# Optimal control problems governed by time-fractional diffusion equations
with control constraint

B.T.Kien††Department of Optimization and Control Theory, Institute of
Mathematics, Vietnam Academy of Science and Technology, 18 Hoang Quoc Viet
road, Hanoi, Vietnam; email<EMAIL_ADDRESS>B.N.Muoi†† Department of
Applied Mathematics, National Sun Yat-sen University, Kaohsiung, 80424,
Taiwan, and Department of Mathematics, Hanoi Pedagogical University 2, Vinh
Phuc, Vietnam; email<EMAIL_ADDRESS>C.F.Wen††Research Center for
Nonlinear Analysis and Optimization, Kaohsiung Medical University; email:
<EMAIL_ADDRESS>and J.C.Yao††Research Center for Nonlinear Analysis and
Optimization, Kaohsiung Medical University; email<EMAIL_ADDRESS>
Abstract. A class of optimal control problems governed by linear fractional
diffusion equation with control constraint is considered. We first establish
some results on the existence of strong solution to the state equation and the
existence of optimal solutions for the optimal control problem. Then we derive
first-and second-order optimality conditions for locally optimal solutions to
the problem.

Key words. Optimal control, Fractional diffusion equation, Solution existence,
First-and second-order optimality conditions.

AMS Subject Classifications. 49K15 $\cdot$ 90C29

##  1 Introduction

Let $\Omega$ be an open and bounded set in $\mathbb{R^{N}}$ with $N=2,3$ and
boundary $\partial\Omega\in C^{2}$, $Q=[0,T]\times\Omega$ and
$\Sigma=[0,T]\times\partial\Omega$, where $T>0$. Let $D:=H^{2}(\Omega)\cap
H_{0}^{1}(\Omega)$. We consider the problem of finding a function $u\in
L^{\infty}(Q)$ and the corresponding state $y\in L^{p}([0,T],D)$ with
$1<p<\infty$, which solve

$\displaystyle\psi(y,u)=\int_{Q}L(t,x,y(t,x),u(t,x))dtdx\to\min$ (1) s.t
$\displaystyle\frac{d^{\alpha}y}{dt^{\alpha}}=\Delta y+u\quad{\rm in}\quad Q$
(2) $\displaystyle y=0\quad\text{on}\quad\Sigma$ (3) $\displaystyle
y(0)=y_{0}\quad\text{in}\quad\Omega$ (4) $\displaystyle a\leq u(t,x)\leq
b\quad\text{a.e.}\quad(t,x)\in Q,$ (5)

where $0<\alpha\leq 1$ and $\frac{d^{\alpha}}{dt^{\alpha}}$ is a fractional
derivative operator in the sense of Caputo, which is given in Section 2,
$L:Q\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is a Caratheodory function,
and $y_{0}\in D$.

The system (2)-(4) is a time-fractional diffusion equation. Note that several
physical systems are not truly modelled with differential equations of
integer-order. So the fractional-order differential equations have been
introduced in order to describe accurately these systems. A motivation for
studying fractional diffusion equations comes from the fact that they describe
efficiently anomalous diffusion on fractals such as physical objects of
fractional dimension, like some amorphous semiconductors or strongly porous
materials (see [2], [22] and references therein).

In contrast with optimal control problems governed by differential equations
with integer order (see [8], [14] and [32]), the class of optimal control
problems governed by fractional differential equations have not been studied
systematically yet. Recently, there have been some mathematicians studying
optimal control problems governed by fractional differential equations. For
the papers which have close connections to the present work, we refer the
reader to [1], [4], [12], [17], [24] and [25]. Among them, O.P. Agrawal [1],
considered optimal control problems governed by fractional differential
equations (FDEs) without constraints. She gave first-order necessary
conditions and some numerical examples for the problem. Kien et al. [17]
studied problems governed by FDEs with control constraint. They gave first-and
second-order optimality conditions for locally optimal solutions to the
problem. G.M. Mophou et al. [12], [24] and [25] considered strongly convex
optimal control problems governed by a time-fractional diffusion equations
with Riemann–Liouville derivative and the cost functions of the type

$J(u)=\frac{1}{2}\|y(u)-y_{0}\|^{2}_{L^{2}(Q)}+\frac{\gamma}{2}\|u\|^{2}_{L^{2}(Q)},\gamma>0$

for cases: control constraints, no constraint and state constraints,
respectively. Then they established results on the existence of optimal
solutions and first-order optimality conditions. Note that, this class of
problems are convex so the second-order optimality conditions are fulfilled
automatically. Recently Bahaa [4] has extended the results of [24] by
considering the problem with control constraint. However, all of them mainly
investigated first-order optimality conditions for the problem.

In this paper we consider problem (1)-(5) governed by time-fractional
diffusion equation with Caputo derivative and with control constraint (5),
where the cost function may not be convex. We shall prove the existence of
solutions to the state equation and the existence of optimal solutions to
(1)-(5). We then derive first- and second-order optimality conditions. We show
that when $p=2$ and $1>\alpha>1/2$, a theory of no-gap second-order optimality
conditions is obtained. Optimality conditions are important for numerical
methods of computing optimal solutions. Based on second-order sufficient
conditions, we can give error estimate for approximate solutions.
Nevertheless, to our best knowledge, so far there have been no result on the
second-order optimality conditions for this class of problems.

The study of optimal control problems governed by fractional diffusion
equations are quite complicated. One of the difficulties is the regularity of
the solution to the state equation is poor. It is known that, if the control
function is Hölder continuous, then the state function is also Höder
continuous. Unfortunately, in optimal control problems, the control variables
are usually belong to $L^{\infty}$-spaces. So the solutions to the state
equation only belong $L^{p}$-spaces. Therefore, the cost function is difficult
to be differentiable with respect to the state variable. Another difficulty
for the problem is about compactness of bounded sequences in function spaces
with Banach space values. To overcome these difficulties, we shall use a
result from [15] in order to study regularity of solutions to the state
equation. We also use a recent result on the compact embedding in [19] which
is a version of the so-call Aubin’s lemma for evolution equations to show that
if a sequence is weakly convergent then it is strongly convergent. Besides,
techniques of optimal control from [7], [16] and [17], nonlinear and
variational analysis are also used to deal with the problem.

The paper is organized as follows. In Section 2, we present some auxiliary
results on the fractional derivatives for functions with values in Banach
space. Section 3 is devoted to results on the existence and regularity of
solutions to the state equation and the existence of optimal solutions to the
problem. Section 4 is destined for establishing first-and second-order
optimality conditions.

##  2 Some auxiliary results on fractional derivatives

Let $X$ be a Banach space. We denote by $L^{p}([0,T],X)$ the space of X-valued
functions $u:[0,T]\to X$ such that
$\|u\|_{p}:=(\int_{0}^{T}\|u(t)\|_{X}^{p}dt)^{1/p}<+\infty$, where $1\leq
p\leq\infty$. Also we denote by $C([0,T],X)$ the space of continuous functions
$v:[0,T]\to X$, by $W^{1,p}([0,T],X)$ the space of absolutely continuous
functions $y:[0,T]\to X$. The norm on $W^{1,p}([0,T],X)$ is defined by

$\|y\|_{1,p}=\|y\|_{0}+\|\dot{y}\|_{p},$

where $\|y\|_{0}=\max_{t\in[0,T]}\|y(t)\|_{X}$ is the norm of $y$ in
$C([0,T],X)$.

Let $\theta>0$ be a real number. The operator $I_{0+}^{\theta}$ defined on
$L^{1}([0,T],X)$, by

$(I_{0+}^{\theta}\phi)(t)=\frac{1}{\Gamma(\theta)}\int_{0}^{t}\frac{\phi(\tau)}{(t-\tau)^{1-\theta}}d\tau$

is called the left Riemann-Liouville fractional integral operator of order
$\theta$. The operator $I_{T-}^{\theta}$ defined on $L^{1}([0,T],\mathbb{R})$,
by

$(I_{T-}^{\theta}\phi)(t)=\frac{1}{\Gamma(\theta)}\int_{t}^{T}\frac{\phi(\tau)}{(\tau-t)^{1-\theta}}d\tau$

is called the right Riemann-Liouville fractional integral operator of order
$\theta$. For the case $\theta=0$, we set $I_{0+}^{0}=I_{T-}^{0}=I$, the
identity operator. Here the integrals are taken in Bochner’s sense. By the
same argument as in the proof of [11, Theorem 2.2], we can show that

$\displaystyle(I_{0+}^{\theta_{1}}(I_{0+}^{\theta_{2}}\phi)(t)=(I_{0+}^{\theta_{1}+\theta_{2}}\phi)(t)\quad\forall\phi\in
L^{1}([0,T],X).$ (6)

Let us denote by $I_{0+}^{\theta}(L^{p}([0,T],X))$ the space of summable
functions $f:[0,T]\to X$ such that $f=I_{0+}^{\theta}\phi$ for some $\phi\in
L^{p}([0,T],X)$. The space $I_{T-}^{\theta}(L^{p})$ is defined similarly. We
have the following characteristic property for
$I_{0+}^{\theta}(L^{p}([0,T],X))$. Its proof is similar to the proof of [17,
Proposition 2.1]

###### Proposition 2.1

$f\in I_{0+}^{\theta}(L^{p}([0,T],X))$ if and only if $I_{0+}^{1-\theta}f\in
W^{1,p}([0,T],X)$.

###### Definition 2.1

Let $0<\alpha<1$ and $\phi\in C([0,T],X)$. Then each of the expressions

$(D_{0+}^{\alpha}\phi)(t):=\frac{d}{dt}I_{0}^{1-\alpha}\phi(t)=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{0}^{t}\frac{\phi(\tau)}{(t-\tau)^{\alpha}}d\tau$

and

$(D_{T-}^{\alpha}\phi)(t):=-\frac{d}{dt}I_{T-}^{1-\alpha}\phi=-\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{t}^{T}\frac{\phi(\tau)}{(\tau-t)^{\alpha}}d\tau$

is called a left Riemann-Liouville fractional derivative and a right Riemann-
Liouville fractional derivative, respectively. The operator $D_{0+}^{\alpha}$
and $D_{T-}^{\alpha}$ is called the left Riemann-Liouville fractional
derivative operator and the right Riemann-Liouville fractional derivative
operator, respectively.

A sufficient condition for the existence of $D_{0+}^{\alpha}(\phi)$ and
$D_{T-}^{\alpha}(\phi)$ is to require $\phi\in W^{1,1}([0,T],X)$ (see [11,
Lemma 2.12]).

###### Proposition 2.2

Let $\alpha\in(0,1)$. Then the following assertions are fulfilled.

