\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 127, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/127\hfil Optimal control problem]
{Optimal control problem for a sixth-order Cahn-Hilliard
equation with nonlinear diffusion}

\author[C. Liu, Z. Wang\hfil EJDE-2012/127\hfilneg]
{Changchun Liu, Zhao Wang}  % in alphabetical order

\address{Changchun Liu \newline
Department of Mathematics, Jilin University,
Changchun 130012, China}
\email{liucc@jlu.edu.cn}

\address{Zhao Wang\hfill\break
Department of Mathematics, Jilin University,
Changchun 130012, China}
\email{wangzhao2717@163.com}

\thanks{Submitted March 1, 2012. Published August 14, 2012.}
\subjclass[2000]{49J20, 35K35, 35K55}
\keywords{Cahn-Hilliard equation; existence; optimal control; optimal solution}

\begin{abstract}
 In this article, we study the initial-boundary-value problem for a
 sixth-order Cahn-Hilliard type equation
 \begin{gather*}
 u_t=D^2\mu, \\
 \mu=\gamma D^4u-a(u)D^2u-\frac{a'(u)}2|D u|^2+f(u)+ku_t,
 \end{gather*}
 which describes the separation properties of oil-water mixtures,
 when a substance enforcing the mixing of the phases is added. The
 optimal control of the sixth order Cahn-Hilliard type equation
 under boundary condition is given and the existence of optimal
 solution to the sixth order Cahn-Hilliard type equation is proved.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

We consider the  equation
\begin{equation} \label{1-1}
u_t=D^2 \big[\gamma D^4u-a(u)D^2u-\frac{a'(u)}2|D u|^2+f(u)+ku_t\big],
\end{equation}
in $\Omega\times(0, T)$, where $\Omega$ is a bounded subset in
$\mathbb{R}$ and $\gamma>0$ with the initial and boundary conditions
\begin{gather} \label{1-2}
u(x, 0)=u_0,\quad\text{in }\Omega,\\
u(x, t)=D^2u(x, t)=D^4u(x,t)=0,\quad\text{on }\partial\Omega.\label{1-3}
\end{gather}
The function $f(u)$ stands for the derivative of a  potential $F(u)$
 with $F(u)$, $a(u)$  approximated, respectively, by a sixth and a
second order polynomial
\begin{gather} \label{1-4}
F(u)=\int_0^u f(s)ds=\gamma_1(u+1)^2(u^2+h_0)(u-1)^2,\\
a(u)=a_2u^2+a_0, \label{1-5}
\end{gather}
where $\gamma_1>0$, $a_2>0$.

The model \eqref{1-1} describes the separation properties of
oil-water mixtures, when a substance enforcing the mixing of the
phases (a surfactant) is added. G. Schimperna et al. \cite{GI} studied  the equation \eqref{1-1}
with logarithmic potential
$$
F(r)=(1-r)\log(1-r)+(1+r)\log(1+r)-\frac\lambda2r^2,\quad \lambda>0.
$$
They investigated the behavior of the solutions to the sixth order
system as the parameter $\gamma$  tended to $0$, the uniqueness and
regularization properties of the solutions have been discussed.


When $k=0$,   equation
\eqref{1-1} is the sixth order equation which describes dynamics of
phase transitions in ternary oil-water-surfactant systems
\cite{GM1, GM2, GJ}. The surfactant which has a character that one part of it
is hydrophilic and the other lipophilic is called amphiphile. In the
system, almost pure oil, almost pure water and microemulsion which
consists of a homogeneous, isotropic mixture of oil and water can
coexist in equilibrium. Paw{\l}ow and Zaj\c{a}czkowski \cite{IW}
proved that the problem \eqref{1-1}-\eqref{1-5} with $k=0$ under
consideration is well posed in the sense that it
admits a unique global smooth solution which depends continuously on the
initial datum. Liu \cite{Liu} studied the equation
$$
\frac{\partial u}{\partial t}-\operatorname{div}
[m(u)(k\nabla\Delta^2 u+\nabla(-a(u)\Delta u-\frac {a'(u)}2 |\nabla u|^2+h(u))]
=0,
$$
and  proved the existence of classical solutions for two dimensions.

In past decades, the optimal control of distributed parameter system
had been received much more attention in academic field. A wide
spectrum of problems in applications can be solved by methods of
optimal control, such as chemical engineering and vehicle dynamics.
Modern optimal control theories and applied models are not only
represented by ODE, but also by PDE.
Kunisch and Volkwein \cite{KS} solved
open-loop and closed-loop optimal control problems for the Burgers
equation. Armaou and Christofides\cite{A} studied the feedback
control of Kuramto-Sivashing equation.

Recently, many authors studied the optimal control problem for the
viscous PDE, such as Tian et al. \cite{CLA}-\cite{LCD}, Zhao and Liu
\cite{ZL}.

In this article, we consider the optimal control problem for the
equation
\begin{align*}
&(u-kD^2u)_t+\frac\gamma kD^4(u-kD^2u)+D^2(a(u)D^2 u
+\frac{a'(u)}2|Du|^2) \\
=&\frac\gamma kD^4u+D^2f(u)+B^*\overline{\omega},
\end{align*}
with \eqref{1-2}-\eqref{1-5}.

