# Control of Kawahara equation using flat outputs

Roberto de A. Capistrano–Filho∗<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Jandeilson Santos da Silva  Departamento de
Matemática, Universidade Federal de Pernambuco, S/N Cidade Universitária,
50740-545, Recife (PE), Brazil

###### Abstract.

In this study we focused on the linear Kawahara equation in a bounded domain,
employing two boundary controls. The controllability of this system has been
previously demonstrated over the past decade using the Hilbert uniqueness
method which involves proving an observability inequality, in general,
demonstrated via Carleman estimates. Here, we extend this understanding by
achieving the exact controllability within a space of analytic functions,
employing the flatness approach which is a new approach for higher-order
dispersive systems.

###### Key words and phrases:

Kawahara equation; controllability to the trajectories; flatness approach;
Gevrey class; smoothing effect.

###### 2020 Mathematics Subject Classification:

37L50, 93B05

∗Corresponding author<EMAIL_ADDRESS>
Capistrano–Filho was partially supported by CAPES/PRINT grant number
88887.311962/2018-00, CAPES/COFECUB grant number 88887.879175/2023-00, CNPq
grant numbers 421573/2023-6, 307808/2021-1 and 401003/2022-1 and Propesqi
(UFPE). Da Silva acknowledges support from CNPq.

##  1\. Introduction

###  1.1. Background and literature review

Many scientists from diverse fields, including hydraulic engineering, fluid
mechanics, physics, and mathematics, have studied water wave models. These
models are generally difficult to derive and complex to analyze for
qualitative information about wave dynamics, making their study interesting
and challenging. Recently, researchers have used appropriate assumptions on
amplitude, wavelength, wave steepness, and other factors to investigate
asymptotic models for water waves and gain insights into the full water wave
system. For rigorous justification of various asymptotic models for surface
and internal waves, see, for instance, [1, 7, 26] and references therein.

In an appropriate non-dimensional form, water waves can be modeled as a free
boundary problem of the incompressible, irrotational Euler equation. This
involves two non-dimensional parameters: $\delta:=\frac{h}{\lambda}$ and
$\varepsilon:=\frac{a}{h}$, where the water depth, the wavelength, and the
amplitude of the free surface are respectively denoted by $h,\lambda$ and $a$.
Additionally, the parameter $\mu$, known as the Bond number, measures the
relative importance of gravitational forces compared to surface tension
forces. Long waves, also known as shallow water waves, are characterized by
the condition $\delta\ll 1$. There are various long-wave approximations
depending on the relationship between $\varepsilon$ and $\delta$.

The discussion above suggests that, instead of relying on models with poor
asymptotic properties, one can rescale the mentioned parameters to find
systems that reveal asymptotic models for surface and internal waves, such as
the Kawahara model. Specifically, by setting $\varepsilon=\delta^{4}\ll 1$,
$\mu=\frac{1}{3}+\nu\varepsilon^{\frac{1}{2}}$, considering the critical Bond
numbe $\mu=\frac{1}{3}$, the Kawahara equation is derived. This equation,
first introduced by Hasimoto and Kawahara [22, 20], takes the form

$\pm 2v_{t}+3vv_{x}-\nu v_{xxx}+\frac{1}{45}v_{xxxxx}=0,$

or, after re-scaling,

$v_{t}+\alpha v_{x}+\beta v_{xxx}-v_{xxxxx}+vv_{x}=0.$

This equation is also known as the fifth-order Korteweg-de Vries (KdV)
equation [8] or the singularly perturbed KdV equation [36]. It describes a
dispersive partial differential equation that encompasses various wave
phenomena, such as magneto-acoustic waves in a cold plasma [21], the
propagation of long waves in a shallow liquid beneath an ice sheet [19], and
gravity waves on the surface of a heavy liquid [15], among others.

Significant efforts in recent decades have aimed to understand this model
within various research frameworks. For instance, numerous studies have
focused on analytical and numerical methods for solving the equation. These
methods include the tanh-function method [5], extended tanh-function method
[6], sine-cosine method [37], Jacobi elliptic functions method [23], direct
algebraic method and numerical simulations [35], decomposition methods [25],
as well as variational iteration and homotopy perturbation methods [24].
Another important research direction is the study of the Kawahara equation
from the perspective of control theory, specifically addressing the boundary
controllability problem [18], which is our motivation.

So, in this context, we are interested in the boundary controllability issue
of the Kawahara equation in a bounded domain. Precisely, we investigate the
linear control problem

(1.5)
$\displaystyle\left\\{\begin{array}[]{lll}u_{t}+u_{x}+u_{xxx}-u_{xxxxx}=0,&(x,t)\text{
in }(-1,0)\times(0,T),\\\ u(0,t)=u_{x}(0,t)=u_{xx}(0,t)=0,&t\text{ in
}(0,T),\\\ u(-1,t)=h_{1}(t),\ \ u_{x}(-1,t)=h_{2}(t),&t\text{ in }(0,T),\\\
u(x,0)=u_{0}(x),&x\text{ in }(-1,0),\end{array}\right.$

where $h_{1},h_{2}$ are the controls input and $u$ is the state function. This
work addresses two main issues:

Problem $\mathcal{A}$: Null controllability. Given an initial data $u_{0}$ in
a suitable space, is it possible to find control functions $h_{1},h_{2}$ such
that the state solution $u$ of the system (1.5) $u(x,T)=0$?

Problem $\mathcal{B}$: Reachable functions. Can we find a space $\mathcal{R}$
with the property that, if the final data $u_{1}\in\mathcal{R}$ then one can
get control functions $h_{1},h_{2}$ such that the solution $u$ of the system
(1.5) with $u_{0}=0$ satisfies $u(\cdot,T)=u_{1}$?

It is important to highlight several results related to control problems
associated with the system (1.5). Here are some of them. Regarding the
analysis of the Kawahara equation in a bounded interval, pioneering work was
done by Silva and Vasconcellos [38, 39]. They studied the stabilization of
global solutions of the linear Kawahara equation in a bounded interval under
the influence of a localized damping mechanism. The second contribution in
this area was made by Capistrano-Filho et al. [2], who considered a
generalized Kawahara equation in a bounded domain.

Glass and Guerrero [18] considered the problem

(1.6) $\left\\{\begin{array}[]{ll}y_{t}+\alpha
y_{5x}=\sum_{k=0}^{3}a_{k}(x,t)\partial_{x}^{k}y+h(x,t),&(x,t)\in(0,1)\times(0,T),\\\
y(0,t)=v_{1}(t),\ \ y(1,t)=v_{2}(t),\ \
y_{x}(0,t)=v_{3}(t),&t\in\times(0,T)\\\ y_{x}(1,t)=v_{4}(t),\ \
y_{xx}(0,t)=v_{5}(t),&t\in\times(0,T),\\\
y(x,0)=y_{0},&x\in(0,1),\end{array}\right.$

where $\alpha>0$ and $y_{0},h,v_{1},...,v_{5}$ are given functions. Using a
Carleman estimate, they demonstrated that the system (1.6) is null
controllable in the energy space $L^{2}(0,1)$ using only the controls on the
right side of the boundary, $v_{2}$ and $v_{4}$, meaning
$v_{1}=v_{3}=v_{5}=0$. However, the authors noted that the controllability
properties might fail if the set of controls is altered. For example, if
$v_{1}=v_{2}=v_{3}=v_{4}=0$, meaning only $v_{5}$ is used as a control input,
controllability does not occur because the adjoint system associated with
(1.6) may have unobservable solutions.

The internal controllability problem for the Kawahara equation with
homogeneous boundary conditions has been addressed by Chen [10]. Using
Carleman estimates associated with the linear operator of the Kawahara
equation with internal observation, a null controllability result was
demonstrated when the internal control is effective in a subdomain
$\omega\subset(0,L)$. In [9], the authors consider the Kawahara equation with
an internal control $f(t,x)$ and homogeneous boundary conditions, the equation
is shown to be exactly controllable in $L^{2}$-weighted Sobolev spaces.
Additionally, it is shown to be controllable by regions in $L^{2}$-Sobolev
space.

###  1.2. Flatness approach and main results

The flatness approach [16, 3, 17], also known as differential flatness,
represents a powerful concept in control theory and nonlinear system analysis.
It characterizes certain dynamical systems where all states and inputs can be
described as algebraic functions of a finite set of independent variables,
referred to as flat outputs, and their derivatives. This approach simplifies
the control and trajectory planning of complex systems. It has found extensive
application across various partial differential equations, including the heat
equation [30, 32, 12], 1-dimensional parabolic equations [31], the
1-dimensional Schrödinger equation [33], the linear KdV equation [34], and
more recently, the linear Zakharov-Kuznetsov equation [11]. Here are the key
aspects of the flatness approach:

  1. (1)

Flat outputs: In a flat system, outputs (flat outputs) describe the system’s
entire state and input trajectories using algebraic relationships involving a
finite number of their derivatives.

  2. (2)

Simplified control design: By utilizing flat outputs, complex nonlinear
control problems can be transformed into simpler linear problems. This
transformation simplifies the design of control laws and facilitates the
generation of desired trajectories.

  3. (3)

Trajectory planning: The flatness approach enables systematic trajectory
planning. Desired trajectories for flat outputs can be planned first, and then
corresponding state and input trajectories can be computed using the flatness
property.

  4. (4)

Real-world applications: The flatness approach has been successfully applied
across various fields including robotics, aerospace, process control, and
automotive systems. For instance, in robotics, it aids in trajectory design
for manipulators and mobile robots.

It is crucial to recognize the advantages of the flatness approach. Typically,
designing controls for nonlinear systems is complex; however, this approach
mitigates this complexity. Furthermore, flatness offers a systematic approach
to generate trajectories and control inputs and, as previously noted, is
applicable across a broad spectrum of dynamical systems.

Nevertheless, it is important to acknowledge the limitations and challenges
associated with this method. Identifying flat outputs for a specific system
can be demanding and necessitates a profound understanding of the system’s
dynamics. Additionally, not all systems exhibit flatness, which restricts the
universal applicability of this approach. In summary, the flatness approach
provides a robust framework for controlling and analyzing nonlinear dynamical
systems through the concept of flat outputs. It simplifies the design of
control laws and trajectory planning, making it a valuable tool across various
engineering applications.

Regarding the primary contribution of this paper, we advance the study of the
control problem for the fifth-order dispersive system (1.5). Unlike recent
works that utilize boundary controls and use the Hilbert uniqueness method,
introduced by Lions [29], to show control properties, this article achieves
the problems $\mathcal{A}$ and $\mathcal{B}$ using two control inputs via the
flatness approach. Let us now present the main results of this work.

Motivated by the smoothing effect exhibited by the equation (1.5) with free
evolution ($h_{1}=h_{2}=0$), we explore the reachable sets within smooth
function spaces, specifically, Gevrey spaces. First, let us introduce the
definition of these spaces.

######  Definition 1 (Gevrey spaces).

Given $s_{1},s_{2}\geq 0$, a function
$u:[a,b]\times[t_{1},t_{2}]\rightarrow\mathbb{R}$ is said to be Gevrey of
order $s_{1}$ on $[a,b]$ and $s_{2}$ on $[t_{1},t_{2}]$ if $u\in
C^{\infty}([a,b]\times[t_{1},t_{2}])$ and there exist positive constants
$C,R_{1}$ and $R_{2}$ such that

$\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}y(x,t)\right|\leq
C\frac{n!^{s_{1}}}{R_{1}^{n}}\frac{m!^{s_{2}}}{R_{2}^{m}},\ \forall n,m\geq
0,\ \forall(x,t)\in[a,b]\times[t_{1},t_{2}].$

The vectorial space of all functions on $[a,b]\times[t_{1},t_{2}]$ which are
Gevrey of order $s_{1}$ in $x$ and $s_{2}$ in $t$ is denoted by
$G^{s_{1},s_{2}}([a,b]\times[t_{1},t_{2}])$.

In this context, we investigate the null controllability problem associated
with (1.5) using the flatness approach to establish controllability
properties. The goal is to identify a set of functions in a Gevrey space that
are null controllable and to demonstrate that, at some intermediate time
$\tau\in(0,T)$, the solutions to the free evolution problem fall into this
set. Specifically, the first main result of this work can be stated as
follows.

######  Theorem 1.1 (Null controllability).

Let $s\in[\frac{5}{2},5)$ and $T>0$ be. Given $u_{0}\in L^{2}(-1,0)$ there
exist control inputs $h_{1},h_{2}\in G^{s}([0,T])$ such that the solution of
(1.5) belongs to the class $u\in C\left([0,T],L^{2}(-1,0)\right)\cap
G^{\frac{s}{5},s}([-1,0]\times[\varepsilon,T]),\ \forall\varepsilon\in(0,T),$
and satisfies $u(\cdot,T)=0$.

The previous result confirms that two flat outputs can be used to achieve null
controllability, thereby addressing Problem $\mathcal{A}$ presented at the
beginning of this work. Now, to present our second main result, we need to
introduce some notations. Given $z_{0}\in\mathbb{C}$ and $R>0$, we denote by
$D(z_{0},R)$ the open disk given by

$\displaystyle D(z_{0},R)=\left\\{z\in\mathbb{C};\ |z-z_{0}|<R\right\\}$

and $H(D(z_{0},R))$ denote the set of holomorphic functions on $D(z_{0},R)$.
Henceforth, we consider the operators

$Pu=\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u,$

where $P^{0}=I_{d}$ and $P^{n}=P\circ P^{n-1},$ when $n\geq 1.$ In this way,
the Kawahara equation can be expressed as

(1.7) $\displaystyle\partial_{t}u+Pu=0.$

Using induction and the fact that $\partial_{t}$ and $P$ commute we see that,
if $u=u(x,t)$ satisfies (1.7) then

(1.8) $\displaystyle\partial_{t}^{n}u+(-1)^{n-1}P^{n}u=0.$

For every $R>1$ we define the set

(1.10) $\displaystyle\mathcal{R}_{R}:=\left\\{u\in
C([-1,0]);\begin{array}[]{c}\exists z\in H(D(0,R));\ u=z|_{[-1,0]}\text{ and
}(\partial_{x}^{j}P^{n}u)(0)=0,\ j=0,1,2\end{array}\right\\},$

now on called a set of reachable states. The following result is the second
main result in this paper, answering the Problem $\mathcal{B}$.

######  Theorem 1.2.

Let $T>0$, $R_{0}:=2\cdot 6^{5^{-1}}\cdot e^{(5e)^{-1}}>1$ and $R>2R_{0}$ be.
Given $u_{1}\in\mathcal{R}_{R}$ there exist control inputs $h_{1},h_{2}\in
G^{5}([0,T])$ for which the solution $u$ of (1.5) with $u_{0}=0$ satisfies
$u(\cdot,T)=u_{1}$ and $u\in G^{1,5}([-1,0]\times[0,T])$.

###  1.3. Further comments and outline

Observe that Theorem 1.2 ensures that for the linear Kawahara equation, any
reachable state can likely be extended as a holomorphic function on some open
set in $\mathbb{C}$. Moreover, the reachable states corresponding to
controllability to the trajectories are in $G^{\frac{1}{2}}([-1,0])$, allowing
them to be extended as functions in $H(\mathbb{C})$. In contrast, the
reachable functions in Theorem 1.2 do not need to be holomorphic over the
entire set $\mathbb{C}$; they can have poles outside $D(0,R)$.

######  Remark 1.

Let us present some comments about our work.

  * i.

Our result is entirely linear and applies only to the linear system (1.5).
Therefore, a natural extension is to consider the nonlinear problem, which
includes the term $uu_{x}$. However, it is required to modify the method used
here. We believe that the strategy used in [27] could be adapted to our work,
though this remains an open problem.

  * ii.

Note that the set defined by (1.10) is an example of reachable functions,
though they are not completely understood. We believe there are other sets for
which Theorem 1.2 remains valid. For instance, in [11], the authors presented
another example for an extension of the KdV equation in a two-dimensional
case. Thus, it remains an open question to verify other sets where Theorem 1.2
holds.

  * iii.

As previously mentioned, several authors have applied the strategy of this
article to different systems. In our case, additional difficulties arise when
dealing with a fifth-order operator in space, specifically
$Pu=\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u$. Consequently,
addressing the smoothing properties is challenging because we are working with
five sets of different regularities and additional terms appear. Furthermore,
the set (1.10), and consequently, Theorem 1.2, can be obtained using the
strategies outlined in [34]. However, the Gevrey space level needs to be
adjusted, and the operator’s order must be carefully adapted to our specific
case.

  * iv.

Finally, observe that our results are verified to the Benney–Lin type
equation:

$\partial_{t}u+\partial_{x}^{3}u+\mu_{0}\partial_{x}^{4}u-\partial_{x}^{5}u=0,\quad(x,t)\in(0,1)\times(0,T),$

with the same boundary conditions as in equation (1.5) and with $\mu_{0}>0$.
This equation describes the evolution of one-dimensional small but finite
amplitude long waves in various physical systems in fluid dynamics (see [4]
and [28]). The coefficient $\mu_{0}>0$ introduces nonconservative dissipative
effects to the dispersive Kawahara equation (1.5) (where $\mu_{0}=0$), and
thus it is sometimes referred to as the strongly dissipative Kawahara equation
[40], the fifth-order Korteweg-de Vries equation [41], or the generalized
Kawahara equation [13]. For more details about the Benney-Lin type equation,
we encourage the reader to see the reference [14].

Let us conclude our introduction with an outline of our work. Section 2
addresses Question $\mathcal{A}$ by presenting the control result, which is a
consequence of the flatness property and the smoothing effect of the Kawahara
equation. In Section 3, we provide an example of a set that can be reached
from $0$ by the system (1.5). This result, stemming from the flatness property
extended to the limit case $s=5$, partially answers Question $\mathcal{B}$.

##  2\. Controllability result

We want to prove the null controllability property for the system (1.5). Our
initial objective is to show the flatness property. This property ensures that
we can parameterize the solution of the system (1.5) using the “flat outputs”
$\partial_{x}^{3}u(0,t)$ and $\partial_{x}^{4}u(0,t)$. To prove it, we will
examine the ill-posed system

(2.4)
$\displaystyle\left\\{\begin{array}[]{lll}u_{t}+\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u=0,&(x,t)\text{
in }(-1,0)\times(0,T),\\\
u(0,t)=\partial_{x}u(0,t)=\partial_{x}^{2}u(0,t)=0,&t\text{ in }(0,T),\\\
\partial_{x}^{3}u(0,t)=y(t),\ \ \partial_{x}^{4}u(0,t)=z(t),&t\text{ in
}(0,T).\end{array}\right.$

Precisely, we will show that the solution of (2.4) belongs to
$G^{\frac{s}{5},s}([-1,0]\times[\varepsilon,T])$ for all $\varepsilon\in(0,T)$
and can be written in the form

(2.5) $\displaystyle
u(x,t)=\sum_{j=0}^{\infty}f_{j}(x)y^{(j)}(t)+\sum_{j=0}^{\infty}g_{j}(x)z^{(j)}(t),\
(x,t)\in[-1,0]\times[\varepsilon,T]$

where $y,z\in G^{s}([0,T])$ for some $s\geq 0$, with $y^{(j)}(T)=z^{(j)}(T)=0$
for every $j\geq 0$. Here, $f_{j}$ and $g_{j}$ are called the generating
functions and are constructed following the ideas introduced in [31]. The
smoothing effect is responsible for ensuring that, from time $\varepsilon$
onward, the solution of the free evolution problem associated to (1.5) is
Gevrey of order $\frac{1}{2}$ in $x$ and $\frac{5}{2}$ in $t$. Finally, a
result of unique continuation is used to ensure that this solution coincides
with the one associated with (2.4), described in (2.5).

