# Boundary control of generalized Korteweg-de Vries-Burgers-Huxley equation:
Well-Posedness, Stabilization and Numerical Studies111 1,2Department of
Mathematics, Indian Institute of Technology Roorkee-IIT Roorkee, Haridwar
Highway, Roorkee, Uttarakhand 247667, INDIA.e-mail: Manil T. Mohan:
<EMAIL_ADDRESS><EMAIL_ADDRESS>e-mail: Shri Lal Raghudev
Ram Singh<EMAIL_ADDRESS>*Corresponding author. Key words:
generalized Korteweg-de Vries-Burgers-Huxley equation; monotonicity; Minty-
Browder theorem; stabilization; boundary control. Mathematics Subject
Classification (2020): 93D15, 93D23, 34H05, 35K51.

Manil T. Mohan1* and Shri Lal Raghudev Ram Singh2

###### Abstract.

A boundary control problem for the following generalized Korteweg-de Vries-
Burgers-Huxley equation:

$u_{t}=\nu u_{xx}-\mu u_{xxx}-\alpha u^{\delta}u_{x}+\beta
u(1-u^{\delta})(u^{\delta}-\gamma),\ x\in[0,1],\ t>0,$

where $\nu,\mu,\alpha,\beta>0,$ $\delta\in[1,\infty)$, $\gamma\in(0,1)$
subject to Neumann boundary conditions is considered in this work. We first
establish the well-posedness of the Neumann boundary value problem by an
application of monotonicity arguments, the Hartman-Stampacchia theorem, the
Minty-Browder theorem, and the Crandall-Liggett theorem. The additional
difficulties caused by the third order linear term is successfully handled by
proving a proper version of the Minty-Browder theorem. By using suitable
feedback boundary controls, we demonstrate $\mathrm{L}^{2}$\- and
$\mathrm{H}^{1}$-stability properties of the closed-loop system for
sufficiently large $\nu>0$. The analytical conclusions from this work are
supported and validated by numerical investigations.

##  1\. Introduction

One of the most basic nonlinear mathematical models that illustrates the
characteristics of both dispersion and dissipation is the Korteweg-de Vries-
Burgers (KdVB) equation ([5, 6, 11, 30]). In this work, we consider a
generalized Korteweg-de Vries-Burgers-Huxley equation which has the
characteristics of _dispersion, dissipation, convection and reaction effects_.

###  1.1. The model

One type of partial differential equation (PDE) for a shallow water wave with
unidirectional motion is the _Korteweg-de Vries (KdV) equation_ ([7]),

$u_{t}=-\mu u_{xxx}-\alpha uu_{x},\ \mu,\alpha>0,$

which is also a common representative of _nonlinear dispersive equations_. The
_viscous Burgers equation_ ([27]),

$u_{t}=\nu u_{xx}-\alpha uu_{x},\ \nu,\alpha>0,$

a common representative of _nonlinear dissipative equations_ or _convection-
diffusion equation_ , is used in a number of applied mathematics fields,
including traffic flow, fluid mechanics, nonlinear acoustics, and gas
dynamics. The case of pure dispersion and pure dissipation is quite uncommon
in many real-world physical problems. The _Korteweg-de Vries-Burgers (KdVB)_
equation is formed when a diffusion term is added to the KdV equation. The
KdVB equations are given by ([5])

$u_{t}=\nu u_{xx}-\mu u_{xxx}-\alpha uu_{x},\ \nu,\mu,\alpha>0.$

It is discovered that this equation also represents a wide range of other
physical phenomena, including lattice waves, the propagation of ion-acoustic
waves in cold plasma, and pressure waves in liquid-gas bubble mixtures, etc.
Actually, in addition to dispersion, several of these phenomena also show
dissipative effects, which makes the so-called KdVB equation a better model
([13]).

A nonlinear PDE that explains the interplay between reaction mechanisms,
convection effects, and diffusion transports is the _Burgers-Huxley equation_
([44]):

$u_{t}=\nu u_{xx}-\alpha uu_{x}+\beta u(1-u)(u-\gamma),\
\nu,\mu,\alpha,\beta>0,\ \gamma\in(0,1).$

The _Korteweg-de Vries-Burgers-Huxley (KdVBH)_ equation is obtained by adding
a reaction term similar to the one found in the Burgers-Huxley equation. We
examine the following generalized version of KdVBH equation in this article:

$\displaystyle u_{t}=\nu u_{xx}-\mu u_{xxx}-\alpha u^{\delta}u_{x}+\beta
u(1-u^{\delta})(u^{\delta}-\gamma),\ x\in(0,1),\ t>0,$ (1.1)

where $\nu,\mu,\alpha,\beta>0,\ \gamma\in(0,1),\delta\in[1,\infty),$ with the
initial condition

$\displaystyle u(x,0)=u_{0}(x),\ x\in[0,1].$ (1.2)

We call $\nu>0$ as the _dissipation coefficient_ , $\alpha>0$ as the
_convection coefficient_ and $\mu>0$ as the _dispersion coefficient._ The
equation (1.1) can be considered as an example of a _convection-diffusion-
reaction equation of dispersive type._

Many authors have actively worked on the KdV and KdVB equations from a variety
of angles ([8, 10, 28, 37], etc., and references therein). The aim of this
work is to analyze a stabilization problem by using a boundary feedback
control for the system (1.1)-(1.2). Similar problems for KdVB equation have
been considered in the works [5, 29].

###  1.2. The boundary control

Control problems for partial differential equations have been a focus of
intense research for the past several decades. Generally speaking, there are
two types of control procedures for partial differential equations: boundary
control and distributed control. It is difficult for engineers to implement
distributed control, which calls for actuators to be positioned at every point
in the spatial domain ([23]). Boundary control is simple and inexpensive to
implement; it only requires applying actuators on the edge of the spatial
region.

Let us now discuss some of the boundary control problems discussed in the
literature for KdVB equation. For the forced KdVB equation, the author in [45]
demonstrated that if the external forcing is time-periodic with small
amplitude, then the problem also admits a unique time-periodic solution with
the same period, and the solution is stable. The boundary stabilization
problem of KdVB equation is studied in [5, 29]. The authors of [5, 29]
demonstrated well-posedness, global exponential stability in $\mathrm{L}^{2}$,
global asymptotic stability in $\mathrm{H}^{1}$, and semi-global exponential
stability in $\mathrm{L}^{2}$ of KdVB equation. Nonlinear boundary
stabilization for a generalized KdVB equation is examined in [38, 39, 40],
etc. By proposing an another class of adaptive controls, the authors in [15]
established well-posedness as well as the $\mathrm{L}^{2}$-global exponential
stability of the solutions of KdV and KdVB equations. By using suitable
boundary controls, the authors in [21, 22] proved that the solutions of the
KdVB equation globally exist and globally exponentially tend to zero as
$t\to\infty$ in a subspace of $\mathrm{H}^{s}$ for $s\in[0,3]$. The authors in
[6] developed two approaches for the stabilization of nonlinear KdV equation
with boundary time-delay feedback. For more interesting problems on the
control and stabilization of KdV and KdVB equations, we refer the interested
readers to [1, 11, 10, 12, 13, 17, 18, 23, 31, 34, 35, 36, 42], etc., and
references therein.

The main aim of this article is to study the well-posedess of a boundary
control problem for (1.1)-(1.2) and establish a feedback stabilization result.
We use two different controls for $\delta=1$ and $\delta=2$. Motivated from
[5], keeping feedback stabilization in mind, the equation (1.1)-(1.2) is
associated with the following boundary conditions for $t\geq 0$:

For $\delta=1$, we consider the control

$\mathopen{}\mathclose{{}\left\\{\begin{aligned} u(0,t)&=0,\\\
u_{x}(1,t)&=-\frac{1}{\nu}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u^{2\delta}(1,t)}\right)u(1,t)=-g_{1}(u(1,t)),\\\
u_{xx}(1,t)&=\frac{\delta}{\nu^{2}}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u^{2\delta}(1,t)}\right)^{2}u(1,t)=g_{2}(u(1,t)),\end{aligned}}\right.$
(1.3)

where $\eta>0$, $g_{1},g_{2}\in\mathrm{C}(\mathbb{R})$ are nondecreasing
functions with $g_{1}(0)=g_{2}(0)=0,$ and $g_{2}(k)k=\delta(g_{1}(k))^{2}$
representing the nonlinear flux feedback controls.

For $\delta=2$, we consider the control

$\mathopen{}\mathclose{{}\left\\{\begin{aligned} u(0,t)&=0,\\\
u_{x}(1,t)&=-\frac{1}{\nu}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha}{(\delta+2)}u^{\delta}(1,t)}\right)u(1,t)=-g_{1}(u(1,t)),\\\
u_{xx}(1,t)&=\frac{\delta}{\nu^{2}}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha}{(\delta+2)}u^{\delta}(1,t)}\right)^{2}u(1,t)=g_{2}(u(1,t)),\end{aligned}}\right.$
(1.4)

where $\eta>0$.

For all $u\in\mathrm{H}^{3}(0,1)$ satisfying (1.3), an integration by parts
yields

$\displaystyle(u_{xxx},u)$
$\displaystyle=-(u_{xx},u_{x})+u_{xx}(1)u(1)-u_{xx}(0)u(0)$
$\displaystyle=-\frac{1}{2}u_{x}^{2}(1)+u_{xx}(1)u(1)=-\frac{1}{2}(g_{1}(u(1)))^{2}+g_{2}(u(1))u(1)=\mathopen{}\mathclose{{}\left(\delta-\frac{1}{2}}\right)(g_{1}(u(1)))^{2}.$
(1.5)

Similarly, for all $u,v\in\mathrm{H}^{3}(0,1)$ satisfying (1.3) and for some
$0<\theta<1$, we obtain

$\displaystyle((u-v)_{xxx},u-v)$
$\displaystyle=-\frac{1}{2}(u_{x}(1)-v_{x}(1))^{2}+(u_{xx}(1)-v_{xx}(1))(u(1)-v(1))$
$\displaystyle=-\frac{1}{2}\mathopen{}\mathclose{{}\left(g_{1}(u(1))-g_{1}(v(1))}\right)^{2}+\mathopen{}\mathclose{{}\left(g_{2}(u(1))-g_{2}(v(1))}\right)(u(1)-v(1))$
$\displaystyle=-\frac{1}{2\nu^{2}}(u(1)-v(1))^{2}\bigg{\\{}\eta^{2}+\frac{2\eta\alpha(\delta+1)}{(\delta+2)}(\theta
u(1)+(1-\theta)v(1))^{\delta}$
$\displaystyle\quad+\frac{\alpha^{2}(\delta+1)^{2}}{(\delta+2)^{2}}(\theta
u(1)+(1-\theta)v(1))^{2\delta}\bigg{\\}}+\frac{\delta}{\nu^{2}}(u(1)-v(1))^{2}\bigg{\\{}\eta^{2}$
$\displaystyle\quad+\frac{2\eta\alpha(\delta+1)}{(\delta+2)}(\theta
u(1)+(1-\theta)v(1))^{\delta}+\frac{\alpha^{2}(2\delta+1)}{(\delta+2)^{2}}(\theta
u(1)+(1-\theta)v(1))^{2\delta}\bigg{\\}}$
$\displaystyle=\frac{1}{\nu^{2}}(u(1)-v(1))^{2}\bigg{\\{}\mathopen{}\mathclose{{}\left(\delta-\frac{1}{2}}\right)\mathopen{}\mathclose{{}\left[\eta^{2}+\frac{2\eta\alpha(\delta+1)}{(\delta+2)}(\theta
u(1)+(1-\theta)v(1))^{\delta}}\right]$
$\displaystyle\quad+\frac{\alpha^{2}(3\delta^{2}-1)}{2(\delta+2)^{2}}(\theta
u(1)+(1-\theta)v(1))^{2\delta}\bigg{\\}}.$ (1.6)

It can be seen that $((u-v)_{xxx},u-v)\geq 0$. A similar calculation can be
performed for $\delta=1$ also.

######  Remark 1.1.

For the case of any $\nu>0$, we are providing the controls for $\delta=1$ and
$\delta=2$ in (1.3) and (1.4), respectively, keeping in mind that one may
extend to other values of $\delta$ also in the future. For
$\delta\in[1,\infty)$ and large values of $\nu$, one can use the control given
in (3.4) below. For small values of $\nu>0,$ we restrict ourselves to
$\delta=1,2$ due to the unavailability of a well-posedness result for other
values of $\delta$ (see Theorem 2.6 below).

Moreover, a control like the following also stabilizes the system (1.1)-(1.2)
([29]):

$\displaystyle u(0,t)=0,\ u_{x}(1,t)=0,\ u_{xx}(1,t)=g(u(1,t)).$ (1.7)

In this case, the condition (1.2) reduces to
$\mathopen{}\mathclose{{}\left(g(u(1))-g(v(1))}\right)(u(1)-v(1))\geq 0$, that
is, $g(\cdot)$ needs to be a monotonicially increasing function. But the rate
of convergence is slower than the controls considered in (1.4) and (1.3).
Examples of such $g(\cdot)$ are

$\displaystyle\frac{1}{\mu}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u^{2\delta}(1,t)}\right)u(1,t)\
(\text{for }\ \delta=1)\ \text{ and }\ \frac{\eta}{\mu}u(1,t)\ (\text{for }\
\delta=2),$ (1.8)

for some $\eta>0$.

Three different adaptive control laws are designed in [41] to show the
$\mathrm{L}^{2}$-global exponential stability for the adaptive control problem
of a forced generalized KdVB equation when either the kinematic viscosity
and/or the dynamic viscosity are unknown. But the controls used in [41] may
not be useful for our case (even for non-adaptive control problem) as the
condition (1.2) may not be satisfied by such controls (see Remark 1.1). The
monotonicity condition (1.2) is crucial in establishing the well-posedness of
the boundary control problem for the generalized KdVBH equation (1.1)-(1.2).

###  1.3. Difficulties, approaches and novelties

The major difficulty of the work lies in establishing the well-posedness of
the generalized KdV-Burgers-Huxley equation (1.1)-(1.2) with the controls
(1.4) and (1.3) for $\delta=1$ and $\delta=2$, respectively. As the problem
(1.1) is of the third order dispersive type, there are limitations in using
the standard Minty-Browder result. The classical result on Minty-Browder
Theorem states that any monotone, hemicontinuous and strongly coercive
operator $T:\mathrm{X}\to\mathrm{X}^{*}$, where $\mathrm{X}$ is a reflexive
Banach space, is onto, that is, $\mathrm{Range}(T)=\mathrm{X}^{*}$ ([16,
Theorem 3.3.1]). By properly defining hemicontinuity and coercivity in our
context (see (2.3) and (2.7) below) and applying the infinite-dimensional
version of the Hartman-Stampacchia theorem ([25, Theorem 1.4, Chapter III]),
we show the same result for $T:\mathrm{X}\to\mathrm{Y}^{*},$ that is,
$\mathrm{Range}(T)=\mathrm{Y}^{*}$ (Theorem 2.3), where $\mathrm{Y}$ is also a
reflexive Banach space such that the embedding $\mathrm{X}\subset\mathrm{Y}$
is dense (Lemma 2.1). We hope that the abstract result obtained in Theorem 2.3
can be used to prove the well-posedness of various systems of the type (1.1).