When $y=u-D^2u$, we take the distributed optimal control problem
\begin{equation}
\begin{gathered}
\min \mathcal {J}(y, \overline{\omega})=
\frac12\|Cy-z\|_S^2+\frac\delta2\|\overline{\omega}\|_{L^2(0, T; Q_0)}^2,
\\
\text{s. t. } y_t+\frac\gamma kD^4y-\frac\gamma kD^4u+D^2(a(u)D^2
u+\frac{a'(u)}2|Du|^2)-D^2f(u)=B^*\overline{\omega},
\\
y(x, 0)=y_0=u_0-D^2u(x, 0),
\\
u(x, t)=D^2u(x, t)=D^4u(x, t)=0.
\end{gathered} \label{1-7}
\end{equation}
For a fixed $T > 0$, we set $\Omega = (0, 1)$ and $Q =\Omega\times(0,T)$.
 Let $Q_0\subset Q$  be an open set with positive measure.

Let $V=H_0^2(0, 1), H=L^2(0, 1), V^*=H^{-2}(0, 1)$ and
$H^*=L^2(0, 1)$ are dual spaces respectively, we have
$V\hookrightarrow H=H^*\hookrightarrow V^*$.
The extension operator $B^*\in L(L^2(0, T;  Q_0), L^2(0, T;  V^*))$
is given by
\begin{equation}
B^*q=\begin{cases}
q,& q\in Q_0,\\
0,& q\in Q/Q_0.
\end{cases}
\label{1-8}
\end{equation}
The space $W(0, T; V)$ is defined by
$$
W(0, T; V)=\{y, y\in L^2(0, T; V), y_t\in L^2(0, T; V^*)\}
$$
which is a Hilbert space endowed with common inner product.

The plan of the paper is as follows. In section 2, we prove the
existence of the weak solution  in a special space. The optimal control
is discussed in section 3, and the
existence of an optimal solution is proved.

\section{Existence of weak solutions}

Consider the  the sixth-order Cahn-Hilliard equation
\begin{equation}
\begin{aligned}
&(u-kD^2u)_t+\frac\gamma kD^4(u-kD^2u)+D^2(a(u)D^2
u+\frac{a'(u)}2|Du|^2)\\
&=\frac\gamma kD^4u+D^2f(u)+B^*\overline{\omega},
\end{aligned}\label{2-1}
\end{equation}
under the initial condition
$$
u(x, 0)=u_0,
$$
and boundary condition
$$
u(x, t)=D^2u(x, t)=D^4u(x, t)=0,
$$
where $B^*\overline{\omega}\in L^2(0, T; V^*)$ and the control item
$\overline{\omega}\in L^2(0, T; Q_0)$.

Let $y=u-kD^2u$. Then the above problem  is rewritten as
\begin{equation}\label{2-2}
\begin{gathered}
 y_t+\frac\gamma kD^4y-\frac\gamma kD^4u+D^2(a(u)D^2
u+\frac{a'(u)}2|Du|^2)-D^2f(u)=B^*\overline{\omega},
\\
y(x, 0)=y_0=u_0-D^2u_0,
\\
u(x, t)=D^2u(x, t)=D^4u(x, t)=0,
\end{gathered}
\end{equation}
with \eqref{1-4}-\eqref{1-5}.

 Now, we give the definition of the weak solution for
problem \eqref{2-2} in the space $W(0, T; V)$.

\begin{definition} \label{def2.1}\rm
A function $y(x, t)\in W(0, T; V)$ is called a weak solution to
problem \eqref{2-2}, if
\begin{align*}
&\frac{d}{dt}(y, \phi)+\frac\gamma k(D^2y,
D^2\phi)-(\frac\gamma kD^2u, D^2\phi)
\\
&+(a(u)D^2u+\frac{a'(u)}2|Du|^2, D^2\phi)-(f(u),
D^2\phi)=(B^*\overline{\omega}, \phi)_{V^*, V},
\end{align*}
for all $\phi\in V$, a.e. $t\in[0, T]$ and $y_0\in H$ are valid.
\end{definition}

\begin{theorem} \label{thm2.1}
Problem \eqref{2-2} admits a weak solution $y(x, t)\in W(0, T;V)$ 
in the interval $[0, T]$, if $B^*\overline{\omega}\in L^2(0, T;
V^*)$, $y_0\in H$ and $u_0\in H^2(\Omega)\cap H_0^1(\Omega)$.
\end{theorem}

\begin{proof}
Employ the standard Galerkin method.
The fourth-order differential operator $A=\partial^4_x$  is a linear
unbounded self-adjoint operator in $H$ with 
$D(A)=\{u|u\in H^4(\Omega), u|_{\partial\Omega}=D^2u|_{\partial\Omega}=0\}$ 
dense in $H$, where $H$ is a Hilbert space with a scalar product $(\cdot, \cdot)$
and norm $\|\cdot\|$.

There exists orthogonal basis $\{\psi_i\}$ of $H$. Let
$\{\psi_i\}_{i=1}^{\infty}$ be the eigenfunctions of the operator
$A=\partial^4_x$ with
$$
A\psi_j=\lambda_j\psi_j, \quad 0<\lambda_1\leq \lambda_2\leq \dots,\quad
\text{as } j\to \infty.
$$
For $n\in \mathbb{N}$, we define the discrete ansatz space by
$$
V_n=\operatorname{span}\{\psi_1, \psi_2,\dots, \psi_n\}\subset V.
$$
Set $y_n(t)=y_n(x, t)=\sum_{i=1}^ny_i^n(t)\psi_i(x)$ require $y_n(0,
\cdot)\mapsto y_0$ in $H$ holds true.

To prove the existence of a unique weak solution to the problem
\eqref{2-2}, we are going to analyze the limiting behavior of
sequences of smooth functions $\{y_n\}$ and $\{u_n\}$.