###  2.1. Flatness property

We need to prove that there exists a one-to-one correspondence between
solutions of (2.4) and a certain space of smooth functions, which we will call
the flatness property. We will often use Stirling’s formula:

$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n},\ \forall n,$

and the inequality

(2.6) $\displaystyle(n+m)!\leq 2^{n+m}n!m!.$

For $n\in\mathbb{N}$, $p\in[1,\infty]$ and $f\in W^{n,p}(-1,0)$ we denote

$\displaystyle\|f\|_{p}=\|f\|_{L^{p}(-1,0)}\text{ \ \ and \ \
}\|f\|_{n,p}=\sum_{i=0}^{n}\|\partial_{x}^{i}f\|_{p}.$

Let us in this part find a solution $u$ for (2.4) in the form

(2.7) $\displaystyle u(x,t)=\sum_{j\geq 0}f_{j}(x)y^{(j)}(t)+\sum_{j\geq
0}g_{j}(x)z^{(j)}(t).$

Here, $y,z,f_{j},g_{j}$ satisfies the following conditions:

  1. $1)$

$y,z\in G^{s}([0,T])$ with $s\in(1,5)$;

  2. $2)$

$f_{j},g_{j}\in L^{\infty}([-1,0])$ with polynomial growths in the form

$\left|f_{j}(x)\right|\leq\frac{|x|^{5j+r}}{(5j+r)!}\text{ \ \ and \ \
}\left|g_{j}(x)\right|\leq\frac{|x|^{5j+r}}{(5j+r)!},\ \forall j\geq 0,\
\forall x\in[-1,0],$

for some $r\in\\{0,1,2,3,4\\}$;

  3. $3)$

$\partial_{x}^{3}(0,t)=y(t)$ and $\partial_{x}^{4}u(0,t)=z(t)$.

Considering $u$ as in (2.7) and assuming that we can derive term by term, we
get that

$\displaystyle u_{t}+\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u=$
$\displaystyle\sum_{j\geq 0}f_{j}(x)y^{(j+1)}(t)+\sum_{j\geq
0}\left(f_{jx}+f_{j3x}-f_{j5x}\right)(x)y^{(j)}(t)$ $\displaystyle+\sum_{j\geq
0}g_{j}(x)z^{(j+1)}(t)+\sum_{j\geq
0}\left(g_{jx}+g_{j3x}-g_{j5x}\right)(x)z^{(j)}(t).$

Considering a new index $l=j+1$ gives us

$\displaystyle\sum_{j\geq 0}f_{j}(x)y^{(j+1)}(t)=\sum_{l\geq
1}f_{l-1}(x)y^{(l)}(t)\text{ \ \ and \ \ }\sum_{j\geq
0}g_{j}(x)z^{(j+1)}(t)=\sum_{l\geq 1}g_{l-1}(x)z^{(l)}(t).$

Thus

$\displaystyle u_{t}+\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u=$
$\displaystyle\sum_{j\geq 1}f_{j-1}(x)y^{(j)}(t)+\sum_{j\geq
1}\left(f_{jx}+f_{j3x}-f_{j5x}\right)(x)y^{(j)}(t)$
$\displaystyle+\left(f_{0x}+f_{03x}-f_{05x}\right)(x)y(t)+\left(g_{0x}+g_{03x}-g_{05x}\right)(x)z(t)$
$\displaystyle+\sum_{j\geq 1}g_{j-1}(x)z^{(j)}(t)+\sum_{j\geq
1}\left(g_{jx}+g_{j3x}-g_{j5x}\right)(x)z^{(j)}(t).$

Hence, if we have

$\displaystyle\begin{cases}&f_{0x}+f_{03x}-f_{05x}=g_{0x}+g_{03x}-g_{05x}=0,\\\
&f_{jx}+f_{j3x}-f_{j5x}=-f_{j-1},\quad\ \forall j\geq 1,\\\
&g_{jx}+g_{j3x}-g_{j5x}=-g_{j-1},\quad\ \forall\ j\geq 1,\end{cases}$

then

$u_{t}+\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u=0.$

Furthermore, imposing the conditions

$\displaystyle\begin{cases}\begin{array}[]{l}f_{0}(0)=f_{0x}(0)=f_{02x}(0)=f_{04x}(0)=0,\quad
f_{03x}(0)=1,\\\
f_{j}(0)=f_{jx}(0)=f_{j2x}(0)=f_{j3x}(0)=f_{j4x}(0)=0,\quad\forall j\geq
1\end{array}\end{cases}$

and

$\displaystyle\begin{cases}\begin{array}[]{l}g_{0}(0)=g_{0x}(0)=g_{02x}(0)=g_{03x}(0)=0,\quad
g_{04x}(0)=1,\\\
g_{j}(0)=g_{jx}(0)=g_{j2x}(0)=g_{j3x}(0)=g_{j4x}(0)=0,\quad\forall j\geq
1,\end{array}\end{cases}$

we obtain,

$\displaystyle\begin{cases}&u(0,t)=\partial_{x}u(0,t)=\partial_{x}^{2}u(0,t)=0,\\\
&\partial_{x}^{3}u(0,t)=y(t),\quad\partial_{x}^{4}u(0,t)=z(t),\end{cases}$

for $t\in(0,T)$. This leads us to define inductively the functions
$\\{f_{j}\\}_{j\geq 0}$ and $\\{g_{j}\\}_{j\geq 0}$ as follows:

  1. $1)$

$f_{0}$ is the solution of the IBVP

(2.10) $\displaystyle\left\\{\begin{array}[]{l}f_{0x}+f_{03x}-f_{05x}=0,\\\
f_{0}(0)=f_{0x}(0)=f_{02x}(0)=f_{04x}(0)=0,\quad
f_{03x}(0)=1.\end{array}\right.$

and, for $j\geq 1$, $f_{j}$ is the solution for the IBVP

(2.13)
$\displaystyle\left\\{\begin{array}[]{l}f_{jx}+f_{j3x}-f_{j5x}=-f_{j-1},\\\
f_{j}(0)=f_{jx}(0)=f_{j2x}(0)=f_{j3x}(0)=f_{j4x}(0)=0.\end{array}\right.$

  2. $2)$

$g_{0}$ is the solution of the IBVP

(2.16) $\displaystyle\left\\{\begin{array}[]{l}g_{0x}+g_{03x}-g_{05x}=0,\\\
g_{0}(0)=g_{0x}(0)=g_{02x}(0)=g_{03x}(0)=0,\quad
g_{04x}(0)=1.\end{array}\right.$

and, for $j\geq 1$, $g_{j}$ is the solution for the IVP

(2.19)
$\displaystyle\left\\{\begin{array}[]{l}g_{jx}+g_{j3x}-g_{j5x}=-g_{j-1},\\\
g_{j}(0)=g_{jx}(0)=g_{j2x}(0)=g_{j3x}(0)=g_{j4x}(0)=0.\end{array}\right.$

We will prove that, given $s\in(1,5)$, there exists a one to one
correspondence between solutions of (2.4) and pairs of functions $(y,z)\in
G^{s}([0,T])\times G^{s}([0,T])$, namely,

$\displaystyle
u\mapsto\left(\partial_{x}^{3}y(0,\cdot),\partial_{x}^{4}z(0,\cdot)\right).$

In the sense of this bijection, we shall say that the system (2.4) is flat.
Observe that, an expression in terms of the families $\\{f_{j}\\}_{j\geq 0}$
and $\\{g_{j}\\}_{j\geq 0}$ for a solution $u$ of (2.4) as in (2.7) must be
unique, that is, if

$\displaystyle u(x,t)=\sum_{j\geq 0}f_{j}(x)y^{(j)}(t)+\sum_{j\geq
0}g_{j}(x)z^{(j)}(t)$

and

$\displaystyle u(x,t)=\sum_{j\geq 0}f_{j}(x)\tilde{y}^{(j)}(t)+\sum_{j\geq
0}g_{j}(x)\tilde{z}^{(j)}(t),$

with $y,\tilde{y},z,\tilde{z}\in G^{s}([0,T])$, then $y=\tilde{y}$ and
$z=\tilde{z}$.

######  Remark 2.

  1. $1)$

Note that considering a toy model, that is,

$\displaystyle\left\\{\begin{array}[]{ll}u_{t}-\partial_{x}^{5}u=0,&(x,t)\in(-1,0)\times(0,T),\\\
u(0,t)=\partial_{x}u(0,t)=\partial_{x}^{2}(0,t)=0,&t\in(0,T),\\\
u(-1,t)=h_{1}(t)\ \ \ \ \partial_{x}u(-1,t)=h_{2}(t),&t\in(0,T),\\\
u(x,0)=u_{0}(x),\end{array}\right.$

the first equation in the systems (2.10)-(2.16) do not have the first and
third derivatives terms. Then, direct computation gives

$\displaystyle f_{j}(x)=\frac{x^{5j+3}}{(5j+3)!}\text{ \ \ and \ \
}g_{j}(x)=\frac{x^{5j+4}}{(5j+4)!},\ \ \forall\ j\geq 0,\ x\in[-1,0].$

  2. $2)$

Returning to the full system, with terms of the first and third derivatives,
we have that the solutions $f_{0}$ and $g_{0}$ of (2.10) and (2.16) are given
by

(2.20) $\displaystyle
f_{0}(x)=\frac{1}{\sqrt{a}(a+b)}\sinh\left(\sqrt{a}x\right)-\frac{1}{\sqrt{b}(a+b)}\sin\left(\sqrt{b}x\right)$

and

(2.21) $\displaystyle
g_{0}(x)=\frac{1}{a(a+b)}\cosh\left(\sqrt{a}x\right)+\frac{1}{b(a+b)}\cos\left(\sqrt{b}x\right)-\frac{1}{a(a+b)}-\frac{1}{b(a+b)},$

respectively, where $a=\frac{\sqrt{5}+1}{2}$ and $b=\frac{\sqrt{5}-1}{2}$.

To conclude that the system (2.4) is flat, it is enough to show that the
solutions of (2.4) can be expressed as in (2.7) with $\\{f_{j}\\}_{j\geq 0}$
and $\\{g_{j}\\}_{j\geq 0}$ given by (2.10)-(2.19). Precisely, we will see
that given $y,z\in G^{s}([0,T])$, then $u$ given by (2.7) is well defined,
belongs to $G^{\frac{s}{5},s}([-1,0]\times[0,T])$ and it solves the problem

(2.25)
$\displaystyle\left\\{\begin{array}[]{lll}u_{t}+\partial_{x}u+\partial_{x}^{3}u-\partial_{x}^{5}u=0&(x,t)\text{
in }(-1,0)\times(0,T)\\\
u(0,t)=\partial_{x}u(0,t)=\partial_{x}^{2}u(0,t)=0,&t\text{ in }(0,T)\\\
\partial_{x}^{3}u(0,t)=y(t),\ \
\partial_{x}^{4}u(0,t)=z(t)&t\in(0,T).\end{array}\right.$

To do this, first, we need to establish estimates to the norms
$\|f_{j}\|_{L^{\infty}(-1,0)}$ and $\|g_{j}\|_{L^{\infty}(-1,0)}$, as
suggested before. At this point, it is very useful to note that, for $j\geq
1$, $f_{j}$ (respectively $g_{j}$) can be written in terms of $f_{0}$ and
$f_{j-1}$ (respectively $g_{0}$ and $g_{j-1}$).

######  Lemma 2.1.

For any $j\geq 1$ and $x\in[-1,0]$ we have

(2.26) $\displaystyle
f_{j}(x)=\int_{0}^{x}\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi dy$

and

(2.27) $\displaystyle g_{j}(x)=\int_{0}^{x}g_{0}(x-\xi)g_{j-1}(\xi)d\xi.$

###### Proof.

Let $x\in[-1,0]$, $y\in[x,0]$ and $\xi\in[y,0]$ be. From (2.13) we have

$\displaystyle
f_{j\xi}(\xi)f_{0}(y-\xi)+f_{j3\xi}(\xi)f_{0}(y-\xi)-f_{j5\xi}(\xi)f_{0}(y-\xi)=-f_{0}(y-\xi)f_{j-1}(\xi).$

Integrating with respect to $\xi$ we obtain

$\displaystyle\int_{0}^{y}f_{j}(\xi)\left(f_{0\xi}(y-\xi)+f_{03\xi}(y-\xi)-f_{05\xi}(y-\xi)\right)d\xi-
f_{j\xi}(y)=-\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$

and by (2.10), after some integration by parts, it follows that

$\displaystyle f_{j}^{\prime}(y)=\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi.$

Integrating from $0$ to $x$ with respect to $y$ we get

$\displaystyle f_{j}(x)=\int_{0}^{x}\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi
dy,\ \forall x\in[-1,0],$

showing (2.26). Similarly (2.27) is verified. ∎

The next lemma will ensure that the series in (2.7) are convergents.

######  Lemma 2.2.

For every $j\geq 0$ we have

(2.28) $\left|f_{j}(x)\right|\leq 2^{j}\frac{|x|^{5j+1}}{(5j+1)!}$

and

(2.29) $\left|g_{j}(x)\right|\leq 2^{j}\frac{|x|^{5j+1}}{(5j+1)!},$

for all $j\geq 0$ and $x\in[-1,0]$.

###### Proof.

Let us start to proving (2.28) by induction on $j$. First, we must verify the
case $j=0$:

$\displaystyle\left|f_{0}(x)\right|\leq|x|,\quad\forall x\in[-1,0].$

Note that $f_{0}(x)\leq 0$ in $[-1,0]$. Indeed, from (2.20), since $\cosh>1$
in $\mathbb{R}\backslash\\{0\\}$ and $\cos\geq 1$ in $\mathbb{R}$,

$\displaystyle
f_{0x}=\frac{1}{a+b}\cosh\left(\sqrt{a}x\right)-\frac{1}{a+b}\cos\left(\sqrt{b}x\right)>\frac{1}{a+b}-\frac{1}{a+b}=0,$

for all $x\in(-\infty,0)$. This implies that $f_{0}$ is increasing, thus
$f_{0}(x)\leq f_{0}(0)=0$. Therefore, it is sufficient to show that

(2.30) $\displaystyle f_{0}(x)\geq x,\quad\ \forall x\in[-1,0].$

Defining $\varphi(x)=f_{0}(x)-x$ we have
$\varphi^{\prime}(x)=f_{0}^{\prime}(x)-1$. On the other hand, we have

$\displaystyle
f_{03x}(x)=\frac{a}{a+b}\cosh\left(\sqrt{a}x\right)+\frac{b}{a+b}\cos\left(\sqrt{b}x\right)>\frac{a-b}{a+b}=\frac{1}{\sqrt{5}}>0,$

for $x\in(-\infty,0)$, and so $f_{02x}$ is increasing in $(-\infty,0]$,
implying that $f_{02x}(x)<f_{02x}(0)=0$ for $x\in(-\infty,0]$. Consequently
$f_{0x}$ is decreasing in $(-\infty,0]$ which implies that

$\displaystyle f_{0x}(x)<f_{0x}(-1)\approxeq 0,54<1,\quad\ \forall\
x\in(-1,0),$

so $\varphi^{\prime}<0$ in $(-1,0)$. Hence $\varphi$ is decreasing on $[-1,0]$
and therefore $\varphi(x)\geq\varphi(0)=0,$ for $x\in[-1,0]$, which implies
(2.30).

For the next step, we need to verify the estimate

(2.31) $\displaystyle\left|f_{03x}(x)\right|\leq 2,$

when $x\in[-1,0].$ Note that we have

$\displaystyle
f_{05x}(x)=\frac{a^{2}}{a+b}\cosh\left(\sqrt{a}x\right)-\frac{b^{2}}{a+b}\cos\left(\sqrt{b}x\right)>\frac{a^{2}-b^{2}}{a+b}=a-b=1,$

for $x\in(-\infty,0)$. This gives us that $f_{04x}$ is increasing in
$(-\infty,0]$, in particular, $f_{04x}(x)<f_{04x}(0)=0.$ Hence $f_{03x}$ is
decreasing in $(-\infty,0]$, and so $0=f_{03x}(0)\leq f_{03x}(x)\leq
f_{03x}(-1)\approxeq 1,59<2,$ for $x\in[-1,0]$, getting (2.31).

Now, suppose that for some $j\geq 1$

$\displaystyle\left|f_{j}(x)\right|\leq
2^{j-1}\frac{|x|^{5(j-1)+1}}{[5(j-1)+1]!},\ \forall x\in[-1,0],$

holds. By Lemma 2.1 we have

$\displaystyle f_{j}(x)=\int_{0}^{x}\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi
dy,\quad\forall x\in[-1,0].$

Using integration by parts (with respect to $\xi$) together with the boundary
conditions in (2.10) we obtain

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{0}(y-\xi)\frac{d}{d\xi}\int_{0}^{\xi}f_{j-1}(\sigma)d\sigma
d\xi$
$\displaystyle=\left[f_{0}(y-\xi)\int_{0}^{\xi}f_{j-1}(\sigma)d\sigma\right]_{0}^{y}+\int_{0}^{y}\int_{0}^{\xi}f_{j-1}(\sigma)d\sigma
f_{0\xi}(y-\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{0\xi}(y-\xi)\int_{0}^{\xi}f_{j-1}(\sigma)d\sigma
d\xi.$

Define $F(\xi)=\int_{0}^{\xi}f_{j-1}(\sigma)d\sigma,$ with this, we can write

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{0\xi}(y-\xi)F(\xi)d\xi=\int_{0}^{y}f_{0\xi}(y-\xi)\frac{d}{d\xi}\int_{0}^{\xi}F(\tau)d\tau
d\xi.$

Integration by parts together with the boundary conditions giving in (2.10) we
get

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\left[f_{0\xi}(y-\xi)\int_{0}^{\xi}F(\tau)d\tau\right]_{0}^{y}+\int_{0}^{y}\int_{0}^{\xi}F(\tau)d\tau
f_{02\xi}(y-\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{02\xi}(y-\xi)\int_{0}^{\xi}F(\tau)d\tau d\xi.$

Setting also $G(\xi)=\int_{0}^{\xi}F(\tau)d\tau,$ and we have

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{02\xi}(y-\xi)G(\xi)d\xi=\int_{0}^{y}f_{02\xi}(y-\xi)\frac{d}{d\xi}\int_{0}^{\xi}G(\rho)d\rho
d\xi.$

Using again integration by parts together with (2.10), it follows that

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\left[f_{02\xi}(y-\xi)\int_{0}^{\xi}G(\rho)d\rho\right]_{0}^{y}+\int_{0}^{y}\int_{0}^{\xi}G(\rho)d\rho
f_{03\xi}(y-\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{03\xi}(y-\xi)\int_{0}^{\xi}G(\rho)d\rho d\xi.$

Therefore, the previous equality ensures that

$\displaystyle\int_{0}^{y}f_{0}(y-\xi)f_{j-1}(\xi)d\xi$
$\displaystyle=\int_{0}^{y}f_{03\xi}(y-\xi)\int_{0}^{\xi}G(\rho)d\rho d\xi$
$\displaystyle=\int_{0}^{y}f_{03\xi}(y-\xi)\int_{0}^{\xi}\int_{0}^{\rho}F(\tau)d\tau
d\rho d\xi$
$\displaystyle=\int_{0}^{y}f_{03\xi}(y-\xi)\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}f_{j-1}(\sigma)d\sigma
d\tau d\rho d\xi$
$\displaystyle=\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}f_{03\xi}(y-\xi)f_{j-1}(\sigma)d\sigma
d\tau d\rho d\xi,$

so

$\displaystyle
f_{j}(x)=\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}f_{03\xi}(y-\xi)f_{j-1}(\sigma)d\sigma
d\tau d\rho d\xi dy,\ \ \forall x\in[-1,0].$

Then, by the induction hypothesis, we get that

$\displaystyle\left|f_{j}(x)\right|$
$\displaystyle\leq\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}2\cdot
2^{j-1}\frac{|\sigma|^{5(j-1)+1}}{[5(j-1)+1]!}d\sigma d\tau d\rho d\xi dy$
$\displaystyle\leq
2^{j}\left|\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}\frac{|\sigma|^{5j-4}}{(5j-4)!}d\sigma
d\tau d\rho d\xi dy\right|$
$\displaystyle=2^{j}\left|\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\int_{0}^{\tau}\frac{(-\sigma)^{5j-4}}{(5j-4)!}d\sigma
d\tau d\rho d\xi dy\right|,$

for $x\in[-1,0]$. Pick $r=-\sigma$, thus $dr=-d\sigma$, that is,

$\displaystyle\int_{0}^{\tau}\frac{(-\sigma)^{5j-4}}{(5j-4)!}d\sigma=-\int_{0}^{-\tau}\frac{r^{5j-4}}{(5j-4)!}dr=\left[-\frac{r^{5j-3}}{(5j-3)(5j-4)!}\right]_{0}^{-\tau}=-\frac{(-\tau)^{5j-3}}{(5j-3)!}$

thus

$\displaystyle\left|f_{j}(x)\right|$ $\displaystyle\leq
2^{j}\left|\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\int_{0}^{\rho}\frac{(-\tau)^{5j-3}}{(5j-3)!}d\tau
d\rho d\xi dy\right|.$

Now set $r=-\tau$ to get

$\displaystyle\int_{0}^{\rho}\frac{(-\tau)^{5j-3}}{(5j-3)!}d\tau=-\int_{0}^{-\rho}\frac{r^{5j-3}}{(5j-3)!}dr=\left[-\frac{r^{5j-2}}{(5j-2)(5j-3)!}\right]_{0}^{-\rho}=-\frac{(-\rho)^{5j-2}}{(5j-2)!},$

and, consequently,

$\displaystyle\left|f_{j}(x)\right|$ $\displaystyle\leq
2^{j}\left|\int_{0}^{x}\int_{0}^{y}\int_{0}^{\xi}\frac{(-\rho)^{5j-2}}{(5j-2)!}d\rho
d\xi dy\right|.$

Proceeding with three more integrations, we obtain

$\displaystyle\left|f_{j}(x)\right|$ $\displaystyle\leq
2^{j}\left|\int_{0}^{x}\int_{0}^{y}\frac{(-\xi)^{5j-1}}{(5j-1)!}d\xi
dy\right|\leq 2^{j}\left|\int_{0}^{x}\frac{(-y)^{5j}}{(5j)!}dy\right|\leq
2^{j}\left|\frac{(-x)^{5j+1}}{(5j+1)!}\right|,$

for all $x\in[-1,0]$, showing (2.28).