Another major difficulty is the restriction of $\delta\in\\{1,2\\}$ for any
$\nu>0$. This restriction is due to the choice of our controls given in (1.3)
and (1.4) and the lack of existence and uniqueness results for the other
values of $\delta$ (see (2.2) in the proof of Theorem 2.6). For
$\delta\in\\{1,2\\}$, we use monotonicity arguments (Lemma 2.5), the Minty-
Browder theorem (Theorem 2.3), and the Crandall-Liggett theorem to prove the
existence and uniqueness of strong solutions for a cutoff problem (see (2.54)
below). Then, by using uniform energy estimates, we show the global
solvability of the problem (1.1)-(1.2) with the controls (1.4) and (1.3) for
$\delta=1$ and $\delta=2$, respectively (Theorem 2.6). The coercivity results
(see (2.4) and (2.4) below) immediately help us to obtain the
$\mathrm{L}^{2}$-exponential stabilization of the problem (1.1)-(1.2) under
the assumption $\nu>\frac{\beta}{4}(1-\gamma)^{2}$ (Theorem 3.1). But under a
restrictive assumption on $\nu$, that is,
$\nu>\frac{\alpha^{2}}{2\beta}(\delta+2)^{-\frac{2(\delta+1)}{\delta+2}}$, we
observe that the convective term can be handled by using the diffusion as well
as reaction terms for any $\delta\in[1,\infty)$ and the well-posedness can be
established. For sufficiently large $\nu$ (see (3.3) below), we are also able
to obtain $\mathrm{L}^{2}$, $\mathrm{H}^{1}$ and pointwise exponential
stabilization results (Theorems 3.2, 3.3 and 3.4).

###  1.4. Organization of the paper

The rest of the paper is organized as follows: The next section deals with the
well-posedness results for the problem (1.1)-(1.2) with the controls (1.4) and
(1.3) for $\delta=1$ and $\delta=2$, respectively. We show the well-posedness
of the Neumann boundary value problem (Theorem 2.6) by an application of
monotonicity arguments, the Minty-Browder theorem (Theorem 2.3), the Hartman-
Stampacchia theorem [25, Theorem 1.4, Chapter III] and the Crandall-Liggett
theorem ([43, Theorems 5.1, 5.2]). In section 3, the
$\mathrm{L}^{2}$-exponential stabilization result for the above problem with
$\nu>\frac{\beta}{4}(1-\gamma)^{2}$ is established (Theorem 3.1). Under
further assumption on $\nu$ (see (3.3) below), $\mathrm{L}^{2}$,
$\mathrm{H}^{1}$ and pointwise exponential stabilization results for the
problem (1.1)-(1.2) with the control (3.4) is obtained in Theorems 3.2, 3.3
and 3.4, respectively. Numerical investigations in section 4 validate and
corroborate the analytical conclusions drawn in section 2. An example is also
provided to show that that controls given in (1.3) and (1.4) converge much
faster than the one given in (1.8).

##  2\. Well-posedness and Stabilization

In order to formulate the problem (1.1)-(1.3) as an abstract initial value
problem, we consider the Hilbert space $\mathrm{H}=\mathrm{L}^{2}(0,1)$ and
the operator
$\mathscr{A}:\mathrm{D}(\mathscr{A})\subset\mathrm{H}\to\mathrm{H}$ as

$\displaystyle\mathscr{A}(v):=-\nu v_{xx}+\mu
v_{xxx}+\frac{\alpha}{\delta+1}(v^{\delta+1})_{x}-\beta
v(1-v^{\delta})(v^{\delta}-\gamma),$ (2.1)

and the domain

$\displaystyle\mathrm{D}(\mathscr{A}):=\mathopen{}\mathclose{{}\left\\{u\in\mathrm{H}^{3}(0,1):u(0)=0,\
u_{x}(1)=-g_{1}(u(1)),\ u_{xx}(1)=g_{2}(u(1))}\right\\}.$ (2.2)

With the above notations, the problem (1.1)-(1.3) can be reformulated as

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
\frac{du(t)}{dt}+\mathscr{A}(u)(t)&=0,\ t>0,\\\
u(0)&=u_{0}.\end{aligned}}\right.$ (2.3)

Performing an integration by parts in (2.1), for each
$v\in\mathrm{D}(\mathscr{A})$, we have

$\displaystyle(\mathscr{A}(v),w)$ $\displaystyle=\nu(v_{x},w_{x})+\nu
g_{1}(v(1))w(1)-\mu(v_{xx},w_{x})+\mu g_{2}(v(1))w(1)$
$\displaystyle\quad+\frac{\alpha}{\delta+1}\mathopen{}\mathclose{{}\left[v^{\delta+1}(1)w(1)-(v^{\delta+1},w_{x})}\right]-\beta(v(1-v^{\delta})(v^{\delta}-\gamma),w),$
(2.4)

for all

$w\in\mathrm{X}:=\mathopen{}\mathclose{{}\left\\{u\in\mathrm{H}^{2}(0,1):u(0)=0,u_{x}(1)=-g_{1}(u(1))}\right\\}.$

Let us define
$\mathrm{Y}:=\mathopen{}\mathclose{{}\left\\{u\in\mathrm{H}^{1}(0,1):u(0)=0}\right\\}$.
Clearly $\mathrm{X}\subset\mathrm{Y}$ and the embedding is continuous. Let us
first prove the following technical result:

######  Lemma 2.1.

The space $\mathrm{X}$ is reflexive and is a dense subspace of $\mathrm{Y}$.

###### Proof.

Remember that closed subspaces of reflexive Banach spaces are reflexive ([24,
Theorem 5.9]). As $\mathrm{H}^{2}(0,1)$ is a Hilbert space, in order to prove
$\mathrm{X}$ is reflexive, it is enough to show that $\mathrm{X}$ is closed.
Claim: _$\mathrm{X}$ is a closed subspace of $\mathrm{H}^{2}(0,1)$._ Let
$u_{n}\in\mathrm{X}$ bs such that $\|u_{n}-u\|_{\mathrm{H}^{2}}\to 0$ as
$n\to\infty$. Clearly $u\in\mathrm{H}^{2}(0,1)$, $u_{n}(0)=0$ and
$u_{n,x}(1)=-g_{1}(u_{n}(1))$. We need to show that $u(0)=0$ and
$u_{x}(1)=-g_{1}(u_{(}1))$. The convergence $\|u_{n}-u\|_{\mathrm{H}^{2}}\to
0$ implies $\|u_{n}-u\|_{\mathrm{H}^{1}}\to 0$, and since
$\mathrm{H}^{1}(0,1)\hookrightarrow\mathrm{C}([0,1])$, we deuce $u_{n}(x)\to
u(x)$ for all $x\in[0,1]$ and $u\in\mathrm{C}([0,1])$ with $u(0)=0$. Moreover,
it is immediate that $u_{n}(1)\to u(1)$ and since $g_{1}$ is continuous
function, we get $g_{1}(u_{n}(1))\to g_{1}(u(1))$ as $n\to\infty$. Since
$\mathrm{H}^{1}(0,1)\hookrightarrow\mathrm{C}^{1}([0,1])$, one can easily
deduce that $u_{n,x}(x)\to u_{x}(x)$ for all $x\in[0,1]$ and by using the
uniqueness of limit, we deduce $u_{x}(1)=-g_{1}(u(1))$. Therefore,
$u\in\mathrm{X}$ and hence $\mathrm{X}$ is a closed subspace of
$\mathrm{H}^{2}(0,1)$.

Claim: _$\mathrm{X}$ is a dense subspace of $\mathrm{Y}$._ Fix any
$v\in\mathrm{Y}$, that is, $v\in\mathrm{H}^{1}(0,1)$ and $v(0)=0$. As a
subspace of $\mathrm{H}^{1}(0,1)$, the space $\mathrm{Y}$ inherits the
$\mathrm{H}^{1}$-norm, but as $v(0)=0$, the $\mathrm{Y}$-norm is equivalent to
$\|v_{x}\|_{\mathrm{L}^{2}}$ for all $v\in\mathrm{Y}$. It is just a simple
application of the Poincaré inequality (see Lemma A.2 below). Since the
embedding $\mathrm{H}^{2}(0,1)\subset\mathrm{H}^{1}(0,1)$ is dense, there
exists a sequence $\\{v_{m}\\}_{m\in\mathbb{N}}\subset\mathrm{H}^{2}(0,1)$
such that $\|v_{m}-v\|_{\mathrm{H}^{1}}\to 0$ as $m\to\infty$. Since
$\mathrm{H}^{2}(0,1)\hookrightarrow\mathrm{C}^{1}([0,1])$, we choose the
sequence $\\{v_{m}\\}_{m\in\mathbb{N}}$ in such a way that $v_{m,x}(1)\to 0$
as $m\to\infty$. By defining $w_{m}(x)=v_{m}(x)-v_{m}(0)$, for all
$x\in[0,1]$, we find $w_{m}(0)=0$ and
$\|w_{m}-v\|_{\mathrm{Y}}=\|(v_{m}-v_{m}(0)-v)_{x}\|_{\mathrm{L}^{2}}=\|v_{m,x}-v_{x}\|_{\mathrm{L}^{2}}\to
0$ as $m\to\infty$. Therefore, the space
$\\{v\in\mathrm{H}^{2}(0,1):v(0)=0\\}$ is dense in $\mathrm{Y}$.

Let us now show that $\mathrm{X}$ is dense in $\mathrm{Y}$. We consider for
$m\geq 2$, $\varphi_{m}(x)=x(1-x)^{m}$. Then, it is clear that
$\varphi_{m}(0)=\varphi_{m}(1)=\varphi_{m,x}(1)=0$ and $\varphi_{m,x}(0)=1$.
Then, by using properties of the beta function, it is immediate that

$\displaystyle\|\varphi_{m}\|_{\mathrm{Y}}^{2}=\|\varphi_{m,x}\|_{\mathrm{L}^{2}}^{2}=\frac{m}{(2m+1)(2m-1)}\to
0\ \text{ as }\ m\to\infty.$

Since $m\geq 2$, it is clear that $\varphi_{m}\in\mathrm{H}^{2}(0,1)$. Let us
define

$\displaystyle
z_{m}(x):=w_{m}(x)+(w_{m,x}(1)+g_{1}(w_{m}(1)))\varphi_{m}(1-x).$

Then $z_{m}\in\mathrm{H}^{2}(0,1)$ for all $m\in\mathbb{N}$ with $z_{m}(0)=0$
and $z_{m,x}(1)=-g_{1}(w_{m}(1))=-g_{1}(z_{m}(1))$, so that
$z_{m}\in\mathrm{X}$. Since
$\mathrm{H}^{1}(0,1)\hookrightarrow\mathrm{C}([0,1])$, the convergence
$\|w_{m}-v\|_{\mathrm{Y}}\to 0$ as $m\to\infty$ implies that $w_{m}(x)\to
v(x)$ as $m\to\infty$ for all $x\in[0,1]$. Since $g_{1}(\cdot)$ is a
continuous function $g_{1}(w_{m}(1))\to g_{1}(v(1))$ as $m\to\infty$.
Therefore $|g_{1}(w_{m}(1))|\leq K_{1}$ for all $m\in\mathbb{N}$. Note that
$w_{m,x}(1)=v_{m,x}(1)\to 0$ as $m\to\infty$. Thus, it can be easily seen that

$\displaystyle\|z_{m}-v\|_{\mathrm{Y}}$
$\displaystyle=\|z_{m,x}-v_{x}\|_{\mathrm{L}^{2}}\leq\|w_{m,x}-v_{x}\|_{\mathrm{L}^{2}}+(|w_{m,x}(1)|+|g_{1}(w_{m}(1))|)\|\varphi_{m,x}\|_{\mathrm{L}^{2}}$
$\displaystyle\to 0\ \text{ as }\ m\to\infty,$

which completes the proof. ∎

######  Remark 2.2.

Since the continuous embedding $\mathrm{X}\subset\mathrm{Y}$ is dense, from
[19, Lemma 2.2.27], we infer that the embedding
$\mathrm{Y}^{*}\subset\mathrm{X}^{*}$ is continuous and the reflexivity of
$\mathrm{X}$ implies that $\mathrm{Y}^{*}$ is dense in $\mathrm{X}^{*}$.

###  2.1. Abstract result

For a mapping $T:\mathrm{X}\to\mathrm{Y}^{*}$, the following result is a
slight modification of the results presented in [3, Corollary 2.1.3.], [16,
Theorem 3.3.1], etc., where the mapping $T:\mathrm{X}\to\mathrm{X}^{*}$.
Therefore, we are providing a proof here (Minty’s theorem).

######  Theorem 2.3.

Let $\mathrm{X}$ and $\mathrm{Y}$ be reflexive Banach spaces such that the
continuous embedding $\mathrm{X}\hookrightarrow\mathrm{Y}$ is dense. Let
$T:\mathrm{X}\to\mathrm{Y}^{*}$ be monotone, that is,

$\displaystyle\langle T(u)-T(v),u-v\rangle\geq 0\ \text{ for all }\
u,v\in\mathrm{X},$ (2.5)

hemicontinuous, that is,

$\displaystyle\lim\limits_{\lambda\to 0}\langle T(u_{1}+\lambda
u_{2}),v\rangle=\langle T(u_{1}),v\rangle,\ \text{ for all }\
u_{1},u_{2}\in\mathrm{X},\ \text{ and }\ v\in\mathrm{Y},$ (2.6)

and coercive, that is,

$\displaystyle\lim\limits_{\|u\|_{\mathrm{Y}\to\infty}}\frac{\langle
T(u),u\rangle}{\|u\|_{\mathrm{Y}}}=\infty,\ \text{ for all }\ u\in\mathrm{X},$
(2.7)

operator. Then $T$ is onto, that is, $\mathrm{Range}(T)=\mathrm{Y}^{*}$.