Performing the Galerkin process for the problem \eqref{2-2}, we
have 
\begin{equation} \label{2-3}
\begin{gathered}
y_{n, t}+\frac\gamma kD^4y_n-\frac\gamma kD^4u_n
+D^2(a(u_n)D^2
u_n+\frac{a'(u_n)}2|Du_n|^2)-D^2f(u_n)=B^*\overline{\omega},
\\
y_n(x, 0)=y_{n, 0}=u_{n,0}-D^2u_n(x, 0),
\\
u_n(x, t)=D^2u_n(x, t)=D^4u_n(x, t)=0.
\end{gathered}
\end{equation}
According to ODE theory, there is a unique solution to \eqref{2-3}
in the interval $[0, t_n]$. We should show that the solution is
uniformly bounded when $t_n\to T$.

First step,  multiplying the first equation of \eqref{2-3} by
$$\mu_n=\gamma D^4u_n-a(u_n)D^2u_n-\frac{a'(u_n)}2|D
u_n|^2+f(u_n)+ku_{n, t},$$ and integrating with respect to $x$, we
obtain 
\begin{equation}\label{2-4}
 \frac{d}{dt} E(u_n)+\|D
\mu_n\|^2+k\|u_{n, t}\|^2=(B^*\overline{\omega}, \mu_n)_{V^*, V},
\end{equation}
where
\begin{gather}\label{2-5} 
E(u_n)=\int_0^1\Big(\frac\gamma2|D^2 u_n|^2+\frac
{a(u_n)}2|D u_n|^2+F(u_n)\Big)dx, \\
\label{2-6} F(u_n)=\gamma_1(u_n^6+(h_0-2)u_n^4+(1-2h_0)u_n^2+h_0).
\end{gather}
Applying a simple calculation, we have
\begin{equation}\label{2-7} 
F(u_n)\geq C_1u_n^6-C_0,
\end{equation}
where $C_1>0$ and $C_0\geq0$.

Since $B^*\overline{\omega}\in L^2(0, T;  V^*)$ is a control item,
we assume
\begin{equation}
\label{2-8} \|B^*\overline{\omega}\|_{V^*}\leq M.
\end{equation}
Taking account \eqref{2-4}, \eqref{2-7}, \eqref{2-8} and \eqref{1-4}
and integrating \eqref{2-4} with respect to time from $0$ to $t$, we
know 
\begin{align*}
&\int_0^1\Big(\frac\gamma2|D^2 u_n|^2+\frac {a_2}2u_n^2|D
u_n|^2+C_1u_n^6\Big)dx
+\int_0^t\|D \mu_n\|^2dt+k\int_0^t\|u_{n, t}\|^2dt
\\
&\leq \int_0^1\frac {|a_0|}2|D u_n|^2dx+E(u_{n, 0})+C_0
+\int_0^t|(B^*\overline{\omega}, \mu_n)_{V^*, V}|dt
\\
&\leq \varepsilon_1\int_0^1\frac {|a_0|}2|D^2
u_n|^2dx+C(\varepsilon_1)\int_0^1u_n^2dx
\\
&\quad+E(u_{n, 0})+C_0+\int_0^t\|B^*\overline{\omega}\|_{V^*}\|\mu_n\|_Vdt
\\
&\leq \varepsilon_1\int_0^1\frac {|a_0|}2|D^2
u_n|^2dx+C(\varepsilon_1)\varepsilon_2\int_0^1u_n^6dx+C(\varepsilon_2)
\\
&\quad +E(u_{n,0})+C_0+C(\varepsilon)\int_0^t\|B^*\overline{\omega}\|_{V^*}^2dt
+\varepsilon\int_0^t\|D^2\mu_n\|^2dt
\\
&= \varepsilon_1\int_0^1\frac {|a_0|}2|D^2 u_n|^2dx+C(\varepsilon_1)\varepsilon_2
\int_0^1u_n^6dx+C(\varepsilon_2)
\\
&\quad +E(u_{n,0})+C_0+C(\varepsilon)\int_0^t\|B^*\overline{\omega}\|_{V^*}^2dt
+\varepsilon \int_0^t\|u_{n, t}\|^2dt,
\end{align*}
where
$$
E(u_{n, 0})=\int_0^1\Big(\frac\gamma2|D^2 u_{n, 0}|^2+\frac
{a(u_{n, 0})}2|D u_{n, 0}|^2+F(u_{n, 0})\Big)dx.
$$
Choosing $\varepsilon_1, \varepsilon_2$ and $\varepsilon$
sufficiently small, from the above inequality and the Poincar\'e
inequality, we have
\begin{gather}
\int_0^1|D^2 u_n|^2dx \leq C,\label{2-9} \\
\int_0^1|D u_n|^2dx \leq  C,\label{2-10}\\
\int_0^1u_n^6dx \leq  C,\label{2-11} \\
\iint_{Q_T}|u_{n, t}|^2\,dx\,dt \leq  C\label{2-12}.
\end{gather}
From \eqref{2-11}, we know that
\begin{equation}\label{2-13}
 \int_0^1u_n^2dx\leq C.
\end{equation}
By  \eqref{2-9}, \eqref{2-10} and \eqref{2-13}, we obtain
\begin{eqnarray}\label{2-14} 
\|u_n\|_{H^2}\leq C.
\end{eqnarray}
By Sobolev's imbedding theorem it follows from \eqref{2-14} that
\begin{equation}\label{2-15} 
\|u_n\|_{L^\infty}\leq C ,\quad \|D u_n\|_{L^\infty}\leq C.
\end{equation}