Now, for (2.29), arguing similarly, we start by checking that
$\left|g_{0}(x)\right|\leq|x|,$ for all $x\in[-1,0]$. Since $\cosh\geq 1$ and
$\cos\geq 1$, from (2.21) we have $g_{0}\geq 0$. Then it is enough to show
that $g_{0}(x)\leq-x$, for $x\in[-1,0]$. Defining $\psi(x)=g_{0}(x)+x$ we have
$\psi^{\prime}(x)=g_{0}^{\prime}(x)+1$. On the other hand,

$\displaystyle
g_{02x}(x)=\frac{1}{a+b}\cosh\left(\sqrt{a}x\right)-\frac{1}{a+b}\cos\left(\sqrt{b}x\right)>0,\quad\forall
x\in(-\infty,0),$

so $g_{0x}$ is increasing in $(-\infty,0]$. Thus
$g_{0x}(x)>g_{0x}(-1)\approxeq-0,18>-1$, for all $x\in(-1,0)$, which implies
that $\psi^{\prime}>0$ in $(-1,0)$ and therefore $\psi$ is increasing in
$[-1,0]$. Consequently

$\displaystyle\psi(x)\leq\psi(0)=0\iff g_{0}(x)\leq-x,\quad\ \forall
x\in[-1,0].$

In the next part, we need the estimate $\left|g_{04x}(x)\right|<2$, for all
$x\in[-1,0]$. This immediately follows from (2.31) and the fact that
$g_{04x}\equiv f_{03x}$.

Now, assume that

$\displaystyle\left|g_{j-1}(x)\right|\leq
2^{j-1}\frac{|x|^{5(j-1)+1}}{[5(j-1)+1]!}\ \ \forall x\in[-1,0],$

holds for some $j\geq 1$. By Lemma 2.1 we have

$\displaystyle g_{j}(x)=\int_{0}^{x}g_{0}(x-\xi)g_{j-1}(\xi)d\xi.$

Proceeding with integration by parts as in the case of $f_{j}$ and using the
boundary conditions in (2.19) we obtain

$\displaystyle
g_{j}(x)=\int_{0}^{x}\int_{0}^{\xi}\int_{0}^{\lambda}\int_{0}^{\rho}\int_{0}^{\tau}g_{04x}(x-\xi)g_{j-1}(\sigma)d\sigma
d\tau d\rho d\lambda d\xi.$

Then, by induction hypothesis we have

$\displaystyle\left|g_{j}(x)\right|$
$\displaystyle\leq\int_{0}^{x}\int_{0}^{\xi}\int_{0}^{\lambda}\int_{0}^{\rho}\int_{0}^{\tau}2\cdot
2^{j-1}\frac{|\sigma|^{5(j-1)+1}}{[5(j-1)+1]!}d\sigma d\tau d\rho d\lambda
d\xi$ $\displaystyle\leq
2^{j}\left|\int_{0}^{x}\int_{0}^{\xi}\int_{0}^{\lambda}\int_{0}^{\rho}\int_{0}^{\tau}\frac{|\sigma|^{5j-4}}{(5j-4)!}d\sigma
d\tau d\rho d\lambda d\xi\right|$
$\displaystyle=2^{j}\left|\int_{0}^{x}\int_{0}^{\xi}\int_{0}^{\lambda}\int_{0}^{\rho}\int_{0}^{\tau}\frac{(-\sigma)^{5j-4}}{(5j-4)!}d\sigma
d\tau d\rho d\lambda d\xi\right|,$

for $x\in[-1,0]$. Using the same process of repeated integration by
substitution as in the case of $f_{j}$ we obtain (2.29), which concludes the
proof. ∎

We are now in a position to solve the system (2.25).

######  Proposition 2.3.

Let $s\in(1,5)$, $T>0$ and $y,z\in G^{s}([0,T])$. The the function $u(x,t)$
defined in (2.7) belongs to $G^{\frac{s}{5},s}([-1,0]\times[0,T])$ and it
solves (2.25). In particular, the corresponding controls $h_{1}(t):=u(-1,t)$
and $h_{2}(t):=\partial_{x}u(-1,t)$ are Gevrey of order $s$ on $[0,T]$.

###### Proof.

Let us briefly explain our strategy to prove this result. Consider $m,n\geq 0$
and $(x,t)\in[-1,0]\times[0,T]$ be given. Our task is first to derivate the
series in (2.7) as follows

(2.32) $\displaystyle\partial_{x}^{n}\partial_{t}^{m}u(x,t)=\sum_{j\geq
0}\partial_{x}^{n}\partial_{t}^{m}\left(f_{j}(x)y^{(j)}(t)\right)+\sum_{j\geq
0}\partial_{x}^{n}\partial_{t}^{m}\left(g_{j}(x)z^{(j)}(t)\right).$

With this, since $m,n$ are arbitrary, we will prove that the series in (2.32)
is uniformly convergent in $[-1,0]\times[0,T]$, which will ensure that $u\in
C^{\infty}([-1,0]\times[0,T])$. Next, to conclude that $u\in
G^{\frac{s}{5},s}([-1,0]\times[0,T])$ we must prove that

$\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}u(x,t)\right|\leq
C\frac{n!^{\frac{s}{5}}}{R_{1}^{n}}\frac{m!^{s}}{R_{2}^{m}}$

for some constants $C,R_{1},R_{2}>0$.

Let us star, choose $s\in(1,5)$ and $y,z\in G^{s}([0,T])$. Then there exists
$M,R>0$ such that

(2.33) $\displaystyle\left|y^{(j)}(t)\right|\leq M\frac{j!^{s}}{R^{j}}\text{ \
\ and \ \ }\left|z^{(j)}(t)\right|\leq M\frac{j!^{s}}{R^{j}}\ \ \forall j\geq
0,\ \forall t\in[0,T].$

For $k\geq 0$, since $\partial_{t}$ and $P$ commute, we have that

$\displaystyle\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)=P^{k}\left(f_{j}(x)\right)y^{(j+m)}(t).$

From (2.10) and (2.13) follows that $Pf_{0}=0$ and $Pf_{j}=-f_{j-1},$ for
$j\geq 1$. We will split into two cases our analysis of $P^{k}(f_{j})$,
namely, $j-k\geq 0$ and $j-k<0$. First, suppose $j-k\geq 0$. In this case,

$\displaystyle
P^{k}(f_{j})=(-1)^{1}P^{k-1}(f_{j-1})=(-1)^{2}P^{k-2}(f_{j-2})=~{}\cdots=(-1)^{k-1}P^{k-(k-1)}\left(f_{j-(k-1)}\right),$

or equivalently,

$\displaystyle
P^{k}(f_{j})=(-1)^{k-1}P(f_{j-k+1})=(-1)^{k-1}(-f_{j-k})=(-1)^{k}f_{j-k}.$

Secondly, assuming $j-k<0$ we have $k-j>0$ so $k-j\geq 1$. Hence

$\displaystyle
P^{k}(f_{j})=(-1)^{1}P^{k-1}(f_{j-1})=(-1)^{2}P^{k-2}(f_{j-2})=~{}\cdots=(-1)^{j}P^{k-j-1}(0)=0.$

Putting both information together we get that

(2.36) $\displaystyle
P^{k}(f_{j})=\left\\{\begin{array}[]{ll}(-1)^{k}f_{j-k},&\text{ if }j-k\geq
0.\\\ 0,&\text{ if }j-k<0,\end{array}\right.$

and consequently

(2.39)
$\displaystyle\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)=\left\\{\begin{array}[]{ll}(-1)^{k}f_{j-k}(x)y^{(j+m)}(t),&\text{
\ if \ }j-k\geq 0,\\\ 0,&\text{ \ if \ }j-k<0.\end{array}\right.$

Assume $j\geq k$, that is, $j-k\geq 0$. Then thanks to the relation (2.39) and
Lemma 2.2, the following estimate holds

$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|\leq
2^{j-k}\frac{|x|^{5(j-k)+1}}{[5(j-k)+1]!}M\frac{(j+m)!^{s}}{R^{j+m}}\leq
M2^{j-k}\frac{(j+m)!^{s}}{R^{j+m}}\frac{1}{\left(5(j-k)+1\right)!}.$

Setting $l=j-k$ we can write the previous inequality as follows

$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq M2^{l}\frac{(l+k+m)!^{s}}{R^{l+k+m}}\frac{1}{(5l+1)!}.$

Furthermore, writing $N=k+m$ yields that

$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq M2^{l}\frac{(l+N)!^{s}}{R^{l+N}}\frac{1}{(5l+1)!}.$

Then using the inequality (2.6) we obtain that

(2.40)
$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq
M2^{l}\frac{(2^{l}2^{N}l!N!)^{s}}{R^{l}R^{N}}\frac{1}{(5l+1)!}=M\frac{2^{(1+s)l}2^{Ns}l!^{s}N!^{s}}{R^{l}R^{N}}\frac{1}{(5l+1)!}.$

Stirling’s formula gives that

$\displaystyle l!\sim\left(\frac{l}{e}\right)^{l}\sqrt{2\pi l}\text{ \ \ and \
\ }(5l)!\sim\left(\frac{5l}{e}\right)^{5l}\sqrt{2\pi 5l}.$

Observer that

$\displaystyle\left(\frac{5l}{e}\right)^{5l}\sqrt{2\pi
5l}=5^{5l+\frac{1}{2}}\left[\left(\frac{l}{e}\right)^{l}\left(\sqrt{2\pi
l}\right)\right]^{5}\left(\sqrt{2\pi l}\right)^{-4}\sim
5^{5l+\frac{1}{2}}\left(\sqrt{2\pi l}\right)^{-4}l!^{5}$

and, consequently, $(5l)!\sim 5^{5l+\frac{1}{2}}\left(\sqrt{2\pi
l}\right)^{-4}l!^{5}.$ Thus, from (2.40), we get the following

$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq\tilde{M}\frac{l!^{s}}{\left[2^{-(1+s)}R\right]^{l}}\frac{N!^{s}}{\left(2^{-s}R\right)^{N}}\frac{1}{(5l+1)5^{5l+\frac{1}{2}}\left(\sqrt{2\pi
l},\right)^{-4}l!^{5}},$

for a suitable constant $\tilde{M}>0$. Since
$\frac{1}{(5l+1)5^{5l+\frac{1}{2}}}\leq 1,$ for $l\geq 0$, it follows that

$\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|\leq\tilde{M}\frac{N!^{s}}{\left(2^{-s}R\right)^{N}}\frac{(2\pi
l)^{2}}{[2^{-(1+s)}R]^{l}l!^{5-s}}.$

Using the inequality (2.6) we get that

$\displaystyle\frac{N!^{s}}{\left(2^{-s}R\right)^{N}}\leq\frac{k!^{s}}{\left(\frac{R}{4^{s}}\right)^{k}}\frac{m!^{s}}{\left(\frac{R}{4^{s}}\right)^{m}}$

and, consequently, the following holds

$\displaystyle\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|\leq\tilde{M}4\pi^{2}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}\frac{l^{2}}{\tilde{R}^{l}l!^{5-s}},$

where $\tilde{R}=2^{-(1+s)}R$. Therefore,

$\displaystyle\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(f_{j}y^{(j)}(t)\right)\right\|_{\infty}\leq
4\tilde{M}\pi^{2}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}\sum_{l\geq
0}\frac{l^{2}}{\tilde{R}^{l}l!^{5-s}},$

for all $m,k\geq 0$. To finish this case, observe that the following series
$\sum_{l\geq 0}\frac{l^{2}}{\tilde{R}^{l}l!^{5-s}}$ is convergent, so by the
Weierstrass M-test the following series

$\displaystyle\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(f_{j}y^{(j)}(t)\right)\right\|_{\infty}$

is uniformly convergent on $[-1,0]\times[0,T]$, for all $m,k\geq 0$.
Furthermore

(2.41) $\displaystyle\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(f_{j}y^{(j)}(t)\right)\right\|_{\infty}\leq\overline{M}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}},$

for $m,k\geq 0$, where $\overline{M}=4\tilde{M}\pi^{2}\sum_{l\geq
0}\frac{l^{2}}{\tilde{R}^{l}l!^{5-s}}.$

Analogously it turns out that

(2.44) $\displaystyle
P^{k}(g_{j})=\left\\{\begin{array}[]{ll}(-1)^{k}g_{j-k},&\text{ if }j-k\geq
0,\\\ 0,&\text{ if }j-k<0,\end{array}\right.$

and

(2.47)
$\displaystyle\partial_{t}^{m}P^{k}\left(g_{j}(x)z^{(j)}(t)\right)=\left\\{\begin{array}[]{ll}(-1)^{k}g_{j-k}(x)z^{(j+m)}(t),&\text{
\ if \ }j-k\geq 0,\\\ 0,&\text{ \ if \ }j-k<0,\end{array}\right.$

so that, using inequalities (2.6), (2.33), Stirling’s formula and Lemma 2.1 we
get

$\displaystyle\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(g_{j}z^{(j)}(t)\right)\right\|_{\infty}\leq
4\tilde{M}\pi^{2}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}\sum_{l\geq
0}\frac{l^{2}}{\tilde{R}^{l}l!^{5-s}},$

for all $m,k\geq 0$. Consequently the series
$\sum\left\|\partial_{t}^{m}P^{k}\left(g_{j}z^{(j)}(t)\right)\right\|_{\infty}$
is uniformly convergent in $[-1,0]\times[0,T]$, for all $m,k\geq 0$, with

(2.48) $\displaystyle\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(g_{j}z^{(j)}(t)\right)\right\|_{\infty}\leq\overline{M}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

Let $K_{3}>0$ be as in Lemma 2.6 for $p=\infty$. From (2.39) and (2.41) we
have

$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(f_{j}y^{(j)}(t)\right)\right\|_{5i,\infty}=K_{3}^{i}\sum_{k=0}^{i}\sum_{j\geq
k}\left\|\partial_{t}^{m}P^{k}\left(f_{j}y^{(j)}(t)\right)\right\|_{\infty}\leq
K_{3}^{i}\sum_{k=0}^{i}\overline{M}\frac{k!^{s}}{\left(4^{-s}R\right)^{k}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

Thus

$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(f_{j}y^{(j)}(t)\right)\right\|_{5i,\infty}$
$\displaystyle\leq
K_{3}^{i}\overline{M}i!^{s}\left(\sum_{k=0}^{i}\frac{1}{\left(4^{-s}R\right)^{k}}\right)\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

Defining $b=\frac{1}{4^{-s}R}$ and noting that

$\displaystyle\sum_{j=0}^{i}\frac{1}{\left(4^{-s}R\right)^{k}}$
$\displaystyle=\sum_{k=0}^{i}b^{k}=1\cdot\frac{b^{i+1}-1}{b-1}\leq\frac{b^{i+1}}{b-1}=\frac{b}{b-1}b^{i},$

yields

$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(f_{j}y^{(j)}(t)\right)\right\|_{5i,\infty}$
$\displaystyle\leq
K_{3}^{i}\overline{M}i!^{s}\frac{b}{b-1}b^{i}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

Hence, defining $\hat{M}=\overline{M}\frac{b}{b-1}$ it follows that

(2.49)
$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(f_{j}y^{(j)}(t)\right)\right\|_{5i,\infty}$
$\displaystyle\leq\hat{M}\frac{i!^{s}}{(K_{3}^{-1}b^{-1})^{i}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}},\
\forall m,i\geq 0,\ t\in[0,T].$

Similarly, from (2.47), (2.48) and Lemma 2.6 we obtain

(2.50)
$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(g_{j}z^{(j)}(t)\right)\right\|_{5i,\infty}$
$\displaystyle\leq\hat{M}\frac{i!^{s}}{(K_{3}^{-1}b^{-1})^{i}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}},\
\forall m,i\geq 0,\ t\in[0,T].$

Given $m,n\geq 0$ consider $i\geq 0$ such that $n\in\left\\{5i-r,\
r=0,1,2,3,4\right\\}.$ Then from (2.49) we get, for
$(x,t)\in[-1,0]\times[0,T]$, that

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq\hat{M}\frac{i!^{s}}{(K_{3}^{-1}b^{-1})^{i}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}},$

and from (2.50)

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(g_{j}(x)z^{(j)}(t)\right)\right|$
$\displaystyle\leq\hat{M}\frac{i!^{s}}{(K_{3}^{-1}b^{-1})^{i}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

By the Weierstrass M-test, it follows these series are uniformly convergent in
$[-1,0]\times[0,T]$ for all $m,n\geq 0$. Thus, defining
$u:[-1,0]\times[0,T]\rightarrow\mathbb{R}$ as in (2.7) we have that
$\partial_{x}^{n}\partial_{t}^{m}u$ is continuous. Consequently $u\in
C^{\infty}([-1,0]\times[0,T])$ and satisfies

(2.51) $\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}u(x,t)\right|\leq
2\hat{M}\frac{i!^{s}}{(K_{3}^{-1}b^{-1})^{i}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}\
\ \forall(x,t)\in[-1,0]\times[0,T].$

As seen before, Stirling’s formula gives us $(5i)!\sim
5^{5i+\frac{1}{2}}\left(\sqrt{2\pi
i}\right)^{-4}i!^{5}=5^{5i+\frac{1}{2}}\left(2\pi i\right)^{-2}i!^{5}$, and so
that $i!^{5}\sim\frac{(2\pi i)^{2}(5i)!}{5^{5i+\frac{1}{2}}}$. Therefore
$i!^{s}\sim\left[\frac{(2\pi
i)^{2}(5i)!}{5^{5i+\frac{1}{2}}}\right]^{\frac{s}{5}}.$ Once we have that
$\frac{i^{2}}{5^{5i+\frac{1}{2}}}\leq 1\ \ \forall i\geq 0$, we can infer that
$i!^{s}\leq M^{*}(4\pi^{2})^{\frac{s}{5}}(5i)!^{\frac{s}{5}},$ for some
constant $M^{*}>0$. Furthermore, $n=5i-r$, that is, $5i=n+r$ with
$r\in\\{0,1,2,3,4\\}$. Then, by (2.6),