$(i)$ If $f\in L^{1}([0,T],X)$, then

$\displaystyle D_{0+}^{\alpha}I_{0+}^{\alpha}f=f\quad{\rm a.e.}$ (7)

$(ii)$ If $f\in I_{0+}^{\alpha}(L^{1})$, then

$\displaystyle I_{0+}^{\alpha}D_{0+}^{\alpha}f=f\quad{\rm a.e.}$ (8)

Proof. $(i)$. Using (6) and the definition of $D_{0+}^{\alpha}$, we have

$\displaystyle
D_{0+}^{\alpha}I_{0+}^{\alpha}f=D^{1}I_{0+}^{1-\alpha}I_{0+}^{\alpha}f=D^{1}I_{0+}^{1}f=f.$

$(ii)$. Let $f=I_{0+}^{\alpha}\phi$ for some $\phi\in L^{1}([0,T],X)$. Then
from (7), we have

$\displaystyle
I_{0+}^{\alpha}D_{0+}^{\alpha}f=I_{0+}^{\alpha}D_{0+}^{\alpha}I^{\alpha}_{0+}\phi=I_{0+}^{\alpha}\phi=f.$

$\Box$

###### Definition 2.2

Let $0<\alpha<1$ and $\phi\in L^{p}([0,T],X)$ with $p\geq 1$. The operator
$\frac{d^{\alpha}}{dt^{\alpha}}$ is defined by

$\frac{d^{\alpha}}{dt^{\alpha}}\phi(t):=D_{0+}^{\alpha}(\phi-\phi(0))=D_{0+}^{\alpha}\phi-\frac{\phi(0)}{(t-0)^{\alpha}\Gamma(1-\alpha)}$

is called the left Caputo differential operator of order $\alpha$, and the
operator $\frac{\hat{d}^{\alpha}}{dt^{\alpha}}$ defined by

$\frac{\hat{d}^{\alpha}}{dt^{\alpha}}\phi(t):=D_{T-}^{\alpha}(\phi-\phi(T))=D_{T-}^{\alpha}\phi-\frac{\phi(T)}{(T-t)^{\alpha}\Gamma(1-\alpha)}$

is called the right Caputo differential operator of order $\alpha$ provided
all terms exist.

From the definition, we see that if $\phi(0)=0$, then
$\frac{d^{\alpha}}{dt^{\alpha}}\phi=D_{0+}^{\alpha}(\phi)$. Also, if
$\phi(T)=0$, then
$\frac{\hat{d}^{\alpha}}{dt^{\alpha}}\phi=D_{T-}^{\alpha}(\phi)$.

###### Lemma 2.1

Suppose that $X=H$ is a Hilbert space, $f\in I_{T-}^{\alpha}(L^{q}([0,T],H))$
and $g\in I_{0+}^{\alpha}(L^{p}([0,T],H)$ with $p,q\geq 1$ and $1/p+1/q\leq
1+\alpha$. Then one has

$\displaystyle\int_{0}^{T}(f,D_{0+}^{\alpha}g)_{H}dt=\int_{0}^{T}(g,D_{T-}^{\alpha}f)_{H}dt$
(9)

Particularly, when $f(T)=0$ and $g(0)=0$, one has

$\displaystyle\int_{0}^{T}(f,\frac{d^{\alpha}g}{dt^{\alpha}})_{H}dt=\int_{0}^{T}(g,\frac{\hat{d}^{\alpha}f}{dt^{\alpha}})_{H}dt.$
(10)

Here $(\cdot,\cdot)_{H}$ denotes the scalar product in $H$.

Proof. We first claim that if $\phi_{1}\in L^{p}([0,T],H)$ and $\phi_{2}\in
L^{q}([0,T],H)$ with $1/p+1/q\leq 1+\alpha$, then

$\displaystyle\int_{0}^{T}(\phi_{2}(t),I_{0+}^{\alpha}\phi_{1}(t))_{H}dt=\int_{0}^{T}(\phi_{1}(t),I_{T-}^{\alpha}\phi_{2}(t))_{H}dt.$
(11)

In fact, from the assumption we see that the integrals in both sides are
defined. By Fubini’s theorem, we have

$\displaystyle\int_{0}^{T}(\phi_{2}(t),I_{0+}^{\alpha}\phi_{1}(t))_{H}dt$
$\displaystyle=\int_{0}^{T}(\phi_{2}(t),\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{1-\alpha}}\phi_{1}(s)ds)_{H}dt$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\int_{0}^{T}\int_{0}^{t}(\phi_{2}(t),\frac{1}{(t-s)^{1-\alpha}}\phi_{1}(s)_{H}dsdt$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\int_{0}^{T}ds\int_{s}^{T}(\phi_{2}(t),\frac{1}{(t-s)^{1-\alpha}}\phi_{1}(s))_{H}dt$
$\displaystyle=\int_{0}^{T}(\phi_{1}(s),I_{T-}^{\alpha}\phi_{2}(s))_{H}ds.$

The claim is justified. We now prove (9). By the assumption, we have
$f=I_{T-}^{\alpha}\phi_{2}$ and $g=I_{0+}^{\alpha}\phi_{1}$ for some
$\phi_{2}\in L^{q}([0,T],H)$ and $\phi_{1}\in L^{p}([0,T],H)$. Then
$\phi_{2}=D_{T-}^{\alpha}(f)$. Using (11), we have

$\displaystyle\int_{0}^{T}(f,D_{0+}^{\alpha}g)_{H}dt=\int_{0}^{T}(I_{T-}^{\alpha}\phi_{2}(t),\phi_{1}(t))_{H}dt=\int_{0}^{T}(\phi_{2}(t),I_{0+}^{\alpha}\phi_{1})_{H}dt=\int_{0}^{T}(D_{T-}^{\alpha}f(t),g(t))_{H}dt.$

We obtain formula (9). $\Box$

###### Proposition 2.3

Suppose that $f\in L^{\infty}([0,T],X)$ and $\alpha\in(0,1)$. Then
$I_{0+}^{\alpha}f\in C^{\alpha}([0,T],X)$-the space of Holder continuous
functions with order $\alpha$. Moreover, one has

$\displaystyle\|I_{0+}^{\alpha}f\|_{C([0,T],X)}\leq\frac{T^{\alpha}}{\alpha\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}.$
(12)

Proof. Take any $t_{0},t\in[0,T]$. Let $t>t_{0}$. Then we have

$\displaystyle(I_{0+}^{\alpha}\phi)(t)-(I_{0+}^{\alpha}f)(t_{0})=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau-\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{0}}\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{0}}[\frac{f(\tau)}{(t-\tau)^{1-\alpha}}-\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}]d\tau+\frac{1}{\Gamma(\alpha)}\int_{t_{0}}^{t}\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau.$

Hence

$\displaystyle\|(I_{0+}^{\alpha}\phi)(t)-(I_{0+}^{\alpha}f)(t_{0})\|_{X}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{0}^{t_{0}}\big{|}\frac{1}{(t-\tau)^{1-\alpha}}-\frac{1}{(t_{0}-\tau)^{1-\alpha}}\big{|}d\tau+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{t_{0}}^{t}\frac{1}{(t-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{0}^{t_{0}}[\frac{1}{(t_{0}-\tau)^{1-\alpha}}-\frac{1}{(t-\tau)^{1-\alpha}}]d\tau+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{t_{0}}^{t}\frac{1}{(t-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}[(t-t_{0})^{\alpha}+t_{0}^{\alpha}-t^{\alpha}]+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t-t_{0})^{\alpha}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t-t_{0})^{\alpha}+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t-t_{0})^{\alpha}$

This implies that

$\|(I_{0+}^{\alpha}\phi)(t)-(I_{0+}^{\alpha}f)(t_{0})\|_{X}\leq
K_{1}|t-t_{0}|^{\alpha}.$

Let $t<t_{0}$. Then we write

$\displaystyle(I_{0+}^{\alpha}\phi)(t_{0})-(I_{0+}^{\alpha}f)(t)$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}d\tau+\frac{1}{\Gamma(\alpha)}\int_{t}^{t_{0}}\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}d\tau-\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}[\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}-\frac{f(\tau)}{(t-\tau)^{1-\alpha}}]d\tau+\frac{1}{\Gamma(\alpha)}\int_{t}^{t_{0}}\frac{f(\tau)}{(t_{0}-\tau)^{1-\alpha}}d\tau.$

Therefore, we have

$\displaystyle\|(I_{0+}^{\alpha}\phi)(t_{0})-(I_{0+}^{\alpha}f)(t)\|_{X}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{0}^{t}\big{|}\frac{1}{(t_{0}-\tau)^{1-\alpha}}-\frac{1}{(t-\tau)^{1-\alpha}}\big{|}d\tau+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{t}^{t_{0}}\frac{1}{(t_{0}-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{0}^{t}\big{[}\frac{1}{(t-\tau)^{1-\alpha}}-\frac{1}{(t_{0}-\tau)^{1-\alpha}}\big{]}d\tau+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{t}^{t_{0}}\frac{1}{(t_{0}-\tau)^{1-\alpha}}d\tau$
$\displaystyle=\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}[t^{\alpha}-t_{0}^{\alpha}+(t_{0}-t)^{\alpha}]+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t_{0}-t)^{\alpha}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t_{0}-t)^{\alpha}+\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\frac{1}{\alpha}(t_{0}-t)^{\alpha}$

This implies that

$\displaystyle\|(I_{0+}^{\alpha}\phi)(t_{0})-(I_{0+}^{\alpha}f)(t)\|_{X}$
$\displaystyle\leq K_{2}|t-t_{0}|^{\alpha}$

Consequently, $I_{0+}^{\alpha}\phi$ is Holder continuous with order $\alpha$.

$\displaystyle\|I_{0+}^{\alpha}\phi(t)\|_{X}\leq\frac{1}{\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}\int_{0}^{t}\frac{1}{(t-s)^{1-\alpha}}ds=\frac{1}{\alpha\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}t^{\alpha}\leq\frac{1}{\alpha\Gamma(\alpha)}\|f\|_{L^{\infty}([0,T],X)}T^{\alpha}$

for all $t\in[0,T]$. Hence (12) is obtained. The proof of the proposition is
completed. $\Box$

Let $J=[0,T]$ and $g:J\to X$ be a $X-$valued function. We consider the
equation of finding $y\in C(J,X)$ such that

$\displaystyle\begin{cases}\frac{d^{\alpha}}{dt^{\alpha}}y(t)=g(t)\quad{\rm
a.e.}\quad t\in J,\\\ y(0)=y_{0}\end{cases}$ (13)

###### Proposition 2.4

Suppose that $g\in L^{1}(J,X)$ and $y\in C(J,X)$. Then $y$ is a solution of
equation (13) if and only if $y$ is a solution of the following integral
equation

$\displaystyle
y(t)=y_{0}+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{g(s)}{(t-s)^{1-\alpha}}ds.$
(14)

Proof. $(\Rightarrow)$. Suppose that $y$ is a solution to (13). Then we have

$g=\frac{d^{\alpha}y}{dt^{\alpha}}=\frac{d}{dt}I_{0+}^{1-\alpha}(y-y(0))\in
L^{1}([0,T],X).$

Hence $I_{0+}^{1-\alpha}(y-y(0))\in W^{1,1}([0,T],X)$. By Proposition 2.1,
$(y-y(0))\in I^{\alpha}_{0+}(L^{1})$. By (8), we have

$I^{\alpha}_{0+}D^{\alpha}_{0+}(y-y(0))=y-y(0)=y-y_{0}.$

We now act $I^{\alpha}_{0+}$ on both sides of (13) to get

$\displaystyle
y-y_{0}=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{g(s)}{(t-s)^{1-\alpha}}ds.$
(15)

Hence $y$ is a solution of equation (14).

$(\Leftarrow)$. Suppose that $y\in C(J,X)$ and satisfies equation (14). Acting
$D_{0+}^{\alpha}$ on both sides of (15) and using (7), we obtain

$\frac{d^{\alpha}y}{dt^{\alpha}}=D_{0+}^{\alpha}(y-y_{0})=D_{0+}^{\alpha}I_{0+}^{\alpha}g=g.$

$\Box$

##  3 Solution existence

In this section, we investigate the existence of solutions to the state
equation (2)-(4) and the existence of optimal solution to problem (1)-(5).