Second step,  multiplying \eqref{1-1} by $D^2u_n$ and integrating
with respect to $x$, we obtain
\begin{equation}\label{2-16}
\begin{split}
& \frac12\frac{d}{dt}\Big(\int_0^1|Du_n|^2dx+k\int_0^1|D^2
u_n|^2dx\Big)+\gamma\int_0^1|D^4u_n|^2dx \\
&= -\int_0^1D^2 f(u_n)D^2u_ndx+\int_0^1a(u_n)D^2 u_nD^4u_ndx\\
&\quad +\int_0^1\frac{a'(u_n)}2|D u_n|^2D^4 u_n
dx-(B^*\overline{\omega}, D^2u_n)_{V^*, V}.
\end{split}
\end{equation}
From a simple calculation, we have
\begin{gather}
a'(u_n)=2a_2u_n,\label{2-17} \\
D^2f(u_n)=f'(u_n)D^2u_n+f''(u_n)(Du_n)^2,\label{2-18}
\end{gather}
where
\begin{gather}\label{2-19}
f'(u_n)=\gamma_1(30u_n^4+12(h_0-2)u_n^2+2(1-2h_0))\geq -C_2,\quad
C_2>0, \\
f''(u)=\gamma_1120u_n^3+24(h_0-2)u_n\label{2-20}.
\end{gather}
Thus it follows from \eqref{2-14}, \eqref{2-18} and\eqref{2-19} that
\begin{equation} \label{2-21}
\begin{aligned}
& \frac12\frac{d}{dt}\Big(\int_0^1|Du_n|^2dx+k\int_0^1|D^2
u_n|^2dx\Big)+\gamma\int_0^1|D^4 u_n|^2dx \\
&\leq -\int_0^1(f'(u_n)D^2u_n+f''(u_n)|Du_n|^2)D^2u_ndx+\int_0^1(a_2u_n^2+a_0)D^2
u_nD^4u_ndx
\\
&\quad +\int_0^1\frac{a'(u_n)}2|D u_n|^2D^4
u_ndx+\|B^*\overline{\omega}\|_{V^*}\|D^2u_n\|_V
\\
&\leq C_2\int_0^1|D^2u_n|^2dx+C(\|u_n\|^3_{L^\infty}+
\|u_n\|_{L^\infty})\|Du_n\|_{L^\infty}\int_0^1|Du_n||D^2u_n|dx
\\
&\quad +|a_2|\|u_n\|^2_{L^\infty}\int_0^1|D^2u_n||D^4u_n|dx+|a_0|\int_0^1|D^2u_n||D^4u_n|dx
\\
&\quad +|a_2|\int_0^1|u_n||D u_n|^2|D^4
u_n|dx+C(\varepsilon)\|B^*\overline{\omega}\|_{V^*}^2+\varepsilon\int_0^1|D^4
u_n|^2dx
\\
&\leq C_2\int_0^1|D^2u_n|^2dx+C(\|u_n\|^3_{L^\infty}+\|u_n\|_{L^\infty})\|Du_n\|_{L^\infty}
\Big(\varepsilon\int_0^1|Du_n|^2 \Big)
\\
&\quad +C(\|u_n\|^3_{L^\infty}+\|u_n\|_{L^\infty})\|Du_n\|_{L^\infty}
\Big(C(\varepsilon)\int_0^1|D^2u_n|^2dx\Big)
\\
&\quad +|a_2|\|u_n\|^2_{L^\infty}\Big(C(\varepsilon)\int_0^1|D^2
u_n|^2dx+\varepsilon\int_0^1|D^4u_n|^2dx\Big)
\\
&\quad +|a_0|\Big(C(\varepsilon)\int_0^1|D^2
u_n|^2dx+\varepsilon\int_0^1|D^4u_n|^2dx\Big)
\\
&\quad +|a_2|\|D u_n\|_{L^\infty}^2\Big(\varepsilon\int_0^1|D^4
u_n|^2dx+C(\varepsilon)\int_0^1u_n^2dx\Big)
\\
&\quad +C(\varepsilon)\|B^*\overline{\omega}\|_{V^*}^2+\varepsilon\int_0^1|D^4
u_n|^2dx
\\
&\leq \frac\gamma2\int_0^1|D^4 u_n|^2dx+C,
\end{aligned}
\end{equation}
where $\varepsilon$ is sufficiently small.

By the Gronwall's inequality,  \eqref{2-21} implies
\begin{equation}
\label{2-22} \iint_{Q_T}|D^4 u_n|^2\,dx\,dt\leq C.
\end{equation}
As we know
\begin{equation} \label{2-23}
\begin{split}
&\int_0^T\int_0^1|D^3u_n|^2\,dx\,dt \\
&\leq \frac12\int_0^T\int_0^1|D^2u_n|^2\,dx\,dt+
\frac12\int_0^T\int_0^1|D^4u_n|^2\,dx\,dt\leq C.
\end{split}
\end{equation}
From a simple calculation, we have
\[
\|y_n\|_{V}=\|u_n-D^2u_n\|_{V}^2
\leq  C(\|u_n\|+\|Du_n\|+\|D^2u_n\|+\|D^3u_n\|+\|D^4u_n\|).
\]
From \eqref{2-14}, \eqref{2-15}, \eqref{2-22} and \eqref{2-23}, we
obtain
\begin{equation}
\|y_n\|_{L^2(0, T; V)}\leq C.
\end{equation}

Third step, from \eqref{2-2}, \eqref{2-14}, \eqref{2-15} and
Sobolev embedding theorem, we have
\begin{align*}
\|y_{n, t}\|_{V^*}
&\leq \|B^*\overline{\omega}\|_{V^*}+\|D^4u_n\|+\|a(u_n)D^2
u_n+\frac{a'(u_n)}2|Du_n|^2\|+\|f(u_n)\|
\\
&\leq \|B^*\overline{\omega}\|_{V^*}+\|D^4u_n\|
+C\|u_n\|^2_{L^\infty}\|D^2u_n\|+C\|u_n\|_{L^\infty}\|Du_n\|^2\\
&\quad + C\|u_n\|^6_{L^\infty}+C
\\
&\leq \|D^4u_n\|+ C.
\end{align*}
Then
$\|y_{n, t}\|_{L^2(0, T; V^*)}\leq C$.
Thus, we have:
\begin{itemize}
\item[(i)] For every $t\in[0, T]$, the sequence $\{y_n\}_{n\in\mathbb{N}}$
is bounded in $L^2(0, T; H)$ as well as in $L^2(0, T; V)$, which is
independent of the dimension of amsatz space $n$.