$\displaystyle i!^{s}\leq
M^{*}\left(4\pi^{2}\right)^{\frac{s}{5}}\left(2^{n}2^{r}n!r!\right)^{\frac{s}{5}}=$
$\displaystyle M^{*}\left(4\pi^{2}\cdot 2^{r}\cdot
r!\right)^{\frac{s}{5}}(2^{n})^{\frac{s}{5}}n!^{\frac{s}{5}}\leq
M^{\prime}\left(2^{\frac{s}{5}}\right)^{n}n!^{\frac{s}{5}},$

where $M^{\prime}=M^{*}\left(4\pi^{2}\cdot 2^{r}\cdot
r!\right)^{\frac{s}{5}}$. With this, returning to inequality (2.51) we obtain

$\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}u(x,t)\right|$
$\displaystyle\leq
2\hat{M}M^{\prime}\left(2^{\frac{s}{5}}\right)^{n}\frac{n!^{\frac{s}{5}}}{\left(K_{3}^{-1}b^{-1}\right)^{\frac{n+r}{5}}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}\leq\frac{2\hat{M}M^{\prime}}{\left(K_{3}^{-1}b^{-1}\right)^{\frac{r}{5}}}\frac{n!^{\frac{s}{5}}}{\left(K_{3}^{-\frac{1}{5}}b^{-\frac{1}{5}}2^{-\frac{s}{5}}\right)^{n}}\frac{m!^{s}}{\left(4^{-s}R\right)^{m}}.$

Finally, defining

$\displaystyle
M^{\prime\prime}:=\max\left\\{\frac{2\hat{M}M^{\prime}}{\left(K_{3}^{-1}b^{-1}\right)^{\frac{r}{5}}};\
r=0,1,2,3,4\right\\},\ \ \
R_{1}:=K_{3}^{-\frac{1}{5}}b^{-\frac{1}{5}}2^{-\frac{s}{5}},\ \text{and}\
R_{2}:=4^{-s}R,$

we have

$\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}u(x,t)\right|$
$\displaystyle\leq
M^{\prime\prime}\frac{n!^{\frac{s}{5}}}{R_{1}^{n}}\frac{m!^{s}}{R_{2}^{m}},\ \
\forall n,m\geq 0,\ \forall(x,t)\in[-1,0]\times[0,T].$

Moreover, $u$ solves (2.25) by construction, the result follows. ∎

###  2.2. Estimates in $W^{5n,p}(-1,0)$-norm

Let us starting remembering that the map

$\displaystyle\begin{array}[]{rcl}\|\cdot\|_{*}:W^{5,p}(-1,0)&\rightarrow&\mathbb{R}_{+}\\\
f&\mapsto&\|f\|_{*}:=\|f\|_{p}+\|Pf\|_{p}\end{array}$

is a norm called the graph norm associated with the operator
$P:W^{5,p}(-1,0)\rightarrow L^{p}(-1,0)$.

######  Lemma 2.4.

Let $p\in[1,\infty]$ be. For all $n\geq 0$ we have

(2.52) $\displaystyle\|P^{n}f\|_{p}\leq 3^{n}\|f\|_{5n,p},\ \forall\ f\in
W^{5n,p}(-1,0).$

###### Proof.

For $n=0$ the inequality is obvious. For $n=1$, given $f\in W^{5n,p}(-1,0)$
follows that

$\displaystyle\|Pf\|_{p}$
$\displaystyle=\|\partial_{x}f+\partial_{x}^{3}f-\partial_{x}^{5}f\|_{p}\leq\|\partial_{x}f\|_{p}+\|\partial_{x}^{3}f\|_{p}+\|\partial_{x}^{5}f\|_{p}\leq
3\|f\|_{5,p}.$

Suppose that (2.52) for $0,1,...,n-1$. Then for $f\in W^{5n,p}(-1,0)$ we get

$\displaystyle\|P^{n}f\|_{p}=\|P^{n-1}Pf\|_{p}$ $\displaystyle\leq
3^{n-1}\|Pf\|_{5n-5,p}\leq
3^{n-1}\left(3\sum_{j=0}^{5n}\|\partial_{x}^{j}f\|_{p}\right)=3^{n}\|f\|_{5n,p},$

so (2.52) is true for $n$, which concludes the proof. ∎

######  Lemma 2.5.

Let $p\in[1,\infty]$ be. There exists a constant $K_{1}=K_{1}(p)>0$ such that

$\displaystyle\|f\|_{5,p}\leq K_{1}\left(\|f\|_{p}+\|Pf\|_{p}\right),\
\forall\ f\in W^{5,p}(-1,0).$

###### Proof.

We know that $\|\cdot\|_{*}$ is a norm in $W^{5,p}(-1,0)$. We will show that
$\left(W^{5,p}(-1,0),\|\cdot\|_{*}\right)$ is a Banach space. To do this,
consider a Cauchy sequence in $\left(W^{5,p}(-1,0),\|\cdot\|_{*}\right)$,
$(f_{n})_{n\geq 0}$. Then,

$\displaystyle\|f_{m}-f_{n}\|_{*}=\|f_{m}-f_{n}\|_{p}+\|Pf_{m}-Pf_{n}\|_{p}\rightarrow
0\text{ \ as \ }m,n\rightarrow\infty.$

Since $L^{p}(-1,0)$ is a Banach space, there exist $f,g\in L^{p}(-1,0)$ such
that

$\displaystyle\|f_{n}-f\|_{p}\rightarrow 0,\ \ \ \|Pf_{n}-g\|_{p}\rightarrow
0\text{ \ as \ }n\rightarrow\infty.$

Given $\varphi\in C_{0}^{\infty}(-1,0)$ we have

$\displaystyle\left|\int_{-1}^{0}(f_{n}-f)\varphi\right|\leq\int_{-1}^{0}|f_{n}-f||\varphi|\leq\left\\{\begin{array}[]{ll}\|f_{n}-f\|_{p}\|\varphi\|_{q},&1<p<\infty,\
\frac{1}{p}+\frac{1}{q}=1\\\ \|f_{n}-f\|_{1}\|\varphi\|_{\infty},&p=1\\\
\|f_{n}-f\|_{\infty}\|\varphi\|_{1},&p=\infty.\end{array}\right.$

Therefore

$\displaystyle\int_{-1}^{0}(f_{n}-f)\varphi\rightarrow 0,\ \ \forall\varphi\in
C_{0}^{\infty}(-1,0)$

which implies that

(2.53) $\displaystyle f_{n}\rightarrow f\text{ \ in \
}\mathcal{D}^{\prime}(-1,0).$

Analogously we infer that $Pf_{n}\rightarrow g$ in
$\mathcal{D}^{\prime}(-1,0)$. But (2.53) implies that

$\displaystyle\partial_{x}^{i}f_{n}\rightarrow\partial_{x}^{i}f\text{ \ in \
}\mathcal{D}^{\prime}(-1,0),\ \forall i\in\mathbb{N}\cup\\{0\\}$

and consequently $Pf_{n}\rightarrow Pf$ in $\mathcal{D}^{\prime}(-1,0)$. By
uniqueness of limit, it follows that

(2.54) $\displaystyle Pf=g\text{ \ in \ }\mathcal{D}^{\prime}(-1,0).$

Consider $T_{1},T_{2}\in D^{\prime}(-1,0)$ given by
$T_{1}=f+\partial_{x}^{2}f-\partial_{x}^{4}f$ and $T_{2}=h_{1}$, where

$h_{1}(x)=\int_{-1}^{x}g(t)dt.$

Note that (2.54) implies that $\partial_{x}T_{1}=g$ in
$\mathcal{D}^{\prime}(-1,0).$ On the other hand, $h_{1}\in L^{p}(-1,0)$ and

$\displaystyle\left\langle\partial_{x}T_{2},\varphi\right\rangle_{\mathcal{D}^{\prime},\mathcal{D}}$
$\displaystyle=-\int_{-1}^{0}h_{1}(x)\varphi^{\prime}(x)dx=-\int_{-1}^{0}\left(\int_{-1}^{x}g(t)dt\right)\varphi^{\prime}(x)dx=-\int_{-1}^{0}\int_{-1}^{x}g(t)\varphi^{\prime}(x)dtdx.$

For $-1\leq t\leq x\leq 0$ the Fubini’s theorem gives us

$\displaystyle\left\langle\partial_{x}T_{2},\varphi\right\rangle_{\mathcal{D}^{\prime},\mathcal{D}}=-\int_{-1}^{0}g(t)(\varphi(0)-\varphi(t))dt=\int_{-1}^{0}g(t)\varphi(t)dt=\left\langle
g,\varphi\right\rangle_{\mathcal{D}^{\prime},\mathcal{D}}$

and therefore $\partial_{x}T_{2}=g=\partial_{x}T_{1}$ in
$\mathcal{D}^{\prime}(-1,0).$ Consequently, there exists a constant
$k_{1}\in\mathbb{R}$ such that $T_{1}=T_{2}+k_{1}$ in
$\mathcal{D}^{\prime}(-1,0),$ that is,

$\displaystyle f+\partial_{x}^{2}f-\partial_{x}^{4}f=h_{1}+k_{1}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0).$

Defining $\tilde{h}_{1}=h_{1}+k_{1}-f$ we have $\tilde{h}_{1}\in L^{p}(-1,0)$
and

(2.55) $\displaystyle\partial_{x}^{2}f-\partial_{x}^{4}f=\tilde{h}_{1}\text{ \
in \ }\mathcal{D}^{\prime}(-1,0).$

Now, consider $T_{3},T_{4}\in\mathcal{D}^{\prime}(-1,0)$ given by
$T_{3}=\partial_{x}f-\partial_{x}^{3}f$ and $T_{4}=h_{2}$, where $h_{2}\in
L^{p}(-1,0)$ is given by

$\displaystyle h_{2}(x)=\int_{1}^{x}\tilde{h}_{1}(t)dt.$

Proceeding as done before we obtain
$\partial_{x}T_{3}=\tilde{h}_{1}=\partial_{x}T_{4}$ in
$\mathcal{D}^{\prime}(-1,0)$, thus there exists a constant
$k_{2}\in\mathbb{R}$ such that $T_{3}=T_{4}+k_{2}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0)$, or equivalently,

$\displaystyle\partial_{x}f-\partial_{x}^{3}f=h_{2}+k_{2}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0).$

Defining $\tilde{h}_{2}:=h_{2}+k_{2}$ we have $\tilde{h}_{2}\in L^{p}(-1,0)$
and

(2.56) $\displaystyle\partial_{x}f-\partial_{x}^{3}f=\tilde{h}_{2}\text{ \ in
\ }\mathcal{D}^{\prime}(-1,0).$

Set $T_{5},T_{6}\in\mathcal{D}^{\prime}(-1,0)$ by $T_{5}=f-\partial_{x}^{2}f$
and $T_{6}=h_{3}$, where $h_{3}\in L^{p}(-1,0)$ is given by

$\displaystyle h_{3}(x)=\int_{-1}^{x}\tilde{h}_{2}(t)dt.$

By the same argument as done before,

$\displaystyle f-\partial_{x}^{2}f=h_{3}+k_{3}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0)$

with $k_{3}\in\mathbb{R}$. Defining $\tilde{h}_{3}=f-h_{3}-k_{3}$ we have
$\tilde{h}_{3}\in L^{p}(-1,0)$ and

(2.57) $\displaystyle\partial_{x}^{2}f=\tilde{h}_{3}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0).$

Now define $T_{7},T_{8}\in\mathcal{D}^{\prime}(-1,0)$ by $T_{7}=\partial_{x}f$
and $T_{8}=h_{4}$, where $h_{4}\in L^{p}(-1,0)$ is given by

$\displaystyle h_{4}(x)=\int_{-1}^{t}\tilde{h}(t)dt.$

Again, the same argument ensures that there exists $k_{4}\in\mathbb{R}$ such
that, defining $\tilde{h}_{4}:=h_{4}+k_{4}$ we have $\tilde{h}_{4}\in
L^{p}(-1,0)$ and

(2.58) $\displaystyle\partial_{x}f=\tilde{h}_{4}\text{ \ in \
}\mathcal{D}^{\prime}(-1,0).$

Since $\tilde{h}_{1}\tilde{h}_{2},\tilde{h}_{3},\tilde{h}_{4}\in L^{p}(-1,0)$,
the equalities (2.54)-(2.58) gives us
$\partial_{x}f,\partial_{x}^{2}f,\partial_{x}^{3}f,\partial_{x}^{4}f$ and
$\partial_{x}^{5}f$ belong to $L^{p}(-1,0)$, so that $f\in W^{5,p}(-1,0)$.
Furthermore,

$\displaystyle\|f_{n}-f\|_{*}=\|f_{n}-f\|_{p}+\|Pf_{n}-Pf\|_{p}=\|f_{n}-f\|_{p}+\|Pf_{n}-g\|_{p}$

which implies that

$\displaystyle\|f_{n}-f\|_{*}\rightarrow 0\text{ \ as \ }n\rightarrow\infty,$

that is, $f_{n}\rightarrow f$ in $\left(W^{5,p}(-1,0),\|\cdot\|_{*}\right)$
showing that $\left(W^{5,p}(-1,0),\|\cdot\|_{*}\right)$ is a Banach space.

Consider the map

$\displaystyle\begin{array}[]{rcl}I:\left(W^{5,p}(-1,0),\|\cdot\|_{5,p}\right)&\rightarrow&\left(W^{5,p}(-1,0),\|\cdot\|_{*}\right)\\\
f&\mapsto&I(f)=f.\end{array}$

Note that $I$ is linear and bijective and $\|I(f)\|_{*}\leq\|f\|_{5,p}$, so
that $I$ is continuous. Thus, $I^{-1}$ is also continuous. Therefore, there
exists $K_{1}>0$ such that

$\displaystyle\|f\|_{5,p}\leq K_{1}\left(\|f\|_{p}+\|Pf\|_{p}\right)\ \ \
\forall f\in W^{5,p}(-1,0),$

showing the result. ∎

######  Remark 3.

As a consequence of the Lemma 2.5, we see that the norms $\|\cdot\|_{*}$ and
$\|\cdot\|_{5,p}$ are equivalents in the space $W^{5,p}(-1,0)$.

######  Lemma 2.6.

Let $p\in[1,\infty]$ be. There exists a constant $K_{2}=K_{2}(p)>0$ such that,
for every $n\geq 1$,

$\displaystyle 2\cdot
3^{-(n+1)}\sum_{i=0}^{n}\|P^{i}f\|_{p}\leq\|f\|_{5n,p}\leq(1+2K_{2})^{n-1}K_{2}\sum_{i=0}^{n}\|P^{i}f\|_{p}\leq\|f\|_{5n,p},\
\forall\ f\in W^{5n,p}(-1,0).$

Consequently,

$\displaystyle 2\cdot
3^{-(n+1)}\sum_{i=0}^{n}\|P^{i}f\|_{p}\leq\|f\|_{5n,p}\leq
K_{3}^{n}\sum_{i=0}^{n}\|P^{i}f\|_{p},\ \forall\ f\in W^{5n,p}(-1,0),$

for every $n\geq 0$, where $K_{3}=K_{3}(p)=1+2K_{2}$.

###### Proof.

Given $n\geq 0$ and $f\in W^{5n,p}(-1,0)$ we have $f\in W^{5i,p}(-1,0)$ and
$\|f\|_{5i,p}\leq\|f\|_{5n,p},$ $0\leq i\leq n.$ Then by Lemma 2.4 we get

$\displaystyle\sum_{i=0}^{n}\|P^{i}f\|_{p}$
$\displaystyle\leq\sum_{i=0}^{n}3^{i}\|f\|_{5i,p}\leq\left(\sum_{i=0}^{n}3^{i}\right)\|f\|_{5n,p}=1\cdot\frac{3^{n+1}-1}{3-1}\|f\|_{5n,p}\leq\frac{3^{n+1}}{2}\|f\|_{5n,p}$

and therefore

$\displaystyle 2\cdot 3^{-(n+1)}\sum_{i=0}^{n}\|P^{i}f\|_{p}\leq\|f\|_{5n,p},\
\ \forall f\in W^{5n,p}(-1,0),\ \ \forall n\geq 0.$

To prove

$\displaystyle\|f\|_{5n,p}\leq(1+2K_{1})^{n-1}K_{1}\sum_{i=0}^{n}\|P^{i}f\|_{p}\
\ \forall f\in W^{5n,p}(-1,0),\ \ \forall n\geq 1$

we use induction on $n$. For $n=1$ this is true since, by Lemma 2.5,

$\displaystyle\|f\|_{5,p}\leq
K_{1}\left(\|f\|_{p}+\|Pf\|_{p}\right)=K_{1}\sum_{i=0}^{1}\|P^{i}f\|_{p}=(1+2K_{1})^{0}K_{1}\sum_{i=0}^{1}\|P^{i}f\|_{p}.$

Assume that, for $n\geq 2$, the inequality is true up to the rank $n-1$. Then,
for any $f\in W^{5n,p}(-1,0)$ we have (setting $j=i-5n+5$)

$\displaystyle\|f\|_{5n,p}$
$\displaystyle=\|f\|_{5n-5,p}+\sum_{j=1}^{5}\|\partial_{x}^{j}\partial_{x}^{5n-5}f\|_{p}\leq\|f\|_{5n-5,p}+\|\partial_{x}^{5n-5}f\|_{5,p}.$

Using Lemma 2.5 and the induction hypothesis we get

$\displaystyle\|f\|_{5n,p}$
$\displaystyle\leq(1+2K_{1})^{(n-1)-1}K_{1}\sum_{i=0}^{n-1}\|P^{i}f\|_{p}+K_{1}\left(\|\partial_{x}^{5n-5}f\|_{p}+\|P\partial_{x}^{5n-5}f\|_{p}\right)$
$\displaystyle\leq(1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n-1}\|P^{i}f\|_{p}+K_{1}\left(\|f\|_{5n-5,p}+\|Pf\|_{5n-5,p}\right).$

Once again induction hypothesis yields

$\displaystyle\|f\|_{5n,p}\leq$
$\displaystyle(1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n-1}\|P^{i}f\|_{p}$
$\displaystyle+K_{1}\left((1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n-1}\|P^{i}f\|_{p}+(1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n-1}\|P^{i}Pf\|_{p}\right).$

Then

$\displaystyle\|f\|_{5n,p}$
$\displaystyle\leq(1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n}\|P^{i}f\|_{p}+2K_{1}(1+2K_{1})^{n-2}K_{1}\sum_{i=0}^{n}\|P^{i}f\|_{p}$
$\displaystyle=(1+2K_{1})^{n-1}K_{1}\sum_{i=0}^{n}\|P^{i}f\|_{p},$

from where the desired follows with $K_{2}=K_{1}$.