###### Proof.

Let $f\in\mathrm{Y}^{*}$. We have for all $\mathrm{X}\ni u\neq 0$,

$\displaystyle\langle
T(u)-f,u\rangle\geq\|u\|_{\mathrm{Y}}\mathopen{}\mathclose{{}\left[\frac{\langle
T(u),u\rangle}{\|u\|_{\mathrm{Y}}}-\|f\|_{\mathrm{Y}^{*}}}\right].$ (2.8)

Since $T$ is coercive in the sense of (2.7), for any $N>0$, there exists an
$R>0$ such that $\frac{\langle T(u),u\rangle}{\|u\|_{\mathrm{Y}}}>N$ for all
$\|u\|_{\mathrm{Y}}\geq R$. Therefore, for $N>\|f\|_{\mathrm{Y}^{*}}$, there
exists an $R>0$ such that

$\displaystyle\langle T(u)-f,u\rangle>0\ \text{ for all }\
\|u\|_{\mathrm{Y}}\geq R.$ (2.9)

Let us define
$\mathbb{K}:=\mathopen{}\mathclose{{}\left\\{u\in\mathrm{X}:\|u\|_{\mathrm{Y}}\leq
R}\right\\}$. Then $\mathbb{K}$ is a closed and convex subset of $\mathrm{X}$.
We further define $\mathbb{K}_{R}=\mathbb{K}\cap\Sigma_{R}=\Sigma_{R}$, where
$\Sigma_{R}:=\mathopen{}\mathclose{{}\left\\{u\in\mathrm{X}:\|u\|_{\mathrm{X}}\leq
R}\right\\}$. Then $\mathbb{K}_{R}$ is not only a closed and convex but also a
bounded subset of $\mathrm{X}$. As $\mathrm{X}$ is reflexive (Lemma 2.1), and
$T$ is monotone and hemicontinuous, we infer from [16, Lemma 3.3.2] that $T$
is demicontinuous. The reflexivity of $\mathrm{X}$ and [32, Remark 5.21] imply
that the demicontinuous operator $T:\mathbb{K}_{R}\to\mathrm{Y}^{*}$ is
continuous on finite dimensional subspaces of $\mathbb{K}_{R},$ that is, for
any finite dimensional subspace $M\subset\mathrm{X}$, the restriction of $T$
to $\mathbb{K}_{R}\cap M$ is weakly continuous, namely, $T:\mathbb{K}_{R}\cap
M\to\mathrm{Y}^{*}$ is weakly continuous. Together with this fact, the
monotonicity of the operator $T:\mathbb{K}_{R}\to\mathrm{Y}^{*}$ and the
Hartman-Stampacchia theorem (see [25, Theorem 1.4, Chapter III]) yield the
existence of a $u_{R}\in\mathbb{K}_{R}$ such that

$\displaystyle\langle T(u_{R})-f,w-u_{R}\rangle\geq 0\ \text{ for all }\
w\in\mathbb{K}_{R}.$ (2.10)

The monotonicity as well as coercivity properties of
$T:\mathbb{K}\to\mathrm{Y}^{*}$ imply (see (2.5) and (2.7))

$\displaystyle\frac{\langle(T(u)-f)-(T(u_{*})-f),u-u_{*}\rangle}{\|u-u_{*}\|_{\mathrm{Y}}}\to\infty\
\text{ as }\ \|u\|_{\mathrm{Y}}\to\infty,\ u\in\mathbb{K},$

for some $u_{*}\in\mathbb{K}_{R}$. For instance, one can choose
$u_{*}=0\in\mathbb{K}_{R}$. Choose, $K>\|T(u_{*})-f\|_{\mathrm{Y}^{*}}$ and
$R>\|u_{*}\|_{\mathrm{Y}}$ such that

$\displaystyle\langle(T(u)-f)-(T(u_{*})-f),u-u_{*}\rangle\geq
K\|u-u_{*}\|_{\mathrm{Y}}\ \text{ for }\ \|u\|_{\mathrm{Y}}\geq R,\
u\in\mathbb{K}.$

Therefore,

$\displaystyle\langle(T(u)-f),u-u_{*}\rangle$ $\displaystyle\geq
K\|u-u_{*}\|_{\mathrm{Y}}+\langle(T(u_{*})-f),u-u_{*}\rangle$
$\displaystyle\geq
K\|u-u_{*}\|_{\mathrm{Y}}-\|T(u_{*})-f\|_{\mathrm{Y}^{*}}\|u-u_{*}\|_{\mathrm{Y}}$
$\displaystyle\geq\mathopen{}\mathclose{{}\left(K-\|T(u_{*})-f\|_{\mathrm{Y}^{*}}}\right)\mathopen{}\mathclose{{}\left(\|u\|_{\mathrm{Y}}-\|u_{*}\|_{\mathrm{Y}}}\right)>0\
\text{ for }\ \|u\|_{\mathrm{Y}}=R.$ (2.11)

Now, let $u_{R}\in\mathbb{K}_{R}$ be a solution of (2.10). Then

$\displaystyle\langle(T(u_{R})-f),u_{R}-u_{*}\rangle=-\langle(T(u_{R})-f),u_{*}-u_{R}\rangle\leq
0,$

and in view of (2.1), we have $\|u_{R}\|_{\mathrm{Y}}\neq R$. Or in other
words, $\|u_{R}\|_{\mathrm{Y}}<R$. Combining the above facts and applying [25,
Theorem 1.7, Chapter III], we deduce the existence of a $u_{0}\in\mathbb{K}$
such that

$\displaystyle\langle T(u_{0})-f,w-u_{0}\rangle\geq 0\ \text{ for all }\
w\in\mathbb{K}.$ (2.12)

In particular, for $w=0$, we have $\langle T(u_{0})-f,u_{0}\rangle\leq 0$,
which by (2.9) implies that

$u_{0}\in\mathopen{}\mathclose{{}\left\\{u\in\mathrm{X}:\|u\|_{\mathrm{Y}}<R}\right\\}.$

Consequently taking $w=u_{0}+\lambda v$ in (2.12) with $\|v\|_{\mathrm{Y}}=1$
and $0<|\lambda|\leq R-\|u_{0}\|_{\mathrm{Y}}$, we obtain

$\displaystyle\langle T(u_{0})-f,v\rangle=0\ \text{ for all }\
v\in\\{u\in\mathrm{X}:\|u\|_{\mathrm{Y}}=1\\}.$ (2.13)

Since the embedding $\mathrm{X}\subset\mathrm{Y}$ is dense, for any
$v\in\mathrm{Y}$ with $\|v\|_{\mathrm{Y}}=1$, we can find a sequence
$\\{v_{n}\\}_{n\in\mathbb{N}}\subset\mathrm{X}$ with
$\|v_{n}\|_{\mathrm{Y}}=1$ such that $\|v_{n}-v\|_{\mathrm{Y}}\to 0$ as
$n\to\infty$ (Lemma 2.1). Therefore, from (2.13), we infer

$\displaystyle\lim\limits_{n\to\infty}\langle
T(u_{0})-f,v_{n}\rangle=0\Rightarrow\langle T(u_{0})-f,v\rangle=0\ \text{ for
all }\ v\in\\{u\in\mathrm{Y}:\|u\|_{\mathrm{Y}}=1\\},$

and hence $T(u_{0})=f\in\mathrm{Y}^{*}.$ ∎

###  2.2. Existence and uniqueness

We use Theorem 2.3 to show the existence and uniqueness of solutions for the
problem (1.1)-(1.2) with the controls (1.4) and (1.3) for $\delta=1$ and
$\delta=2$, respectively. We provide a cut-off function for the function
$f(v)=\frac{1}{\delta+1}v^{\delta+1}$ in order to obtain monotonicity of the
nonlinear map $\mathscr{A}$ defined in (2) due to the lack of global Lipschitz
continuity.

Let us define the modified nonlinearity $f_{\rho}:\mathrm{Y}\to\mathrm{Y}^{*}$
by

$\displaystyle
f_{\rho}(y):=\mathopen{}\mathclose{{}\left\\{\begin{array}[]{cc}f(y)&\ \text{
if }\ |y|\leq\rho,\\\
\mathopen{}\mathclose{{}\left(\frac{\rho}{\|y\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(y)&\
\text{ if }\ |y|>\rho,\end{array}}\right.$ (2.16)

for $\rho\in\mathbb{N}$. Since each $y\in\mathrm{Y}$ is a continuous function
over $[0,1]$, $\|y\|_{\mathrm{L}^{\infty}}=\max\limits_{x\in[0,1]}|y(x)|.$

Let us define the nonlinear map
$\mathscr{A}_{\rho}:\mathrm{X}\subset\mathrm{Y}\to\mathrm{Y}^{*}$ by

$\displaystyle\langle\mathscr{A}_{\rho}(v),w\rangle$
$\displaystyle=\nu(v_{x},w_{x})+\nu g_{1}(v(1))w(1)-\mu(v_{xx},w_{x})+\mu
g_{2}(v(1))w(1)$
$\displaystyle\quad+\alpha\mathopen{}\mathclose{{}\left[f_{\rho}(v(1))w(1)-(f_{\rho}(v),w_{x})}\right]-\beta(v(1-v^{\delta})(v^{\delta}-\gamma),w),$
(2.17)

for all $v\in\mathrm{X}$ and $w\in\mathrm{Y}$. Therefore, for each
$v\in\mathrm{X}$ and $w\in\mathrm{Y}$, we have

$\displaystyle|\langle\mathscr{A}_{\rho}(v),w\rangle|$ $\displaystyle\leq
C\big{[}\nu(\|v_{x}\|_{\mathrm{L}^{2}}+|g_{1}(v(1))|)+\mu(\|v_{xx}\|_{\mathrm{L}^{2}}+|g_{2}(v(1))|)+\alpha(\|f_{\rho}(v)\|_{\mathrm{L}^{2}}+|f_{\rho}(v(1))|)$
$\displaystyle\quad+\beta((1+\gamma)\|v\|_{\mathrm{L}^{2(\delta+1)}}^{\delta+1}+\gamma\|v\|_{\mathrm{L}^{2}}+\|v\|_{\mathrm{L}^{2(\delta+1)}}^{2\delta+1})\big{]}\|w\|_{\mathrm{H}^{1}},$
(2.18)

where we have used the continuous Sobolev embedding
$\mathrm{H}^{1}(0,1)\hookrightarrow\mathrm{C}([0,1])\hookrightarrow\mathrm{L}^{p}(0,1)$
for all $1\leq p<\infty$. From (2.2), it is immediate that the operator
$\|\mathscr{A}_{\rho}(v)\|_{\mathrm{Y}^{*}}$ is bounded for each
$v\in\mathrm{X}$.

######  Remark 2.4.

Note that if one takes $w=v$ in (2.2), then we have for all $v\in\mathrm{X}$:

$\displaystyle\langle\mathscr{A}_{\rho}(v),v\rangle$
$\displaystyle=\nu\|v_{x}\|_{\mathrm{L}^{2}}^{2}+\nu
g_{1}(v(1))v(1)-\frac{\mu}{2}(g_{1}(v(1)))^{2}+\frac{\mu}{2}(v_{x}(0))^{2}+\mu
g_{2}(v(1))v(1)$
$\displaystyle\quad+\alpha\mathopen{}\mathclose{{}\left[f_{\rho}(v(1))v(1)-(f_{\rho}(v),v_{x})}\right]-\beta(1+\gamma)(v^{\delta+1},v)+\beta\gamma\|v\|_{\mathrm{L}^{2}}^{2}+\beta\|v\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}.$
(2.19)

For $\delta=1$ and $|v|\leq\rho$, using the control given in (1.3), we find

$\displaystyle\langle\mathscr{A}_{\rho}(v),v\rangle$
$\displaystyle\geq\nu\|v_{x}\|_{\mathrm{L}^{2}}^{2}+\eta
v^{2}(1)+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}v^{2(\delta+1)}(1)+\frac{\mu}{2}(g_{1}(v(1)))^{2}+\frac{\alpha}{\delta+2}v^{\delta+2}(1)$
$\displaystyle\quad-\frac{\beta}{4}(1-\gamma)^{2}\|v\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\geq\nu\|v_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{\eta}{2}v^{2}(1)+\frac{\alpha^{2}}{2\eta(\delta+2)^{2}}v^{2(\delta+1)}(1)+\frac{\mu}{2}(g_{1}(v(1)))^{2}-\frac{\beta}{4}(1-\gamma)^{2}\|v\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\geq\mathopen{}\mathclose{{}\left(\nu-\frac{\beta}{4}(1-\gamma)^{2}}\right)\|v\|_{\mathrm{L}^{2}}^{2}\geq
0,$ (2.20)

provided $\nu\geq\frac{\beta}{4}(1-\gamma)^{2}$. For $\delta=2$ and
$|v|\leq\rho$, using the control given in (1.4), we deduce

$\displaystyle\langle\mathscr{A}_{\rho}(v),v\rangle$
$\displaystyle\geq\nu\|v_{x}\|_{\mathrm{L}^{2}}^{2}+\eta
v^{2}(1)+\frac{2\alpha}{(\delta+2)}v^{\delta+2}(1)+\frac{\mu}{2}(g_{1}(v(1)))^{2}-\frac{\beta}{4}(1-\gamma)^{2}\|v\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\geq\mathopen{}\mathclose{{}\left(\nu-\frac{\beta}{4}(1-\gamma)^{2}}\right)\|v\|_{\mathrm{L}^{2}}^{2}\geq
0,$ (2.21)

provided $\nu\geq\frac{\beta}{4}(1-\gamma)^{2}$. As
$1>\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}$, the same results hold true for
$|v|>\rho$ also.

The following result shows that the operator
$\mathscr{B}_{\rho,\omega}:=\mathscr{A}_{\rho}+\omega\mathrm{I}:\mathrm{X}\subset\mathrm{Y}\to\mathrm{Y}^{*}$
is monotone, hemicontinuous and coercive for sufficiently large $\omega$ (in
the sense of Theorem 2.3).

######  Lemma 2.5.

Let $g_{1},g_{2}\in\mathrm{C}(\mathbb{R})$ be non-decreasing functions,
$f\in\mathrm{C}^{1}(\mathbb{R})$, $\nu,\mu,\alpha,\beta,\rho>0$,
$\delta\in\\{1,2\\}$, $\gamma\in(0,1)$ be constants, and $\mathscr{A}_{\rho}$
be defined by (2.2). Then for all $\omega\geq\omega_{\rho}$, with
$\omega_{\rho}$ defined by

$\displaystyle\omega_{\rho}=\mathopen{}\mathclose{{}\left(\frac{\alpha
L_{\rho}}{\nu}+2^{2\delta-1}\beta(1+\gamma)^{2}(\delta+1)^{2}}\right),$ (2.22)

the nonlinear map
$\mathscr{B}_{\rho,\omega}:=\mathscr{A}_{\rho}+\omega\mathrm{I}:\mathrm{X}\to\mathrm{Y}^{*}$
is _monotone_ in the sense that

$\displaystyle\langle\mathscr{B}_{\rho,\omega}(v)-\mathscr{B}_{\rho,\omega}(w),v-w\rangle\geq
0\ \text{ for all }\ v,w\in\mathrm{X}.$ (2.23)

Moreover, $\mathscr{B}_{\rho,\omega}$ is _hemicontinuous_ in the sense that
for all $v_{1},v_{2}\in\mathrm{X}$ and $w\in\mathrm{Y}$,

$\displaystyle\lim\limits_{\lambda\to
0}\langle\mathscr{B}_{\rho,\omega}(v_{1}+\lambda
v_{2}),w\rangle=\langle\mathscr{B}_{\rho,\omega}(v_{1}),w\rangle.$ (2.24)

Finally, $\mathscr{B}_{\rho,\omega}$ is _coercive_ in the sense that

$\displaystyle\lim\limits_{\|v\|_{\mathrm{Y}\to\infty}}\frac{\langle\mathscr{B}_{\rho,\omega}(v),v\rangle}{\|v\|_{\mathrm{Y}}}=\infty,$
(2.25)

for all $v\in\mathrm{X}$.