\item[(ii)] For every $t\in[0, T]$, the sequence 
$\{y_{n, t}\}_{n\in\mathbb{N}}$ is bounded in $L^2(0, T; V^*)$,  which is
independent of the dimension of amsatz space $n$.
\end{itemize}
By theorem we get $\{y_{n, t}\}_{n\in\mathbb{N}}\subset W(0, T; V)$
and $W(0, T; V)$ is continuously embedded into $C(0, T; H)$. $\{y_{n,
t}\}_{n\in\mathbb{N}}$ weak in $W(0, T; V)$, weak star in
$L^{\infty}(0, T; H)$ and strong in $L^2(0, T; H)$ to a function
$y(x, t)\in W(0, T; V)$. Obviously, the uniqueness of solution is
easy to obtained \cite{MS}. We omit it here.
\end{proof}

To ensure that the norm of weak solution in the space $W(0, T; V)$
can be controlled by initial value and control item, we need the
following theorem.

\begin{theorem} \label{thm2.2}
If $B^*\overline{\omega}\in L^2(0, T;  V^*)$, $u_0\in H^2(\Omega)\cap H_0^1(\Omega)$ 
and $y_0\in H$, then there exists a constant $C_3>0$ and $C_4>0$, such that
\[
\|y\|_{W(0, T; V)}^2\leq
C_3\Big(\|y_0\|_H^2+\|\overline{\omega}\|_{L^2(0, T;Q_0)}^2\Big)+C_4.
\]
\end{theorem}

\begin{proof}
As in the proof of  Theorem \ref{thm2.1}, we obtain
\begin{equation}\label{2-26} 
\|u\|\leq C,\quad \|Du\|\leq C,\quad \|u\|_V\leq C,\quad \|D^3u\|\leq C.
\end{equation}
Multiplying this equation by $y$ and integrating the equation with
respect to $x$,  we obtain
\begin{equation} \label{2-27}
\begin{split}
&\frac12\frac{d}{dt}\|y\|_H^2+\frac\gamma k\|D^2y\|_H^2\\
&= \int_0^1\frac\gamma kD^2yD^2udx-\int_0^1(a(u)D^2
u+\frac{a'(u)}2|Du|^2)D^2ydx\\
&\quad -\int_0^1D y D f(u)dx+(B^*\overline{\omega}, y)_{V^*, V}.
\end{split}
\end{equation}
From H\"older and Young inequalities, we have
\begin{equation} \label{2-28}
\int_0^1\frac\gamma kD^2yD^2udx
\leq C(\varepsilon)\|D^2u\|^2+\varepsilon\|D^2y\|^2.
\end{equation}
From \eqref{2-26}, we have
\begin{equation} \label{2-29}
\begin{aligned}
&\int_0^1(a(u)D^2 u+\frac{a'(u)}2|Du|^2)D^2y\,dx \\
&\leq \|u\|_{L^\infty}^3\|D^2y\|\|D^2u\|+|a_2|\|u\|_{L^\infty}
 \|Du\|_{L^\infty}^2\|D^2y\| \\
&\leq C(\epsilon)C\|D^2u\|^2+\varepsilon\|D^2y\|^2
 +CC(\epsilon)+\varepsilon\|D^2y\|^2 \\
&\leq \varepsilon\|D^2y\|^2+C,
\end{aligned}
\end{equation}
and
\begin{equation} \label{2-30}
-\int_0^1D y D f(u)dx\leq C_2\int_0^1|Dy|dx\leq C\|Dy\|+C\leq C.
\end{equation}
Note that
\begin{equation} \label{2-31}
 (B^*\overline{\omega}, y)_{V^*, V}\leq \|B^*\overline{\omega}\|_{V^*}\|y\|_{V}.
\end{equation}
From \eqref{2-27}-\eqref{2-31}, we have
\begin{equation} \label{2-32}
\frac12\frac{d}{dt}\|y\|_H^2+\frac\gamma k\|D^2y\|_H^2
\leq \varepsilon\|D^2y\|_H^2+C\|B^*\overline{\omega}\|_{V^*}^2+C.
\end{equation}
Integrating the above inequality with respect to $t$ yields
\begin{equation} \label{2-33}
\|y\|_H^2\leq \|y_0\|_H^2+C\|B^*\overline{\omega}\|_{L^2(0, T; V^*)}^2+C.
\end{equation}
By \eqref{2-33}, \eqref{2-2} and \eqref{2-26}, we deduce that
\begin{equation} \label{2-34}
\begin{split}
\|y_t\|_{V^*}^2
&\leq \|B^*\overline{\omega}\|_{V^*}^2+\frac\gamma
k\|y\|_V^2+\frac\gamma k\|D^2u\|
+\|(a(u)D^2 u+\frac{a'(u)}2|Du|^2)\|+\|f(u)\|
\\
&\leq \|B^*\overline{\omega}\|_{V^*}^2+C\|y\|_V^2+C\\
&\leq \|y_0\|_H^2+C\|B^*\overline{\omega}\|_{L^2(0, T; V^*)}^2+C.
\end{split}
\end{equation}
From \eqref{2-33} and \eqref{2-34}, we have
\begin{align*}
\|y\|_{W(0, T; V)}
&=\|y\|_{L^2(0, T; V)}+\|y_t\|_{L^2(0, T; V^*)} \\
&\leq  C_3\left(\|y_0\|_H^2+\|\overline{\omega}\|_{L^2(0, T;Q_0)}^2\right)+C_4.
\end{align*}
The proof is completed.
\end{proof}

\section{Optimal problem}

In this section, we will study the distributed optimal control of
the viscous generalized Cahn-Hilliard equation and the existence of
optimal solution is obtained based on Lions' theory.