Finally, consider $K_{3}=1+2K_{2}$ and $f\in W^{5n,p}(-1,0)$. For $n=0$, we
see that $\|f\|_{5n,p}=K_{3}^{n}\sum_{i=0}^{n}\|P^{i}f\|_{p}.$ Now, For $n\geq
1$, we have that

$\displaystyle\|f\|_{5n,p}$
$\displaystyle\leq(1+2K_{2})^{n-1}K_{2}\sum_{i=0}^{n}\|P^{i}f\|_{p}\leq(1+2K_{2})^{n-1}(1+2K_{2})\sum_{i=0}^{n}\|P^{i}f\|_{p}=K_{3}^{n}\sum_{i=0}^{n}\|P^{i}f\|_{p}.$

∎

###  2.3. Smoothing property

With the previous section in hand, let us show that for any $u_{0}\in
L^{2}(-1,0)$, the solution of (1.5) with $h_{1}=h_{2}=0$ is a Gevrey unction
of order $\frac{1}{2}$ in the variable $x$ and $\frac{5}{2}$ in the variable
$t$. Precisely, we will prove the existence of positive constants $M,R_{1}$,
and $R_{2}$ such that

(2.59)
$\displaystyle\left|\partial_{x}^{p}\partial_{t}^{n}u(x,t)\right|\leq\frac{M}{t^{\frac{5n+p+5}{2}}}\frac{p!^{\frac{1}{2}}}{R_{1}^{p}}\frac{n!^{\frac{5}{2}}}{R_{2}^{n}},$

for all $t\in(0,T]$ and for all $x\in[-1,0].$

To do this, using the inequality (2.59) on intervals of length one we can
assume, without loss of generality, $T=1$. Consider the operator given by
$Au=-Pu=-\partial_{x}u-\partial_{x}^{3}u+\partial_{x}^{5}u,$ with
$D(A)=\left\\{u\in H^{5}(-1,0);\
u(-1)=u(0)=u_{x}(-1)=u_{x}(0)=u_{xx}(0)=0\right\\}.$ It has been proven in [2]
that $A$ generates a semigroup of contractions $\\{S(t)\\}_{t\geq 0}$ in
$L^{2}(-1,0)$, moreover, that smoothing effect is verified, that is, for
$u_{0}\in L^{2}(-1,0)$, the mild solution $u(\cdot,t)=S(t)u_{0}$ of (1.5) with
$h_{1}=h_{2}=0$ satisfies $u\in C([0,1],L^{2}(-1,0))\cap
L^{2}(0,1,H^{2}(-1,0))$ and

(2.60)
$\displaystyle\|u\|_{L^{2}(0,T,H^{2}(-1,0))}\leq\sqrt{3}\|u_{0}\|_{L^{2}(-1,0)}.$

Now on, we will denote the norm $\|\cdot\|_{L^{2}(-1,0)}$ by simplicity for
$\|\cdot\|_{L^{2}}$ and the spaces

$\displaystyle X_{0}$ $\displaystyle=L^{2}(-1,0)\quad
X_{1}=H_{0}^{1}(-1,0),\quad X_{2}=H_{0}^{2}(-1,0),$ $\displaystyle X_{3}$
$\displaystyle=\left\\{u\in H^{3}(-1,0);\
u(-1)=u(0)=u_{x}(-1)=u_{x}(0)=u_{xx}(0)=0\right\\},$ $\displaystyle X_{4}$
$\displaystyle=\left\\{u\in H^{4}(-1,0);\
u(-1)=u(0)=u_{x}(-1)=u_{x}(0)=u_{xx}(0)=0\right\\},$ $\displaystyle X_{5}$
$\displaystyle=D(A).$

For any $m\in\\{1,2,3,4,5\\},(X_{m},\|\cdot\|_{H^{m}})$ is a Banach space. The
next propositions are paramount in our analysis and given several estimates in
the $X_{s}$-spaces.

######  Proposition 2.7.

For any $t\in(0,1]$ the map $S(t):L^{2}(-1,0)\rightarrow H^{2}(-1,0)$ is
continuous. More precisely, there exists a positive constant $C_{1}>0$ such
that $\|u(\cdot,t)\|_{H^{2}}\leq C_{1}\|u_{0}\|_{L^{2}},$ where
$u(\cdot,t)=S(t)u_{0}$.

###### Proof.

Given $u_{0}\in D(A)$, the semigroup theory ensures that $S(t)u_{0}\in D(A)$
with

(2.61) $\displaystyle\frac{d}{dt}S(t)u_{0}=AS(t)u_{0}=S(t)Au_{0}$

and $u=S(\cdot)u_{0}\in C\left([0,\infty),D(A)\right).$ From (2.61) we obtain
$\|AS(t)u_{0}\|_{L^{2}}=\|S(t)Au_{0}\|_{L^{2}}$ and, therefore,
$\|S(t)u_{0}\|_{D(A)}\leq\|u_{0}\|_{D(A)}.$

Using the Lemma 2.5 (see also Remark 3) we obtain a positive constant
$C^{\prime}>0$ (which does not depend on $t$) such that
$\|S(t)u_{0}\|_{H^{5}}\leq C_{1}^{\prime}\|u_{0}\|_{H^{5}}.$ Then we see that
the map
$S(t):\left(D(A),\|\cdot\|_{H^{5}}\right)\to\left(D(A),\|\cdot\|_{H^{5}}\right)$
is continuous. Since $S(t):L^{2}(-1,0)\to L^{2}(-1,0)$ is also continuous and,
by interpolation argument, $\left[X_{0},X_{5}\right]_{\frac{2}{5}}=X_{2}$, we
have that
$S(t):\left(X_{2},\|\cdot\|_{H^{2}}\right)\to\left(X_{2},\|\cdot\|_{H^{2}}\right)$
is continuous with $\|u(\cdot,t)\|_{H^{2}}=\|S(t)u_{0}\|_{H^{2}}\leq
C_{1}^{\prime\prime}\|u_{0}\|_{H^{2}},$ for $u_{0}\in X_{2}$ and $t\in(0,1]$,
where $C_{1}^{\prime\prime}=\max\\{C_{1}^{\prime},1\\}$.

In this way, given $u_{0}\in X_{2}$, for any $t\in(0,1]$ we have

$\displaystyle\|u(\cdot,t)\|_{H^{2}}^{2}\leq(C_{1}^{\prime\prime})^{2}\|u(\cdot,s)\|_{H^{2}}^{2},$

for all $s\in(0,t].$ Integrating with respect to $s$ from $0$ to $t$ we get

$\displaystyle
t\|u(\cdot,t)\|_{H^{2}}^{2}\leq(C_{1}^{\prime\prime})^{2}\int_{0}^{t}\|u(\cdot,s)\|_{H^{2}}^{2}ds\leq(C_{1}^{\prime\prime})^{2}\|u\|_{L^{2}(0,1,H^{2}(-1,0))}^{2}.$

Using (2.60) we obtain

$\displaystyle\|u(\cdot,t)\|_{H^{2}}\leq\frac{C_{1}^{\prime\prime}\sqrt{3}}{\sqrt{t}}\|u_{0}\|_{L^{2}}^{2},$

and the result is achieved. ∎

######  Proposition 2.8.

For any $t\in(0,1]$ the map $S(t):D(A)\rightarrow H^{7}(-1,0)$ is continuous,
that is, there exists a positive constant $C_{2}>0$ such that
$\|u(\cdot,t)\|_{H^{7}}\leq C_{2}\|u_{0}\|_{H^{5}},$ for all $u_{0}\in D(A)$.

###### Proof.

First, we need to check that $S(t)u_{0}\in H^{7}(-1,0)$ whenever $u_{0}\in
D(A)$. Indeed, given $u_{0}\in D(A)$ we have $u(\cdot,t)=S(t)u_{0}\in D(A)$
and (2.61) holds. Then $Au(\cdot,t)=AS(t)u_{0}=S(t)Au_{0}\in H^{2}(-1,0)$
which implies that $\partial_{x}^{5}u(\cdot,t)\in H^{2}(-1,0)$ and
consequently $u(\cdot,t)\in H^{7}(-1,0).$ Furthermore, Proposition 2.7 gives
us

(2.62)
$\displaystyle\|u(\cdot,t)\|_{H^{2}}+\left\|Au(\cdot,t)\right\|_{H^{2}}\leq\frac{C_{1}}{\sqrt{t}}\left(\|u_{0}\|_{L^{2}}+\|Au_{0}\|_{L^{2}}\right).$

Now, let $v\in H^{7}(-1,0)$. Observe that
$\partial_{x}^{6}v=\partial_{x}^{2}v+\partial_{x}^{4}v-\partial_{x}Pv$ and
$\partial_{x}^{7}v=\partial_{x}^{3}v+\partial_{x}^{5}v-\partial_{x}^{2}Pv.$
Then,

$\displaystyle\|v\|_{H^{7}}\leq
C_{2}^{\prime}\left(\|v\|_{H^{5}}+\|Pv\|_{H^{2}}\right),$

for some positive constant $C_{2}^{\prime}$. Thus, using the Lemma 2.5, we
obtain

$\displaystyle\|v\|_{H_{7}}\leq
C_{2}^{\prime}\left(\|v\|_{H^{2}}+\|Av\|_{H^{2}}\right).$

With this and (2.62), we get that

$\displaystyle\|u(\cdot,t)\|_{H^{7}}\leq\frac{C_{2}^{\prime}C_{1}}{\sqrt{t}}\left(\|u_{0}\|_{L^{2}}+\|Au_{0}\|_{L^{2}}\right)=\frac{C_{2}^{\prime}C_{1}}{\sqrt{t}}\|u_{0}\|_{D(A)}\leq\frac{C_{2}}{\sqrt{t}}\|u_{0}\|_{H^{5}},$

showing the proposition. ∎

######  Proposition 2.9.

For every $t\in(0,1]$ and $m\in\\{1,2,3,4,5\\}$ the map $S(t):H^{m}\rightarrow
H^{m+2}$ is continuous; there exists a constant $C_{3}>1$ such that
$\|u(\cdot,t)\|_{H^{m+2}}\leq C_{3}\|u_{0}\|_{H^{m}},$ for all $u_{0}\in
X_{m}.$

###### Proof.

By Propositions 2.7 and 2.8, the linear maps $S(t):L^{2}(-1,0)\rightarrow
L^{2}(-1,0)$ and $S(t):D(A)\rightarrow H^{7}(-1,0)$ are continuous. Moreover,
there exists $C_{3}>1$ such that

(2.63) $\begin{cases}\begin{aligned}
&\|S(t)u_{0}\|_{H^{2}}\leq\frac{C_{3}}{\sqrt{t}}\|u_{0}\|_{L^{2}},\ \forall\
u_{0}\in L^{2}(-1,0),\\\
&\|S(t)u_{0}\|_{H^{7}}\leq\frac{C_{3}}{\sqrt{t}}\|u_{0}\|_{H^{5}},\ \forall\
u_{0}\in D(A).\end{aligned}\end{cases}$

For $m=1,2,3,4$, thanks to the interpolation arguments, it follows that
$S(t)\left(\left[X_{0},X_{5}\right]_{\frac{m}{5}}\right)\subset\left[H^{2},H^{7}\right]_{\frac{m}{5}}$,
and the maps
$S(t):\left[X_{0},X_{5}\right]_{\frac{m}{5}}\rightarrow\left[H^{2},H^{7}\right]_{\frac{m}{5}}$
are continuous with

$\displaystyle\|S(t)u_{0}\|_{\left[H^{2},H^{7}\right]_{\frac{m}{5}}}\leq\frac{C_{3}}{\sqrt{t}}\|u_{0}\|_{\left[X_{0},X_{5}\right]_{\frac{m}{5}}},\
\forall\ u_{0}\in\left[X_{0},X_{5}\right]_{\frac{m}{5}}.$

Since $\left[H^{2},H^{7}\right]_{\frac{m}{5}}=H^{(1-\frac{m}{5})\cdot
2+\frac{m}{2}\cdot 7}=H^{m+2}$ and
$\left[X_{0},X_{5}\right]_{\frac{m}{5}}=X_{m}$, it follows that the maps
$S(t):X_{m}\rightarrow H^{m+2}$ for $m=1,2,3,4$, are continuous with

(2.64)
$\displaystyle\|S(t)u_{0}\|_{H^{m+2}}\leq\frac{C_{3}}{\sqrt{t}}\|u_{0}\|_{H^{m}},\
\forall\ u_{0}\in H^{m}.$

From (2.63) and (2.64) we obtain the desired. ∎

It is convenient to remember that for $n\in\mathbb{N}$, we define inductively
by

$\displaystyle D(A^{n})$ $\displaystyle=\left\\{v\in L^{2}(-1,0);\ v\in
D(A^{n-1})\text{ and }Av\in D(A^{n-1})\right\\},\ \ A^{n}v=A^{n-1}(Av).$

So, with this in hand, we have the following result.

######  Proposition 2.10.

Let $u_{0}\in L^{2}(-1,0)$ and $t\in(0,1]$. We have for $u=S(\cdot)u_{0}$
that:

  1. $(i)$

$u(\cdot,t)\in D(A)$ and

$\displaystyle\|Au(\cdot,t)\|_{L^{2}}\leq\frac{C_{4}}{t^{\frac{3}{2}}}\|u_{0}\|_{L^{2}}$

for some positive constant $C_{4}>1$ (which does not depend on $t$).

  2. $(ii)$

$u(\cdot,t)\in D(A^{n})$ for every $n\in\mathbb{N}$ and

$\displaystyle\|A^{n}u(\cdot,t)\|_{L^{2}}\leq\frac{C_{5}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|u_{0}\|_{L^{2}},$

where $C_{5}>\max\\{1,C_{4}\\}$ is a constant which does not depend on $t$.

  3. $(iii)$

$u\in C((0,1],D(A^{n}))$, for every $n\in\mathbb{N}\cup\\{0\\}$.

###### Proof.

(i) Assume $u_{0}\in D(A)$. Splitting $[0,t]$ into
$[0,t/3]\cup[t/3,2t/3]\cup[2t/3,t]$, using Proposition 2.9 and take in mind
that
$Au(\cdot,t)=-\partial_{x}u(\cdot,t)-\partial_{x}^{3}u(\cdot,t)+\partial_{x}^{5}u(\cdot,t),$
we get that

(2.65)
$\displaystyle\|Au(\cdot,t)\|_{L^{2}}\leq\frac{\tilde{C}_{4}}{t^{\frac{3}{2}}}\|u_{0}\|_{L^{2}},\
\forall\ u_{0}\in D(A),$

where $\tilde{C}_{4}=\frac{3C_{3}^{3}}{\left(\sqrt{\frac{1}{3}}\right)^{3}}$
and $\tilde{C}_{4}>1$.

Note that, thanks to the inequality (2.65), the linear operator
$S(t):(D(A),\|\cdot\|_{L^{2}})\rightarrow(D(A),\|\cdot\|_{D(A)})$ is a bounded
linear, and holds that

$\displaystyle\|S(t)u_{0}\|_{D(A)}\leq\frac{2\tilde{C}_{4}}{t^{\frac{3}{2}}}\|u_{0}\|_{L^{2}},$

for $u_{0}\in D(A)$. Since $D(A)$ is dense in $L^{2}(-1,0)$, there exists a
bounded linear operator
$\Lambda_{t}:(L^{2}(-1,0),\|\cdot\|_{L^{2}})\rightarrow(D(A),\|\cdot\|_{D(A)})$
such that

(2.66) $\displaystyle\Lambda_{t}\big{|}_{D(A)}=S(t)\text{ \ \ and \ \
}\|\Lambda_{t}v\|_{D(A)}\leq\frac{2\tilde{C}_{4}}{t^{\frac{3}{2}}}\|v\|_{L^{2}},\
\ \forall\ v\in L^{2}(-1,0).$

Now, given $u_{0}\in L^{2}(-1,0)$, there exists $(u_{k})\subset D(A)$ such
that $u_{k}\rightarrow u_{0}$ in $L^{2}(-1,0)$. Then from (2.66),

$\displaystyle\|\Lambda_{t}u_{0}-S(t)u_{0}\|_{L^{2}}\leq\frac{2\tilde{C}_{4}}{t^{\frac{3}{2}}}\|u_{0}-u_{k}\|_{L^{2}}+\|u_{0}-u_{k}\|_{L^{2}}.$

Making $k\rightarrow\infty$ we obtain $\Lambda_{t}u_{0}=S(t)u_{0}.$ Therefore,
$S(t)u_{0}\in D(A),$ for $u_{0}\in L^{2}(-1,0)$, and by (2.66)

$\displaystyle\|Au(\cdot,t)\|_{L^{2}}=\|AS(t)u_{0}\|_{L^{2}}\leq\|S(t)u_{0}\|_{D(A)}=\|\Lambda_{t}u_{0}\|_{D(A)}\leq\frac{C_{4}}{t^{\frac{3}{2}}}\|u_{0}\|_{L^{2}},$

where $C_{4}=2\tilde{C}_{4}$, giving the iten (i).

(ii) Assume $u_{0}\in D(A^{n})$. From the demigroup theory we have
$S(t)u_{0}\in D(A^{n})$ and $A^{n}u(\cdot,t)=AS(t)A^{n-1}u_{0}$. Using the
item $(i)$ we get

(2.67)
$\displaystyle\|A^{n}u(\cdot,t)\|_{L^{2}}=\|AS(t)A^{n-1}u_{0}\|_{L^{2}}\leq\frac{C_{4}}{t^{\frac{3}{2}}}\left\|A^{n-1}u_{0}\right\|_{L^{2}}.$

Splitting $[0,t]$ into $[0,t]=[0,t/n]\cup[t/n,2t/n]\cup\cdots\cup[(n-1)t/n,t]$
and using (2.67) several times we obtain

$\displaystyle\|A^{n}u(\cdot,t)\|_{L^{2}}\leq\frac{C_{4}}{t^{\frac{3}{2}}}\left\|A^{n-1}u(\cdot,(n-1)t/n)\right\|_{L^{2}}\leq\cdots$
$\displaystyle\leq\frac{C_{4}}{t^{\frac{3}{2}}}\frac{C_{4}}{\left(\frac{n-1}{n}t\right)^{\frac{3}{2}}}\frac{C_{4}}{\left(\frac{n-2}{n}t\right)^{\frac{3}{2}}}\cdots\frac{C_{4}}{\left(\frac{2t}{n}\right)^{\frac{3}{2}}}\frac{C_{4}}{\left(\frac{t}{n}\right)^{\frac{3}{2}}}\|u_{0}\|_{L^{2}}$
$\displaystyle\leq\left(\frac{C_{4}}{\left(\frac{t}{n}\right)^{\frac{3}{2}}}\right)^{n}\|u_{0}\|_{L^{2}},$

thus, for $u_{0}\in D(A^{n})$, holds that

$\displaystyle\|A^{n}u(\cdot,t)\|_{L^{2}}$
$\displaystyle\leq\frac{C_{4}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|u_{0}\|_{L^{2}}.$

Now, remark that
$S(t):\left(D(A^{n}),\|\cdot\|_{L^{2}}\right)\rightarrow\left(D(A^{n}),\|\cdot\|_{D(A^{n})}\right)$
is a bounded linear operator, since, for $u_{0}\in D(A^{n})$, the previous
estimate ensures that

$\displaystyle\|S(t)u_{0}\|_{D(A^{n})}=\|S(t)u_{0}\|_{L^{2}}+\|A^{n}S(t)u_{0}\|_{L^{2}}\leq\left(1+\frac{C_{4}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\right)\|u_{0}\|_{L^{2}}.$

Since $D(A^{n})$ is dense in $L^{2}(-1,0)$, there exists a bounded linear
operator
$\Lambda_{t,n}:\left(L^{2}(-1,0),\|\cdot\|_{L^{2}}\right)\rightarrow\left(D(A^{n}),\|\cdot\|_{D(A^{n})}\right)$,
such that

(2.68) $\displaystyle\Lambda_{t,n}\big{|}_{D(A^{n})}=S(t)\text{ \ \ and \ \
}\|\Lambda_{t,n}v\|_{D(A^{n})}\leq\frac{C_{5}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|v\|_{L^{2}},\
\forall v\in L^{2}(-1,0)$

for some constant $C_{5}>\max\\{1,C_{4}\\}$ which does not depend on $t$.