In the sequel, we will assume that $J=[0,T]$, $D=H^{2}(\Omega)\cap
H_{0}^{1}(\Omega)$, $V=H^{1}_{0}(\Omega)$ and $H=L^{2}(\Omega)$. Then we have

$D\hookrightarrow V\hookrightarrow H.$

Besides, since $n=2,3$, the embedded theorem implies that $D\hookrightarrow
C(\overline{\Omega})$ is compact. Given functions $f$ and $g$ on $J$, the
convolution of $f$ and $g$ is denoted by $f\ast g$ which is defined by

$(f\ast g)(t)=\int_{0}^{t}f(t-s)g(s)ds.$

We say $y$ is a strong solution to (2)-(4) if $y\in C(J,H)$ and $\Delta y+u\in
L^{p}(J,H)$ with $p\geq 1$ such that

$\displaystyle\begin{cases}\frac{d^{\alpha}y}{dt^{\alpha}}=\Delta y+u\quad{\rm
a.e.}\quad t\in J,\\\ y(0)=y_{0}.\end{cases}$ (16)

Here $y_{0}\in D$ and $u\in L^{p}(J,H)$ are given.

Let us assume that $y\in C(J,H)$ is a solution to (16). Then $\Delta y+u\in
L^{p}(J,H)$ with $p\geq 1$. By Proposition 2.4, $y$ satisfies the integral
equation

$\displaystyle y(t)=y_{0}+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{\Delta
y(s)}{(t-s)^{1-\alpha}}ds+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}\frac{u(s)}{(t-s)^{1-\alpha}}ds$
(17)

or equivalently

$\displaystyle y(t)=y_{0}+k_{1-\alpha}\ast(\Delta y+u),$ (18)

where

$\displaystyle k_{1-\alpha}(t):=\frac{1}{\Gamma(\alpha)t^{1-\alpha}}.$ (19)

Each solution of (18) is called a mild solution to (2)-(4). It is clear that
any strong solution is a mild solution. We now establish formula representing
solution of (18).

It is know that $\Delta:D\subset H\to H$ is a sectorial operator. Namely, the
resolvent set $\rho(\Delta)$ contains the sector

$S_{\theta}=\\{\lambda\in C:\lambda\neq 0,|{\rm
arg}\lambda|<\theta\\},\pi/2<\theta<\pi,$

and

$\|R(\lambda,\Delta)\|_{L(L^{2}(\Omega))}\leq\frac{M}{\lambda}\quad\forall\lambda\in
S_{\theta}.$

Hence $\Delta$ is a generator of a uniformly bounded analytic semigroup
$\\{S(t),t\geq 0\\}$ (see Theorem 2.4.1 and Proposition 2.4.2 in [21], [20,
Theorem 3.1.2] and [26, Theorem 2.7, p. 211]). Besides, the following formula
for the resolvent operator is valid:

$\displaystyle R(\lambda,A)=(\lambda I-A)^{-1}=\int_{0}^{+\infty}e^{-\lambda
t}S(t)dt\quad\forall t>0.$ (20)

In other words, $R(\lambda,\Delta)$ is the Laplace transform of $S(t)$.

Let us denote by $z$ the Laplace transform of $y$, that is

$z(s)=\mathcal{L}(y)(s):=\int_{0}^{+\infty}e^{-st}y(t)dt.$

By taking Laplace transform both sides of (18) and using the property of
Laplace transform for convolution (see [18, Theorem 1, p. 232]), we get

$\displaystyle z(\lambda)$
$\displaystyle=\frac{y_{0}}{\lambda}+\frac{1}{\lambda^{\alpha}}\Delta
z(\lambda)+\frac{1}{\lambda^{\alpha}}\mathcal{L}(u)(\lambda).$

This implies that

$\displaystyle
z(\lambda)=(\lambda^{\alpha}I-\Delta)^{-1}[\lambda^{\alpha-1}y_{0}]+(\lambda^{\alpha}I-\Delta)^{-1}[\mathcal{L}(u)(\lambda)].$
(21)

By taking the inverse Laplace transform, we obtain

$\displaystyle y(t)=F(t)y_{0}+\int_{0}^{t}E(t-s)u(s)ds,$

where

$\displaystyle F(t)=\frac{1}{2\pi i}\int_{\Gamma_{\theta,\delta}}e^{\lambda
t}\lambda^{\alpha-1}(\lambda^{\alpha}I+\Delta)^{-1}d\lambda,$ $\displaystyle
E(t)=\frac{1}{2\pi i}\int_{\Gamma_{\theta,\delta}}e^{\lambda
t}(\lambda^{\alpha}I+\Delta)^{-1}d\lambda.$

Here $\Gamma_{\theta,\delta}$ is a Hankel contour, oriented with an increasing
imaginary part. Namely,

$\Gamma_{\theta,\delta}=\\{\lambda\in C:|\lambda|=\delta,|{\rm
arg}\lambda|<\theta\\}\cup\\{\lambda\in C:\lambda=\rho e^{\pm
i\theta},\rho\geq\delta\\}.$

We refer the reader to [15, Theorem 6.4] for properties of $F$ and $E$.

There is another way to represent solution $y(t)$ of (18). Let $\phi_{\alpha}$
be the one-sided probability density function such that

$\int_{0}^{\infty}\phi_{\alpha}(\tau)d\tau=1\quad\text{and}\quad\int_{0}^{\infty}e^{-\lambda\tau}\phi_{\alpha}(\tau)d\tau=e^{-\lambda^{\alpha}}.$

For the explicit formula of $\phi_{\alpha}$, we refer the reader to [23] and
[29]. Moreover, from the explicit formula of $\phi_{\alpha}$ in [23], we can
show (see [27]) that

$\int_{0}^{\infty}t^{\mu}\phi_{\alpha}(t)dt=\frac{\Gamma(-\mu/\alpha)}{\alpha\Gamma(-\mu)}.$
(22)

Let us define

$\zeta_{\alpha}(\tau)=\frac{1}{\alpha\tau^{1+1/\alpha}}\phi_{\alpha}(\frac{1}{\tau^{1/\alpha}}).$

Then we can check that $\int_{0}^{\infty}\zeta_{\alpha}(\tau)d\tau=1$.
Moreover, using (22), we can show that

$\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)d\theta=\frac{1}{\alpha\Gamma(\alpha)}.$

Using (21), we have

$\displaystyle z(\lambda)$
$\displaystyle=(\lambda^{\alpha}I-A)^{-1}[h(\lambda)]=\int_{0}^{\infty}e^{-\lambda^{\alpha}s}S(s)h(\lambda)ds$
$\displaystyle=\int_{0}^{\infty}e^{-(\lambda
s^{1/\alpha})^{\alpha}}S(s)h(\lambda)ds$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}e^{-\lambda
s^{1/\alpha}\tau}\phi_{\alpha}(\tau)d\tau S(s)h(\lambda)ds$
$\displaystyle=\int_{0}^{\infty}\int_{0}^{\infty}e^{-\lambda
t}\frac{1}{s^{1/\alpha}}\phi_{\alpha}(\frac{t}{s^{1/\alpha}})dtS(s)h(\lambda)ds$
$\displaystyle=\int_{0}^{\infty}e^{-\lambda
t}[\int_{0}^{\infty}\frac{1}{s^{1/\alpha}}\phi_{\alpha}(\frac{t}{s^{1/\alpha}})S(s)h(\lambda)ds]dt.$

Here $h(\lambda):=\lambda^{\alpha-1}y_{0}+\mathcal{L}(u)(\lambda)$. By
changing variable $s=\theta t^{\alpha}$, we have

$\displaystyle z(\lambda)=\int_{0}^{\infty}e^{-\lambda
t}[\int_{0}^{\infty}\alpha t^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta
t^{\alpha})h(\lambda)d\theta]dt.$

By defining

$\displaystyle P(t)=\int_{0}^{\infty}\alpha
t^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta t^{\alpha})d\theta,$ (23)

we get

$\displaystyle z(\lambda)$
$\displaystyle=\mathcal{L}(P)(\lambda)[\lambda^{\alpha-1}y_{0}+\mathcal{L}(u)(\lambda)]$
$\displaystyle=\mathcal{L}((\frac{1}{\Gamma(1-\alpha)t^{\alpha}})\ast
P)(\lambda)y_{0}+\mathcal{L}(P\ast u)(\lambda).$

By acting the converse Laplace transform on both sides and using the
convolution property of Laplace transform, we obtain

$\displaystyle
y(t)=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}P(s)y_{0}ds+\int_{0}^{t}P(t-s)u(s)ds$
(24)
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}\int_{0}^{\infty}\alpha
s^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta s^{\alpha})y_{0}d\theta
ds+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta(t-s)^{\alpha})d\theta
u(s)ds$ (25)
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{1}\frac{1}{(1-\tau)^{\alpha}\tau^{1-\alpha}}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta(t\tau)^{\alpha})y_{0}d\theta
d\tau$
$\displaystyle+t^{\alpha}\int_{0}^{1}(1-\tau)^{\alpha-1}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta
t^{\alpha}(1-\tau)^{\alpha})u(t\tau)d\theta d\tau.$ (26)

Based on the representation, we have

###### Proposition 3.1

Let $y_{0}\in D=H^{2}(\Omega)\cap H_{0}^{1}(\Omega)$. The following assertions
are fulfilled:

$(i)$ If $u\in L^{p}(J,H)$ with $p>1/\alpha$, then equation (18) has a unique
mild solution $y\in C(J,H)$ which is given by (26) and there exist constants
$K>0$ and $K^{\prime}>0$ such that

$\displaystyle\|y\|_{C(J,H)}\leq
K\|y_{0}\|_{H}+K^{\prime}T^{\alpha-1/p}\|u\|_{L^{p}(J,H)}.$ (27)

Moreover, $y(t)\in D$ for a.e. $t\in[0,T]$.

$(ii)$ If $u\in C^{\gamma}(J,X)$ with $0<\gamma<1$, then equation (18) has a
unique solution $y\in C(J,D)\cap C^{\gamma}(J,X)$ and there exist positive
constants $K,K_{1}$ and $K_{2}$ such that

$\displaystyle\|y\|_{C(J,D)}\leq
K\|y_{0}\|_{D}+K_{1}\|u\|_{C^{\gamma}(J,X)}+K_{2}\|u\|_{C(J,X)}.$ (28)

In this case $y$ is also a strong solution to (2)-(4).