We study the following Problem when $\overline{\omega}\in L^2(0, T;Q_0)$
\begin{gather*}
\min \mathcal {J}(y, \overline{\omega})=
\frac12\|Cy-z\|_S^2+\frac\delta2\|\overline{\omega}\|_{L^2(0, T;
Q_0)}^2
\\
\text{s. t. } y_t+\frac\gamma kD^4y-\frac\gamma kD^4u+D^2(a(u)D^2
u+\frac{a'(u)}2|Du|^2)-D^2f(u)=B^*\overline{\omega},\\
y(x, 0)=y_0=u_0-D^2u(x, 0), \\
u(x, t)=D^2u(x, t)=D^4u(x, t)=0,
\end{gather*}
where $y=u-D^2u$.

As we know that there exists a weak solution $y$ to the equation
\eqref{2-2}, due to $u=(1-\partial_x^2)^{-1}y$, we know that there
exists a weak solution $u$ to the equation \eqref{2-1}. Given an
observation operator $C\in L(W(0, T; V), S)$,  in which $S$ is a
real Hilbert space and $C$ is continuous.

We choose the performance index of tracking type
\begin{equation} \label{3-1}
\mathcal {J}(y, \overline{\omega})=
\frac12\|Cy-z\|_S^2+\frac\delta2\|\overline{\omega}\|_{L^2(0, T;Q_0)}^2,
\end{equation}
where $z\in S$ is a desired state and $\delta>0$ is fixed.

The optimal control problem about the sixth-order Cahn-hilliard equation
is
\begin{equation} \label{3-2} 
\min \mathcal {J}(y, \overline{\omega}),
\end{equation}
where $(y, \overline{\omega})$ satisfies  \eqref{2-2}.

Let $X=W(0, T; V)\times L^2(0, T;  Q_0)$ and $Y=L^2(0, T; V)\times
H$
We define an operator $e=e(e_1, e_2): X\to Y$ by
\begin{align*}
e(y, \overline{\omega})=e(e_1(y, \overline{\omega}), e_2(y,\overline{\omega})),
\end{align*}
where
\begin{gather*}
\begin{aligned}
e_1(y, \overline{\omega})
&=(\Delta^2)^{-1}\Big(y_t+\frac\gamma kD^4y-\frac\gamma kD^4u+D^2(a(u)D^2
u+\frac{a'(u)}2|Du|^2)\\
&\quad -D^2f(u)-B^*\overline{\omega}\Big),
\end{aligned}\\
e_2=y(x, 0)-y_0,
\end{gather*}
and $\Delta^2$ is an operator from $H^2(0, 1)$ to $H^{-2}(0, 1)$.

Then equation \eqref{3-2} is rewritten as
$$
\min \mathcal {J}(y, \overline{\omega})\quad 
\text{subject to } e=e(y, \overline{\omega})=0.
$$
Now, we have the following theorem.

\begin{theorem} \label{thm3.1}
There exists an optimal control solution to the above problem.
\end{theorem}

\begin{proof}
Let $(y, \overline{\omega})\in X$ satisfies the equation 
$e=e(y, \overline{\omega})=0$. In view of  \eqref{3-1}, we have
$$
\mathcal {J}(y,
\overline{\omega})\geq\frac\delta2\|
\overline{\omega}\|^2_{L^2(0, T;Q_0)}.
$$
From  Theorem \ref{thm2.2}, we have
$$
\|y\|_{W(0, T;V)}\to\infty\quad \text{yields}\quad
\|\overline{\omega}\|_{L^2(0,T;  Q_0)}\to\infty.
$$
Hence
\begin{equation}\label{3-3} 
\mathcal {J}(y,\overline{\omega})\to+\infty\quad \text{when }
\|y,\overline{\omega}\|_X\to\infty.
\end{equation}
As the norm is weakly lowered semi-continuous \cite{A2}, we achieve
that $\mathcal {J}$ is weakly lowered semi-continuous.

Since $\mathcal {J}(y, \overline{\omega})\geq0$ for all $(y,
\overline{\omega})\in X$ holds, there exist
$$
\eta=\inf\{\mathcal {J}(y, \overline{\omega})|(y,
\overline{\omega})\in X \text{ such that } e(y,\overline{\omega})=0\},
$$
which means that there exists  a minimizing sequence 
$\{(y^n, \overline{\omega^n})\}_{n\in \mathbb{N}}$ in $X$ such that
$$
\eta=\lim_{n\to\infty}\mathcal {J}(y^n,
\overline{\omega}^n)\quad \text{and}\quad 
e=e(y^n, \overline{\omega}^n)=0,\quad \forall n\in \mathbb{N}.
$$
From \eqref{3-3}, there exists an element $(y^*,\overline{\omega}^*)\in X$ 
such that
\begin{gather}
y^n\rightharpoonup y^*,\quad y\in W(0, T; V)\label{3-4},\\
\overline{\omega^n}\rightharpoonup\overline{\omega}^*,\quad 
\overline{\omega}\in L^2(0, T;  Q_0), \label{3-5}
\end{gather}
when $n\to\infty$.
From  \eqref{3-4}, we have
\begin{align*}
\lim_{n\to\infty}\int_0^T(y^n(t)-y^*(t), \phi(t))_{V^*,
V}dt=0,\quad \forall\phi\in L^2(0, T;  V).
\end{align*}

