrix}\right]=0,\ t>0,\quad 0<\alpha<\alpha^{*},\\\ $ (3.5)

subject to the initial conditions

$\begin{cases}u(0)=\varphi\in\mathcal{H}^{\frac{2}{3}}\\\
v(0)=\dfrac{1}{(\varUpsilon_{1}^{\alpha})^{3}-(\varUpsilon_{0}^{\alpha})^{3}}\Big{[}\varUpsilon_{1}^{\alpha}(\varUpsilon_{2}^{\alpha}-\varUpsilon_{0}^{\alpha})A^{\frac{1}{3}}\varphi-(\varUpsilon_{2}^{\alpha}+3\varUpsilon_{0}^{\alpha}\varUpsilon_{1}^{\alpha})A^{\frac{1-\alpha}{3}}\psi-\varUpsilon_{1}^{\alpha}A^{\frac{1-2\alpha}{3}}\xi\Big{]}\in\mathcal{H}^{\frac{1}{3}},\\\
w(0)=\dfrac{1}{(\varUpsilon_{2}^{\alpha})^{3}-(\varUpsilon_{1}^{\alpha})^{3}}\Big{[}\varUpsilon_{2}^{\alpha}(\varUpsilon_{0}^{\alpha}-\varUpsilon_{1}^{\alpha})A^{\frac{2}{3}}\varphi+(\varUpsilon_{1}^{\alpha}+3\varUpsilon_{0}^{\alpha}\varUpsilon_{2}^{\alpha})A^{\frac{2-\alpha}{3}}\psi-\varUpsilon_{2}^{\alpha}A^{\frac{1-2\alpha}{3}}\xi\Big{]}\in\mathcal{H},\end{cases}$

where $\mathbb{A}^{\alpha}$ is given by (2.15) and $\varUpsilon_{j}^{\alpha}$
are defined by (2.16) $0<\alpha<\alpha^{*}$. Obviously, (3.5) is equivalent to
the following system

$\begin{cases}u^{\prime}(t)+\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}u(t)-\varUpsilon_{2}^{\alpha}A^{\frac{\alpha-1}{3}}v(t)+\varUpsilon_{1}^{\alpha}A^{\frac{\alpha-2}{3}}w(t)=0\\\
v^{\prime}(t)-\varUpsilon_{1}^{\alpha}A^{\frac{\alpha+1}{3}}u(t)-\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}v(t)-\varUpsilon_{2}^{\alpha}A^{\frac{\alpha-1}{3}}w(t)=0\\\
w^{\prime}(t)-\varUpsilon_{2}^{\alpha}A^{\frac{\alpha+2}{3}}u(t)-\varUpsilon_{1}^{\alpha}A^{\frac{\alpha+1}{3}}v(t)+\varUpsilon_{0}^{\alpha}A^{\frac{\alpha}{3}}w(t)=0.\end{cases}$
(3.6)

By straightforward calculus, we obtain (3.2) subject to (3.3). The existence
and the uniqueness result follows from Proposition 2.9 and [21, Corollary 1.5,
Chap 4.]. Also, we have

$\left[\begin{smallmatrix}u(t)\\\ v(t)\\\
w(t)\end{smallmatrix}\right]\in\mathcal{D}(\mathbb{A}^{\alpha})=\mathcal{H}^{\frac{\alpha+2}{3}}\times\mathcal{H}^{\frac{\alpha+1}{3}}\times\mathcal{H}^{\frac{\alpha}{3}},\quad
t\geq 0,\quad 0<\alpha<\alpha^{*}.$

Consequently, by (3.6), we get

$u(t)\in\mathcal{H}^{\frac{\alpha+2}{3}},u^{\prime}(t)\in\mathcal{H}^{\frac{\alpha}{3}},u^{\prime\prime}(t)\in\mathcal{H}^{\frac{2-\alpha}{3}},\quad
t\geq 0,\quad 0<\alpha<\alpha^{*}.$

On the other hand, (3.2) it can be written as the equivalent Cauchy problem
for the system of first order differential equations

$\begin{gathered}u^{\prime}(t)-aBu(t)=f(t),\\\ f^{\prime}(t)-bBf(t)=g(t),\\\
g^{\prime}(t)+Bg(t)=0,\quad t>0,\end{gathered}$ (3.7)

where $B=A^{\frac{\alpha}{3}}$, $a=e^{i\pi{\frac{3-2\alpha}{3}}}$ and
$b=e^{-i\pi{\frac{3+2\alpha}{3}}}$. Since the operators $aB$, $bB$ and $-B$
are generators of strongly continuous analytic semigroups, then by integrating
these equations, we can write

$\begin{gathered}g(t)=e^{-tB}g(0),\\\
f(t)=e^{tbB}f(0)+\int_{0}^{t}e^{(t-s)bB}g(s)ds,\\\
u(t)=e^{taB}u(0)+\int_{0}^{t}e^{(t-s)aB}f(s)ds.\end{gathered}$ (3.8)

Applying system of equations (3.7), we obtain

$\displaystyle f(0)=u^{\prime}(0)-aBu(0),$ $\displaystyle
g(0)=v^{\prime}(0)-bBv(0)=u^{\prime\prime}(0)-(a+b)Bu^{\prime}(0)+abB^{2}u(0).$

The fact that
$u(0)\in\mathcal{H}^{\frac{2}{3}}\subset\mathcal{H}^{\frac{\alpha}{3}}$,
$u^{\prime}(0)\in\mathcal{H}^{\frac{2-\alpha}{3}}\subset\mathcal{H}^{\frac{\alpha}{3}}$
and
$u^{\prime\prime}(0)\in\mathcal{H}^{\frac{2-\alpha}{3}}\subset\mathcal{H}^{\frac{2\alpha}{3}}$,
then, we have $g(0)\in\mathcal{H}^{\frac{2\alpha}{3}}$,
$f(0)\in\mathcal{H}^{\frac{\alpha}{3}}$ and

$g(t)=e^{-tB}[u^{\prime\prime}(0)-(a+b)Bu^{\prime}(0)+abB^{2}u(0)].$ (3.9)

Using formulas (3.8), (3.9), we obtain

$\displaystyle f(t)$ $\displaystyle=e^{tbB}(u^{\prime}(0)-aBu(0))$ (3.10)
$\displaystyle\quad+\frac{1}{1+b}B^{-1}\Big{(}e^{tbB}-e^{-tB}\Big{)}(u^{\prime\prime}(0)-(a+b)Bu^{\prime}(0)+abB^{2}u(0)).$

Using formulas (3.8), (3.10), we obtain by integration,

$\displaystyle
u(t)=e^{taB}\varphi+\dfrac{1}{b-a}B^{-1}\Big{[}e^{tbB}-e^{taB}\Big{]}(\psi-
aB\varphi)$ (3.11)
$\displaystyle\dfrac{1}{1+b}B^{-2}\Big{[}\dfrac{1}{b-a}e^{tbB}-\dfrac{1+b}{(1+a)(b-a)}e^{taB}+\dfrac{1}{1+a}e^{-tB}\Big{]}(\xi-(a+b)B\psi+abB\varphi),$

Now, the formula (3.4) holds after some manipulations. ∎

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