Proof. $(i)$ Let us represent $y$ by (24)-(26). We now show that $\lim_{t\to
0+}y(t)=y_{0}$. In fact, it is clear that

$\displaystyle\lim_{t\to 0+}y(t)$
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{1}\frac{1}{(1-\tau)^{\alpha}\tau^{1-\alpha}}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(0)y_{0}d\theta
d\tau+0$
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{1}\frac{1}{(1-\tau)^{\alpha}\tau^{1-\alpha}}d\tau\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)y_{0}d\theta$
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha)}B(\alpha,1-\alpha)\frac{1}{\alpha\Gamma(\alpha)}y_{0}$
$\displaystyle=\frac{\alpha}{\Gamma(1-\alpha)}\frac{\Gamma(\alpha)\Gamma(1-\alpha)}{\Gamma(1)}\frac{1}{\alpha\Gamma(\alpha)}y_{0}=y_{0}.$

Here the beta function is defined by

$B(a,b)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},0<a,b<+\infty.$

Note that $S(\tau)$ is uniform bounded. Hence there exists a constant $K>0$
such that

$\displaystyle\|P(t)\|_{H}\leq Kt^{\alpha-1}\quad\forall t>0.$

We now claim that $y\in C([0,T],H)$. Take any $t_{0},t\in[0,T]$ and assume
that $t_{0}<t$. Then we have

$\displaystyle\|y(t)-y(t_{0})\|_{H}=\big{\|}\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{1}\frac{1}{(1-\tau)^{\alpha}\tau^{1-\alpha}}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)[S(\theta(t\tau)^{\alpha})-S(\theta(t_{0}\tau)^{\alpha}))]y_{0}d\theta
d\tau$
$\displaystyle+t^{\alpha}\int_{0}^{1}(1-\tau)^{\alpha-1}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta
t^{\alpha}(1-\tau)^{\alpha})u(t\tau)d\theta d\tau$ $\displaystyle-
t_{0}^{\alpha}\int_{0}^{1}(1-\tau)^{\alpha-1}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta
t_{0}^{\alpha}(1-\tau)^{\alpha})u(t_{0}\tau)d\theta d\tau\big{\|}.$
$\displaystyle\leq\frac{\alpha}{\Gamma(1-\alpha)}\int_{0}^{1}\frac{1}{(1-\tau)^{\alpha}\tau^{1-\alpha}}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)\|S(\theta(t\tau)^{\alpha})-S(\theta(t_{0}\tau)^{\alpha}))\|\|y_{0}\|_{H}d\theta
d\tau$ $\displaystyle+$
$\displaystyle\big{\|}t^{\alpha}\int_{0}^{1}(1-\tau)^{\alpha-1}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta
t^{\alpha}(1-\tau)^{\alpha})u(t\tau)d\theta d\tau$ $\displaystyle-
t_{0}^{\alpha}\int_{0}^{1}(1-\tau)^{\alpha-1}\int_{0}^{\infty}\theta\zeta_{\alpha}(\theta)S(\theta
t_{0}^{\alpha}(1-\tau)^{\alpha})u(t_{0}\tau)d\theta d\tau\big{\|}$
$\displaystyle:=I(t)+II(t).$ (29)

Since $S(t)$ is continuous and uniformly bounded, the dominated convergence
theorem implies that $I(t)\to 0$ as $t\to t_{0}$. To estimate $II(t)$, we
write

$\displaystyle
II(t)=\big{\|}\int_{0}^{t}P(t-s)u(s)ds-\int_{0}^{t_{0}}P(t_{0}-s)u(s)ds\big{\|}_{H}$
$\displaystyle=\big{\|}\int_{0}^{t_{0}}(P(t-s)-P(t_{0}-s))u(s)ds+\int_{t_{0}}^{t}P(t-s)u(s)ds\big{\|}_{H}$
$\displaystyle=\big{\|}\int_{0}^{t_{0}}[\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta(t-s)^{\alpha})d\theta-\int_{0}^{\infty}\alpha(t_{0}-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta(t_{0}-s)^{\alpha})d\theta]u(s)ds$
$\displaystyle+\int_{t_{0}}^{t}P(t-s)u(s)ds\big{\|}_{H}$
$\displaystyle=\big{\|}\int_{0}^{t_{0}}[\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)(S(\theta(t-s)^{\alpha})-S(\theta(t_{0}-s)^{\alpha})d\theta$
$\displaystyle+\int_{0}^{t_{0}}\int_{0}^{\infty}\alpha((t-s)^{\alpha-1}-(t_{0}-s)^{\alpha-1})\theta\zeta_{\alpha}(\theta)S(\theta(t_{0}-s)^{\alpha})d\theta]u(s)ds$
$\displaystyle+\int_{t_{0}}^{t}P(t-s)u(s)ds\big{\|}_{H}$
$\displaystyle\leq\|u\|_{L^{p}(J,H)}\big{[}\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|S(\theta(t-s)^{\alpha})-S(\theta(t_{0}-s)^{\alpha})\|d\theta\big{)}^{q}ds\big{]}^{1/q}$
$\displaystyle+\|u\|_{L^{p}(J,H)}\big{[}\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}|(t-s)^{\alpha-1}-(t_{0}-s)^{\alpha-1}|\theta\zeta_{\alpha}(\theta)\|S(\theta(t_{0}-s)^{\alpha})\|d\theta\big{)}^{q}ds\big{]}^{1/q}$
$\displaystyle+\|u\|_{L^{p}(J,H)}\big{[}\int_{t_{0}}^{t}\|P(t-s)\|^{q}ds\big{]}^{1/q}$
$\displaystyle\leq\|u\|_{L^{p}(J,H)}\big{[}\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|S(\theta(t-s)^{\alpha})-S(\theta(t_{0}-s)^{\alpha})\|d\theta\big{)}^{q}ds\big{]}^{1/q}$
$\displaystyle+\|u\|_{L^{p}(J,H)}\big{[}\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}|(t-s)^{\alpha-1}-(t_{0}-s)^{\alpha-1}|\theta\zeta_{\alpha}(\theta)\|S(\theta(t_{0}-s)^{\alpha})\|d\theta\big{)}^{q}ds\big{]}^{1/q}$
$\displaystyle+\|u\|_{L^{p}(J,H)}K(t-t_{0})^{(q(\alpha-1)+1)/q}$
$\displaystyle:=II_{1}(t)+II_{2}(t)+II_{3}(t).$

Since $p>1/\alpha$, $q(\alpha-1)+1>0$. Hence $II_{3}(t)\to 0$ as $t\to
t_{0}+$. By the continuity of $S$,

$\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|S(\theta(t-s)^{\alpha})-S(\theta(t_{0}-s)^{\alpha})\|\to
0\quad{\rm as}\quad t\to t_{0}+$

and

$\displaystyle\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|S(\theta(t-s)^{\alpha})-S(\theta(t_{0}-s)^{\alpha})\|d\theta\big{)}^{q}ds$
$\displaystyle\leq
2K\int_{0}^{t_{0}}\big{(}\int_{0}^{\infty}\alpha(t_{0}-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)d\theta\big{)}^{q}ds$
$\displaystyle=\frac{(2K)^{q}}{\Gamma(\alpha+1)^{q}}t_{0}^{q(\alpha-1)+1}.$

The dominated implies that $II_{1}(t)\to 0$ when $t\to t_{0}+$. By the same
argument, we can show that $II_{2}(t)\to 0$ as $t\to t_{0}+$. Combining these
we conclude that

$\lim_{t\to t_{0}+}\|y(t)-y(t_{0})\|_{X}=0.$

Repeating the procedure, we also have

$\lim_{t\to t_{0}-}\|y(t)-y(t_{0})\|_{X}=0.$

Hence $y$ is continuous at $t_{0}$. Since $t_{0}$ is arbitrary in $[0,T]$, we
have $y\in C([0,T],H)$. To prove (27), we use (25) and the fact that
$\|S(t)\|\leq K$ for all $t>0$. Then we have

$\displaystyle\|y(t)\|_{H}$
$\displaystyle\leq\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}(t-s)^{-\alpha}s^{\alpha-1}\int_{0}^{\infty}\alpha\theta\zeta_{\alpha}(\theta)\|S(\theta
s^{\alpha})\|\|y_{0}\|_{H}d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|S(\theta(t-s)^{\alpha})\|\|u(s)\|_{H}d\theta
ds$ $\displaystyle\leq
K\|y_{0}\|_{H}+\frac{K}{\Gamma(\alpha)}[\int_{0}^{t}(t-s)^{q(\alpha-1)}ds]^{1/q}\|u\|_{L^{p}(J,H)}$
$\displaystyle\leq
K\|y_{0}\|_{H}+\frac{K}{\Gamma(\alpha)}T^{\alpha-1/p}\|u\|_{L^{p}(J,H)}$

This leads to (27).

$(ii)$. Note that $S^{\prime}(t)=\Delta S(t)$. Since $y_{0}\in D$, $\Delta
S(t)y_{0}=S(t)\Delta y_{0}$. Therefore, we have from (25) that

$\displaystyle\Delta y(t)$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}\int_{0}^{\infty}\alpha
s^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta s^{\alpha})\Delta
y_{0}d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\Delta
S(\theta(t-s)^{\alpha})u(s)d\theta ds$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}\int_{0}^{\infty}\alpha
s^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta s^{\alpha})\Delta
y_{0}d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\Delta
S(\theta(t-s)^{\alpha})[u(s)-u(t)]d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\Delta
S(\theta(t-s)^{\alpha})u(t)d\theta dt$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}\int_{0}^{\infty}\alpha
s^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta s^{\alpha})\Delta
y_{0}d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\Delta
S(\theta(t-s)^{\alpha})[u(s)-u(t)]d\theta ds$
$\displaystyle+\int_{0}^{\infty}\zeta_{\alpha}(\theta)\int_{0}^{t}\frac{d}{ds}[S(\theta(t-s)^{\alpha})]u(t)dsd\theta$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\int_{0}^{t}\frac{1}{(t-s)^{\alpha}}\int_{0}^{\infty}\alpha
s^{\alpha-1}\theta\zeta_{\alpha}(\theta)S(\theta s^{\alpha})\Delta
y_{0}d\theta ds$
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\Delta
S(\theta(t-s)^{\alpha})[u(s)-u(t)]d\theta ds$
$\displaystyle+u(t)+\int_{0}^{\infty}\zeta_{\alpha}(\theta)S(\theta
t^{\alpha})u(t)d\theta.$

It follows that

$\displaystyle\|\Delta y(t)\|_{H}$ $\displaystyle\leq K\|\Delta
y_{0}\|_{H}+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{\alpha-1}\theta\zeta_{\alpha}(\theta)\|\Delta
S(\theta(t-s)^{\alpha})\|\|u(s)-u(t)\|_{H}d\theta ds+(1+K)\|u(t)\|_{H}$
$\displaystyle\leq K\|\Delta
y_{0}\|_{H}+\int_{0}^{t}\int_{0}^{\infty}\alpha(t-s)^{-1}\zeta_{\alpha}(\theta)K^{\prime}\|u(s)-u(t)\|_{H}d\theta
ds+(1+K)\|u(t)\|_{H}$ $\displaystyle\leq K\|\Delta
y_{0}\|_{H}+\int_{0}^{t}\alpha(t-s)^{-1}K^{\prime}\|u\|_{C^{\gamma}([0,T],H)}|t-s|^{\gamma}ds+(1+K)\|u(t)\|_{H}$
$\displaystyle=K\|y_{0}\|_{D}+K^{\prime}\|u\|_{C^{\gamma}([0,T],H)}\frac{t^{\gamma}}{\gamma}+(1+K)\|u(t)\|_{H}.$

This implies that

$\displaystyle\|y\|_{C(J,D)}\leq
K\|y_{0}\|_{D}+K^{\prime}\|u\|_{C^{\gamma}([0,T],H)}\frac{T^{\gamma}}{\gamma}+(1+K)\|u\|_{C(J,H)}.$

Hence (28) is obtained. Since $\Delta y\in C(J,H)$, we have $\Delta y+u\in
L^{\infty}(J,H)$. By Proposition 2.3, $y\in C^{\alpha}(J,H)$. By Proposition
2.4, $y$ is a strong solution to (2)-(4). $\Box$

###### Proposition 3.2

Suppose that $y_{0}\in D$ and $u\in L^{p^{\prime}}([0,T],H)$ with
$1<p^{\prime}<\infty$. Then equation (18) has a unique solution $y\in
L^{p^{\prime}}([0,T],D)$ such that $\frac{d^{\alpha}y}{dt^{\alpha}}\in
L^{p^{\prime}}([0,T],H)$ and

$\displaystyle\|y\|_{L^{p^{\prime}}([0,T],D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}y\|_{L^{p^{\prime}}([0,T],H)}\leq
C_{1}\|u\|_{L^{p^{\prime}}([0,T],H)}+C_{2}\|\Delta y_{0}\|_{H},$ (30)

where $C_{1}$ and $C_{2}$ are positive constants which are independent of
$y,f$ and $y_{0}$. In addition if $p^{\prime}>1/\alpha$, then $y\in
C([0,T],H)\cap L^{p^{\prime}}([0,T],D)$. In this case $y$ is a strong solution
to (2)-(4).