###### Proof.

The proof is divided into the following steps: Step 1: _Monotonicity._ Let us
first prove the monotoniciy property. For all $v,w\in\mathrm{X}$, since
$f_{\rho}$ satisfies a monotone condition (see (2.30) below), $g_{1}$ is non-
decreasing and (1.2) holds, we have

$\displaystyle\langle\mathscr{A}_{\rho}(v)-\mathscr{A}_{\rho}(w),v-w\rangle$
$\displaystyle=\nu\|v_{x}-w_{x}\|_{\mathrm{L}^{2}}^{2}-\frac{\mu}{2}(g_{1}(v(1))-g_{1}(w(1)))^{2}+\frac{\mu}{2}(v_{x}(0)-w_{x}(0))^{2}-\alpha(f_{\rho}(v)-f_{\rho}(w),v_{x}-w_{x})$
$\displaystyle\quad-\beta(1+\gamma)(v^{\delta+1}-w^{\delta+1},v-w)+\beta\gamma\|v-w\|_{\mathrm{L}^{2}}^{2}+\beta(v^{2\delta+1}-w^{2\delta+1},v-w)$
$\displaystyle\quad+\nu(g_{1}(v(1))-g_{1}(w(1)))(v(1)-w(1))+\mu(g_{2}(v(1))-g_{2}(w(1)))(v(1)-w(1))$
$\displaystyle\quad+\alpha(f_{\rho}(v(1))-f_{\rho}(w(1)))(v(1)-w(1))$
$\displaystyle\geq\frac{\nu}{2}\|v_{x}-w_{x}\|_{\mathrm{L}^{2}}^{2}-\frac{\alpha
L_{\rho}}{\nu}\|v-w\|_{\mathrm{L}^{2}}^{2}+\beta\gamma\|v-w\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\quad-\beta(1+\gamma)(v^{\delta+1}-w^{\delta+1},v-w)+\beta(v^{2\delta+1}-w^{2\delta+1},v-w).$
(2.26)

The estimate of the convective term can be justified in the following way:
Without loss of generality, we may assume that $|u|\leq|v|$ for all
$u,v\in\mathrm{Y}$. Then for $|u|,|v|\leq\rho,$ we have

$\displaystyle|\langle f_{\rho,x}(u)-f_{\rho,x}(v),u-v\rangle|$
$\displaystyle=|(f_{\rho}(u(1))-f_{\rho}(v(1)))(u(1)-v(1))-\langle
f_{\rho}(u)-f_{\rho}(v),u_{x}-v_{x}\rangle|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\frac{1}{\delta+1}\mathopen{}\mathclose{{}\left(u^{\delta+1}(1)-v^{\delta+1}(1)}\right)(u(1)-v(1))-\frac{1}{\delta+1}\langle
u^{\delta+1}-v^{\delta+1},u_{x}-v_{x}\rangle}\right|$
$\displaystyle\leq\frac{1}{\delta+1}\|u^{\delta+1}-v^{\delta+1}\|_{\mathrm{L}^{\infty}}\|u-v\|_{\mathrm{L}^{\infty}}+\|(\theta
u+(1-\theta)v)^{\delta}(u-v)\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\leq\|(\theta
u+(1-\theta)v)^{\delta}\|_{\mathrm{L}^{\infty}}\mathopen{}\mathclose{{}\left[\|u-v\|_{\mathrm{L}^{\infty}}^{2}+\|u-v\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}}\right]$
$\displaystyle\leq
2^{\delta+1}\mathopen{}\mathclose{{}\left(\|u\|_{\mathrm{L}^{\infty}}^{\delta}+\|v\|_{\mathrm{L}^{\infty}}^{\delta}}\right)\|u-v\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\leq\frac{\nu}{2}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{2^{4\delta+3}\rho^{2\delta}}{\nu}\|u-v\|_{\mathrm{L}^{2}}^{2},$
(2.27)

where we have used Hölder’s, Agmon’s and Young’s inequalities. For
$|u|,|v|>\rho,$ we have

$\displaystyle|\langle f_{\rho,x}(u)-f_{\rho,x}(v),u-v\rangle|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{|u(1)|}}\right)^{\delta+1}f(u(1))-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}f(v(1))}\right](u(1)-v(1))}\right.$
$\displaystyle\quad\mathopen{}\mathclose{{}\left.-\mathopen{}\mathclose{{}\left<\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(u)-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(v),u_{x}-v_{x}}\right>}\right|$
$\displaystyle=\Bigg{|}\mathopen{}\mathclose{{}\left(\frac{\rho}{|u(1)|}}\right)^{\delta+1}\frac{1}{\delta+1}(u^{\delta+1}(1)-v^{\delta+1}(1))(u(1)-v(1))$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{|u(1)|}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}}\right]\frac{1}{\delta+1}v^{\delta+1}(1)$
$\displaystyle\quad-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\frac{1}{\delta+1}\langle
u^{\delta+1}-v^{\delta+1},v_{x}-u_{x}\rangle$
$\displaystyle\quad-\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}}\right]\frac{1}{\delta+1}\langle
v^{\delta+1},u_{x}-v_{x}\rangle\Bigg{|}$
$\displaystyle=\Bigg{|}\mathopen{}\mathclose{{}\left(\frac{\rho}{|u(1)|}}\right)^{\delta+1}(\theta
u(1)+(1-\theta)v(1))^{\delta}(u(1)-v(1))^{2}$
$\displaystyle\quad+\frac{\rho^{\delta+1}}{|u(1)|^{\delta+1}|v(1)|^{\delta+1}}(\theta|v(1)|+(1-\theta)|u(1)|)^{\delta}\mathopen{}\mathclose{{}\left(|v(1)|-|u(1)|}\right)(u(1)-v(1))$
$\displaystyle\quad-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\langle(\theta
u+(1-\theta v))^{\delta}(u-v),u_{x}-v_{x}\rangle$
$\displaystyle\quad-\frac{\rho^{\delta+1}}{\|u\|_{\mathrm{L}^{\infty}}^{\delta+1}\|v\|_{\mathrm{L}^{\infty}}^{\delta+1}}\mathopen{}\mathclose{{}\left(\theta\|v\|_{\mathrm{L}^{\infty}}+(1-\theta)\|u\|_{\mathrm{L}^{\infty}}}\right)^{\delta}\mathopen{}\mathclose{{}\left(\|v\|_{\mathrm{L}^{\infty}}-\|u\|_{\mathrm{L}^{\infty}}}\right)\langle
v^{\delta+1},u_{x}-v_{x}\rangle\Bigg{|}$ $\displaystyle\leq
2^{2\delta}\rho^{\delta}\mathopen{}\mathclose{{}\left[\|u-v\|_{\mathrm{L}^{\infty}}^{2}+\|u-v\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}+\|u-v\|_{\mathrm{L}^{\infty}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}}\right]$
$\displaystyle\leq\frac{\nu}{2}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{L_{\rho}}{\nu}\|u-v\|_{\mathrm{L}^{2}}^{2},$
(2.28)

where we have used Hölder’s, Agmon’s, Sobolev’s and Young’s inequalities.
Finally, for $|u|\leq\rho$ and $|v|>\rho$, we deduce

$\displaystyle|\langle f_{\rho,x}(u)-f_{\rho,x}(v),u-v\rangle|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left[f(u(1))-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}f(v(1))}\right](u(1)-v(1))-\mathopen{}\mathclose{{}\left<f(u)-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(v),u_{x}-v_{x}}\right>}\right|$
$\displaystyle=\Bigg{|}\mathopen{}\mathclose{{}\left[1-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}}\right]f(u(1))(u(1)-v(1))+\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}(f(u(1))-f(v(1)))(u(1)-v(1))$
$\displaystyle\quad-\mathopen{}\mathclose{{}\left[1-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}}\right]\langle
f(u),u_{x}-v_{x}\rangle-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\mathopen{}\mathclose{{}\left<(f(u)-f(v)),u_{x}-v_{x}}\right>\Bigg{|}$
$\displaystyle\leq\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{|u(1)|}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}}\right]|f(u(1))||u(1)-v(1)|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}|(f(u(1))-f(v(1)))||u(1)-v(1)|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}}\right]\|f(u)\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\|f(u)-f(v)\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\leq\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}(\theta|v(1)|+(1-\theta)|u(1)|)^{\delta}(|v(1)|-|u(1)|)|u(1)-v(1)|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}|\theta
u(1)+(1-\theta)v(1)|^{\delta}|u(1)-v(1)|^{2}$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}(\theta\|v\|_{\mathrm{L}^{\infty}}+(1-\theta)\|u\|_{\mathrm{L}^{\infty}})^{\delta}(\|v\|_{\mathrm{L}^{\infty}}-\|u\|_{\mathrm{L}^{\infty}})\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\|\theta
u+(1-\theta)v\|_{\mathrm{L}^{\infty}}^{\delta}\|u-v\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\leq
2^{2\delta}\rho^{\delta}\mathopen{}\mathclose{{}\left[\|u-v\|_{\mathrm{L}^{\infty}}^{2}+\|u-v\|_{\mathrm{L}^{\infty}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}+\|u-v\|_{\mathrm{L}^{2}}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}}\right]$
$\displaystyle\leq\frac{\nu}{2}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{L_{\rho}}{\nu}\|u-v\|_{\mathrm{L}^{2}}^{2}.$
(2.29)

From the above calculations, one can conclude that

$\displaystyle|\langle
f_{\rho,x}(u)-f_{\rho,x}(v),u-v\rangle|\leq\frac{\nu}{2}\|u_{x}-v_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{L_{\rho}}{\nu}\|u-v\|_{\mathrm{L}^{2}}^{2},$
(2.30)

for all $u,v\in\mathrm{Y}$.

Let us now estimate the term $-\beta(v^{2\delta+1}-w^{2\delta+1},v-w)$ from
(2.2) as

$\displaystyle\beta(v^{2\delta+1}-w^{2\delta+1},v-w)$
$\displaystyle=\beta(v^{2\delta},(v-w)^{2})+\beta(w^{2\delta},(v-w)^{2})+\beta(v^{2\delta}w-w^{2\delta}v,v-w)$
$\displaystyle=\beta\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\beta\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\beta(vw,v^{2\delta}+w^{2\delta})-\beta(v^{2},w^{2\delta})-\beta(w^{2},v^{2\delta})$
$\displaystyle=\frac{\beta}{2}\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{2}\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{2}(v^{2\delta}-w^{2\delta},v^{2}-w^{2})$
$\displaystyle\geq\frac{\beta}{2}\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{2}\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2},$
(2.31)

since $(v^{2\delta}-w^{2\delta},v^{2}-w^{2})\geq 0$. Using Taylor’s formula,
Hölder’s and Young’s inequalities, we estimate the term
$\beta(1+\gamma)(v^{\delta+1}-w^{\delta+1},v-w)$ from (2.2) as

$\displaystyle\beta(1+\gamma)(v^{\delta+1}-w^{\delta+1},v-w)$
$\displaystyle=\beta(1+\gamma)(\delta+1)((\theta
v+(1-\theta)w)^{\delta}(v-w),v-w)$
$\displaystyle\leq\beta(1+\gamma)(\delta+1)2^{\delta-1}(\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}+\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}})\|v-w\|_{\mathrm{L}^{2}}$
$\displaystyle\leq\frac{\beta}{4}\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{4}\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{2}2^{2\delta}(1+\gamma)^{2}(\delta+1)^{2}\|v-w\|_{\mathrm{L}^{2}}^{2}.$
(2.32)

It should also be noted that for all $v,w\in\mathrm{L}^{2(\delta+1)}(0,1)$,

$\displaystyle\|v-w\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}$
$\displaystyle=\int_{0}^{1}|v(x)-w(x)|^{2\delta}|v(x)-w(x)|^{2}dx$
$\displaystyle\leq
2^{2\delta-1}\mathopen{}\mathclose{{}\left[\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}}\right].$
(2.33)

Combining (2.2)-(2.2) and substituting it in (2.2), we obtain

$\displaystyle\langle\mathscr{A}_{\rho}(v)-\mathscr{A}_{\rho}(w),v-w\rangle$
$\displaystyle\geq\frac{\nu}{2}\|v_{x}-w_{x}\|_{\mathrm{L}^{2}}^{2}-\frac{\alpha
L_{\rho}}{\nu}\|v-w\|_{\mathrm{L}^{2}}^{2}+\beta\gamma\|v-w\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\quad+\frac{\beta}{4}\|v^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta}{4}\|w^{\delta}(v-w)\|_{\mathrm{L}^{2}}^{2}-2^{2\delta-1}\beta(1+\gamma)^{2}(\delta+1)^{2}\|v-w\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\geq\max\mathopen{}\mathclose{{}\left\\{\frac{\nu}{2},\beta\gamma}\right\\}\|v-w\|_{\mathrm{H}^{1}}^{2}+\frac{\beta}{2^{2\delta+1}}\|v-w\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}-\omega_{\rho}\|v-w\|_{\mathrm{L}^{2}}^{2},$
(2.34)

where $\omega_{\rho}$ is defined in (2.22). Therefore for all
$\omega\geq\omega_{\rho}$, we deduce from (2.2) that

$\displaystyle\langle\mathscr{A}_{\rho}(v)-\mathscr{A}_{\rho}(w),v-w\rangle+\omega\|v-w\|_{\mathrm{L}^{2}}^{2}\geq\max\mathopen{}\mathclose{{}\left\\{\frac{\nu}{2},\beta\gamma}\right\\}\|u-v\|_{\mathrm{H}^{1}}^{2},$
(2.35)

for all $v,w\in\mathrm{X}$, so that the monotonicity of the operator
$\mathscr{B}_{\rho,\omega}=\mathscr{A}_{\rho}+\omega\mathrm{I}$ follows.

Step 2: _Hemicontinuity._ Note that for reflexive Banach spaces, the
demicontinuity property implies hemicontinuity [16, Section 3.3]. As
$\mathrm{X}$ is a reflexive Banach space (Lemma 2.1), it is enough to show
that $\mathscr{B}_{\rho,\omega}$ is demicontinuous in the sense that $v_{n}\to
v$ in $\mathrm{X}$ implies
$\mathscr{B}_{\rho,\omega}(v_{n})\xrightharpoonup{w}\mathscr{B}_{\rho,\omega}(v)$
in $\mathrm{Y}^{*}$ as $n\to\infty$. Claim: _The operator
$\mathscr{B}_{\rho,\omega}$ is demicontinuous._ In order to show the
demicontinuity of the operator $\mathscr{B}_{\rho,\omega}$, it is enough to
prove the demicontinuity of the operator $\mathscr{A}_{\rho}(\cdot)$ defined
in (2.2), that is, we need to show that if $v_{n}\to v$ in $\mathrm{X}$
implies $\mathscr{A}_{\rho}(v_{n})\xrightharpoonup{w}\mathscr{A}_{\rho}(v)$ in
$\mathrm{Y}^{*}$ as $n\to\infty$. We first consider the case
$|v_{n}|,|v|\leq\rho$. We choose a sequence
$\\{v_{n}\\}_{n\in\mathbb{N}}\in\mathrm{X}$ such that $v_{n}\to v$ in
$\mathrm{X}$ and $|v_{n}|,|v|\leq\rho$. Since
$\mathrm{H}^{1}(0,1)\hookrightarrow\mathrm{C}([0,1])$, the convergence
$v_{n}\to v$ in $\mathrm{X}$ implies $v_{n}(x)\to v(x)$ for all $x\in[0,1]$.
For any $w\in\mathrm{Y}$, we consider

$\displaystyle\mathopen{}\mathclose{{}\left|\langle\mathscr{A}(v_{n})-\mathscr{A}(v),w\rangle}\right|$
$\displaystyle\leq\nu\big{|}\big{(}(v_{n,x}-v_{x}),w_{x}\big{)}\big{|}+\nu|(g_{1}(v_{n}(1))-g_{1}(v(1)))w(1)|$
$\displaystyle\quad+\mu|((v_{n,xx}-v_{xx}),w_{x})|+\mu|(g_{2}(v_{n}(1))-g_{2}(v(1)))w(1)|$
$\displaystyle\quad+\alpha|(f(v_{n})-f(v),w_{x})|+\alpha|(f(v_{n}(1))-f(v(1)))w(1)|$
$\displaystyle\quad+\beta(1+\gamma)|(v_{n}^{\delta+1}-v^{\delta+1},w)|+\beta|(v_{n}^{2\delta+1}-v^{2\delta+1},w)|+\beta\gamma|(v_{n}-v,w)|$
$\displaystyle\leq\nu\|v_{n,x}-v_{x}\|_{\mathrm{L}^{2}}\|w_{x}\|_{\mathrm{L}^{2}}+\eta|v_{n}(1)-v(1)||w(1)|$
$\displaystyle\quad+\frac{\alpha(\delta+1)}{(\delta+2)}|\theta
v_{n}(1)+(1-\theta)v(1)|^{\delta}|v_{n}(1)-v(1)||w(1)|+\mu\|v_{n,xx}-v_{xx}\|_{\mathrm{L}^{2}}\|w_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\quad+\frac{\delta\mu}{\nu^{2}}\bigg{[}\eta^{2}|v_{n}(1)-v(1)|+\frac{2\alpha\eta(\delta+1)}{(\delta+2)}\theta
v_{n}(1)+(1-\theta)v(1)|^{\delta}|v_{n}(1)-v(1)|$
$\displaystyle\quad+\frac{\alpha^{2}(2\delta+1)}{(\delta+2)^{2}}|\theta
v_{n}(1)+(1-\theta)v(1)|^{2\delta}|v_{n}(1)-v(1)|\bigg{]}|w(1)|$
$\displaystyle\quad+\alpha\|\theta
v_{n}+(1-\theta)v\|_{\mathrm{L}^{\infty}}^{\delta}\|v_{n}-v\|_{\mathrm{L}^{2}}\|w_{x}\|_{\mathrm{L}^{2}}+\alpha|\theta
v_{n}(1)+(1-\theta)v(1)|^{\delta}|v_{n}(1)-v(1)||w(1)|$
$\displaystyle\quad+\beta\Big{[}(1+\gamma)(\delta+1)\|\theta
v_{n}+(1-\theta)v\|_{\mathrm{L}^{\infty}}^{\delta}+(2\delta+1)\|\theta
v_{n}+(1-\theta)v\|_{\mathrm{L}^{\infty}}^{2\delta}+1\Big{]}\|v_{n}-v\|_{\mathrm{L}^{2}}\|w\|_{\mathrm{L}^{2}}$
$\displaystyle\to 0\ \text{ as }\ n\to\infty,$ (2.36)

for $\delta=2$. Similarly, one can show for the case $\delta=1$ also. For the
other cases, the proof follows in a similar way except for the convective term
with cutoff function. For $|v_{n}|,|v|>\rho$, we have