Since $W(0, T; V)$ is compactly embedded into $L^2(0, T;
L^{\infty})$, we have $y^n\to y^*$ strongly in $L^2(0, T;  L^{\infty})$. 
Then we also derive that $D^2u^n\to D^2u^*$ strongly in $L^2(0, T; L^\infty)$.
On the other hand, by \eqref{2-26} and \eqref{2-34}, we know that 
$u_n\in L^\infty(0,T; V)$ and $y_{nt}\in L^2(0, T; V^*)$. 
Hence by \cite[Lemma 4]{Si}
we have $u^n \to u^*$ strongly in $C(0, T; L^\infty)$, $Du^n \to Du^*$ strongly 
in $C(0, T; H)$,  as $n\to\infty$.

As the sequence $\{y^n\}_{n\in\mathbb{N}}$ converges weakly, then
$\|y^n\|_{W(0, T; V)}$ is bounded. And 
$\|y^n\|_{L^2(0, T; L^{\infty})}$ is also bounded based on the embedding theorem.

Since $y^n\to y^*$ strongly in $L^2(0, T;  L^{\infty})$,
then we derive that $\|y^*\|_{L^2(0, T;  L^{\infty})}$,
$\|u^*\|_{L^2(0, T; L^{\infty})}$ and $\|D^2u^*\|_{L^2(0, T;
L^{\infty})}$ are bounded.