Proof. Define $v(t)=y(t)-y_{0}$. Then $v(0)=0$, $y(t)=v(t)+y_{0}$ and
$D_{0+}^{\alpha}v=\frac{d^{\alpha}v}{dt^{\alpha}}$. By a simple computation,
we have $\frac{d^{\alpha}y}{dt}=\frac{d^{\alpha}v}{dt^{\alpha}}$. Then
equation (16) becomes

$\displaystyle\begin{cases}\frac{d^{\alpha}v}{dt^{\alpha}}=\Delta v+(u+\Delta
y_{0})\\\ v(0)=0.\end{cases}$ (31)

Since $u\in L^{p^{\prime}}(J,H)$, we have $u+\Delta y_{0}\in
L^{p^{\prime}}(J,H)$. By Theorem 6.11 in [15], equation (31) has a unique mild
solution $v\in L^{p^{\prime}}(J,D)$ with $\frac{d^{\alpha}v}{dt^{\alpha}}\in
L^{p^{\prime}}(J,H)$ and

$\displaystyle\|v\|_{L^{p^{\prime}}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}v\|_{L^{p^{\prime}}(J,H)}\leq
C\|f+\Delta y_{0}\|_{L^{p^{\prime}}(J,H)}.$

It follows that equation (17) has a unique solution $y\in L^{p^{\prime}}(J,D)$
and $\frac{d^{\alpha}y}{dt^{\alpha}}\in L^{p^{\prime}}(J,H)$ such that

$\displaystyle\|y\|_{L^{p^{\prime}}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}y\|_{L^{p^{\prime}}(J,H)}\leq
C\|f\|_{L^{p^{\prime}}(J,H)}+(1+c)T^{\frac{1}{p^{\prime}}}\|\Delta
y_{0}\|_{H},$

where $c$ is a positive constant which is independent of $f$ and $y_{0}$. If
$p^{\prime}>1/\alpha$, then Proposition 3.1 implies that $y\in C(J,H)$.
Finally, since $\Delta y+u\in L^{p^{\prime}}(J,H)$, Proposition 2.4 implies
that $y$ is a strong solution to (2)-(4). The proof of the proposition is
completed. $\Box$

In the sequel, we shall denote by $\Phi$ the feasible set of problem (1)-(5).
In order to establish a result for the existence of optimal solution to
problem (1)-(5), we make the following assumption

$(A1)$ $L:[0,T]\times\Omega\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is a
Carathéodory function, that is, for each $y,u\in\mathbb{R}$,
$L(\cdot,\cdot,y,u)$ is integrable in $(t,x)$ and for each
$(t,x)\in[0,T]\times\Omega$, $L(t,x,\cdot,\cdot)$ is continuous. Moreover,
there exist functions $\phi_{1},\phi_{2}\in L_{1}(Q)$, nonnegative numbers
$a_{1},b_{1},\beta_{1},\beta_{2}$ and positive numbers $a_{2},b_{2}$ such that

$\displaystyle\phi_{1}(t,x)-a_{1}|y|^{p}-b_{1}|u|^{\beta_{1}}\leq
L(t,x,y,u)\leq\phi_{2}(t,x)+a_{2}|y|^{p}+b_{2}|u|^{\beta_{2}}$ (32)

for all $(t,x,y,u)\in[0,T]\times\Omega\times\mathbb{R}\times\mathbb{R}.$

$(A2)$ $L(t,x,\cdot,\cdot)$ is convex in $(y,u)$ for each fixed
$(t,x)\in[0,T]\times\Omega$.

$(A3)$ $L(t,x,y,\cdot)$ is convex in $u$ for each fixed
$(t,x,y)\in[0,T]\times\Omega\times\mathbb{R}$, $p>1/\alpha$ and $p\leq 6$
whenever $N=3$.

###### Theorem 3.1

Suppose that $(A1)$ and $(A2)$ are satisfied or $(A1)$ and $(A3)$ are
satisfied. Then problem (1)-(5) has an optimal solution.

Proof. Let $\xi=\inf_{(y,u)\in\Phi}\psi(y,u)$. Then for each $(y,u)\in\Phi$,
$(y,u)$ satisfies equation (16) and inequality (30) is valid. It follows that

$\displaystyle\|y\|_{L^{p}(Q)}+\|\frac{d^{\alpha}}{dt^{\alpha}}y\|_{L^{p}(J,H)}$
$\displaystyle\leq\|y\|_{L^{p}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}y\|_{L^{p}(J,H)}$
$\displaystyle\leq C_{1}\|u\|_{L^{p}(J,H)}+C_{2}\|\Delta y_{0}\|_{H}$
$\displaystyle\leq C^{\prime}\|u\|_{L^{\infty}(Q)}+C_{2}^{\prime}\|\Delta
y_{0}\|_{H}.$

This and (32) imply that $\xi$ is finite, that is, $\xi\in\mathbb{R}$.

Let $(y_{n},u_{n})\in\Phi$ such that $V=\lim_{n\to\infty}\psi(y_{n},u_{n})$.
Since $a\leq u_{n}\leq b$, $\\{u_{n}\\}$ is bounded in $L^{\infty}(\Omega)$.
Therefore, we may assume that $u_{n}$ converges in
$\sigma(L^{\infty}(Q),L^{1}(Q))$, the weak star topology, to a function
$\overline{u}\in L^{\infty}(Q)$. From the above inequality, we see that
$\\{y_{n}\\}$ is bounded in $L^{p}(J,D)$, $\\{\Delta y_{n}\\}$ and
$\\{\frac{d^{\alpha}}{dt^{\alpha}}y_{n}\\}$ is bounded in $L^{p}(J,H)$.
Therefore, we may assume that $y_{n}$ converges weakly to $\overline{y}$ in
$L^{p}(J,D)$ and so $\Delta y_{n}\rightharpoonup\Delta y$ in $L^{p}(J,H)$. Let
us claim that $(\overline{y},\overline{u})\in\Phi$. Since
$(y_{n},u_{n})\in\Phi$, from (18) we have

$y_{n}=y_{0}+k_{1-\alpha}\ast(\Delta y_{n}+u_{n}).$

Taking any $\phi\in L^{q}(J,H)$, we have

$\displaystyle(y_{n}(t),\phi(t))_{H}=(y_{0},\phi(t))_{H}+\frac{1}{\Gamma(\alpha)}\int_{0}^{t}((t-s)^{\alpha-1}\phi(t),\Delta
y_{n}(s)+u_{n}(s))_{H}ds.$ (33)

Integrating both sides of (33) on $[0,T]$ and using Fubini’s theorem, we have

$\displaystyle\int_{0}^{T}(y_{n}(t),\phi(t))_{H}dt=\int_{0}^{T}(y_{0},\phi(t))_{H}dt+\frac{1}{\Gamma(\alpha)}\int_{0}^{T}\int_{0}^{t}((t-s)^{\alpha-1}\phi(t),\Delta
y_{n}(s)+u_{n}(s))_{H}dsdt$
$\displaystyle=\int_{0}^{T}(y_{0},\phi(t))_{H}dt+\frac{1}{\Gamma(\alpha)}\int_{0}^{T}ds\int_{s}^{T}((t-s)^{\alpha-1}\phi(t),\Delta
y_{n}(s)+u_{n}(s))_{H}dt$
$\displaystyle=\int_{0}^{T}(y_{0},\phi(t))_{H}dt+\frac{1}{\Gamma(\alpha)}\int_{0}^{T}(\int_{s}^{T}(t-s)^{\alpha-1}\phi(t)dt,\Delta
y_{n}(s)+u_{n}(s))_{H}ds$
$\displaystyle=\int_{0}^{T}(y_{0},\phi(t))_{H}dt+\frac{1}{\Gamma(\alpha)}\int_{0}^{T}(\int_{s^{\prime}+T}^{T}(t-s^{\prime}-T)^{\alpha-1}\phi(t)dt,\Delta
y_{n}(s^{\prime}+T)+u_{n}(s^{\prime}+T))_{H}ds^{\prime}$
$\displaystyle=\int_{0}^{T}(y_{0},\phi(t))_{H}dt-\frac{1}{\Gamma(\alpha)}\int_{0}^{T}(\int_{0}^{s^{\prime}}(t^{\prime}-s^{\prime})^{\alpha-1}\phi(t^{\prime}+T)dt^{\prime},\Delta
y_{n}(s^{\prime}+T)+u_{n}(s^{\prime}+T))_{H}ds^{\prime}$
$\displaystyle=\int_{0}^{T}(y_{0},\phi(t))_{H}dt-\int_{0}^{T}(k_{1-\alpha}\ast\phi(\cdot+T)(s^{\prime}),\Delta
y_{n}(s^{\prime}+T)+u_{n}(s^{\prime}+T))_{H}ds^{\prime}.$

By a property for convolution, we see that the function $s^{\prime}\mapsto
k_{1-\alpha}\ast\phi(\cdot+T)(s^{\prime})$ belongs to $L^{q}(J,H)$. Thus we
have shown that

$\int_{0}^{T}(y_{n}(t),\phi(t))_{H}dt=\int_{0}^{T}(y_{0},\phi(t))_{H}dt-\int_{0}^{T}(k_{1-\alpha}\ast\phi(\cdot+T)(s^{\prime}),\Delta
y_{n}(s^{\prime}+T)+u_{n}(s^{\prime}+T))_{H}ds^{\prime}.$

By letting $n\to\infty$, we obtain

$\int_{0}^{T}(\overline{y}(t),\phi(t))_{H}dt=\int_{0}^{T}(y_{0},\phi(t))_{H}dt-\int_{0}^{T}(k_{1-\alpha}\ast\phi(\cdot+T)(s^{\prime}),\Delta\overline{y}(s^{\prime}+T)+\overline{u}(s^{\prime}+T))_{H}ds^{\prime}.$

This implies that

$\overline{y}=y_{0}+k_{1-\alpha}\ast(\Delta\overline{y}+\overline{u}).$

It remains to show that $a\leq\overline{u}(t,x)\leq b$ for a.e. $(t,x)\in Q$.
In fact, the set

$K:=\\{u\in L^{p}(Q):a\leq\overline{u}(t,x)\leq b,{\rm a.e.}\\}$

is a closed convex set in $L^{p}(Q)$. Hence it is a weakly closed set. Since
$u_{n}\in K$ and $u_{n}\rightharpoonup\overline{u}$ in $L^{p}(Q)$, we have
$\overline{u}\in K$. In a summary, we have showed that
$(y_{n},u_{n})\rightharpoonup(\overline{y},\overline{u})$ in $L^{p}(J,D)\times
L^{p}(Q)$ and $(\overline{y},\overline{u})\in\Phi.$

$(i)$ If $(A1)$ and $(A2)$ are valid, then the functional $\psi(y,u)$ is
sequentially lower semicontinuous on $L^{p}(Q)\times L^{p}(Q)$. Since
$L^{p}(J,D)\times L^{p}(Q)\hookrightarrow L^{p}(Q)\times L^{p}(Q)$, we see
that $(y_{n},u_{n})\rightharpoonup(\overline{y},\overline{u})$ in
$L^{p}(Q)\times L^{p}(Q)$. It follows that

$\xi=\lim_{n\to\infty}\psi(y_{n},u_{n})\geq\psi(\overline{y},\overline{u}).$

Hence $(\overline{y},\overline{u})$ is an optimal solution to problem (1)-(5).