Given $u_{0}\in L^{2}(-1,0)$, there exists $(u_{k})\subset D(A^{n})$ such that
$u_{k}\rightarrow u_{0}$ in $L^{2}(-1,0)$. Thus, (2.68) gives that

$\displaystyle\|\Lambda_{t,n}u_{0}-S(t)u_{0}\|_{L^{2}}$
$\displaystyle\leq\|\Lambda_{t,n}u_{0}-\Lambda_{t,n}u_{k}\|_{D(A^{n})}+\|S(t)u_{k}-S(t)u_{0}\|_{L^{2}}$
$\displaystyle\leq\frac{C_{5}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|u_{0}-u_{k}\|_{L^{2}}+\|u_{k}-u_{0}\|_{L^{2}}.$

Making $k\rightarrow\infty$ we obtain $S(t)u_{0}=\Lambda_{t,n}u_{0}$.
Therefore, $S(t)u_{0}\in D(A),$ for $u_{0}\in L^{2}(-1,0)$, and, due to
(2.68),

$\displaystyle\|A^{n}u(\cdot,t)\|_{L^{2}}\leq\|S(t)u_{0}\|_{D(A^{n})}=\|\Lambda_{t,n}u_{0}\|_{D(A^{n})}\leq\frac{C_{5}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|u_{0}\|_{L^{2}},$

and item (ii) holds.

(iii) Let $n\in\mathbb{N}$ and $\varepsilon\in(0,1)$ be. By the item $(ii)$ we
have $S(t)u_{0}\in D(A^{n})$. In particular, $S(\varepsilon)u_{0}\in D(A^{n})$
so, from the semigroup theory, we obtain
$S(\cdot)u_{0}\in\bigcap_{j=0}^{n}C^{n-j}([\varepsilon,1];D(A^{j}))$. Taking
$j=n$ we get $S(\cdot)u_{0}\in C([\varepsilon,1];D(A^{n}))$, and as
$\varepsilon\in(0,1)$ is arbitrary it follows that $S(\cdot)u_{0}\in
C((0,1];D(A^{n}))$, showing the item (iii), and the proof is finished. ∎

The last lemma will be useful to prove the main result of this section.

######  Lemma 2.11.

For every $n\in\mathbb{N}\cap\\{0\\}$ we have $D(A^{n})\subset H^{5n}(-1,0).$

###### Proof.

The result is obvious for $n=0$ and $n=1$. For $n=2$, given $v\in D(A^{2})$
there exists $g\in D(A)\subset H^{5}$ such that $Av=g$, that is,
$\partial_{x}^{5}v=g+\partial_{x}v+\partial_{x}^{3}v\in H^{2}$ and therefore
$v\in H^{7}$. Thus,
$\partial_{x}^{7}v=\partial_{x}^{2}g+\partial_{x}^{3}v+\partial_{x}^{5}v\in
H^{2}$, so $v\in H^{9}$. Deriving, again, we get
$\partial_{x}^{9}v=\partial_{x}^{4}g+\partial_{x}^{5}v+\partial_{x}^{7}v\in
H^{1}$, and so $v\in H^{10}$. Therefore $D(A^{2})\subset H^{10}.$

To conclude the result for any $n\in\mathbb{N}$ we proceed by induction.
Suppose that for some $n\geq 1$ we have $D(A^{n})\subset H^{5n}(-1,0).$ Let
$v\in D(A^{n+1})$, by induction hypothesis, and using the same procedure, we
have $v,Av\in H^{5n}$ which implies that there exists $f\in H^{5n}$ with
$Av=f$, that is, $\partial_{x}^{5}v\in H^{5n-3}$ and so $v\in H^{5n+2}$. By
deriving $Av=f$ twice we obtain $v\in H^{5n+4}$. Deriving again twice it
follows that $v\in H^{5n+5}$ and therefore $D(A^{n+1})\subset H^{5(n+1)}$,
which concludes the proof. ∎

The previous results ensure directly the following one.

######  Proposition 2.12.

For any $u_{0}\in L^{2}(-1,0)$ the solution $u(\cdot,t)=S(t)u_{0}$ of (1.5)
with $h_{1}=h_{2}=0$ satisfies $u\in C^{\infty}([-1,0]\times(0,1]).$

###### Proof.

Consider $n\in\mathbb{N}$, $t\in(0,1]$ and $u_{0}\in L^{2}(-1,0)$. From
Proposition 2.10 we have $u(\cdot,t)\in D(A^{n})$. Using Lemma 2.11 we obtain
$u(\cdot,t)\in H^{5n}$. The Sobolev embedding $H^{5n}\hookrightarrow
C^{5n-1}([-1,0])$ provides $u(\cdot,t)\in C^{5n-1}([-1,0])$. Since
$n\in\mathbb{N}$ is arbitrary, it follows that $u(\cdot,t)\in
C^{\infty}([-1,0]).$ By the other hand, given $\varepsilon>0$ and
$n\in\mathbb{N}$, Proposition 2.10 yields that $u(\cdot,\varepsilon)\in
D(A^{n+1})$ so, the semigroup theory, follows that
$u\in\bigcap_{j=0}^{n+1}C^{n+1-j}\left([\varepsilon,1],D(A^{j})\right).$ In
particular, taking $j=1$, we have that $u\in C^{n}([\varepsilon,1],D(A)).$
Using the Lemma 2.5 and the embedding $H^{5}\hookrightarrow C^{4}([-1,0])$ we
obtain that $\partial_{t}^{i}u(x,\cdot)$ is continuous at $t_{0}$ and, as
$t_{0}\in[\varepsilon,1]$ is arbitrary we conclude that
$\partial_{t}^{i}u(x,\cdot)$ is continuous for $i=0,1,...,n$. Since
$n\in\mathbb{N}$ and $x\in[-1,0]$ are arbitrary we get $u(x,\cdot)\in
C^{\infty}([\varepsilon,1]$. Furthermore, $u(x,\cdot)\in C^{\infty}((0,1]),$
and the result holds. ∎

We are now in a position to prove a smooth property for the solutions of
(1.5). With these auxiliary results in hand, the main result of this section
is the following one.

######  Proposition 2.13.

Let $u_{0}\in L^{2}(-1,0)$ and $h_{1}(t)=h_{2}(t)=0$ for $t\in[0,1]$. Then the
corresponding solution $u$ of (1.5) satisfies $u\in
G^{\frac{1}{2},\frac{5}{2}}\left([-1,0]\times[\varepsilon,1]\right)$, for all
$\varepsilon\in(0,1)$, that is, we can find positive constants $M,R_{1},R_{2}$
such that

$\displaystyle\left|\partial_{x}^{p}\partial_{t}^{n}u(x,t)\right|\leq\frac{M}{t^{\frac{5n+p+5}{2}}}\frac{p!^{\frac{1}{2}}}{R_{1}^{p}}\frac{n!^{\frac{5}{2}}}{R_{2}^{n}}.$

###### Proof.

Consider $p\in\mathbb{N}\cup\\{0\\}$ and choose $n\in\mathbb{N}$ such that
$5n-5\leq p\leq 5n-1$. Then, the Sobolev embedding, Lemma 2.6 and Proposition
2.10 ensures that

(2.69) $\begin{split}\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}&\leq
C_{6}K_{3}^{n}\sum_{i=0}^{n}\left\|P^{i}u(\cdot,t)\right\|_{L^{2}}\\\
&=C_{6}K_{3}^{n}\left(\|u(\cdot,t)\|_{L^{2}}+\sum_{i=1}^{n}\left\|A^{i}u(\cdot,t)\right\|_{L^{2}}\right)\\\
&\leq
C_{6}K_{3}^{n}\left(1+\sum_{i=1}^{n}\frac{C_{5}^{i}}{t^{\frac{3i}{2}}}i^{\frac{3i}{2}}\right)\|u_{0}\|_{L^{2}}\\\
&\leq
C_{6}K_{3}^{n}(n+1)\frac{C_{5}^{n}}{t^{\frac{3n}{2}}}n^{\frac{3n}{2}}\|u_{0}\|_{L^{2}}\\\
&=C_{6}K_{3}^{n}(n+1)\frac{C_{5}^{n}}{t^{\frac{5n}{2}}}\frac{(5n)^{\frac{5n}{2}}}{5^{\frac{5n}{2}}}\|u_{0}\|_{L^{2}}.\end{split}$

Using the Stirling’s formula we get

$\displaystyle(5n)^{5n}\sim\frac{e^{5n}(5n)!}{(2\pi)^{\frac{1}{2}}(5n)^{\frac{1}{2}}}\iff(5n)^{\frac{5n}{2}}\sim\frac{e^{\frac{5n}{2}}(5n)!^{\frac{1}{2}}}{(2\pi)^{\frac{1}{4}}(5n)^{\frac{1}{4}}}\iff\frac{(5n)^{\frac{5n}{2}}}{5^{\frac{5n}{2}}}\sim\frac{1}{(2\pi)^{\frac{1}{4}}(5n)^{\frac{1}{4}}}\left(\frac{e}{5}\right)^{\frac{5n}{2}}(5n)!^{\frac{1}{2}},$

which implies that

$\displaystyle\frac{(5n)^{\frac{5n}{2}}}{5^{\frac{5n}{2}}}\leq
C_{7}\frac{1}{(2\pi)^{\frac{1}{4}}(5n)^{\frac{1}{4}}}\left(\frac{e}{5}\right)^{\frac{5n}{2}}(5n)!^{\frac{1}{2}}\leq
C_{7}\left(\frac{e}{5}\right)^{\frac{5n}{2}}(5n)!^{\frac{1}{2}}$

for some constant $C_{7}>0$. Since $(n+1)\leq e^{\frac{5n}{2}},$ for
$n\in\mathbb{N},$ the inequality (2.69) gives us

$\displaystyle\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}$
$\displaystyle\leq
C_{6}K_{3}^{n}e^{\frac{5n}{2}}\frac{C_{5}^{n}}{t^{\frac{5n}{2}}}C_{7}\left(\frac{e}{5}\right)^{\frac{5n}{2}}(5n)!^{\frac{1}{2}}\|u_{0}\|_{L^{2}}=C_{6}C_{7}\frac{C_{5}^{n}}{t^{\frac{5n}{2}}}K_{3}^{n}\left(\frac{e^{2}}{5}\right)^{\frac{5n}{2}}(5n)!^{\frac{1}{2}}\|u_{0}\|_{L^{2}}.$

Remember that $5n-5\leq p\leq 5n-1$ which allow us write $p=5n-r$ with
$r\in\\{1,2,3,4,5\\}$, that is, $5n=p+r,$ with $r\in\\{1,2,3,4,5\\}.$ Using
(2.6) it follows that

$\displaystyle\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}$
$\displaystyle\leq
C_{6}C_{7}\frac{C_{5}^{\frac{p+r}{5}}}{t^{\frac{p+r}{2}}}K_{3}^{\frac{p+r}{5}}\left(\frac{e^{2}}{5}\right)^{\frac{p+r}{2}}(p+r)!^{\frac{1}{2}}\|u_{0}\|_{L^{2}}$
$\displaystyle\leq
C_{6}C_{7}C_{5}^{\frac{r}{5}}K_{3}^{\frac{r}{5}}\left(\frac{e^{2}}{5}\right)^{\frac{r}{2}}\frac{C_{5}^{\frac{p}{5}}}{t^{\frac{p+r}{2}}}K_{3}^{\frac{p}{5}}\left(\frac{e^{2}}{5}\right)^{\frac{p}{2}}\left(2^{p}2^{r}p!r!\right)^{\frac{1}{2}}\|u_{0}\|_{L^{2}}$
$\displaystyle=C_{6}C_{7}C_{5}^{\frac{r}{5}}K_{3}^{\frac{r}{5}}\left(\frac{e^{2}}{5}\right)^{\frac{r}{2}}2^{\frac{r}{2}}r!^{\frac{1}{2}}\|u_{0}\|_{L^{2}}\frac{1}{t^{\frac{p+r}{2}}}\left(\frac{C_{5}^{\frac{1}{5}}K_{3}^{\frac{1}{5}}e2^{\frac{1}{2}}}{5^{\frac{1}{2}}}\right)^{p}p!^{\frac{1}{2}}.$

Consequently

$\displaystyle\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}$
$\displaystyle\leq\frac{C_{8}}{t^{\frac{p+r}{2}}}\left(\frac{C_{5}^{\frac{1}{5}}K_{0}^{\frac{1}{5}}e2^{\frac{1}{2}}}{5^{\frac{1}{2}}}\right)^{p}p!^{\frac{1}{2}}=\frac{C_{8}}{t^{\frac{p+r}{2}}}\frac{p!^{\frac{1}{2}}}{\left(5^{\frac{1}{2}}C_{5}^{-\frac{1}{5}}K_{0}^{-\frac{1}{5}}e^{-1}2^{-\frac{1}{2}}\right)^{p}},$

with

$C_{8}:=C_{6}C_{7}C_{5}^{\frac{r}{5}}K_{3}^{\frac{r}{5}}\left(\frac{e^{2}}{5}\right)^{\frac{r}{2}}2^{\frac{r}{2}}r!^{\frac{1}{2}}\|u_{0}\|_{L^{2}}\text{
\ \ and \ \ }K_{0}=K_{3}+1.$

Defining
$R=5^{\frac{1}{2}}C_{5}^{-\frac{1}{5}}K_{0}^{-\frac{1}{5}}e^{-1}2^{-\frac{1}{2}}$,
follows that $R\in(0,1)$ and

$\displaystyle\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}$
$\displaystyle\leq\frac{C_{8}}{t^{\frac{p+r}{2}}}\frac{p!^{\frac{1}{2}}}{R^{p}}.$

Here, $p\geq 0$, $r\in\\{1,2,3,4,5\\}$ and $t\in(0,1]$. Observe that, as
$t\in(0,1]$ and $0\leq r\leq 5$ holds that

(2.70) $\displaystyle\left\|\partial_{x}^{p}u(\cdot,t)\right\|_{\infty}$
$\displaystyle\leq\frac{C_{8}}{t^{\frac{p+5}{2}}}\frac{p!^{\frac{1}{2}}}{R^{p}},\
\forall\ p\geq 0,\ \forall t\in(0,1].$

Finally, for every $n,p\geq 0$ from (1.8) we have

$\displaystyle\partial_{t}^{n}\partial_{x}^{p}u$
$\displaystyle=\partial_{x}^{p}\partial_{t}^{n}u=\partial_{x}^{p}\left(-(-1)^{n-1}P^{n}u\right)=\partial_{x}^{p}\left((-1)^{n}P^{n}u\right)=(-1)^{n}P^{n}\partial_{x}^{p}u.$

Using Newton’s Binomial theorem we have that

$\displaystyle P^{n}\partial_{x}^{p}u$
$\displaystyle=\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}(-\partial_{x}^{5})^{j}(\partial_{x}^{3})^{q-j}\partial_{x}^{n-q}\partial_{x}^{p}u$
$\displaystyle=\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}(-1)^{j}\partial_{x}^{5j}\partial_{x}^{3q-3j}\partial_{x}^{n-q}\partial_{x}^{p}u.$

Thus, from (2.70), for $(x,t)\in[-1,0]\times(0,1]$,

$\displaystyle\left|\partial_{t}^{n}\partial_{x}^{p}u(x,t)\right|$
$\displaystyle\leq\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}\left|\partial_{x}^{n+2q+2j+p}u(x,t)\right|$
$\displaystyle\leq\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}\frac{C_{8}}{t^{\frac{n+2q+2j+p+5}{2}}}\frac{(n+2q+2j+p)!^{\frac{1}{2}}}{R^{n+2q+2j+p}}.$

Once that $t\in(0,1]$, $R<1$ and $n+2q+2j+p\leq 5n+p$ it follows that

$\displaystyle\left|\partial_{t}^{n}\partial_{x}^{p}u(x,t)\right|$
$\displaystyle\leq\frac{C_{8}}{t^{\frac{5n+p+5}{2}}}\frac{(5n+p)!^{\frac{1}{2}}}{R^{5n+p}}\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}.$

Noting that

$\displaystyle 2^{q}=(1+1)^{q}=\sum_{j=0}^{q}\binom{q}{j}1^{j}\cdot
1^{q-j}=\sum_{j=0}^{q}\binom{q}{j}$

and

$\displaystyle\sum_{q=0}^{n}\binom{n}{q}\sum_{j=0}^{q}\binom{q}{j}=\sum_{q=0}^{n}\binom{n}{q}2^{q}=\sum_{q=0}^{n}\binom{n}{q}2^{q}\cdot
1^{n-q}=(2+1)^{n}=3^{n},$

using (2.6) one more time, yields

$\displaystyle\left|\partial_{t}^{n}\partial_{x}^{p}u(x,t)\right|$
$\displaystyle\leq\frac{C_{8}}{t^{\frac{5n+p+5}{2}}}\frac{3^{n}\cdot
2^{\frac{5n}{2}}2^{\frac{2}{p}}(5n)!^{\frac{1}{2}}p!^{\frac{1}{2}}}{R^{5n}R^{p}}.$

The Stirling’s formula gives us

(2.71) $\displaystyle(5n)!\sim\left(\frac{5n}{e}\right)^{5n}\sqrt{2\pi
5n}=(10\pi)^{\frac{1}{2}}\left(\frac{5n}{e}\right)^{5n}n^{\frac{1}{2}}.$

Moreover, we also have $n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi
n}=(2\pi)^{\frac{1}{2}}\left(\frac{n}{e}\right)^{n}n^{\frac{1}{2}},$ then

$\displaystyle\frac{5^{5n}n!^{5}}{(2\pi)^{\frac{5}{2}}n^{\frac{5}{2}}}\sim\left(\frac{5n}{e}\right)^{5n}.$

This previous relation together with (2.71) leads us

$\displaystyle(5n)!\sim(10\pi)^{\frac{1}{2}}\frac{5^{5n}n!^{5}}{(2\pi)^{\frac{5}{2}}n^{\frac{5}{2}}}n^{\frac{1}{2}}=\frac{(10\pi)^{\frac{1}{2}}}{(2\pi)^{\frac{5}{2}}}\frac{5^{5n}n!^{5}}{n^{2}}\sim\frac{(10\pi)^{\frac{1}{4}}}{(2\pi)^{\frac{5}{4}}}\frac{5^{\frac{5n}{2}}n!^{\frac{5}{2}}}{n}.$

Thus, there exists a constant $C_{9}>0$ such that $(5n)!^{\frac{1}{2}}\leq
C_{9}5^{\frac{5n}{2}}n!^{\frac{5}{2}}.$ Hence

$\displaystyle\left|\partial_{t}^{n}\partial_{x}^{p}u(x,t)\right|$
$\displaystyle\leq\frac{C_{8}C_{9}}{t^{\frac{5n+p+5}{2}}}\left(\frac{3\cdot
2^{\frac{5}{2}}\cdot
5^{\frac{5}{2}}}{R^{5}}\right)^{n}n!^{\frac{5}{2}}\left(\frac{2^{\frac{1}{2}}}{R}\right)^{p}p!^{\frac{1}{2}}=\frac{C_{8}C_{9}}{t^{\frac{5n+p+5}{2}}}\frac{n!^{\frac{5}{2}}}{\left(3^{-1}\cdot
10^{-\frac{5}{2}}\cdot
R^{5}\right)^{n}}\frac{p!^{\frac{1}{2}}}{\left(2^{-\frac{1}{2}}R\right)^{p}}$

and the result is achieved with $M=C_{8}C_{9},$ $R_{1}=2^{-\frac{1}{2}}R$ and
$R_{2}=3^{-1}\cdot 10^{-\frac{5}{2}}\cdot R^{5}.$ ∎

###  2.4. Null controllability results

Let us now prove the null controllability result. Precisely, employing two
flat output controls, the solution of (1.5) satisfies $u(\cdot,T)=0$.