Notice that 
\begin{align*}
&\big|\int_0^T\int_0^1(D^2f(u^n)-D^2f(u^*))\psi \,dx\,dt\big|\\
&=\big|\int_0^T\int_0^1(f(u^n)-f(u^*))D^2\psi \,dx\,dt\big|\\
&\leq \int_0^T\int_0^1|(u^n-u^*)\Big(6\gamma_1((u^n)^4+(u^n)^2(u^*)^2+(u^n)^3u^*
 +u^n(u^*)^3+(u^*)^4)\\
&\quad +4(h_0-2)((u^2)^2+(u^*)^2+u^nu^*)+2(1-2h_0)\Big)|D^2\psi|\,dx\,dt
\\
&\leq  \big( \| u^n \|^4_{L^8(0,T;L^\infty)} + \| u^* \|^4_{L^8(0,T;L^\infty)}\big)
  % Referee suggestin &\leq (\|u^n\|_{L^2(0, T; L^\infty)}^4+\|u^*\|_{L^2(0, T;L^\infty)}^4)
\|u^n-u^*\|_{C(0, T; H)}\|D^2\psi\|_{L^2(0, T; H)}
\\
&\to 0, \quad \forall\psi\in L^2(0, T;  V).
\end{align*}
As we know
\begin{align*}
&\Big|\int_0^T\int_0^1\Big(D^2(a(u^n)D^2 u^n+\frac{a'(u^n)}2|Du^n|^2)
\\
&\quad -D^2(a(u^*)D^2u^*+\frac{a'(u^*)}2|Du^*|^2)\Big)\psi \,dx\,dt\Big|
\\
=&\Big|\int_0^T\int_0^1D^2(a(u^n)D^2 u^n-a(u^*)D^2 u^*)\psi\,dx\,dt
\\
&\quad +\int_0^T\int_0^1D^2(\frac{a'(u^n)}2|Du^n|^2-\frac{a'(u^*)}2|Du^*|^2)\psi
\,dx\,dt\Big|
\\
&=|I_1+I_2|.
\end{align*}
Note that
\begin{align*}
|I_1|
&=\Big|\int_0^T\int_0^1D^2(a(u^n)D^2 u^n-a(u^*)D^2 u^*)\psi \,dx\,dt\Big|
\\
&=\Big|\int_0^T\int_0^1(a(u^n)D^2 u^n-a(u^*)D^2 u^*)D^2\psi\,dx\,dt\Big|
\\
&=\Big|\int_0^T\int_0^1((a_2(u^n)^2+a_0)D^2
u^n-(a_2(u^*)^2+a_0)D^2 u^*)D^2\psi \,dx\,dt\Big|
\\
&=\Big|\int_0^T\int_0^1(a_2(u^n)^2D^2 u^n-a_2(u^*)^2D^2u^*)D^2\psi dx dt
\\
&\quad+\int_0^T\int_0^1(a_0D^2 u^n-a_0D^2 u^*)D^2\psi\,dx\,dt\Big|
\\
&=|I^1_1+I_1^2|.
\end{align*}
Now, we deal with $I^1_1$ and $I_1^2$,
\begin{align*}
I^1_1
&=\int_0^T\int_0^1(a_2(u^n)^2D^2 u^n-a_2(u^*)^2D^2 u^*)D^2\psi\,dx\,dt
\\
&=\int_0^T\int_0^1\Big(a_2(u^n)^2D^2 u^n-a_2(u^n)^2D^2 u^*
\\
&\quad+a_2(u^n)^2D^2 u^*-a_2(u^*)^2D^2 u^*\Big)D^2\psi\,dx\,dt
\\
&= \int_0^T\int_0^1(a_2(u^n)^2D^2 u^n-a_2(u^n)^2D^2 u^*)D^2\psi \,dx\,dt
\\
&\quad +\int_0^T\int_0^1(a_2(u^n)^2D^2 u^*-a_2(u^*)^2D^2 u^*)D^2\psi\,dx\,dt
\\
&\leq \int_0^T\int_0^1a_2(u^n)^2(D^2 u^n-D^2 u^*)D^2\psi \,dx\,dt
\\
&\quad +\int_0^T\int_0^1(a_2(u^n)^2-a_2(u^*)^2)D^2 u^*D^2\psi \,dx\,dt
\\
&\leq \int_0^Ta_2\|u^n\|_{L^4}^2\|D^2 u^n-D^2 u^*\|_{L^\infty}\|D^2\psi\|_H dt
\\
&\quad +\int_0^Ta_2\|(u^n)^2-(u^*)^2\|_H\|D^2
u^*\|_{L^{\infty}}\|D^2\psi\|_H dt
\\
&\leq a_2\|u^n\|_{C(0, T;  L^4)}^2\|D^2 u^n-D^2
u^*\|_{L^2(0, T; L^\infty)}\|D^2\psi\|_{L^2(0, T;  H)}
\\
&\quad +a_2\|u^n-u^*\|_{C(0, T; H)}(\|u^n\|_{C(0, T; L^\infty)}
 +\|u^*\|_{C(0, T; L^\infty)})
\\
&\quad\cdot\|D^2 u^*\|_{L^2(0, T; L^{\infty})}\|D^2\psi\|_{L^2(0, T;  H)}
\to0,\quad  \forall\psi\in L^2(0, T;  V),
\end{align*}
and
\begin{align*}
I^2_1
&=\int_0^T\int_0^1(a_0D^2u^n-a_0D^2u^*)D^2\psi \,dx\,dt
\\
&\leq \int_0^T|a_0|\|D^2u^n-D^2u^*\|_H\|D^2\psi\|_H dt
\\
&\leq |a_0|\|D^2u^n-D^2u^*\|_{L^2(0, T; H)}\|D^2\psi\|_{L^2(0, T;  H)}
\\
&\to 0,\quad \forall\psi\in L^2(0, T;  V).
\end{align*}
Further,
\begin{align*}
I_2
&=\int_0^T\int_0^1D^2(a_2u^n|Du^n|^2-a_2u^*|Du^*|^2)\psi \,dx\,dt
\\
&=\int_0^T\int_0^1(a_2u^n|Du^n|^2-a_2u^*|Du^*|^2)D^2\psi \,dx\,dt
\\
&=\int_0^T\int_0^1(a_2u^n|Du^n|^2-a_2u^n|Du^*|^2)D^2\psi \,dx\,dt
\\
&\quad +\int_0^T\int_0^1(a_2u^n|Du^*|^2-a_2u^*|Du^*|^2)D^2\psi \,dx\,dt
\\
&\leq \int_0^Ta_2\|u^n\|_{L^\infty}\|Du^n-Du^*\|_H(\|Du^n\|_{L^\infty}
 +\|Du^*\|_{L^\infty})\|D^2\psi\|_H
\\
&\quad + \int_0^Ta_2\|u^n-u^*\|_{L^\infty}\||Du^*|^2\|_{H}\|D^2\psi\|_H dt
\\
&\leq a_2\|u^n\|_{C(0, T;L^{\infty})}\|Du^n\|_{L^2(0, T; L^\infty)}
\|Du^n-Du^*\|_{C(0, T; H)}\|D^2\psi\|_{L^2(0,T;  H)}
\\
&\quad +a_2\|u^n\|_{C(0, T;L^\infty)}\|Du^*\|_{L^2(0, T; L^\infty)}
\|Du^n-Du^*\|_{C(0, T; H)}\|D^2\psi\|_{L^2(0,T;  H)}
\\
& \quad +a_2\|u^n-u^*\|_{L^2(0, T; L^{\infty})}\||Du^*|^2\|_{C(0, T;
H)}\|D^2\psi\|_{L^2(0, T; H)}
\\
&\to 0,\quad  \forall\psi\in L^2(0, T;  V).
\end{align*}
From \eqref{3-5}, we have
\[
\Big|\int_0^T\int_0^1(B^*\overline{\omega}^n-B^*\overline{\omega}^*)\psi\Big|
\to0,\quad \forall\psi\in L^2(0, T;  V).
\]
In view of the above discussion, we can conclude that
$$
e_1(y^*, \overline{\omega}^*)=0,\quad \forall n\in\mathbb{N}.
$$
Since $y^*\in W(0, T; V)$, we have $y^*(0)\in H$. From
$y^n\rightharpoonup y^*$ in $W(0, T; V)$, we  can infer that
$y^n(0)\rightharpoonup y^*(0)$. Thus we obtain
$$
(y^n(0)-y^*(0), \psi)\to0,\quad\forall\psi\in H,
$$
which means that $e_2(y^*, \overline{\omega}^*)=0$, for all
$n\in\mathbb{N}$.
Hence, we can derive that $e(y^*, \overline{\omega}^*)=0$, for all
$n\in\mathbb{N}$.

In conclusion, there exists an optimal solution 
$(y^*, \overline{\omega}^*)$  to the problem. And  we can infer that there
exists an optimal solution $(y^*, \overline{\omega}^*)$ to the
viscous generalized Cahn-Hilliard equation due to
$u=(1-\partial_x^2)^{-1}y$.
\end{proof}

\subsection*{Acknowledgments}
The authors would like to express their deep thanks to the anonymous referees
for their valuable suggestions for the revision of this article.

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\end{document}