$(ii)$ Suppose that $(A1)$ and $(A3)$ are valid. We claim that
$y_{n}\to\overline{y}$ strongly in $L^{p}(Q)$. In fact, it is known that the
embeddings

$D\hookrightarrow V\hookrightarrow H$

are compact. Since $1/p<\alpha<1$, we have

$\displaystyle\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\|y_{n}(s)\|_{D}ds$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}(\int_{0}^{t}(t-s)^{(\alpha-1)q}ds)^{1/q}\|y_{n}\|_{L^{p}(J,D)}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\frac{1}{((\alpha-1)q+1)^{1/q}}t^{(\alpha-1)+1/q}\|y_{n}\|_{L^{p}(J,D)}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\frac{1}{((\alpha-1)q)^{1/q}}T^{\alpha-1/p}M:=M_{1}.$

Hence

$\sup_{t\in(0,T)}\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\|y_{n}(s)\|_{D}ds\leq
M_{1}.$

Besides, we have $p>1$ and $p>\frac{p}{1+p\alpha}$, and
$\\{D_{0+}^{\alpha}y_{n}\\}=\\{\frac{d^{\alpha}}{dt^{\alpha}}y_{n}\\}$ is
bounded in $L^{2}(J,H)$. By Theorem 4.1 in [19], we see that $\\{y_{n}\\}$ is
a relatively compact set in $L^{p}(J,V)$. Therefore, we may assume that
$z_{k}\to z$ strongly in $L^{p}(J,V)$. If $N=2$, then $V\hookrightarrow
L^{\infty}(\Omega)$. If $N=3$, then $V\hookrightarrow L^{6}(\Omega)$. Since
$1<p\leq 6$, we have $L^{p}(J,V)\hookrightarrow
L^{p}(J,L^{6}(\Omega))\hookrightarrow L^{p}(Q)$. Thus we have shown that
$y_{n}\to\overline{y}$ strongly in $L^{p}(Q)$ and
$u_{n}\rightharpoonup\overline{u}$ in $L^{p}(Q)$. By $(A3)$, the functional
$\psi(y,u)$ is sequentially lower semicontinuous (see Theorem 3.23 in [9]).
Hence we have

$\xi=\lim_{n\to\infty}\psi(y_{n},u_{n})\geq\psi(\overline{y},\overline{u})$

and so $(\overline{y},\overline{u})$ is an optimal solution to problem
(1)-(5). The proof of the theorem is completed. $\Box$

##  4 Optimality conditions

We now impose the following assumption for $L$ which makes sure that $\psi$ is
of class $C^{2}$ around a feasible point $(\overline{y},\overline{u})$. Given
$1<p<+\infty$, we denote by $q$ the conjugate number of $p$, that is, $q>1$
and $1/p+1/q=1$.

$(A4)$ $L:[0,T]\times\Omega\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is a
Carathéodory function of class $C^{2}$ with respect to variable $(y,u)$ and
satisfies the property that for each $M>0$, there exist a positive number
$k_{LM}$ and a function $r_{M}\in L^{\infty}(Q)$ such that

$\displaystyle|L_{y}(t,x,y,u)|+|L_{u}(t,x,y,u)|\leq
k_{LM}\left(|y|^{p-1}+|u|\right)+r_{M}(t,x),$
$\displaystyle|L_{u}(t,x,y_{1},u_{1})-L_{u}(t,x,y_{2},u_{2})|\leq
k_{LM}(|y_{1}-y_{2}|+|u_{1}-u_{2}|),$
$\displaystyle|L_{y}(t,x,y_{1},u)-L_{y}(t,x,y_{2},u)|\leq
k_{LM}\sum_{j=0}^{[p-1]}|y_{1}-y_{2}|^{p-1-j}|y_{2}|^{j}$

for all $u,u_{1},u_{2}\in\mathbb{R}$ satisfying $|u|,|u_{i}|\leq M$ and any
$y_{1},y_{2}\in\mathbb{R}$. Also for each $M>0$, there is a number $k_{LM}>0$
such that

$\displaystyle\big{|}L_{uu}(t,x,y_{1},u_{1})-L_{uu}(t,x,y_{2},u_{2})\big{|}\leq
k_{LM}(|y_{1}-y_{2}|+|u_{1}-u_{2}|),$
$\displaystyle\big{|}L_{yu}(t,x,y_{1},u_{1})-L_{yu}(t,x,y_{2},u_{2})\big{|}\leq
k_{LM}(\varepsilon|y_{1}-y_{2}|^{p-1}+|u_{1}-u_{2}|)$

( with $\varepsilon=0$ if $p=2$ and $\varepsilon=1$ if $p>2$ ) and

$\displaystyle|L_{yy}(t,x,y_{1},u)-L_{yy}(t,x,y_{2},u)|\begin{cases}=0&\text{if}\quad
p=2,\\\ \leq
k_{LM}\sum_{j=0}^{[p-2]}|y_{1}-y_{2}|^{p-2-j}|y_{2}|^{j}&\text{if}\quad p>2\\\
\end{cases}$ (34)

for a.e. $x\in\Omega$, for all $u,u_{i},y_{i}\in\mathbb{R}$ with
$|u|,|u_{i}|\leq M$ and $i=1,2,$ where $[p-2]$ is the integer part of $p-2$.
Moreover, we require that $L(t,x,0,0)\in L^{1}(\Omega)$ and
$L_{yy}(t,x,0,u)\in L^{\infty}(Q)$.

Note that $(A4)$ makes sure that the functional $\psi$ is of class $C^{2}$.
The following example showing that $L$ satisfies $(A4)$.

###### Example 4.1

Let $p\geq 2$ and $p\in\mathbb{N}$. Then

$L(t,x,y,u):=(y-y_{0}(t,x))^{p}+y^{p-1}+\cdots+y^{2}+yu+u^{2}+u^{4}+\cdots+u^{2p}$

satisfies assumption $(A4)$, where $y_{0}\in L^{p}(Q)$ is given.

Let us define the mapping $F:L^{p}(J,D)\times L^{\infty}(Q)\to L^{p}(J,X)$ by
setting

$\displaystyle F(y,u)=y-y_{0}-k_{1-\alpha}\ast(\Delta y+u),$ (35)

where $k_{1-\alpha}$ is given by (19). Put

$\displaystyle K_{\infty}:$ $\displaystyle=\\{u\in L^{\infty}(Q)|a\leq
u(t,x)\leq b\quad{\rm a.e.}\ t\in Q\\}$ (36) $\displaystyle K_{p}:$
$\displaystyle=\\{u\in L^{p}(Q)|a\leq u(t,x)\leq b\quad{\rm a.e.}\ t\in Q\\}.$
(37)

Obviously, $K_{\infty}=K_{p}$. Besides, we can check that $\overline{y}$ is a
strong solution to (2)-(4) corresponding to $\overline{u}\in L^{\infty}(Q)$ if
and only if $\overline{y}\in L^{p}(J,D)$ and $F(\overline{y},\overline{u})=0$.
Therefore, problem (1)-(5) can be formulated in the following form: find
$(\overline{y},\overline{u})\in L^{p}(J,D)\times L^{\infty}(Q)$ which solves

$\displaystyle\psi(y,u)\to{\rm inf}$ (38) $\displaystyle{\rm s.t.}$
$\displaystyle F(y,u)=0,$ (39) $\displaystyle u\in K_{\infty}.$ (40)

In this section we assume that $(\overline{y},\overline{u})\in\Phi$, that is,
$F(\overline{y},\overline{u})=0$ and $\overline{u}\in K_{\infty}$. The symbols

$L[t,x],L_{y}[t,x],L_{u}[t,x],L_{yy}[t,x],L_{uu}[t,x],L_{yu}[t,x]$

stand for

$\displaystyle
L(t,x,\overline{y}(t,x),\overline{u}(t,x)),L_{y}(t,x,\overline{y}(t,x),\overline{u}(t,x)),L_{u}(t,x,\overline{y}(t,x),\overline{u}(t,x))$

and so on, respectively.

###### Proposition 4.1

Suppose that assumption $(A4)$ is valid. Then the mappings $\psi$ and $F$ are
of class $C^{2}$ on $L^{p}(J,D)\times L^{\infty}(Q)$. Moreover, the following
formulae are valid:

$\displaystyle\langle\psi_{y}(\overline{y},\overline{u}),y\rangle=\int_{Q}L_{y}[t,x]y(t,x)dtdx,\
\langle\psi_{u}(\overline{y},\overline{u}),u\rangle=\int_{Q}L_{u}[t,x]u(t,x)dtdx,$
$\displaystyle\langle\psi_{yy}(\overline{y},\overline{u})y_{1},y_{2}\rangle=\int_{Q}L_{yy}[t,x]y_{1}(t,x)y_{2}(t,x)dtdx,\
\langle\psi_{yu}(\overline{y},\overline{u})y,u\rangle=\int_{Q}L_{yu}[t,x]y(t,x)u(t,x)dtdx,$
$\displaystyle\langle\psi_{uu}(\overline{y},\overline{u})u,v\rangle=\int_{Q}L_{uu}[t,x]u(t,x)v(t,x)dtdx$

and

$\displaystyle F_{y}(\overline{y},\overline{u})y=y-k_{1-\alpha}\ast(\Delta
y),$ $\displaystyle F_{u}(\overline{y},\overline{u})u=-k_{1-\alpha}\ast(u),$
$\displaystyle F_{yy}(\overline{y},\overline{u})[y_{1},y_{2}]=0,\
F_{uu}(\overline{y},\overline{u})[u,v]=0,\
F_{yu}(\overline{y},\overline{u})[y,u]=0,$

for all $y\in L^{p}(J,D)$ and $u,v\in L^{\infty}(Q)$.

Proof. By the same argument as in the proof of Proposition A in [16], we can
show that $\psi$ is of class $C^{2}$ on $L^{p}(Q)\times L^{\infty}(Q)$. Since
$L^{p}(J,D)\hookrightarrow L^{p}(Q)$, we see that $\psi$ is of class $C^{2}$
on $L^{p}(J,D)\times L^{\infty}(Q)$. The differentiability of $F$ follows from
the linear property of $F$. $\Box$

Fixing any $u\in L^{\infty}(Q)$. By Proposition 3.2, the equation $F(y,u)=0$
has a unique solution $y=y(u)\in L^{p}(J,D)$. Let us define the mapping
$G:L^{\infty}(Q)\to L^{p}(J,D)$ by setting $G(u)=y(u)$. The mapping $G$ is
said to be a control-to-state mapping. By (24), we have

$G(u)=k_{\alpha}\ast Py_{0}+P\ast u.$

It is clear that $G(u)$ is linear and so it is of class $C^{\infty}$. Namely,
we have $G^{\prime}(u)v=P\ast v$ and $G^{\prime\prime}(u)[v,v]=0$.

Let us define the mapping $f(u)=\psi(G(u),u)$. Then by a simple argument, we
can show that $(\overline{y},\overline{u})$ is a locally optimal solution of
(38)-(40) if and only if $\overline{u}$ is a locally optimal solution of the
problem:

$\displaystyle\begin{cases}&f(u)=\psi(G(u),u)\to{\rm min}\\\ &u\in
K_{\infty}.\end{cases}$ (41)

The following proposition gives some properties of control-to-state mapping
$G$.

###### Proposition 4.2

The mapping $G:L^{\infty}(Q)\to L^{p}(J,D)$ has the following properties.