###### Proof of Theorem 1.1.

Consider $u_{0}\in L^{2}(-1,0)$ and denote by $\overline{u}$ the solution of
(1.5) for $h_{1}=h_{2}=0$. From Proposition 2.13 we have, for
$\varepsilon\in(0,T)$ , that $\overline{u}\in
G^{\frac{1}{2},\frac{5}{2}}([-1,0]\times[\varepsilon,T]).$ In particular,
$\partial_{x}^{3}\overline{u}(0,t),\partial_{x}^{4}\overline{u}(0,t)\in
G^{\frac{5}{2}}([\varepsilon,T])$ for any $\varepsilon\in(0,T)$. Choose
$\tau\in(0,T)$ and define

$\displaystyle
y(t)=\phi_{s}\left(\frac{t-\tau}{T-\tau}\right)\partial_{x}^{3}\overline{u}(0,t)\text{
\ \ and \ \
}z(t)=\phi_{s}\left(\frac{t-\tau}{T-\tau}\right)\partial_{x}^{4}\overline{u}(0,t),$

where $\phi_{s}$ is the step function given by

$\displaystyle\phi_{s}(r)=\left\\{\begin{array}[]{ll}1,&\text{ if }r\leq 0,\\\
0,&\text{ if }r\geq 1,\\\
\frac{e^{-\frac{K}{(1-r)^{\sigma}}}}{e^{-\frac{K}{r^{\sigma}}}+e^{-\frac{K}{(1-r)^{\sigma}}}},&\text{
if }r\in(0,1),\end{array}\right.$

with $K>0$ and $\sigma:=(s-1)^{-1}$. As $\phi_{s}$ is Gevrey of order $s$
(see, for example, [33]) and $s\geq\frac{5}{2}$ we infer that $y,z\in
G^{s}\left([\varepsilon,T]\right),\ \forall\ \varepsilon\in(0,T).$ Then,
defining $u:[-1,0]\times(0,T]\rightarrow\mathbb{R}$ by

$\displaystyle u(x,t)$
$\displaystyle=\left\\{\begin{array}[]{ll}u_{0}(x),&\text{ if }t=0,\\\
\sum_{i\geq 0}f_{i}(x)y^{(i)}(t)+\sum_{i\geq 0}g_{i}(x)z^{(i)}(t),&\text{ if
}t\in(0,T],\end{array}\right.$

the Proposition 2.3 gives us that $u$ satisfies (2.25) with $u\in
G^{\frac{s}{5},s}\left([-1,0]\times[\varepsilon,T]\right)$ for all
$\varepsilon\in(0,T)$. In particular

$\partial_{x}^{m}u(0,t)=0,\ m=0,1,2,\ \ \partial_{x}^{3}u(0,t)=y(t)\text{ \ \
and \ \ }\partial_{x}^{4}u(0,t)=z(t).$

Furthermore, by construction

$\displaystyle\partial_{x}^{m}\overline{u}(0,t)=0,\ m=0,1,2$

and, for $t\in(0,\tau)$

$\displaystyle
y(t)=\underbrace{\phi_{s}\left(\frac{t-\tau}{T-\tau}\right)}_{=1}\partial_{x}^{3}\overline{u}(0,t)=\partial_{x}^{3}\overline{u}(0,t),$

and

$\displaystyle
z(t)=\underbrace{\phi_{s}\left(\frac{t-\tau}{T-\tau}\right)}_{=1}\partial_{x}^{4}\overline{u}(0,t)=\partial_{x}^{4}\overline{u}(0,t).$

Therefore $\partial_{x}^{m}u(0,t)=\partial_{x}^{m}\overline{u}(0,t)$, for
$m=0,1,2,3,4$ and $t\in(0,\tau)$. Thanks to the Holmgren theorem, we conclude
that $u(x,t)=\overline{u}(x,t)$, for all $(x,t)\in[-1,0]\times(0,\tau)$.
Hence, $u\in C([0,T],L^{2}(-1,0))$ and it solves the system (1.5) with

$\displaystyle h_{1}(t)=\sum_{i\geq 0}f_{i}(-1)y^{(i)}(t)+\sum_{i\geq
0}g_{i}(-1)z^{(i)}(t)$

and

$\displaystyle h_{2}(t)=\sum_{i\geq 0}f_{ix}(-1)y^{(i)}(t)+\sum_{i\geq
0}g_{ix}(-1)z^{(i)}(t).$

Observe that $h_{1},h_{2}\in G^{s}([0,T])$ and $h_{1}(t)=h_{2}(t)=0$ for
$0<t<\tau$ since

$\displaystyle\begin{cases}\begin{array}[]{l}h_{1}(t)=u(-1,t)=\overline{u}(-1,t)=0,\\\
h_{2}(t)=u_{x}(-1,t)=\overline{u}_{x}(-1,t)=0,\end{array}\ \forall\
t\in(0,\tau).\end{cases}$

Finally, once we have
$\operatorname{{supp}}y^{(i)}\subset\operatorname{{supp}}y\subset(-\infty,T)$
and
$\operatorname{{supp}}z^{(i)}\subset\operatorname{{supp}}z\subset(-\infty,T)$,
follows that $y^{(i)}(T)=0$ and $z^{(i)}(T)=0$, for every $i\geq 0$, so that
$u(\cdot,T)=0$, and the first main result is showed. ∎

##  3\. A class of reachable functions

In this section, we will establish a class of sets that can be reachable from
$0$ by the system (1.5). Our goal is to prove that, given
$u_{1}\in\mathcal{R}_{R}$ (see the definition in (1.10)), one can find control
inputs $h_{1}$ and $h_{2}$ for which the solution of (1.5), with $u_{0}=0$,
satisfies $u(x,T)=u_{1}(x)$.

###  3.1. Auxiliary results

To prove what we mentioned before we need auxiliary results. The first
establishes the flatness property for the limit case $s=5$.

######  Proposition 3.1.

Let $R>1$ and $y,z\in G^{5}([0,T])$ with

(3.1) $\displaystyle|y^{(j)}(t)|,|z^{(j)}(t)|\leq M\frac{(5j)!}{R^{5j}},\
\forall j\geq 0,\ \forall t\in[0,T].$

Then, defining $u(x,t)$ as in (2.7) we have $u\in G^{1,5}([-1,0]\times[0,T])$
and it solves (2.25).

###### Proof.

Let $m,k\geq 0$. For $j\geq k$ we have, from (3.1) and Lemma 2.2, that

$\displaystyle\sum_{j=0}^{\infty}\left|f_{j-k}(x)y^{(j+m)}(t)\right|\leq\sum_{j\geq
k}2^{j-k}\frac{|x|^{5(j-k)+1}}{\left[5(j-k)+1\right]!}M\frac{[5(j+m)]!}{R^{5(j+m)}}.$

Pick $l=5(j-k)$ and $N=5(k+m)$ so $l+N=5(j+m)$, gives that

$\displaystyle\sum_{j=0}^{\infty}\left|f_{j-k}(x)y^{(j+m)}(t)\right|\leq
M\sum_{l\geq 0}2^{\frac{l}{5}}\frac{1}{(l+1)!}\frac{(l+N)!}{R^{l+N}}.$

If $N\leq 1$ then

$\displaystyle\sum_{j=0}^{\infty}\left|f_{j-k}(x)y^{(j+m)}(t)\right|\leq\frac{M}{R^{N}}\sum_{l\geq
0}\left(\frac{1}{R\cdot 2^{-\frac{1}{5}}}\right)^{l}<\infty.$

Assume $N>1$, note that $(l+N)!=(l+N)(l+N-1)\cdots(l+2)(l+1)!$, so that

$\displaystyle\sum_{j=0}^{\infty}\left|f_{j-k}(x)y^{(j+m)}(t)\right|$
$\displaystyle\leq M\sum_{l\geq
0}2^{\frac{l}{5}}\frac{(l+N)(l+N-1)\cdots(l+2)}{R^{l+N}}$ $\displaystyle\leq
M\sum_{q\geq 0}\sum_{qN\leq
l<(q+1)N}\frac{2^{\frac{l+N}{5}}(l+N)^{N-1}}{R^{l+N}}$
$\displaystyle=M\sum_{q\geq 0}\sum_{qN\leq
l<(q+1)N}\frac{[(q+2)N]^{N-1}}{\left(2^{-\frac{1}{5}}R\right)^{l+N}}.$

For $l>qN$ we have $l+N\geq(q+1)N$ and, since $2^{-\frac{1}{5}}R>1$, follows
that

$\frac{1}{\left(2^{-\frac{1}{5}}R\right)^{l+N}}\leq\frac{1}{\left(2^{-\frac{1}{5}}R\right)^{(q+1)N}}.$

Thus, thanks to the relation (2.39), the following estimate

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq M\sum_{q\geq 0}\sum_{qN\leq
l<(q+1)N}\frac{(q+2)^{N-1}N^{N-1}}{\left(2^{-\frac{1}{5}}R\right)^{(q+1)N}}$
$\displaystyle\leq MN^{N}\sum_{q\geq
0}\left(\frac{q+2}{\tilde{R}^{q+1}}\right)^{N},$

is verified with $\tilde{R}=2^{-\frac{1}{5}}R$. Pick any $\sigma\in(0,1)$ and
define $f_{\sigma}(x)=\frac{x+2}{\left(\tilde{R}^{1-\sigma}\right)^{x+1}}.$
Note that
$\lim_{x\rightarrow\infty}\left(\tilde{R}^{1-\sigma}\right)^{x+1}=\infty$,
since $\tilde{R}>1$. The L’Hospital rule ensures that

$\displaystyle\lim_{x\rightarrow\infty}f_{\sigma}(x)=\lim_{x\rightarrow\infty}\frac{1}{\ln(\tilde{R}^{1-\sigma})(\tilde{R}^{1-\sigma})^{x+1}}=0,$

and so,

$\displaystyle\frac{q+2}{\left(\tilde{R}^{1-\sigma}\right)^{q+1}}\rightarrow
0,\text{ \ as \ }q\rightarrow\infty.$

Defining $a:=\sup_{q\geq
0}\frac{q+2}{\left(\tilde{R}^{1-\sigma}\right)^{q+1}}$ holds that

$\displaystyle\sum_{q\geq
0}\left(\frac{q+2}{\tilde{R}^{q+1}}\right)^{N}\leq\frac{a^{N}}{\tilde{R}^{N\sigma}}\sum_{q\geq
0}\frac{1}{\tilde{R}^{N\sigma
q}}=\frac{a^{N}}{\tilde{R}^{N\sigma}}\left(\frac{1}{1-\frac{1}{\tilde{R}^{N\sigma}}}\right).$

Once we have the following convergence

$\displaystyle\alpha_{N}:=\frac{1}{1-\frac{1}{\tilde{R}^{N\sigma}}}\rightarrow
1,\text{ \ as \ }N\rightarrow\infty,$

we can define $\displaystyle\tilde{M}:=\sup_{N>1}\alpha_{N}$ to get

$\displaystyle\sum_{q\geq
0}\left(\frac{q+2}{\tilde{R}^{q+1}}\right)^{N}\leq\tilde{M}\frac{a^{N}}{\tilde{R}^{N\sigma}}.$

Consequently

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq MN^{N}\tilde{M}\frac{a^{N}}{\tilde{R}^{N\sigma}},$

and using Stirling’s formula we get that

$\displaystyle\frac{a^{N}N^{N}}{\tilde{R}^{N\sigma}}\sim\frac{1}{\sqrt{2\pi}}\frac{a^{N}e^{N}}{\tilde{R}^{N\sigma}}\frac{N!}{N^{\frac{1}{2}}},$

which ensures the following estimate

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq
M^{\prime}\left(\frac{ae}{\tilde{R}^{\sigma}}\right)^{N}\frac{N!}{N^{\frac{1}{2}}},$

for some constant $M^{\prime}>0$. Moreover, noting that $N!=(5k+5m)!\leq
2^{5k}2^{5m}(5k)!(5m)!$ and using Stirling’s formula again, namely, $(5m)!\sim
5^{5m+\frac{1}{2}}\left(\sqrt{2\pi m}\right)^{-4}m!^{5}$, follows that

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq\sqrt{5}M^{\prime\prime}\left(\frac{2ae}{\tilde{R}^{\sigma}}\right)^{5k}(5k)!\left(\frac{10ae}{\tilde{R}^{\sigma}}\right)^{5m}m!^{5}\frac{1}{(k+1)^{\frac{1}{2}}}.$

Now, define $R_{1}=(2ae)^{-1}\tilde{R}^{\sigma}$,
$R_{2}=\left[(10ae)^{-1}\tilde{R}\right]^{5}$ and
$M^{\prime\prime\prime}=\sqrt{5}M^{\prime\prime}$, it follows that

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{t}^{m}P^{k}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq
M^{\prime\prime\prime}\frac{(5k)!}{R_{1}^{5k}}\frac{m!^{5}}{R_{2}^{m}}\frac{1}{(k+1)^{\frac{1}{2}}}.$

Observe that we can assume $R_{1}<1$. Let $K_{3}>0$ as in Lemma 2.6 for
$p=\infty$. Then, the previous inequality yields that

$\displaystyle\sum_{j=0}^{\infty}\left\|\partial_{t}^{m}\left(f_{j}y^{(j)}(t)\right)\right\|_{5i,\infty}$
$\displaystyle\leq
M^{\prime\prime\prime}K_{3}^{i}\frac{m!^{5}}{R_{2}^{m}}\frac{(5i)!}{R_{1}^{5i}}\sum_{k=0}^{i}\frac{1}{(k+1)^{\frac{1}{2}}},$

for all $i\geq 0$. Given $m,n\geq 0$ consider $i\geq 0$ such that
$n\in\\{5i-r,\ r=0,1,2,3,4\\}$. Thus,

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(f_{j}(x)y^{(j)}(t)\right)\right|$
$\displaystyle\leq
M^{\prime\prime\prime}K_{3}^{i}\frac{m!^{5}}{R_{2}^{m}}\frac{(5i)!}{R_{1}^{5i}}\sum_{k=0}^{i}\frac{1}{(k+1)^{\frac{1}{2}}},$

for $(x,t)\in[-1,0]\times[0,T]$. Analogously one can see that

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(g_{j}(x)z^{(j)}(t)\right)\right|$
$\displaystyle\leq
M^{\prime\prime\prime}K_{3}^{i}\frac{m!^{5}}{R_{2}^{m}}\frac{(5i)!}{R_{1}^{5i}}\sum_{k=0}^{i}\frac{1}{(k+1)^{\frac{1}{2}}}.$

Therefore these series are uniformly convergent on $[-1,0]\times[0,T]$, for
all $m,n\geq 0$ so that, the function $u$ defined by (2.7) satisfies $u\in
C^{\infty}([-1,0]\times[0,T])$. Furthermore,

$\displaystyle\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(f_{j}(x)y^{(j)}(t)\right)\right|+\sum_{j=0}^{\infty}\left|\partial_{x}^{n}\partial_{t}^{m}\left(g_{j}(x)z^{(j)}(t)\right)\right|$
$\displaystyle\leq
M^{\prime\prime\prime}K_{3}^{i}\frac{m!^{5}}{R_{2}^{m}}\frac{(5i)!}{R_{1}^{5i}}\sum_{k=0}^{i}\frac{1}{(k+1)^{\frac{1}{2}}}.$

Since $n=5i-r$ with $r\in\\{0,1,2,3,4\\}$ we have

$\displaystyle\frac{K_{3}^{i}}{R_{1}^{5i}}(5i)!\leq\frac{K_{3}^{\frac{r}{5}}\cdot
2^{r}\cdot r!}{R_{1}^{r}}\cdot\frac{n!}{\left(K_{3}^{-\frac{1}{5}}\cdot
2^{-1}\cdot R_{1}\right)^{n}}.$

Defining

$\displaystyle\hat{M}=2M^{\prime\prime\prime}\left(\max_{0\leq r\leq
4}\frac{K_{3}^{\frac{r}{5}}\cdot 2^{r}\cdot
r!}{R_{1}^{r}}\right)\sum_{k=0}^{i}\frac{1}{(k+1)^{\frac{1}{2}}}\text{ \ \ and
\ \ }R_{1}^{\prime}=K_{3}^{-\frac{1}{5}}\cdot 2^{-1}\cdot R_{1}$

it follows that

$\displaystyle\left|\partial_{x}^{n}\partial_{t}^{m}u(x,t)\right|$
$\displaystyle\leq\hat{M}\frac{n!}{(R_{1}^{\prime})^{n}}\frac{m!^{5}}{R_{2}^{m}}\
\ \forall\ n,m\geq 0,\ \ \forall(x,t)\in[-1,0]\times[0,T],$

which concludes the proof. ∎

The next result is a particular case of [32, Proposition 3.6] with $a_{0}=1$
and $a_{p}=[5p(5p-1)(5p-2)(5p-3)(5p-4)]^{-1}$, for $p\geq 1$.

######  Proposition 3.2.

Let $(d_{q})_{q\geq 0}$ be a sequence of real numbers satisfying $|d_{q}|\leq
CH^{q}(5q)!$, for all $q\geq 0$ and for some constants $H>0$ and $C>0$. Then,
for each $\tilde{H}>e^{e^{-1}}H$, one can find a function $f\in
C^{\infty}(\mathbb{R})$ such that $f^{(q)}(0)=d_{q}$, for all $q\geq 0$, and

$\displaystyle\left|f^{(q)}(x)\right|\leq C\tilde{H}^{q}(5q)!,\quad\forall
q\geq 0,\ \forall x\in\mathbb{R}.$

The next lemma is a consequence of the theory of analytic functions.

######  Lemma 3.3.

Let $\psi\in G^{1}([-1,0])$ be such that $\partial_{x}^{j}P^{n}\psi(0)=0,\
\forall n\geq 0,\ j=0,1,2,3,4.$ Then $\psi\equiv 0$.

###### Proof.

First, remember (from the proof of Proposition 2.13) that

$\displaystyle P^{n}$
$\displaystyle=\sum_{q=0}^{n}\binom{n}{q}\sum_{k=0}^{q}\binom{q}{k}(-1)^{k}\partial_{x}^{n+2q+2k}\
\ \ \forall n\geq 0$

so

$\displaystyle\partial_{x}^{j}P^{n}=\sum_{q=0}^{n}\binom{n}{q}\sum_{k=0}^{q}\binom{q}{k}(-1)^{k}\partial_{x}^{n+2q+2k+j}\
\ \ \forall n,j\geq 0.$

We claim that for any $n\geq 0$

(3.2) $\displaystyle\partial_{x}^{j}\psi(0)=0\ \ \forall
j\in\\{0,1,...,5n+4\\}.$

To prove this, we use induction on $n$. For $n=0$ this immediately follows
from the hypothesis, since

(3.3) $\displaystyle\partial_{x}^{j}\psi(0)=\partial_{x}^{j}P^{0}\psi(0)=0,\ \
\ j=0,1,2,3,4.$

Let us also analyze the case $n=1$. Using the hypothesis and (3.3) we obtain

$\displaystyle\begin{cases}&P\psi(0)=0\Rightarrow\partial_{x}^{5}\psi(0)=0,\quad\partial_{x}P\psi(0)=0\partial_{x}^{6}\psi(0)=0,\quad\partial_{x}^{2}P\psi(0)=0\Rightarrow\partial_{x}^{7}\psi(0)=0,\\\
&\partial_{x}^{3}P\psi(0)=0\Rightarrow\partial_{x}^{8}\psi(0)=0,\quad\partial_{x}^{4}P\psi(0)=0\Rightarrow\partial_{x}^{9}\psi(0)=0.\end{cases}$

Combining this with (3.3) we get (3.2), for $n=1$. Now, suppose that (3.2)
holds for some $n\geq 1$ and let us show that

$\displaystyle\partial_{x}^{j}\psi(0)=0\ \ \forall
j\in\\{0,1,...,5(n+1)+4\\}.$

By the induction hypothesis, it is sufficient to show that
$\partial_{x}^{j}\psi(0)=0$ for $j=5(n+1)+r$ with $r=0,1,2,3,4$. From
hypothesis $P^{n+1}\psi(0)=0$, that is,

$\displaystyle\sum_{q=0}^{n+1}\binom{n+1}{q}\sum_{k=0}^{q}\binom{q}{k}(-1)^{k}\partial_{x}^{n+1+2q+2k}\psi(0)=0.$

Thus,

$\displaystyle\sum_{q=0}^{n}\binom{n+1}{q}\sum_{k=0}^{q}\binom{q}{k}(-1)^{k}\partial_{x}^{n+1+2q+2k}\psi(0)+\binom{n+1}{n+1}\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}\partial_{x}^{n+1+2(n+1)+2k}\psi(0)=0.$

Note that $n+1+2q+2k\leq 5n+1$, for $0\leq k\leq q\leq n$. So, from induction
hypothesis it follows that $\partial_{x}^{n+1+2q+2k}\psi(0)=0$ and the last
equality becomes

$\displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^{k}\partial_{x}^{3(n+1)+2k}\psi(0)=0,$

or, equivalently,

$\displaystyle\sum_{k=0}^{n}\binom{n+1}{k}(-1)^{k}\partial_{x}^{3(n+1)+2k}\psi(0)+\binom{n+1}{n+1}(-1)^{n+1}\partial_{x}^{5(n+1)}\psi(0)=0.$

But, for $0\leq k\leq n$ we have, $3(n+1)+2k\leq 5n+3$, and the induction
hypothesis gives us $\partial_{x}^{3(n+1)+2k}\psi(0)=0$ and therefore, from
the last equality we concludes $\partial_{x}^{5(n+1)}\psi(0)=0$. In a similar
way,

$\displaystyle\partial_{x}^{r}P^{n+1}\psi(0)=0\implies\partial_{x}^{5(n+1)+r}\psi(0)=0,\
\ \ r=1,2,3,4,$

concluding the proof of (3.2), and we conclude that
$\partial_{x}^{j}\psi(0)=0$, for all $j\geq 0.$ Since $\psi$ is analytic in
$[-1,0]$, it follows that $\psi\equiv 0$. ∎

###  3.2. Reachable states

We are now in a position to prove the second main result of the article.