$\displaystyle|(f_{\rho}(v_{n})-f_{\rho}(v),w_{x})|+|(f_{\rho}(v_{n}(1))-f_{\rho}(v(1)))w(1)|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v_{n}\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(v_{n})-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(v),w_{x}}\right)}\right|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{|v_{n}(1)|}}\right)^{\delta+1}f(v_{n}(1))-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}f(v(1))}\right]w(1)}\right|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v_{n}\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}}\right]f(v_{n})+\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}(f(v_{n})-f(v)),w_{x}}\right)}\right|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left\\{\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\frac{\rho}{|v_{n}(1)|}}\right)^{\delta+1}-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}}\right]f(v_{n}(1))+\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}(f(v_{n}(1))-f(v(1)))}\right\\}w(1)}\right|$
$\displaystyle\leq\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}\Big{[}(\theta\|v\|_{\mathrm{L}^{\infty}}+(1-\theta)\|v_{n}\|_{\mathrm{L}^{\infty}})^{\delta}\|v_{n}-v\|_{\mathrm{L}^{\infty}}$
$\displaystyle\quad+\|(\theta
v_{n}+(1-\theta)v)^{\delta}\|_{\mathrm{L}^{\infty}}\|v_{n}-v\|_{\mathrm{L}^{2}}\Big{]}\|w_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\quad+2\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}\Big{[}(\theta|v_{n}(1)|+(1-\theta)|v(1))^{\delta}|v_{n}(1)-v(1)|\Big{]}|w(1)|$
$\displaystyle\leq
2^{\delta+1}(\|v_{n}\|_{\mathrm{L}^{\infty}}^{\delta}+\|v\|_{\mathrm{L}^{\infty}}^{\delta})\|v_{n}-v\|_{\mathrm{L}^{\infty}}\|w\|_{\mathrm{Y}}$
$\displaystyle\to 0\ \text{ as }\ n\to\infty.$ (2.37)

For the case $|v_{n}|\leq\rho$ and $|v|>\rho$, we infer

$\displaystyle|(f_{\rho}(v_{n})-f_{\rho}(v),w_{x})|+|(f_{\rho}(v_{n}(1))-f_{\rho}(v(1)))w(1)|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left(f(v_{n})-\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}f(v),w_{x}}\right)}\right|+\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left[f(v_{n}(1))-\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}f(v(1))}\right]w(1)}\right|$
$\displaystyle=\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left(\frac{\|v\|_{\mathrm{L}^{\infty}}^{\delta+1}-\rho^{\delta+1}}{\|v\|_{\mathrm{L}^{\infty}}^{\delta+1}}}\right)(f(v_{n}),w_{x})+\mathopen{}\mathclose{{}\left(\frac{\rho}{\|v\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}(f(v_{n})-f(v),w_{x})}\right|$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left|\mathopen{}\mathclose{{}\left(\frac{|v(1)|^{\delta+1}-\rho^{\delta+1}}{|v(1)|^{\delta+1}}}\right)f(v_{n}(1))w(1)+\mathopen{}\mathclose{{}\left(\frac{\rho}{|v(1)|}}\right)^{\delta+1}(f(v_{n}(1))-f(v(1)))w(1)}\right|$
$\displaystyle\leq\mathopen{}\mathclose{{}\left[\mathopen{}\mathclose{{}\left(\theta\|v\|_{\mathrm{L}^{\infty}}+(1-\theta)\|v_{n}\|_{\mathrm{L}^{\infty}}}\right)^{\delta}\|v_{n}-v\|_{\mathrm{L}^{\infty}}+\|\theta
v_{n}+(1-\theta)v\|_{\mathrm{L}^{\infty}}^{\delta}\|v_{n}-v\|_{\mathrm{L}^{2}}}\right]\|w_{x}\|_{\mathrm{L}^{2}}$
$\displaystyle\quad+\mathopen{}\mathclose{{}\left[(\theta|v(1)|^{\delta}+(1-\theta)|v_{n}|^{\delta})+|\theta
v_{n}(1)+(1-\theta)v(1)|^{\delta}}\right]|v_{n}(1)-v(1)||w(1)|$
$\displaystyle\to 0\ \text{ as }\ n\to\infty.$ (2.38)

A similar calculation holds for the case, $|v_{n}|>\rho$ and $|v|\leq\rho$
also. Thus the operator $\mathscr{A}(\cdot)$ is demicontinuous and hence
hemicontinuous also.

Step 3: _Coercivity._ By taking $w=0$ in (2.2), one can easily see that

$\displaystyle\frac{\langle\mathscr{B}_{\rho,\omega}(v),v\rangle}{\|v\|_{\mathrm{H}^{1}}}\geq\frac{1}{2}\max\mathopen{}\mathclose{{}\left\\{\frac{\nu}{2},\beta\gamma}\right\\}\|v\|_{\mathrm{H}^{1}}-\|\mathscr{A}_{\rho}(0)\|_{\mathrm{Y}^{*}}\to\infty$
(2.39)

for all $v\in\mathrm{X}$ such that $\|v\|_{\mathrm{H}^{1}}\to\infty$, and the
coercivity of $\mathscr{B}_{\rho,\omega}(\cdot)$ follows. ∎

Finally, we have the following result on the existence and uniqueness of
solutions for the problem (1.1)-(1.2) with the controls (1.4) and (1.3) for
$\delta=1$ and $\delta=2$, respectively.

######  Theorem 2.6.

For any initial data $u_{0}\in\mathrm{D}(\mathscr{A})$, the system (2.3)
possesses a _unique strong solution_
$u\in\mathrm{C}^{1}([0,\infty);\mathrm{L}^{2}(0,1))\cap\mathrm{C}([0,\infty);\mathrm{H}^{3}(0,1))$.

###### Proof.

The proof is divided into the following steps: Step 1: _Strong solution of the
approximate problem:_ We use the well-known Crandall-Liggett and Minty
Theorems to establish this result. Since $\mathscr{B}_{\omega,\rho}$ is
monotone, hemicontinuous and coercive from $\mathrm{X}$ to $\mathrm{Y}^{*}$,
then by an application of Theorem 2.3 (Minty’s Theorem) yields that
$\mathscr{B}_{\omega,\rho}$ is onto, that is,
$\mathrm{Range}(\mathscr{B}_{\omega,\rho})=\mathrm{Y}^{*}$. Since
$\mathscr{B}_{\rho,\omega}:\mathrm{X}\to\mathrm{Y}^{*}$ is monotone,
hemicontinuous and coercive, a further application of [9, Example 2.3.7]
implies that $\mathscr{B}_{\rho,\omega}$ (the restriction of
$\mathscr{B}_{\rho,\omega}$ to $\mathrm{H}$) is maximal monotone in
$\mathrm{H}$ with
$\mathrm{D}(\mathscr{A}_{\rho})=\mathrm{D}(\mathscr{B}_{\omega,\rho})$. Since
$\mathrm{H}^{*}=\mathrm{H}$, from [2, Theorem 3.1], we infer that the maximal
monotone and $m$-accretive sets coincide. Therefore, for sufficiently large
$\omega$, $\mathscr{B}_{\omega,\rho}$ is $m$-accretive with domain
$\mathrm{D}(\mathscr{A}_{\rho})=\mathrm{D}(\mathscr{B}_{\omega,\rho})$. The
above arguments can be justified in the following way: Since
$\mathrm{Range}(\mathscr{B}_{\omega,\rho})=\mathrm{Y}^{*}$, for each
$\xi\in\mathrm{Y}^{*}$, there exists a unique $\varphi\in\mathrm{X}$ such that
$\mathscr{B}_{\omega,\rho}(\varphi)=\xi$ in the distributional sense, which
means

$\displaystyle\langle\mathscr{B}_{\omega,\rho}(\varphi),v\rangle=\langle\xi,v\rangle,\
\text{ for all }\ v\in\mathrm{Y}.$ (2.40)

Furthermore, if $\xi\in\mathrm{L}^{2}(0,1),$ then
$\varphi\in\mathrm{H}^{3}(0,1)$ and it satisfies (cf. [20])

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
\omega\varphi(x)-\nu\varphi_{xx}(x)+\mu\varphi_{xxx}(x)+\alpha(f_{\rho}(\varphi(x)))_{x}-\beta\varphi(x)(1-\varphi^{\delta}(x))(\varphi^{\delta}(x)-\gamma)&=\xi(x),\\\
\varphi(0)=0,\ \varphi_{x}(1)+g_{1}(\varphi(1))=0,\
\varphi_{xx}(1)-g_{2}(\varphi(1))&=0,\end{aligned}}\right.$ (2.41)

where the first equation in (2.41) holds for a.e. $x\in[0,1]$. In fact, taking
the inner product with $\varphi$ to the first equation in (2.41), we find

$\displaystyle\omega\|\varphi\|_{\mathrm{L}^{2}}^{2}+\langle\mathscr{A}_{\rho}(\varphi),\varphi\rangle$
$\displaystyle=\langle\xi,\varphi\rangle.$ (2.42)

Applying (2.4) and (2.4) in (2.42), we deduce

$\displaystyle\mathopen{}\mathclose{{}\left(\omega-\frac{\beta}{4\theta}(1-2(2\theta-1)\gamma+\gamma^{2})^{2}}\right)\|\varphi\|_{\mathrm{L}^{2}}^{2}+\frac{\nu}{2}\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\beta\theta\|\varphi\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}\leq\frac{1}{2\nu}\|\xi\|_{\mathrm{Y}^{*}}^{2},$
(2.43)

for some $0<\theta<1$ and sufficiently large $\omega$. An application of
Agmon’s inequality (Lemma A.2) yields

$\displaystyle\max_{x\in[0,1]}|\varphi(x)|\leq\sqrt{2}\|\varphi\|_{\mathrm{L}^{2}}^{1/2}\|\varphi_{x}\|_{\mathrm{L}^{2}}^{1/2}\leq\frac{1}{\sqrt{\nu}}\|\xi\|_{\mathrm{Y}^{*}}.$
(2.44)

Taking the inner product with $-\varphi_{xx}$ to the first equation in (2.41),
we obtain

$\displaystyle\omega\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+(\mathscr{A}_{\rho}(\varphi),-\varphi_{xx})=(\xi,-\varphi_{xx}).$
(2.45)

Since $\frac{\rho}{\|\varphi\|_{\mathrm{L}^{\infty}}}<1$ for $|\varphi|>\rho$,
we need to consider the case $|\varphi|\leq\rho$ only as the calculations for
the other case be performed in a similar way. For $|\varphi|\leq\rho$, we have

$\displaystyle(\mathscr{A}(\varphi),-\varphi_{xx})$
$\displaystyle=\nu\|\varphi_{xx}\|_{\mathrm{L}^{2}}^{2}-\frac{\mu}{2}g_{2}^{2}(\varphi(1))+\frac{\mu}{2}\varphi_{xx}^{2}(0)-\alpha(\varphi^{\delta}\varphi_{x},\varphi_{xx})+\beta(1+\gamma)(\varphi^{\delta+1},\varphi_{xx})$
$\displaystyle\quad+\beta\gamma\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\beta\gamma
g_{1}(\varphi(1))\varphi(1)+\beta(2\delta+1)\|\varphi^{\delta}\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\beta
g_{1}(\varphi(1))\varphi^{2\delta+1}(1).$ (2.46)

We estimate $\alpha|(\varphi^{\delta}\varphi_{x},\varphi_{xx})|$ as

$\displaystyle\alpha|(\varphi^{\delta}\varphi_{x},\varphi_{xx})|\leq\alpha\|\varphi_{xx}\|_{\mathrm{L}^{2}}\|\varphi^{\delta}\|_{\mathrm{L}^{\infty}}\|\varphi_{x}\|_{\mathrm{L}^{2}}\leq\frac{\nu}{4}\|\varphi_{xx}\|_{\mathrm{L}^{2}}^{2}+\frac{\alpha^{2}}{\nu}\|\varphi\|_{\mathrm{L}^{\infty}}^{2\delta}\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}.$
(2.47)

Similarly, we estimate $\beta(1+\gamma)|(\varphi^{\delta+1},\varphi_{xx})|$ as

$\displaystyle\beta(1+\gamma)|(\varphi^{\delta+1},\varphi_{xx})|\leq\beta(1+\gamma)\|\varphi^{\delta+1}\|_{\mathrm{L}^{2}}\|\varphi_{xx}\|_{\mathrm{L}^{2}}\leq\frac{\nu}{4}\|\varphi_{xx}\|_{\mathrm{L}^{2}}^{2}+\frac{\beta^{2}(1+\gamma)^{2}}{\nu}\|\varphi\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}.$
(2.48)

Substituting (2.2)-(2.48) in (2.45) leads to

$\displaystyle\omega\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{\nu}{4}\|\varphi_{xx}\|_{\mathrm{L}^{2}}^{2}+\beta(2\delta+1)\|\varphi^{\delta}\varphi\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\leq\frac{1}{\nu}\|\xi\|_{\mathrm{L}^{2}}^{2}+\frac{\mu}{2}g_{2}^{2}(\varphi)+\frac{\alpha^{2}}{\nu}\|\varphi\|_{\mathrm{L}^{\infty}}^{2\delta}\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{\beta^{2}(1+\gamma)^{2}}{\nu}\|\varphi\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}.$
(2.49)

From the definition of controls given in (1.3) and (1.4), using Agmon’s
inequality (Lemma A.2) and the estimate (2.44), we deduce from (2.2) that

$\displaystyle\omega\|\varphi_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{\nu}{4}\|\varphi_{xx}\|_{\mathrm{L}^{2}}^{2}+\beta(2\delta+1)\|\varphi^{\delta}\varphi\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\leq
C_{\nu,\mu,\alpha,\beta,\gamma,\delta}\mathopen{}\mathclose{{}\left(1+\|\xi\|_{\mathrm{L}^{2}}^{8\delta}}\right),$
(2.50)

where we have used (2.43). Taking the inner product with $\varphi_{xxx}$ to
the first equation in (2.41), we get

$\displaystyle(\mathscr{A}_{\rho}(\varphi),\varphi_{xxx})=(\xi,\varphi_{xxx})-\omega(\varphi,\varphi_{xxx}).$
(2.51)

An integration by parts and application of Hölder’s and Young’s inequalities
yield