$(i)$ If $v\in L^{\infty}(Q)$ and $z_{v}=G^{\prime}(\overline{u})v$, then
$z_{v}$ is a unique solution of two equivalent equations:

$\displaystyle z_{v}=k_{1-\alpha}\ast(\Delta z_{v}+v)$ (42)

and

$\displaystyle\frac{d^{\alpha}z_{v}}{dt}=\Delta z_{v}+v,\quad z_{v}(0)=0.$
(43)

$(ii)$ For every $v_{1},v_{2}\in L^{\infty}$,
$G^{\prime\prime}(\overline{u})[v_{1},v_{2}]=0$

Proof. For all $u\in L^{\infty}(Q)$, we have $F(G(u),u)=0$. By differentiating
both sides in $u$, we get

$F_{y}(\overline{y},\overline{u})G^{\prime}(\overline{u})+F_{u}(\overline{y},\overline{u})=0.$

This means

$F_{y}(\overline{y},\overline{u})G^{\prime}(\overline{u})v+F_{u}(\overline{y},\overline{u})v=0\quad\forall
v\in L^{\infty}(Q).$

By Proposition 4.1, we get

$\displaystyle z_{v}-k_{1-\alpha}\ast(\Delta z_{v})-k_{1-\alpha}\ast v=0.$

Hence $z_{v}$ is a solution of (42) and (43). Assertion $(ii)$ follows from
linear property of $G(u)$. $\Box$

###### Proposition 4.3

The function $f$ is of class $C^{2}$ around $\overline{u}$ and for any
$v,v_{1},v_{2}\in L^{\infty}(Q)$ the following formulae are valid:

$\displaystyle\langle
f^{\prime}(\overline{u}),v\rangle=\int_{Q}(L_{u}[t,x]-\varphi(t,x))v(t,x)dxdt$
(44)

and

$\displaystyle f^{\prime\prime}(\overline{u})[v_{1},v_{2}]$
$\displaystyle=\int_{Q}(L_{yy}[t,x]z_{v_{1}}z_{v_{2}}+L_{yu}[t,x]z_{v_{1}}v_{2}+L_{uy}[t,x]z_{v_{1}}v_{2}+L_{uu}[t,x]v_{1}v_{2})dtdx,$
(45)

where $\varphi\in L^{q}(J,D)$ is a unique mild solution of the adjoint
equation

$\displaystyle\frac{\hat{d}^{\alpha}\varphi}{dt^{\alpha}}=\Delta\varphi-
L_{y}[\cdot,\cdot],\ \varphi(T)=0.$ (46)

Proof. We first show that equation (46) has a unique solution. Indeed, by
changing variable $\zeta(t)=\varphi(T-t)$, equation (46) is equivalent to the
equation

$\displaystyle\frac{{\hat{d}}^{\alpha}}{dt^{\alpha}}\varphi(T-t)=\Delta\zeta(t)-L_{y}[T-t,\cdot],\
\zeta(0)=0.$ (47)

Since $\varphi(T)=0$, we have
$\frac{\hat{d}^{\alpha}\varphi}{dt^{\alpha}}=D_{T-}^{\alpha}\varphi$ and

$\displaystyle D_{T-}^{\alpha}\varphi(T-t)$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{T-t}^{T}\frac{\varphi(\tau)}{(\tau-T+t)^{\alpha}}d\tau=-\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{t}^{0}\frac{\varphi(T-s)}{(t-s)^{\alpha}}ds$
$\displaystyle=\frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{0}^{t}\frac{\varphi(T-s)}{(t-s)^{\alpha}}ds=D_{0+}^{\alpha}\zeta(t)=\frac{d^{\alpha}}{dt^{\alpha}}\zeta(t).$

Hence equation (47) is equivalent to

$\displaystyle\frac{d^{\alpha}}{dt^{\alpha}}\zeta(t)=\Delta\zeta(t)-L_{y}[T-t,\cdot],\
\zeta(0)=0.$ (48)

By assumption $(H1)$, $L_{y}[\cdot,\cdot]\in
L^{q}(J,L^{\infty}(\Omega))\hookrightarrow L^{q}(J,H)$. By Proposition 3.2,
equation (48) has a unique mild solution $\zeta\in L^{q}(J,D)$ with
$\frac{d^{\alpha}}{dt^{\alpha}}\zeta\in L^{q}(J,H)$. Consequently, equation
(46) has a unique mild solution $\varphi(t)=\zeta(T-t)\in L^{q}(J,D)$. We now
have $f^{\prime}(u)=\psi_{y}(G(u),u)G^{\prime}(u)+\psi_{u}(G(u),u)$. Hence

$\displaystyle\langle
f^{\prime}(u),v\rangle=\psi_{y}(G(u),u)G^{\prime}(u)v+\psi_{u}(G(u),u)v.$ (49)

By taking $u=\overline{u}$ and putting $z=z_{v}=G^{\prime}(\overline{u})v$, we
get

$\displaystyle\langle
f^{\prime}(\overline{u}),v\rangle=\psi_{y}(\overline{y},\overline{u})G^{\prime}(\overline{u})v+\psi_{u}(\overline{y},\overline{u})v=\int_{Q}(L_{y}[t,x]z(t,x)+L_{u}[t,x]v(t,x))dtdx.$
(50)

Note that $z=z_{v}$ satisfies (43) and $\varphi$ is a mild solution of (46).
Hence

$\displaystyle\varphi=I_{T-}^{\alpha}(\Delta\varphi-L_{y}[\cdot,\cdot]).$ (51)

Since $\Delta\varphi-L_{y}[\cdot,\cdot]\in L^{q}(J,H)$, we see that
$\varphi\in I_{T-}^{\alpha}(L^{q})$. Hence Lemma 2.1 is applicable for
$\varphi$ and $z=z_{v}\in I_{0+}^{\alpha}(L^{p})$. By applying operator
$D_{T-}^{\alpha}$ on both sides of (51), we get

$D_{T-}^{\alpha}\varphi=\Delta\varphi-L_{y}[\cdot,\cdot]$

and so

$L_{y}[\cdot,\cdot]=\Delta\varphi-D_{T-}^{\alpha}\varphi.$

Besides, for each $t\in J$, $\varphi(t),z(t,\cdot)\in H^{2}(\Omega)\cap
H_{0}^{1}(\Omega)$. By Lemma 1.5.3.2 in [13], we have

$\int_{\Omega}\Delta\varphi(t,x)z(t,x)dx=\int_{\Omega}\varphi(t,x)\Delta
z(t,x)dx.$

Using this, (43), (46) and (9), we obtain

$\displaystyle\langle f^{\prime}(\overline{u}),v\rangle$
$\displaystyle=\int_{0}^{T}\int_{\Omega}(-D_{T-}^{\alpha}\varphi+\Delta\varphi)z(t,x)+L_{u}[t,x]v(t,x))dtdx$
$\displaystyle=\int_{0}^{T}\int_{\Omega}(-D_{T-}^{\alpha}\varphi
z(t,x)dxdt+\int_{0}^{T}\int_{\Omega}\Delta\varphi(t,x)z(t,x)dxdt+\int_{0}^{T}\int_{\Omega}L_{u}[t,x]v(t,x)dxdt$
$\displaystyle=-\int_{0}^{T}(-D_{T-}^{\alpha}\varphi(t),z(t,\cdot))_{H}dt+\int_{0}^{T}\int_{\Omega}\varphi(t,x)\Delta
z(t,x)dxdt+\int_{0}^{T}\int_{\Omega}L_{u}[t,x]v(t,x)dxdt$
$\displaystyle=-\int_{0}^{T}(D_{0+}^{\alpha}z(t),\varphi(t,\cdot))_{H}dt+\int_{0}^{T}\int_{\Omega}\varphi(t,x)\Delta
z(t,x)dxdt+\int_{0}^{T}\int_{\Omega}L_{u}[t,x]v(t,x)dxdt$
$\displaystyle=\int_{0}^{T}(-\frac{d^{\alpha}z(t,\cdot)}{dt^{\alpha}}+\Delta
z(t,\cdot),\varphi(t,\cdot))_{H}dt+\int_{0}^{T}\int_{\Omega}L_{u}[t,x]v(t,x)dxdt$
$\displaystyle=\int_{0}^{T}(-v,\varphi)_{H}dt+\int_{0}^{T}\int_{\Omega}L_{u}[t,x]v(t,x)dxdt$
$\displaystyle=\int_{Q}(L_{u}[t,x]-\varphi(t,x))v(t,x)dxdt.$

Hence (44) is obtained. Next, we have from (49) that

$\displaystyle f^{\prime\prime}(u)v$
$\displaystyle=(\psi_{yy}(G(u),u)G^{\prime}(u)+\psi_{yu}(G(u),u))G^{\prime}(u)v+\psi_{y}(G(u),u)G^{\prime\prime}(u)v$
$\displaystyle+(\psi_{uy}(G(u),u)G^{\prime}(u)+\psi_{uu}(G(u),u))v.$

Replacing $u=\overline{u}$, $v=v_{1}$ and acting on $v_{2}$, we obtain

$\displaystyle
f^{\prime\prime}(u)[v_{1},v_{2}]=\psi_{yy}(\overline{y},\overline{u})[z_{v_{1}},z_{v_{2}}]+\psi_{yu}(\overline{y},\overline{u})[z_{v_{1}},v_{2}]+0$
$\displaystyle+\psi_{uy}(\overline{y},\overline{u})[z_{v_{1}},v_{2}]+\psi_{uu}(\overline{y},\overline{u})[v_{1},v_{2}]$
$\displaystyle=\int_{Q}(L_{yy}[t,x]z_{v_{1}}z_{v_{2}}+L_{yu}[t,x]z_{v_{1}}v_{2}+L_{uy}[t,x]z_{v_{1}}v_{2}+L_{uu}[t,x]v_{1}v_{2})dtdx$

which is (45). The proof of the proposition is complete. $\Box$

In the sequel, we shall use the following sets.

$\displaystyle Q_{a}=\\{(t,x)\in Q:\overline{u}(t,x)=a\\},$ $\displaystyle
Q_{b}=\\{(t,x)\in Q:\overline{u}(t,x)=b\\},$ $\displaystyle Q_{ab}=\\{(t,x)\in
Q:a<\overline{u}(t,x)<b\\}.$

The following theorem is a result on the first-order necessary optimality
conditions.