###### Proof of Theorem 1.2.

Let $R>2R_{0}$ and $u_{1}\in\mathcal{R}_{R}$. Later on, it will be shown that
$u_{1}$ can be written in the form

(3.4) $\displaystyle u_{1}(x)=\sum_{i\geq 0}c_{i}f_{i}(x)+\sum_{i\geq
0}b_{i}g_{i}(x),\ \forall x\in[-1,0].$

Assume for a moment that (3.4) holds with a convergence in
$W^{n,\infty}(-1,0)$ for all $n\geq 0$. Then, using (2.36) and (2.44) we
obtain

$\displaystyle P^{n}u_{1}(x)=(-1)^{n}\sum_{i\geq
n}c_{i}f_{i-n}(x)+(-1)^{n}\sum_{i\geq n}b_{i}g_{i-n}(x)$

and

$\displaystyle\partial_{x}^{j}P^{n}u_{1}(x)=(-1)^{n}\sum_{i\geq
n}c_{i}\partial_{x}^{j}f_{i-n}(x)+(-1)^{n}\sum_{i\geq
n}b_{i}\partial_{x}^{j}g_{i-n}(x)\ \ \forall j\geq 0.$

From (2.10)-(2.19) it follows that

$\displaystyle\partial_{x}^{3}P^{n}u_{1}(0)=(-1)^{n}c_{n}\partial_{x}^{3}f_{0}(0)+(-1)^{n}\sum_{i>n}c_{i}\partial_{x}^{3}f_{i-n}(0)+(-1)^{n}\sum_{i\geq
n}b_{i}\partial_{x}^{3}g_{i-n}(0)=(-1)^{n}c_{n}$

and

$\displaystyle\partial_{x}^{4}P^{n}u_{1}(0)=(-1)^{n}\sum_{i\geq
n}c_{i}\partial_{x}^{4}f_{i-n}(0)+(-1)^{n}b_{n}\partial_{x}^{4}g_{0}(0)+(-1)^{n}\sum_{i>n}b_{i}\partial_{x}^{4}g_{i-n}(0)=(-1)^{n}b_{n}.$

This leads us to define

(3.5) $\displaystyle c_{n}=(-1)^{n}\partial_{x}^{3}P^{n}u_{1}(0)\text{ \ \ and
\ \ }b_{n}=(-1)^{n}\partial_{x}^{4}P^{n}u_{1}(0),\ \forall\ n\geq 0.$

Claim. There exist $r\in(R_{0},R)$ and a constant $K=K(r)>0$ such that

$\displaystyle\left|\partial_{x}^{n}u_{1}(x)\right|\leq K\frac{n!}{r^{n}},\
\forall\ x\in[-1,0].$

Indeed, since $R>2R_{0}$ we can write $R=2R_{0}+\alpha$ with $\alpha>0$. Then
$R>R_{0}+R_{0}+\frac{\alpha}{2}$, taking $r=R_{0}+\frac{\alpha}{2}$ we have
$r\in(R_{0},R)$ and $R_{0}+r<R$. Consequently
$\overline{D(w,r)}\subset\overline{D(0,R_{0}+r)}\subset D(0,R),$ for all
$w\in\overline{D(0,R_{0})}$. Define $K:=\max\left\\{|z(w)|;\
w\in\overline{D(0,R_{0}+r)}\right\\},$ where $z\in H(D(0,R))$ is such that
$z|_{[-1,0]}=u_{1}$. So, under the hypothesis of $u_{1}\in\mathcal{R}$,
follows that

$\displaystyle\left|\partial_{x}^{n}u_{1}(x)\right|\leq K\frac{n!}{r^{n}},\
\forall\ x\in[-1,0],$

as desired, showing the claim.

Using Lemma 2.6, with $p=\infty$, the claim and (2.6) we get that

$\displaystyle|c_{n}|\leq\sum_{i=0}^{n}\|P^{i}\partial_{x}^{3}u_{1}\|_{\infty}$
$\displaystyle\leq\frac{3^{n+1}}{2}\|\partial_{x}^{3}u_{1}\|_{5n,\infty}\leq\frac{3^{n+1}}{2r^{3}}K\cdot
2^{5n}\cdot 2^{3}\cdot(5n)!\cdot
3!\cdot\frac{\frac{1}{r}}{\frac{1}{r}-1}\cdot\frac{1}{r^{5n}}.$

and analogously

$\displaystyle|b_{n}|\leq\frac{3^{n+1}}{2r^{4}}K\cdot 2^{5n}\cdot
2^{4}\cdot(5n)!\cdot
4!\cdot\frac{\frac{1}{r}}{\frac{1}{r}-1}\cdot\frac{1}{r^{5n}}.$

Therefore

$\displaystyle|c_{n}|,|b_{n}|$ $\displaystyle\leq K^{\prime}\left(\frac{3\cdot
2^{5}}{r^{5}}\right)^{n}(5n)!,\ \forall\ n\geq 0,$

for some positive constant $K^{\prime}>0$.

Define $H=\frac{3\cdot 2^{5}}{r^{5}}$ and observe that $r>R_{0}$ implies that
$He^{e^{-1}}<\frac{1}{2}$. Choose
$\tilde{H}\in\left(He^{e^{-1}},\frac{1}{2}\right)$. Then, from Proposition
3.2, there exist functions $\tilde{f},\tilde{g}\in C^{\infty}(\mathbb{R})$
such that

$\displaystyle\tilde{f}^{(n)}(0)=c_{n},\ \ \
\tilde{g}^{(n)}(0)=b_{n},\quad\text{and}\quad\left|\tilde{f}^{(n)}(t)\right|,\left|\tilde{g}^{(n)}(t)\right|\leq
K^{\prime}\tilde{H}^{n}(5n)!\ \ \forall t\in\mathbb{R},$

for every $n\geq 0$. Define $f(t)=\tilde{f}(t-T)$ and $g(t)=\tilde{g}(t-T)$.
Then $f,g\in C^{\infty}(\mathbb{R})$ with

$f^{(n)}(t)=\tilde{f}^{(n)}(t-T)\text{ \ \ and \ \
}g^{(n)}(t)=\tilde{g}^{(n)}(t-T)\ \ \forall n\geq 0,\ \forall t\in\mathbb{R}.$

Moreover, $f^{(n)}(T)=c_{n}$, $g^{(n)}(T)=b_{n}$, and

$\left|f^{(n)}(t)\right|,\left|g^{(n)}(t)\right|\leq
K^{\prime}\tilde{H}^{n}(5n)!\ \ \ \forall n\geq 0,\ \forall t\in\mathbb{R}.$

From the last inequality, we have

(3.6) $\displaystyle\left|f^{(n)}(t)\right|,\left|g^{(n)}(t)\right|\leq
K^{\prime}\frac{(5n)!}{R_{3}^{5n}}\ \ \forall n\geq 0,\ \forall
t\in\mathbb{R},$

where $R_{3}=\tilde{H}^{-\frac{1}{5}}>2^{\frac{1}{5}}$. Note that (3.6)
implies that $f,g\in G^{5}([0,T])$. Indeed, from Stirling’s formula we have

$\displaystyle(5n)!\sim 5^{5n+\frac{1}{2}}\left(\sqrt{2\pi
n}\right)^{-4}n!^{5}=\frac{5^{\frac{1}{2}}(2\pi)^{-2}}{n^{2}}5^{5n}n!^{5}.$

Hence, for some positive constant $K^{\prime\prime}>0$ we have

(3.7) $\displaystyle\left|f^{(n)}(t)\right|,\left|g^{(n)}(t)\right|\leq
K^{\prime}\frac{(5n)!}{R_{3}^{5n}}\leq
K^{\prime\prime}\frac{5^{5n}}{R_{3}^{5n}}n!^{5}=K^{\prime\prime}\frac{n!^{5}}{\left(5^{-5}R_{3}^{5}\right)^{n}}.$

Pick any $\tau\in(0,T)$ and let

$\displaystyle\beta(t)=1-\phi_{2}\left(\frac{t-\tau}{T-\tau}\right),\ \ \
t\in[0,T].$

Observe that

$\displaystyle\beta^{(i)}(t)=\left(-\frac{1}{T-\tau}\right)^{i}\phi_{2}^{(i)}\left(\frac{t-\tau}{T-\tau}\right),\
i\geq 1.$

By definition we have $\operatorname{{supp}}\phi_{2}\subset\subset(-\infty,1)$
so $\beta(T)=1$ and $\beta^{(i)}(T)=0,$ for all $i\geq 1.$ Define

(3.8) $\displaystyle y(t)=f(t)\beta(t)\text{ \ \ and \ \ }z(t)=g(t)\beta(t),\
\ t\in[0,T].$

From the Leibniz rule, we have

$\displaystyle
y^{(n)}(t)=\sum_{j=0}^{n}\binom{n}{j}f^{(j)}(t)\beta^{(n-j)}(t)\text{ \ \ \
and \ \ \ }z^{(n)}(t)=\sum_{j=0}^{n}\binom{n}{j}g^{(j)}(t)\beta^{(n-j)}(t).$

Then

(3.9) $\displaystyle
y^{(n)}(T)=\sum_{j=0}^{n}\binom{n}{j}f^{(j)}(T)\beta^{(n-j)}(T)=f^{(n)}(T)\beta(T)=c_{n}$

and

(3.10) $\displaystyle
z^{(n)}(T)=\sum_{j=0}^{n}\binom{n}{j}g^{(j)}(T)\beta^{(n-j)}(T)=g^{(n)}(T)\beta(T)=b_{n}.$

On the other hand, by definition $\phi_{2}\equiv 1$ in $(-\infty,0]$ so that
$\beta\equiv 0$ in $(-\infty,\tau)$. Consequently $y^{(n)}\equiv z^{(n)}\equiv
0$ in $(-\infty,\tau)$ for any $n\geq 0$. In particular

(3.11) $\displaystyle y^{(n)}(0)=z^{(n)}(0)=0,\ \ \forall n\geq 0.$

Since $\beta\in G^{2}([0,T])$ and $f,g\in G^{5}([0,T])$ we have $y,z\in
G^{5}([0,T])$. Moreover, from [32, Lemma 3.7], the same constant “$R$ ” of $f$
and $g$ in the definition of $G^{5}([0,T])$ works for $y$ and $z$. Hence, from
(3.7), there exists $K^{\prime\prime\prime}>0$ such that

$\displaystyle\left|y^{(n)}(t)\right|,\left|z^{(n)}(t)\right|\leq
K^{\prime\prime\prime}\frac{n!^{5}}{\left(5^{-5}R_{3}^{5}\right)^{n}},\ \ \
\forall n\geq 0,\ \forall t\in[0,T].$

Using Stirling’s formula we obtain $(5n)!\sim
5^{5n+\frac{1}{2}}\left(\sqrt{2\pi n}\right)^{-4}n!^{5}$, and so

$\displaystyle n!^{5}\sim 5^{-\left(5n+\frac{1}{2}\right)}\left(\sqrt{2\pi
n}\right)^{4}(5n)!=\frac{(2\pi)^{2}}{5^{\frac{1}{2}}}n^{2}\frac{(5n)!}{5^{5n}}.$

Then, there exists a constant $\overline{K}>0$ such that

$\displaystyle\left|y^{(n)}(t)\right|,\left|z^{(n)}(t)\right|\leq
K^{\prime\prime\prime}\frac{1}{\left(5^{-5}R_{3}^{5}\right)^{n}}\overline{K}n^{2}\frac{(5n)!}{5^{5n}}=K^{\prime\prime\prime}\overline{K}\cdot\frac{n^{2}(5n)!}{R_{3}^{5n}}.$

Note that, since $R_{3}>2^{\frac{1}{5}}$ we can pick
$\rho\in\left(1,R_{3}2^{-\frac{1}{5}}\right)$, so that the sequence
$\left(\frac{n^{2}}{\rho^{5n}}\right)_{n\in\mathbb{N}}$ is bounded. Indeed,
using the L’Hospital rule we see that

$\displaystyle\lim_{x\rightarrow\infty}\frac{x^{2}}{\rho^{5x}}=\lim_{x\rightarrow\infty}\frac{2x}{(5\ln\rho)\rho^{5x}}=\lim_{x\rightarrow\infty}\frac{2}{(5\ln\rho)^{2}\rho^{5x}}=0.$

Hence, there exists $\overline{K}^{\prime}>0$ such that
$n^{2}\leq\overline{K}^{\prime}\rho^{5n}$, and therefore

$\displaystyle\left|y^{(n)}(t)\right|,\left|z^{(n)}(t)\right|\leq
K^{\prime\prime\prime}\overline{K}\cdot\overline{K}^{\prime}\cdot\frac{\rho^{5n}(5n)!}{R_{3}^{5n}}=K^{\prime\prime\prime}\overline{K}\cdot\overline{K}^{\prime}\cdot\frac{(5n)!}{\left(R_{3}\rho^{-1}\right)^{5n}}.$

Defining
$K^{\prime\prime\prime\prime}=K^{\prime\prime\prime}\overline{K}\cdot\overline{K}^{\prime}$
and $R_{3}^{\prime}=\frac{R_{3}}{\rho}$ we have
$R_{3}^{\prime}>2^{\frac{1}{5}}$ and

(3.12) $\displaystyle\left|y^{(n)}(t)\right|,\left|z^{(n)}(t)\right|\leq
K^{\prime\prime\prime\prime}\frac{(5n)!}{\left(R_{3}^{\prime}\right)^{5n}},\ \
\forall n\geq 0,\forall t\in[0,T].$

Let $u$ be as in (2.7) corresponding to $y$ and $z$ given in (3.8). From
(3.12) and by Proposition 3.1 we have that $y\in G^{1,5}([-1,0]\times[0,T])$
and it solves (2.25). Furthermore, (3.11) gives us

$\displaystyle u(x,0)=\sum_{j\geq 0}f_{j}(x)y^{(j)}(0)+\sum_{j\geq
0}g_{j}(x)z^{(j)}(0)=0.$

Setting $h_{1}=u(-1,t)$ and $h_{2}=u_{x}(-1,t)$, we get $h_{1},h_{2}\in
G^{5}([0,T])$ and therefore $u$ solves (1.5) with $h_{1}$ and $h_{2}$ as
control inputs and $y_{0}=0$ as initial data. From the proof of Proposition
3.1 we know that for all $n,m\in\mathbb{N}$ the sequence of the series

$\sum_{j\geq
0}\partial_{t}^{m}\partial_{x}^{n}\left(f_{j}(x)y^{(j)}(t)+g_{j}(x)z^{(j)}(t)\right)$

converges uniformly on $[-1,0]\times[0,T]$ to
$\partial_{t}^{m}\partial_{x}^{n}u$ and consequently, for all $n,i\geq 0$,

$\displaystyle P^{n}u(x,t)$ $\displaystyle=\sum_{j\geq
0}P^{n}f_{j}(x)y^{(j)}(t)+\sum_{j\geq 0}P^{n}g_{j}(x)z^{(j)}(t)$
$\displaystyle=(-1)^{n}\sum_{j\geq n}f_{j-n}(x)y^{(j)}(t)+(-1)^{n}\sum_{j\geq
n}g_{j-n}(x)z^{(j)}(t)$

and

$\displaystyle\partial_{x}^{i}P^{n}u(x,t)$ $\displaystyle=(-1)^{n}\sum_{j\geq
n}\partial_{x}^{i}f_{j-n}(x)y^{(j)}(t)+(-1)^{n}\sum_{j\geq
n}\partial_{x}^{i}g_{j-n}(x)z^{(j)}(t),$

for every $(x,t)\in[-1,0]\times[0,T]$. In particular, (3.9) and (3.10) gives
us that

$\displaystyle\partial_{x}^{i}P^{n}u(x,T)=(-1)^{n}\sum_{j\geq
n}c_{j}\partial_{x}^{i}f_{j-n}(x)+(-1)^{n}\sum_{j\geq
n}b_{j}\partial_{x}^{i}g_{j-n}(x)\ \ \forall i,n\geq 0,\ \forall x\in[-1,0].$

Thus, (2.10)-(2.19) provide us

$\begin{cases}\begin{aligned}
P^{n}u(0,T)&=\partial_{x}P^{n}u(0,T)=\partial_{x}^{2}P^{n}u(0,T)=0,\\\
\partial_{x}^{3}P^{n}u(0,T)&=(-1)^{n}\sum_{j\geq
n}c_{j}\partial_{x}^{3}f_{j-n}(0)+(-1)^{n}\sum_{j\geq
n}b_{j}\partial_{x}^{3}g_{j-n}(0)=(-1)^{n}c_{n},\\\
\partial_{x}^{4}P^{n}u(0,T)&=(-1)^{n}\sum_{j\geq
n}c_{j}\partial_{x}^{4}f_{j-n}(0)+(-1)^{n}\sum_{j\geq
n}b_{j}\partial_{x}^{4}g_{j-n}(0)=(-1)^{n}b_{n},\end{aligned}\end{cases}$

and therefore, using (3.5),

(3.13) $\begin{cases}\begin{aligned} &\partial_{x}^{j}P^{n}u(0,T)=0,\
j=0,1,2,\\\
&\partial_{x}^{3}P^{n}u(0,T)=\partial_{x}^{3}P^{n}u_{1}(0),\quad\partial_{x}^{4}P^{n}u(0,T)=\partial_{x}^{4}P^{n}u_{1}(0).\end{aligned}\end{cases}$

Define $\psi\in G^{1}([-1,0])$ by $\psi(x)=u(x,T)-u_{1}(x),$ for all
$x\in[-1,0]$, and using the fact that $u_{1}\in\mathcal{R}_{R}$ together with
(3.13), holds that $\partial_{x}^{j}P^{n}\psi(0)=0,$ for $j=0,1,2,3,4.$ From
Lemma 3.3 it follows that $\psi\equiv 0$, that is, $u(x,T)=u_{1}(x)$, which
concludes the proof. ∎

### Acknowledgments.

This work is part of the Ph.D. thesis of da Silva at the Department of
Mathematics of the Universidade Federal de Pernambuco.

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