$\displaystyle\mu\|\varphi_{xxx}\|_{\mathrm{L}^{2}}^{2}+\frac{\nu}{2}\varphi_{xx}^{2}(0)$
$\displaystyle=(\xi,\varphi_{xxx})-(\omega+\beta\gamma)(\varphi,\varphi_{xxx})-\alpha(\varphi^{\delta}\varphi_{x},\varphi_{xxx})+\beta(1+\gamma)(\varphi^{\delta+1},\varphi_{xxx})-\beta(\varphi^{2\delta+1},\varphi_{xxx})$
$\displaystyle\leq\frac{3\mu}{4}\|\varphi_{xxx}\|_{\mathrm{L}^{2}}^{2}+\frac{1}{\mu}\|\xi\|_{\mathrm{L}^{2}}^{2}+\frac{2(\omega+\beta\gamma)^{2}}{\mu}\|\varphi\|_{\mathrm{L}^{2}}^{2}+\frac{2\alpha^{2}}{\mu}\|\varphi^{\delta}\varphi\|_{\mathrm{L}^{2}}^{2}+\frac{2\beta^{2}(1+\gamma)^{2}}{\mu}\|\varphi\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}$
$\displaystyle\quad+\frac{2\beta^{2}}{\mu}\|\varphi\|_{\mathrm{L}^{4\delta+2}}^{4\delta+2}.$
(2.52)

Using (2.50) in (2.2), we finally arrive at

$\displaystyle\frac{\mu}{4}\|\varphi_{xxx}\|_{\mathrm{L}^{2}}^{2}\leq
C_{\nu,\mu,\alpha,\beta,\gamma,\delta,\omega}\mathopen{}\mathclose{{}\left(1+\|\xi\|_{\mathrm{L}^{2}}^{8\delta}}\right).$
(2.53)

Combining (2.43), (2.50) and (2.53), one can easily seen that

$\|\mathscr{B}_{\omega,\rho}(\varphi)\|_{\mathrm{L}^{2}}\leq
C_{\nu,\mu,\alpha,\beta,\gamma,\delta,\omega}\mathopen{}\mathclose{{}\left(1+\|\xi\|_{\mathrm{L}^{2}}^{4\delta}}\right),$

that is,
$\mathrm{H}\subseteq\mathrm{Range}(\mathscr{B}_{\omega,\rho}(\varphi))$.
Taking the inner product with $\xi$ to the first equation in (2.41), we infer

$\displaystyle\|\xi\|_{\mathrm{L}^{2}}\leq\|\mathscr{B}_{\omega,\rho}(\varphi)\|_{\mathrm{L}^{2}},$

so that
$\mathrm{Range}(\mathscr{B}_{\omega,\rho}(\varphi))\subseteq\mathrm{H}$.
Therefore, it is immediate that
$\mathrm{Range}(\mathscr{B}_{\omega,\rho}(\varphi))=\mathrm{H}$. Thus, for
each $\lambda\in(0,\frac{1}{\omega_{\rho}})$, there exists a
$\varphi\in\mathrm{X}$ such that
$\mathscr{B}_{\rho,\frac{1}{\lambda}}(\varphi)=\frac{\xi}{\lambda}$, which
yields $(\mathrm{I}+\lambda\mathscr{A}_{\rho})(\varphi)=\xi$, so that
$\mathrm{Range}(\mathrm{I}+\lambda\mathscr{A}_{\rho})=\mathrm{H}$.

Then by an application of the Crandall-Liggett Theorem ([3, Theorem 4.1.3,
4.1.4], [14], [43, Theorems 5.1, 5.2]), we have that the problem

$\mathopen{}\mathclose{{}\left\\{\begin{aligned}
\frac{du_{\rho}(t)}{dt}+\mathscr{A}_{\rho}(u_{\rho})(t)&=0,\ t>0,\\\
u_{\rho}(0)&=u_{0},\end{aligned}}\right.$ (2.54)

has a _unique strong solution_
$u_{\rho}\in\mathrm{C}\mathopen{}\mathclose{{}\left([0,\infty);\mathrm{D}(\mathscr{A}_{\rho})}\right)\cap\mathrm{C}^{1}\mathopen{}\mathclose{{}\left((0,\infty);\mathrm{L}^{2}(0,1)}\right)\subset\mathrm{C}\mathopen{}\mathclose{{}\left([0,\infty);\mathrm{H}^{3}(0,1)}\right)\cap\mathrm{C}^{1}\mathopen{}\mathclose{{}\left((0,\infty);\mathrm{L}^{2}(0,1)}\right)$
for all $\rho>0$ and $u_{0}\in\mathrm{D}(\mathscr{A})$. Moreover, (2.54) has a
unique mild solution

$u_{\rho}(t)=\mathrm{S}_{\rho}(t)(u_{0})=\lim\limits_{n\to\infty}\mathopen{}\mathclose{{}\left(\mathrm{I}+\frac{t}{n}\mathscr{A}_{\rho}}\right)^{-n}(u_{0}),$

which is a limit in $\mathrm{C}([0,T];\mathrm{L}^{2}(0,1))$ of a sequence of
strong solutions ([3, Theorem 4.1.3]).

Step 2: _Uniform bounds for the solutions:_ Let us now show the existence and
uniqueness of solutions for the problem (2.3). We first establish the uniform
boundedness of the sequence of solutions $\\{u_{\rho}\\}_{\rho>0}$. Taking the
inner product with $u_{\rho}$ in (2.54) and using a calculation similar to
(2.4), one obtains

$\displaystyle\|u_{\rho}(t)\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle=\|u_{0}\|_{\mathrm{L}^{2}}^{2}-2\int_{0}^{t}\langle\mathscr{A}_{\rho}(u_{\rho})(s),u_{\rho}(s)\rangle
ds$
$\displaystyle\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}-2\nu\int_{0}^{t}\|u_{\rho,x}(s)\|_{\mathrm{L}^{2}}^{2}ds-\eta\int_{0}^{t}u_{\rho}^{2}(1,s)ds-\frac{\alpha^{2}}{\eta(\delta+2)^{2}}\int_{0}^{t}u_{\rho}^{2(\delta+1)}(1,s)ds$
$\displaystyle\quad-\mu\int_{0}^{t}(g_{1}(u_{\rho}(1,s)))^{2}ds-\beta\int_{0}^{t}\|u_{\rho}(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds+\beta(1+\gamma^{2})\int_{0}^{t}\|u_{\rho}(s)\|_{\mathrm{L}^{2}}^{2}ds,$
(2.55)

for all $t\in[0,T]$ and $|u_{\rho}|\leq\rho$. A similar expression involving
$\mathopen{}\mathclose{{}\left(\frac{\rho}{\|u_{\rho}\|_{\mathrm{L}^{\infty}}}}\right)^{\delta+1}$
holds for $|u_{\rho}|>\rho$ also. Since
$\frac{\rho}{\|u_{\rho}\|_{\mathrm{L}^{\infty}}}<1$ for $|u_{\rho}|>\rho$, we
need to consider the case $|u_{\rho}|\leq\rho$ only. An application of
Gronwall’s inequality in (2.2) yields

$\displaystyle\|u_{\rho}(t)\|_{\mathrm{L}^{2}}^{2}+2\nu\int_{0}^{t}\|u_{\rho,x}(s)\|_{\mathrm{L}^{2}}^{2}ds+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}\int_{0}^{t}u_{\rho}^{2(\delta+1)}(1,s)ds$
$\displaystyle\quad+\frac{\mu}{\delta}\int_{0}^{t}g_{2}(u_{\rho}(1,s))u_{\rho}(1,s)ds+\beta\int_{0}^{t}\|u_{\rho}(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}e^{\beta
T(1+\gamma^{2})},$ (2.56)

for all $t\in[0,T]$, and the right hand side is independent of $\rho$.

Taking the inner product with $-u_{\rho,xx}$ in (2.54), we find for a.e.
$t\in[0,T]$

$\displaystyle-\mathopen{}\mathclose{{}\left(u_{\rho,t}(t),u_{\rho,xx}(t)}\right)=(\mathscr{A}_{\rho}u_{\rho}(t),u_{\rho,xx}(t)).$
(2.57)

We first consider the case $\delta=1$. Note that

$\displaystyle-$ $\displaystyle\int_{0}^{1}u_{\rho,t}(t,x)u_{\rho,xx}(t,x)dx$
$\displaystyle=-u_{\rho,t}(t,x)u_{\rho,x}(t,x)\big{|}_{0}^{1}+\int_{0}^{1}u_{\rho,xt}(t,x)u_{\rho,x}(t,x)dx$
$\displaystyle=\frac{1}{2}\frac{d}{dt}\|u_{\rho,x}(t)\|_{\mathrm{L}^{2}}^{2}+\frac{1}{\nu}u_{\rho,t}(1,t)\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u_{\rho}^{2\delta}(1,t)}\right)u_{\rho}(1,t)$
$\displaystyle=\frac{1}{2}\frac{d}{dt}\mathopen{}\mathclose{{}\left[\|u_{\rho,x}(t)\|_{\mathrm{L}^{2}}^{2}+\frac{\eta}{\nu}u_{\rho}^{2}(1,t)+\frac{\alpha^{2}}{\nu\eta(\delta+2)^{2}(\delta+1)}u_{\rho}^{2(\delta+1)}(1,t)}\right],$
(2.58)

since $u_{\rho,t}(0,t)=0$. Substituting the calculations (2.2)-(2.48) and
(2.2) in (2.57), and then integrating from $0$ to $t$, we deduce

$\displaystyle\|u_{\rho,x}(t)\|_{\mathrm{L}^{2}}^{2}+\frac{\eta}{\nu}u_{\rho}^{2}(1,t)+\frac{\alpha^{2}}{\nu\eta(\delta+2)^{2}(\delta+1)}u_{\rho}^{2(\delta+1)}(1,t)+\nu\int_{0}^{t}\|u_{\rho,xx}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\quad+2\beta(2\delta+1)\int_{0}^{t}\|u_{\rho}^{\delta}(s)u_{\rho,x}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\leq\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+\frac{\eta}{\nu}u_{0}^{2}(1)+\frac{\alpha^{2}}{\nu\eta(\delta+2)^{2}(\delta+1)}u_{0}^{2(\delta+1)}(1)$
$\displaystyle\quad+\mu\int_{0}^{t}\mathopen{}\mathclose{{}\left[\frac{\delta}{\nu^{2}}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u_{\rho}^{2\delta}(1,s)}\right)^{2}u(1,s)}\right]^{2}ds+\frac{2\alpha^{2}}{\nu}\int_{0}^{t}\|u_{\rho}(s)\|_{\mathrm{L}^{\infty}}^{2\delta}\|u_{\rho,x}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\quad+\frac{2\beta^{2}(1+\gamma)^{2}}{\nu}\int_{0}^{t}\|u_{\rho}(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds,$
(2.59)

for all $t\in[0,T]$. Note that for the time dependent problem, the estimate
(2.47) is valid for $1\leq\delta\leq 2$ only. Application of Gronwall’s and
Agmon’s inequalities in (2.2) yield

$\displaystyle\|u_{\rho,x}(t)\|_{\mathrm{L}^{2}}^{2}+\frac{\eta}{\nu}u_{\rho}^{2}(1,t)+\frac{\alpha^{2}}{\nu\eta(\delta+2)^{2}(\delta+1)}u_{\rho}^{2(\delta+1)}(1,t)+\nu\int_{0}^{t}\|u_{\rho,xx}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\quad+2\beta(2\delta+1)\int_{0}^{t}\|u_{\rho}^{\delta}(s)u_{\rho,x}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\leq\mathopen{}\mathclose{{}\left\\{\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+\frac{1}{\nu(\delta+1)}\mathopen{}\mathclose{{}\left[u_{0}^{2}(1)+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u_{0}^{2(\delta+1)}(1)}\right]+\frac{2\beta^{2}(1+\gamma)^{2}}{\nu}\int_{0}^{T}\|u_{\rho}(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds}\right\\}$
$\displaystyle\quad\times\exp\mathopen{}\mathclose{{}\left\\{\frac{\delta(\delta+1)}{\nu}\int_{0}^{T}\mathopen{}\mathclose{{}\left(\eta
u_{\rho}^{2}(1,s)+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u_{\rho}^{2\delta+2}(1,s)}\right)ds+\frac{2\alpha^{2}}{\nu}\int_{0}^{T}\|u_{\rho}(s)\|_{\mathrm{L}^{\infty}}^{2\delta}ds}\right\\}$
$\displaystyle\leq
C\mathopen{}\mathclose{{}\left(\|u_{0}\|_{\mathrm{L}^{2}},\alpha,\beta,\gamma,\delta,\mu,\nu,T}\right),$
(2.60)

for all $t\in[0,T]$ and $\delta=1$. Once again Agmon’s inequality leads to

$\displaystyle\sup_{t\in[0,T]}\|u_{\rho}(t)\|_{\mathrm{L}^{\infty}}\leq\sup_{t\in[0,T]}\|u_{\rho}(t)\|_{\mathrm{H}^{1}}^{1/2}\sup_{t\in[0,T]}\|u_{\rho}(t)\|_{\mathrm{L}^{2}}^{1/2}\leq
C,$ (2.61)

where $C$ is independent of $\rho$. Therefore, for $\rho\geq C$,
$f_{\rho}(y)=f(y)$, and hence $u=u_{\rho}$ is a solution to (2.3) on $[0,T]$.
In this way, one can find the standard existence result
$u\in\mathrm{C}\mathopen{}\mathclose{{}\left([0,\infty);\mathrm{D}(\mathscr{A}_{\rho})}\right)\cap\mathrm{C}^{1}\mathopen{}\mathclose{{}\left((0,\infty);\mathrm{L}^{2}(0,1)}\right)\subset\mathrm{C}\mathopen{}\mathclose{{}\left([0,\infty);\mathrm{H}^{3}(0,1)}\right)\cap\mathrm{C}^{1}\mathopen{}\mathclose{{}\left((0,\infty);\mathrm{L}^{2}(0,1)}\right)$
for the solution of the generalized Korteweg-de Vries-Burgers-Huxley equation
(1.1)-(1.2) with the boundary control (1.3). The case of $\delta=2$ can be
established in a similar way.

Let us now prove the uniqueness for $\delta=1,2$. Let $u_{1}$ and $u_{2}$ be
two solutions of (2.3) on $[0,T]$. Then $u=u_{1}-u_{2}$ satisfies the
following energy equality:

$\|u(t)\|_{\mathrm{L}^{2}}^{2}=\|u_{0}\|_{\mathrm{L}^{2}}^{2}-2\int_{0}^{t}\langle\mathscr{A}(u_{1})(s)-\mathscr{A}(u_{2})(s),u_{1}(s)-u_{2}(s)\rangle
ds,$ (2.62)

for all $t\in[0,T]$. The term
$\langle\mathscr{A}(u_{1})-\mathscr{A}(u_{2}),u_{1}-u_{2}\rangle$ can be
estimated similar way as in (2.2) except for the term
$\langle(f(u_{1}))_{x}-(f(u_{2}))_{x},u_{1}-u_{2}\rangle$. We estimate it
using Hölder’s, Agmon’s and Young’s inequalities as

$\displaystyle\langle(f(u_{1}))_{x}-(f(u_{2}))_{x},u_{1}-u_{2}\rangle$
$\displaystyle=\langle
u_{1}^{\delta}u_{1,x}-u_{2}^{\delta}u_{2,x},u_{1}-u_{2}\rangle$
$\displaystyle=\langle(u_{1}^{\delta}-u_{2}^{\delta})u_{1,x},u_{1}-u_{2}\rangle+\langle
u_{2}^{\delta}(u_{1,x}-u_{2,x}),u_{1}-u_{2}\rangle$
$\displaystyle\leq\delta\|(\theta
u_{1}+(1-\theta)u_{2})^{\delta-1}\|_{\mathrm{L}^{2}}\|u_{1,x}\|_{\mathrm{L}^{2}}\|u_{1}-u_{2}\|_{\mathrm{L}^{\infty}}^{2}$
$\displaystyle\quad+\|u_{2}\|_{\mathrm{L}^{\infty}}^{\delta}\|u_{1,x}-u_{2,x}\|_{\mathrm{L}^{2}}\|u_{1}-u_{2}\|_{\mathrm{L}^{2}}$
$\displaystyle\leq\frac{\nu}{2}\|u_{x}\|_{\mathrm{L}^{2}}^{2}+\frac{\delta^{2}}{\nu}\mathopen{}\mathclose{{}\left(\|u_{1}\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}+\|u_{2}\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}}\right)\|u_{1,x}\|_{\mathrm{L}^{2}}^{2}\|u\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\quad+\frac{1}{\nu}\|u_{2}\|_{\mathrm{L}^{\infty}}^{2\delta}\|u\|_{\mathrm{L}^{2}}^{2}.$
(2.63)

Therefore, from (2.62), we infer

$\displaystyle\|u(t)\|_{\mathrm{L}^{2}}^{2}+\nu\int_{0}^{t}\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds+\frac{\beta}{2^{2\delta}}\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds$
$\displaystyle\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}+2^{2\delta}\beta(1+\gamma)^{2}(\delta+1)^{2}\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2}}^{2}ds+\frac{2}{\nu}\int_{0}^{t}\|u_{2}(s)\|_{\mathrm{L}^{\infty}}^{2\delta}\|u(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\quad+\frac{2\delta^{2}}{\nu}\int_{0}^{t}\mathopen{}\mathclose{{}\left(\|u_{1}(s)\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}+\|u_{2}(s)\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}}\right)\|u_{1,x}(s)\|_{\mathrm{L}^{2}}^{2}\|u(s)\|_{\mathrm{L}^{2}}^{2}ds,$
(2.64)

for all $t\in[0,T]$. An application of Gronwall’s inequality in (2.2) provides

$\displaystyle\|u(t)\|_{\mathrm{L}^{2}}^{2}+\nu\int_{0}^{t}\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}e^{2^{2\delta}\beta(1+\gamma)^{2}(\delta+1)^{2}T}\exp\mathopen{}\mathclose{{}\left(\frac{2}{\nu}\sup_{t\in[0,T]}\|u_{2}(t)\|_{\mathrm{L}^{2}}^{2\delta}\int_{0}^{T}\|u_{2}(t)\|_{\mathrm{H}^{1}}^{2\delta}dt}\right)$
$\displaystyle\quad\times\exp\mathopen{}\mathclose{{}\left(\frac{2\delta^{2}}{\nu}\sup_{t\in[0,T]}\mathopen{}\mathclose{{}\left(\|u_{1}(t)\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}+\|u_{2}(t)\|_{\mathrm{L}^{2(\delta-1)}}^{2(\delta-1)}}\right)\int_{0}^{T}\|u_{1,x}(t)\|_{\mathrm{L}^{2}}^{2}dt}\right),$
(2.65)

for all $t\in[0,T]$. Since $\delta=1,2$, the right hand side of (2.2) is
finite even for $u_{0}\in\mathrm{L}^{2}(0,1)$ and the uniqueness of solutions
follows since $u_{0}=0$. ∎

##  3\. Stabilization

The aim of this section is to establish the $\mathrm{L}^{2}$-,
$\mathrm{H}^{1}$\- and pointwise exponential stabilization results for the
generalized KdV-Burgers-Huxley equation (1.1)-(1.2) subject to the controls
(1.4) and (1.3).

######  Theorem 3.1.

For

$\nu>\frac{\beta}{4}(1-\gamma)^{2},$

the generalized KdV-Burgers-Huxley equation (1.1)-(1.2) is exponentially
stable in $\mathrm{L}^{2}(0,1)$ under the control laws (1.3), (1.4),
respectively for $\delta=1,2$.

###### Proof.

Let $\mathcal{V}(\cdot)$ be a Lyapunov function defined by

$\displaystyle\mathcal{V}(t)=\frac{1}{2}\int_{0}^{1}u^{2}(x,t)dx,\ t\geq 0.$
(3.1)

Taking the derivative with respect to $t$, performing an integration by parts
and then using (2.4) and (2.4), for $\delta=1,2$, respectively, we deduce

$\displaystyle\dot{\mathcal{V}}(t)$
$\displaystyle=\int_{0}^{1}u(x,t)u_{t}(x,t)dx$
$\displaystyle=\int_{0}^{1}u(x,t)[\nu u_{xx}(x,t)-\mu u_{xxx}-\alpha
u^{\delta}(x,t)u_{x}(x,t)+\beta
u(x,t)(1-u^{\delta}(x,t))(u^{\delta}(x,t)-\gamma)]dx$
$\displaystyle=-\nu\int_{0}^{1}u_{x}^{2}(x,t)dx+\nu u(1,t)u_{x}(1,t)-\mu
u(1,t)u_{xx}(1,t)+\frac{\mu}{2}u_{x}^{2}(1,t)-\frac{\mu}{2}u_{x}^{2}(0,t)$
$\displaystyle\quad-\frac{\alpha}{\delta+2}u^{\delta+2}(1,t)+\beta(1+\gamma)\int_{0}^{1}u^{\delta+2}(x,t)dx-\beta\gamma\int_{0}^{1}u^{2}(x,t)dx-\beta\int_{0}^{1}u^{2(\delta+1)}(x,t)dx$
$\displaystyle\leq-2\mathopen{}\mathclose{{}\left(\nu-\frac{\beta}{4}(1-\gamma)^{2}}\right)\mathcal{V}(t),$
(3.2)

for all $t>0$. Thus, we deduce
$\|u(t)\|_{\mathrm{L}^{2}}\leq\|u_{0}\|_{\mathrm{L}^{2}}e^{-\zeta t},$ for all
$t\geq 0$ and since
$\zeta=\mathopen{}\mathclose{{}\left(\nu-\frac{\beta}{4}(1-\gamma)^{2}}\right)>0$,
$\|u(t)\|_{\mathrm{L}^{2}}$ converges to zero exponentially as $t\to\infty$. ∎

By controlling the convective term using diffusion and reaction terms, under
further restrictions on $\nu$, for any $\delta\in[1,\infty)$, one can obtain
$\mathrm{L}^{2}$-stabilization of generalized KdV-Burgers-Huxley equation by
using an another control law.

######  Theorem 3.2.

For any $\delta\in[1,\infty)$ and for

$\displaystyle\nu>\max\mathopen{}\mathclose{{}\left\\{\frac{\alpha^{2}}{2\beta(2\delta+1)},\frac{\beta}{2}(1+\gamma^{2})+\frac{\alpha^{2}}{2\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}}\right\\},$
(3.3)

the generalized KdV-Burgers-Huxley equation (1.1)-(1.2) is exponentially
stable in $\mathrm{L}^{2}(0,1)$ under the control law

$u(0,t)=0,\ u_{x}(1,t)=-u(1,t),\ u_{xx}(1,t)=u(1,t).$ (3.4)

###### Proof.

For $\nu>\frac{\alpha^{2}}{2\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}$,
the well-posedness of the problem (1.1)-(1.2) for any $\delta\in[1,\infty)$
under the control law (3.4) is not difficult. Note that the control given in
(3.4) does not cause any additional difficulty in establishing the estimate
(2.61). Using Lemma A.3, one can estimate
$-\frac{\alpha}{\delta+2}u^{\delta+2}(1)$ as

$\displaystyle-\frac{\alpha}{\delta+2}u^{\delta+2}(1)$
$\displaystyle\leq\frac{\alpha}{\delta+2}\|u\|_{\mathrm{L}^{\infty}}^{\delta+2}\leq\frac{\alpha}{\delta+2}\|u_{x}\|_{\mathrm{L}^{2}}\|u\|_{\mathrm{L}^{2(\delta+1)}}^{\delta+1}$
$\displaystyle\leq\frac{\theta\beta}{2}\|u\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}+\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}\|u_{x}\|_{\mathrm{L}^{2}}^{2},$
(3.5)

for some $0<\theta\leq 1$. But we know that

$\displaystyle\beta(1+\gamma)(u^{\delta+1},u)\leq\beta(1+\gamma)\|u\|_{\mathrm{L}^{2(\delta+1)}}^{\delta+1}\|u\|_{\mathrm{L}^{2}}\leq\frac{\theta\beta}{2}\|u\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}+\frac{\beta(1+\gamma)^{2}}{2\theta}\|u\|_{\mathrm{L}^{2}}^{2}.$
(3.6)

In fact, for the control given in (3.4), the equality (2.4) yields

$\displaystyle\langle\mathscr{A}(u),u\rangle\geq\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}}\right)\|u_{x}\|_{\mathrm{L}^{2}}^{2}+\mathopen{}\mathclose{{}\left(\nu+\frac{\mu}{2}}\right)u^{2}(1)-\frac{\beta(1+\gamma^{2})}{2}\|u\|_{\mathrm{L}^{2}}^{2}.$
(3.7)

Therefore, a calculation similar to (2.2) yields

$\displaystyle\|u(t)\|_{\mathrm{L}^{2}}^{2}+2\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}}\right)\int_{0}^{t}\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds+2\mathopen{}\mathclose{{}\left(\nu+\frac{\mu}{2}}\right)\int_{0}^{t}u^{2}(1,s)ds$
$\displaystyle\quad+2\beta(1-\theta)\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds$
$\displaystyle\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}+\frac{\beta(1+\gamma^{2})}{\theta}\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2}}^{2}ds,$
(3.8)

for all $t\in[0,T]$ and some $0<\theta<1$. Therefore, for
$\nu>\frac{\alpha^{2}}{2\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}$, an
application of Gronwall’s inequality yields

$\displaystyle\|u(t)\|_{\mathrm{L}^{2}}^{2}+2\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}}\right)\int_{0}^{t}\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds+2\mathopen{}\mathclose{{}\left(\nu+\frac{\mu}{2}}\right)\int_{0}^{t}u^{2}(1,s)ds$
$\displaystyle\quad+2\beta(1-\theta)\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2}e^{\frac{\beta(1+\gamma^{2})T}{\theta}},$
(3.9)

for all $t\in[0,T]$ and some $0<\theta<1$.

Let us now establish an estimate similar to (2.2). Calculations similar to
(2.2) and (2.2) provide

$\displaystyle-\int_{0}^{1}u_{t}(t,x)$ $\displaystyle
u_{xx}(t,x)dx=\frac{1}{2}\frac{d}{dt}\mathopen{}\mathclose{{}\left[\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}+u^{2}(1,t)}\right],$

and

$\displaystyle(\mathscr{A}(u),-u_{xx})$
$\displaystyle\geq\nu\|u_{xx}\|_{\mathrm{L}^{2}}^{2}-\frac{\mu}{2}u^{2}(1)-\alpha(u^{\delta}u_{x},u_{xx})+\beta(1+\gamma)(u^{\delta+1},u_{xx})$
$\displaystyle\quad+\beta\gamma\|u_{x}\|_{\mathrm{L}^{2}}^{2}+\beta
u^{2}(1)+\beta(2\delta+1)\|u^{\delta}u_{x}\|_{\mathrm{L}^{2}}^{2}+\beta
u^{2\delta+2}(1).$

We calculate the terms $-\alpha(u^{\delta}u_{x},u_{xx})$ and
$\beta(1+\gamma)(u^{\delta+1},u_{xx})$ as

$\displaystyle-\alpha(u^{\delta}u_{x},u_{xx})$
$\displaystyle\leq\alpha\|u^{\delta}u_{x}\|_{\mathrm{L}^{2}}\|u_{xx}\|_{\mathrm{L}^{2}}\leq\frac{\theta\nu}{2}\|u_{xx}\|_{\mathrm{L}^{2}}^{2}+\frac{\alpha^{2}}{2\theta\nu}\|u^{\delta}u_{x}\|_{\mathrm{L}^{2}}^{2},$
$\displaystyle\beta(1+\gamma)(u^{\delta+1},u_{xx})$
$\displaystyle\leq\beta(1+\gamma)\|u\|_{\mathrm{L}^{2(\delta+1)}}^{\delta+1}\|u_{xx}\|_{\mathrm{L}^{2}}\leq\frac{\theta\nu}{2}\|u_{xx}\|_{\mathrm{L}^{2}}^{2}+\frac{\beta^{2}(1+\gamma)^{2}}{2\theta\nu}\|u\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)},$

for some $0<\theta\leq 1$. Substituting the above estimates in (2.57), we
deduce

$\displaystyle\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}+u^{2}(1,t)+2(1-\theta)\nu\int_{0}^{t}\|u_{xx}(s)\|_{\mathrm{L}^{2}}^{2}ds+2\beta\int_{0}^{t}u^{2(\delta+1)}(1,s)ds$
$\displaystyle\quad+2\mathopen{}\mathclose{{}\left(\beta(2\delta+1)-\frac{\alpha^{2}}{2\theta\nu}}\right)\int_{0}^{t}\|u^{\delta}(s)u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds$
$\displaystyle\leq\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+u^{2}_{0}(1)+\mu\int_{0}^{t}u^{2}(1,s)ds+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}\int_{0}^{t}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds,$
(3.10)

for all $t\in[0,T]$ and some $0<\theta<1$. For
$\nu>\frac{\alpha^{2}}{2\beta(2\delta+1)}$, an application of Gronwall’s and
Agmon’s inequalities in (3) yields

$\displaystyle\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}\leq\mathopen{}\mathclose{{}\left\\{\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+2\|u_{0}\|_{\mathrm{L}^{2}}\|u_{0,x}\|_{\mathrm{L}^{2}}+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}\int_{0}^{T}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds,}\right\\}e^{\mu
T},$ (3.11)

for all $t\in[0,T]$. Therefore an application of (3) and Agmon’s inequality
imply (2.61) holds for all $\delta\in[1,\infty)$ and the proof of existence is
completed. The uniqueness of strong solutions follow immediately from the
estimate (2.2) as $u\in\mathrm{L}^{\infty}((0,T)\times(0,1))$.

Finally, for the control law given in (3.4), we infer from (3) under the
condition (3.3) that

$\displaystyle\dot{\mathcal{V}}(t)$
$\displaystyle\leq-2\mathopen{}\mathclose{{}\left(\nu-\frac{\beta}{2\theta}(1+\gamma^{2})-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}}\right)\mathcal{V}(t),$
(3.12)

for all $t>0$ and some $0<\theta<1$. Therefore, one can obtain the exponential
stabilization in $\mathrm{L}^{2}(0,1)$ for any $\delta\in[1,\infty)$. ∎

Under the condition (3.3), let us establish the $\mathrm{H}^{1}$-stabilization
of generalized KdV-Burgers-Huxley equation with the control law (3.4).

######  Theorem 3.3.

Under the condition (3.3), global exponential stability in the
$\mathrm{H}^{1}$-norm sense holds for the generalized KdV-Burgers-Huxley
equation (1.1)-(1.2) under the control law (3.4), that is,

$\displaystyle\|u(t)\|_{\mathrm{H}^{1}}^{2}\leq
2\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}\leq
2\mathopen{}\mathclose{{}\left\\{\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+u_{0}^{2}(1)+\mathopen{}\mathclose{{}\left[\frac{1}{\varrho}(\varrho+2(\mu+1))+\frac{\beta(1+\gamma)^{2}}{2\theta(1-\theta)\nu}}\right]\|u_{0}\|_{\mathrm{L}^{2}}^{2}}\right\\}e^{-\frac{\varrho}{2}t},$
(3.13)

for all $t\geq 0$ and $u_{0}\in\mathrm{H}^{1}(0,1)$, where

$\displaystyle\varrho=\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}-\frac{\beta(1+\gamma^{2})}{2\theta}}\right)>0,$
(3.14)

for some $0<\theta<1$.

###### Proof.

A calculation similar to (3) yields

$\displaystyle\frac{d}{dt}\Psi(t)+2(1-\theta)\nu\|u_{xx}(t)\|_{\mathrm{L}^{2}}^{2}+2\beta
u^{2(\delta+1)}(1,t)+2\mathopen{}\mathclose{{}\left(\beta(2\delta+1)-\frac{\alpha^{2}}{2\theta\nu}}\right)\|u^{\delta}(t)u_{x}(t)\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\leq\mu
u^{2}(1,t)+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}\|u(t)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)},$

where $\Psi(t)=\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}+u^{2}(1,t)$ for a.e.
$t\in[0,T]$. Let us multiply the above equation by $e^{\frac{\varrho}{2}t}$ to
deduce

$\displaystyle\frac{d}{dt}[e^{\frac{\varrho}{2}t}\Psi(t)]+2(1-\theta)\nu
e^{\frac{\varrho}{2}t}\|u_{xx}(t)\|_{\mathrm{L}^{2}}^{2}+2\beta
e^{\frac{\varrho}{2}t}u^{2(\delta+1)}(1,t)$
$\displaystyle\quad+2\mathopen{}\mathclose{{}\left(\beta(2\delta+1)-\frac{\alpha^{2}}{2\theta\nu}}\right)e^{\frac{\varrho}{2}t}\|u^{\delta}(t)u_{x}(t)\|_{\mathrm{L}^{2}}^{2}$
$\displaystyle\leq\frac{\varrho}{2}e^{\frac{\varrho}{2}t}\Psi(t)+\mu
e^{\frac{\varrho}{2}t}u^{2}(1,t)+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}e^{\frac{\varrho}{2}t}\|u(t)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}$
$\displaystyle\leq
e^{\frac{\varrho}{2}t}\mathopen{}\mathclose{{}\left[(\varrho+2(\mu+1))\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}\|u(t)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}}\right],$

where $\varrho$ is defined in (3.14), and we have used the fact that
$u^{2}(1)\leq\|u\|_{\mathrm{L}^{\infty}}^{2}\leq
2\|u\|_{\mathrm{L}^{2}}\|u_{x}\|_{\mathrm{L}^{2}}\leq
2\|u_{x}\|_{\mathrm{L}^{2}}^{2}$. Integrating the above inequality from $0$ to
$t$, we find

$\displaystyle
e^{\frac{\varrho}{2}t}\Psi(t)\leq\Psi(0)+\int_{0}^{t}e^{\frac{\varrho}{2}s}\mathopen{}\mathclose{{}\left[(\varrho+2(\mu+1))\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}+\frac{\beta^{2}(1+\gamma)^{2}}{\theta\nu}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}}\right]ds,$
(3.15)

for all $t\geq 0$.

A calculation similar to (3) yields

$\displaystyle\frac{d}{dt}\|u(t)\|_{\mathrm{L}^{2}}^{2}+2\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}-\frac{\beta(1+\gamma^{2})}{2\theta}}\right)\|u_{x}(t)\|_{\mathrm{L}^{2}}^{2}+2\beta(1-\theta)\|u(t)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}\leq
0,$

for a.e. $t\in[0,T]$. Therefore, by an application of the Poincaré inequality
and variation of constants formula in the above inequality implies, for
$\varrho=\mathopen{}\mathclose{{}\left(\nu-\frac{\alpha^{2}}{2\theta\beta(\delta+2)^{\frac{2(\delta+1)}{\delta+2}}}-\frac{\beta(1+\gamma^{2})}{2\theta}}\right)>0$,

$\displaystyle\|u(t)\|_{\mathrm{L}^{2}}^{2}+\varrho\int_{0}^{t}e^{\varrho
s}\|u_{x}(s)\|_{\mathrm{L}^{2}}^{2}ds+2\beta(1-\theta)\int_{0}^{t}e^{\varrho
s}\|u(s)\|_{\mathrm{L}^{2(\delta+1)}}^{2(\delta+1)}ds\leq\|u_{0}\|_{\mathrm{L}^{2}}^{2},$
(3.16)

for all $t\geq 0$. Substituting (3.16) in (3.15), we obtain our required
result as

$\displaystyle\Psi(t)\leq\mathopen{}\mathclose{{}\left\\{\Psi(0)+\mathopen{}\mathclose{{}\left[\frac{1}{\varrho}(\varrho+2(\mu+1))+\frac{\beta(1+\gamma)^{2}}{2\theta(1-\theta)\nu}}\right]\|u_{0}\|_{\mathrm{L}^{2}}^{2}}\right\\}e^{-\frac{\varrho}{2}t},$
(3.17)

for some $0<\theta<1$. Thus the exponential stabilization result in
$\mathrm{H}^{1}(0,1)$ (see (3.13)) follows immediately. ∎

Finally, an application of Agmon’s inequality (Lemma A.2) yields the following
result on the pointwise exponential convergence:

######  Theorem 3.4.

Consider the generalized KdV-Burgers-Huxley equation (1.1)-(1.2) under the
control law (3.4). If $u_{0}\in\mathrm{H}^{1}(0,1)$ and $\nu$ satisfies the
condition (3.3), then we have

$\displaystyle\max_{x\in[0,1]}|u(x,t)|\leq
4\mathopen{}\mathclose{{}\left\\{\|u_{0,x}\|_{\mathrm{L}^{2}}^{2}+u_{0}^{2}(1)+\mathopen{}\mathclose{{}\left[\frac{1}{\varrho}(\varrho+2(\mu+1))+\frac{\beta(1+\gamma)^{2}}{2\theta(1-\theta)\nu}}\right]\|u_{0}\|_{\mathrm{L}^{2}}^{2}}\right\\}e^{-\frac{\varrho}{2}t},$
(3.18)

for all $t\geq 0$, where $\varrho$ is defined in (3.14).

##  4\. Numerical Results

In our numerical studies of the third-order differential system (1.1)-(1.2),
we implement a modified Chebyshev Collocation Method given by Kosloff and Tal-
Ezer [26] to ensure stability in time steps. This method derives modified
grids, which are then transformed to fit within our desired $[0,1]$ interval.
Consequently, we compute the corresponding modified Chebyshev differentiation
matrices, following the approach discussed in [33]. For spatial
discretization, these matrices are applied over the transformed Chebyshev
nodes, and we utilize the backward Euler method for time discretization. A
MATLAB program is developed in accordance with these methodologies and
effectively solved both uncontrolled and controlled versions of the
Generalized Korteweg-de Vries-Burgers-Huxley (GKdVBH) equation (1.1)-(1.2) and
controls outlined in (1.3), (1.4), (3.4) and Remark 1.1. Over transformed
nodes, we apply the Gauss-Lobatto quadrature for numerical integration, and
thus, the $L^{2}$-norm of the solution $u(\cdot,t)$ at each time step is
plotted against time $t$.

Figure 1. Time evolution of uncontrolled GKdVBH equation for
$\nu=1,\alpha=1,\beta=1,\delta=1,\gamma=0.5$ and $u_{0}(x)=\sin(\pi x)$.

Figure 2. Time evolution of uncontrolled GKdVBH equation for
$\nu=1,\alpha=1,\beta=1,\delta=2,\gamma=0.5$ and $u_{0}(x)=\sin(\pi x)$.

Figure 3. Time evolution of uncontrolled GKdVBH equation for
$\nu=1,\alpha=1,\beta=1,\delta=3,\gamma=0.5$ and $u_{0}(x)=\sin(\pi x)$.

Figure 1, 2 and 3 depicts the temporal evolution of the solution $u(x,t)$ in
the absence of control, characterized by the parameters $\alpha=1$, $\beta=1$,
$\gamma=0.5$, $\nu=1$, $\mu=0.1$, and the initial condition $u(x,0)=\sin(\pi
x)$ for $\delta=1$, $\delta=2$ and $\delta=3$ respectively. When the control
delineated in (1.3) ($\delta=1$) is implemented with a parameter $\eta=1$,
figure 4 elucidates that the solution $u(x,t)$ converges towards the zero
state. Similarly, when a control law (1.4) ($\delta=2$) is applied with
$\eta=1$, figure 5 demonstrates that $u(x,t)$ approaches the desired quiescent
state. These observations validate Theorem 3.1.

Figure 4. Time evolution of controlled GKdVBH equation for
$\nu=1,\alpha=1,\beta=1,\delta=1,\gamma=0.5,\eta=1,\mu=0.1$ and
$u_{0}(x)=\sin(\pi x)$.

Figure 5. Time evolution of uncontrolled GKdVBH equation for
$\nu=1,\alpha=1,\beta=1,\delta=2,\gamma=0.5,\eta=1,\mu=0.1$ and
$u_{0}(x)=\sin(\pi x)$.

Figures 6 and 7 describe the asymptotic stability achieved by implementing the
controls as outlined in Remark 1.1. Specifically, along with condition (1.7),
defined by the function $g(\cdot)$ are
$\frac{1}{\mu}\mathopen{}\mathclose{{}\left(\eta+\frac{\alpha^{2}}{\eta(\delta+2)^{2}}u^{2\delta}(1,t)}\right)u(1,t)\
(\text{for }\ \delta=1)\ \text{ and }\ \frac{\eta}{\mu}u(1,t)\ (\text{for }\
\delta=2)$ respectively.

Figure 8 validates Theorem 3.2, which shows exponential stability of the
control defined in (3.4) $(\text{for }\ \delta=3)$.

Moreover, in studying the convergence rates for controls, the logarithms of
the norms plotted over time offer enhanced visual clarity. Linear segments in
these plots are indicative of exponential decay, with the slope of these
segments directly corresponding to the rate of decay. A less steep slope
signifies a slower rate of decay. Note that we have taken $\nu=1$ so that
condition (3.3) can be met along with the condition in Theorem 3.1, thus,
facilitating proper comparison. Figure 9 and 10 conclusively demonstrates that
the convergence rates for control (1.7) is notably slower than controls (3.4)
and (1.3) $\text{for }\ \delta=1\ \text{and }\ \delta=2,$ respectively,
corroborating the observations made in Remark 1.1. Figure 11 shows that
$\max\limits_{x\in[0,1]}|u(x,t)|$ decays to zero, as in accordance with
Theorem 3.4.

Figure 6. Time evolution of GKdVBH equation under control defined in (1.7) for
$\nu=1,\alpha=1,\beta=1,\delta=1,\gamma=0.5,\eta=1,\mu=0.1$ and
$u_{0}(x)=\sin(\pi x)$.

Figure 7. Time evolution of GKdVBH equation under control defined in (1.7) for
$\nu=1,\alpha=1,\beta=1,\delta=2,\gamma=0.5,\eta=1,\mu=0.1$ and
$u_{0}(x)=\sin(\pi x)$.

Figure 8. Time evolution of GKdVBH equation under control defined in (3.4) for
$\nu=1,\alpha=1,\beta=1,\delta=3,\gamma=0.5,\eta=1,\mu=0.1$ and
$u_{0}(x)=\sin(\pi x)$.

Figure 9. Time evolution of the log $\|u(\cdot,t)\|_{L^{2}}$ under the control
(1.3), (1.7) and (3.4) for $\delta=1$.

Figure 10. Time evolution of the log $\|u(\cdot,t)\|_{L^{2}}$ under the
control (1.4), (1.7) and (3.4) for $\delta=2$.

Figure 11. Time evolution of the $\max_{x\in[0,1]}|u(x,t)|$ under the control
(3.4) for $\delta=1$.

##  Appendix A Some Useful Inequalities

######  Lemma A.1 (Poincaré’s inequality).

For any $w\in\mathrm{C}^{1}([0,1])$, we have

$\displaystyle\int_{0}^{1}w^{2}(x)dx\leq\mathopen{}\mathclose{{}\left(\int_{0}^{1}w_{x}^{2}(x)dx}\right).$
(A.1)

###### Proof.

For any $w\in\mathrm{C}^{1}[0,1]$, we know that

$\displaystyle w(x)=\int_{0}^{x}w_{x}(\zeta)d\zeta,$

so that

$\displaystyle
w^{2}(x)=\mathopen{}\mathclose{{}\left(\int_{0}^{x}w_{x}(\zeta)d\zeta}\right)^{2}\leq\mathopen{}\mathclose{{}\left(\int_{0}^{1}|w_{x}(\zeta)|^{2}d\zeta}\right).$

Ingratiating the above inequality from $0$ to $1$, we have the Poincaré
inequality (A.1). By a density argument, the result (A.1) holds true for all
$w\in\mathrm{H}^{1}(0,1)$. ∎

######  Lemma A.2 (Agmon’s inequality).

For any $w\in\mathrm{C}^{1}([0,1])$, the following inequality holds:

$\displaystyle\max_{x\in[0,1]}|w(x)|\leq\sqrt{2}\|w\|_{\mathrm{L}^{2}}^{1/2}\|w_{x}\|_{\mathrm{L}^{2}}^{1/2}.$
(A.2)

###### Proof.

An application of Hölder’s inequality yields

$\displaystyle w^{2}(x)$
$\displaystyle=2\int_{0}^{x}w(\zeta)w_{x}(\zeta)d\zeta\leq
2\mathopen{}\mathclose{{}\left(\int_{0}^{1}|w(\zeta)|^{2}d\zeta}\right)^{1/2}\mathopen{}\mathclose{{}\left(\int_{0}^{1}|w_{x}(\zeta)|^{2}d\zeta}\right)^{1/2},$
(A.3)

for all $x\in[0,1]$. Therefore, from (A.3), we infer

$\displaystyle\max\limits_{x\in[0,1]}|w(x)|^{2}\leq
2\|w\|_{\mathrm{L}^{2}}\|w_{x}\|_{\mathrm{L}^{2}},$ (A.4)

and (A.2) follows. By a density argument, the result (A.2) holds true for all
$w\in\mathrm{H}^{1}(0,1)$. ∎

######  Lemma A.3.

For any $w\in\mathrm{C}^{1}([0,1])$, the following inequality holds:

$\displaystyle\max_{x\in[0,1]}|w(x)|\leq(\delta+2)^{\frac{1}{\delta+2}}\|w\|_{\mathrm{L}^{2(\delta+1)}}^{\frac{\delta+1}{\delta+2}}\|w_{x}\|_{\mathrm{L}^{2}}^{\frac{1}{\delta+2}}.$
(A.5)

###### Proof.

An application of Hölder’s inequality yields

$\displaystyle w^{\delta+2}(x)$
$\displaystyle=\int_{0}^{x}\partial_{\zeta}w^{\delta+2}(\zeta)d\zeta=(\delta+2)\int_{0}^{x}w^{\delta+1}(\zeta)w_{x}(\zeta)d\zeta$
$\displaystyle\leq(\delta+2)\mathopen{}\mathclose{{}\left(\int_{0}^{1}|w(\zeta)|^{2(\delta+1)}d\zeta}\right)^{1/2}\mathopen{}\mathclose{{}\left(\int_{0}^{1}|w_{x}(\zeta)|^{2}d\zeta}\right)^{1/2},$
(A.6)

for all $x\in[0,1]$. Therefore, from (A), we infer

$\displaystyle\max\limits_{x\in[0,1]}|w(x)|^{\delta+2}\leq(\delta+2)\|w\|_{\mathrm{L}^{2(\delta+1)}}^{\delta+1}\|w_{x}\|_{\mathrm{L}^{2}},$
(A.7)

and (A.5) follows. By a density argument, the result (A.5) holds true for all
$w\in\mathrm{H}^{1}(0,1)$. ∎

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