###### Theorem 4.1

Suppose that assumption $(A4)$ is valid. If
$(\overline{y},\overline{u})\in\Phi$ is a locally optimal solution of problem
(1)-(5), then there exist a function $\varphi\in L^{q}(J,D)$ with
$\frac{\hat{d}^{\alpha}\varphi}{dt^{\alpha}}\in L^{q}(J,H)$ and a function
$e\in L^{q}(J,L^{\infty}(\Omega))$ such that the following conditions are
fulfilled:

$(i)$ $\varphi$ is a mild solution of the adjoint equation

$\displaystyle\frac{\hat{d}^{\alpha}\varphi}{dt^{\alpha}}=\Delta\varphi-
L_{y}[\cdot,\cdot],\ \varphi(T,\cdot)=0;$ (52)

$(ii)$ (the stationary condition in $u$)

$\displaystyle L_{u}[t,x]-\varphi(t,x)=e(t,x)\quad{\rm a.e.}\quad(t,x)\in Q;$

$(iii)$ (the complimentary condition) $e(t,x)$ has the property that

$\displaystyle e(t,x)\begin{cases}\geq 0\quad&{\rm a.e.}\quad(t,x)\in
Q_{a},\\\ \leq 0\quad&{\rm a.e.}\quad(t,x)\in Q_{b},\\\ 0\quad&{\rm
a.e.}\quad(t,x)\in Q_{ab}.\end{cases}$

Proof. It is clear that $(\overline{y},\overline{u})$ is a locally optimal
solution of (1)-(5) if and only if $\overline{u}$ is a locally optimality
solution of problem (41). By definition, there exists an open ball
$B(\overline{u},\epsilon)\subset L^{\infty}(Q)$ so that

$f(u)\geq f(\overline{u})\quad\forall u\in B(\overline{u},\epsilon)\cap
K_{\infty}.$

Fix any $v\in K_{\infty}$. Then for $\lambda>0$ small enough, we have
$v_{\lambda}=\overline{u}+\lambda(v-\overline{u})\in
B(\overline{u},\epsilon)\cap K_{\infty}$. Hence

$f(\overline{u}+\lambda(v-\overline{u}))\geq f(\overline{u}).$

This implies that

$\frac{f(\overline{u}+\lambda(v-\overline{u}))-f(\overline{u})}{\lambda}\geq
0.$

By letting $\lambda\to 0+$, we obtain

$\displaystyle\langle f^{\prime}(\overline{u}),v-\overline{u}\rangle\geq
0\quad\forall v\in K_{\infty}.$ (53)

By Proposition 3.2, equation (52) has a unique solution $\varphi\in C(J,H)\cap
L^{q}(J,D)$. Hence $(i)$ is valid. From Proposition 4.3 and (53), we have

$\displaystyle\int_{Q}(L_{u}[t,x]-\varphi(t,x))(v(t,x)-\overline{u}(t,x))dtdx\geq
0\quad\forall v\in K_{\infty}.$ (54)

By putting $L_{u}[t,x]-\varphi(t,x)=e(t,x)$, we see that $e\in
L^{q}(J,L^{\infty}(\Omega))$ which satisfies $(ii)$ and (54) becomes

$\displaystyle\int_{Q}e(t,x)(v(t,x)-\overline{u}(t,x))dtdx\geq 0\quad\forall
v\in K_{\infty}.$ (55)

Using variational arguments as in the proof of [17, Theorem 4.1], we can show
that $e$ satisfies $(iii)$. The theorem is proved. $\Box$

To deal with second-order optimality conditions, we need the so-called
critical cone as follows.

###### Definition 4.1

A couple $(y,v)\in L^{p}(J,D)\times L^{p}(Q)$ is said to be a critical
direction to problem (1)-(5) at $(\overline{y},\overline{u})$ if the following
conditions are fulfilled:

$(c_{1})$ $\int_{Q}(L_{y}[t,x]y(t,x)+L_{u}[t,x]v(t,x))dtdx\leq 0$;

$(c_{2})$ $y=k_{1-\alpha}\ast(\Delta y+v)$;

$(c_{3})$ $v(t,x)\geq 0$ for a.e. $(t,x)\in Q_{a}$ and $v(t,x)\leq 0$ for a.e.
$(t,x)\in Q_{b}$.

The set of critical directions at $(\overline{y},\overline{u})$ is denoted by
$\mathcal{C}_{p}[(\overline{y},\overline{u})]$ which is a closed convex cone
containing $(0,0)$.

We now take a couple $(y,v)\in\mathcal{C}(\overline{y},\overline{u})$ such
that $v\in L^{\infty}(Q)$. Then $(c_{2})$ is equivalent to saying that
$y=z_{v}=G^{\prime}(\overline{u})v$. Moreover, if
$(\overline{y},\overline{u})$ is a locally optimal solution, then

$\int_{Q}(L_{y}[t,x]y(t,x)+L_{u}[t,x]v(t,x))dtdx=0.$

Indeed, it is clear that $e(t,x)v(t,x)\geq 0$ for a.e. $(t,x)\in Q$. Since
$y=z_{v}$ we have from Proposition 4.3 and (50) that

$\displaystyle
0\geq\int_{Q}(L_{y}[t,x]y(t,x)+L_{u}[t,x]v(t,x))dtdx=\int_{Q}(L_{u}[t,x]-\varphi(t,x))v(t,x)dtdx=\int_{Q}e(t,x)v(t,x)dt\geq
0.$ (56)

Let $\mathcal{L}(y,u,\varphi,e)$ be the Lagrange function associated with the
problem, which is given by

$\displaystyle\mathcal{L}(y,u,\varphi,e)=\int_{Q}[L(t,x,y,u)+y(\frac{\hat{d}^{\alpha}\varphi}{dt^{\alpha}}-\Delta\varphi)-\varphi
u-eu]dtdx.$ (57)

Then conditions $(i)$ and $(ii)$ of Theorem 4.1 is equivalent to the condition

$\displaystyle\nabla_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)=0.$
(58)

We have the following result on second-order necessary optimality conditions.

###### Theorem 4.2

Suppose that $(A4)$ is valid, $(\overline{y},\overline{u})$ is a locally
optimal solution of (1)-(5) and $(\varphi,e)$ satisfies the conclusion of
Theorem 4.1. Then

$\displaystyle\nabla^{2}_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)[(y,v),(y,v)]=\int_{Q}(L_{yy}[t,x]y^{2}+L_{yu}[t,x]yv+L_{uy}[t,x]yv+L_{uu}[t,x]v^{2})dtdx\geq
0$ (59)

for all $(y,v)\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$, where $p\geq
2$.

Proof. We divide the proof into two steps.

${\it Step1}$. Proving (59) for the case
$(y,v)\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$ where $v\in
L^{\infty}(Q)$.

Take any $(y,v)\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$ with $v\in
L^{\infty}(Q)$. By $(c_{2})$, $y=z_{v}$. For each $0<\rho<b-a$, we define

$\displaystyle v_{\rho}(t,x)=\begin{cases}0\quad&{\rm if}\quad
a<\overline{u}(t,x)<a+\rho\quad{\rm or}\quad b-\rho<\overline{u}(t,x)<b\\\
v(t)\quad&{\rm otherwise}\end{cases}$

and $y_{\rho}=z_{v_{\rho}}$. Then by the same argument as in the proof of [17,
Theorem 4.3], we can show that
$(y_{\rho},u_{\rho})\in\mathcal{C}_{p}[(\overline{y},\overline{u})]$ and
$v_{\rho}\to v$ strongly in $L^{p}(Q)$. For
$\lambda\in(0,\rho/\|v\|_{\infty})$, we have $\overline{u}+\lambda v_{\rho}\in
K_{\infty}$. Since $\overline{u}$ is an optimal solution, for $\lambda>0$
small enough, we get

$\displaystyle 0\leq\frac{f(\overline{u}+\lambda
v_{\rho})-f(\overline{u})}{\lambda}=\lambda
f^{\prime}(\overline{u})v_{\rho}+\frac{\lambda^{2}}{2}f^{\prime\prime}(\overline{u}+\theta\lambda
v_{\rho})[v_{\rho},v_{\rho}],$

where $0\leq\theta\leq 1$. By Proposition 4.3,

$f^{\prime}(\overline{u})v_{\rho}=\int_{Q}e(t,x)v_{\rho}(t,x)dtdx\leq 0.$

It follows that

$\displaystyle 0\leq f^{\prime\prime}(\overline{u}+\theta\lambda
v_{\rho})[v_{\rho},v_{\rho}].$

By letting $\lambda\to 0+$, we obtain $0\leq
f^{\prime\prime}(\overline{u})[v_{\rho},v_{\rho}].$ This and Proposition 4.3
imply

$\displaystyle
0\leq\int_{Q}[L_{yy}[t,x]y_{\rho}^{2}+2L_{yu}[t,x]y_{\rho}v_{\rho}+L_{uu}[t,x]v_{\rho}^{2}]dtdx$
(60)

Now we have

$\displaystyle y=k_{1-\alpha}\ast(\Delta y+v)$ $\displaystyle
y_{\rho}=k_{1-\alpha}\ast(\Delta y_{n}+v_{\rho})$

Hence

$\displaystyle y-y_{\rho}=k_{1-\alpha}\ast(\Delta(y-y_{\rho})+(v-v_{\rho})).$

By Proposition 3.2, there exist positive constants $C$ and $C^{\prime}$ such
that

$\displaystyle\|y_{\rho}-y\|_{L^{p}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}(y_{\rho}-y)\|_{L^{p}(J,H)}\leq
C\|v-v_{\rho}\|_{L^{p}(J,H)}\leq C^{\prime}\|v-v_{\rho}\|_{L^{p}(Q)}.$ (61)

This implies that $\|y_{\rho}-y\|_{L^{p}(J,D)}\to 0$ as $\rho\to 0+$. Since
$L^{p}(J,D)\hookrightarrow L^{p}(J,L^{\infty}(\Omega)$, we also have
$\|y_{\rho}-y\|_{L^{p}(J,L^{\infty}(\Omega))}\to 0$. From assumption $(A4)$,
we see that $L_{yy}[\cdot,\cdot],L_{yu}[\cdot,\cdot],L_{uy}[\cdot,\cdot]\in
L^{q}(J,L^{\infty})$. We now claim that

$\int_{Q}[L_{yy}[t,x]y_{\rho}^{2}\to\int_{Q}L_{yy}[t,x]y^{2}dtdx.$

In fact, using (34) with $y_{1}=0$ and $y_{2}=\overline{y}$, we have

$\displaystyle\big{|}\int_{Q}[L_{yy}[t,x]y_{\rho}^{2}-\int_{Q}L_{yy}[t,x]y^{2}dtdx\big{|}=\big{|}\int_{Q}L_{yy}[t,x](y_{\rho}-y)(y_{\rho}+y)dtdx\big{|}$
$\displaystyle\leq
C^{\prime\prime}\int_{0}^{T}\|L_{yy}[\cdot,\cdot]\|_{\infty}\|y_{\rho}-y\|_{\infty}\|y_{\rho}+y\|_{\infty}dt$
$\displaystyle\leq
C^{\prime\prime}[\int_{0}^{T}\|L_{yy}[\cdot,\cdot]\|^{\frac{p}{p-2}}_{\infty}\|dt]^{1-2/p}[\int_{0}^{T}\|y_{\rho}-y\|_{\infty}^{p}]dt]^{1/p}[\int_{0}^{T}\|y_{\rho}+y\|_{\infty}^{p}]dt]^{1/p}$
$\displaystyle\leq
C^{\prime\prime}[\int_{0}^{T}(\|\overline{y}\|^{p-2}_{\infty}+\|L_{yy}[\cdot,\cdot]\|_{L^{\infty}(Q)})^{\frac{p}{p-2}}\|dt]^{1-2/p}[\int_{0}^{T}\|y_{\rho}-y\|_{\infty}^{p}]dt]^{1/p}[\int_{0}^{T}\|y_{\rho}+y\|_{\infty}^{p}]dt]^{1/p}$
$\displaystyle\leq
M\|y_{\rho}-y\|_{L^{p}(J,L^{\infty}(\Omega))}\|y_{\rho}+y\|_{L^{p}(J,L^{\infty}(\Omega))}\to
0.$

The claim is justified. Similarly, we also have

$\int_{Q}[2L_{yu}[t,x]y_{\rho}v_{\rho}+L_{uu}[t,x]v_{\rho}^{2}]dtdx\to\int_{Q}[2L_{yu}[t,x]yv+L_{uu}[t,x]v^{2}]dtdx.$

By letting $\rho\to 0+$, we obtain from (60) that

$0\leq\int_{Q}[L_{yy}[t,x]y^{2}+2L_{yu}[t,x]yv+L_{uu}[t,x]v^{2}]dtdx.$

Hence (59) is valid.

Step 2. Proving (59) for any
$(y,v)\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$.

Next we take $(y,v)\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$. Then from
$(c_{2})$ of $\mathcal{C}_{p}[(\overline{x},\overline{u})]$, we have
$y=z_{v}$. For each $n\geq 1$, we define

$v_{k}(t,x)=P_{[-n,n]}(v(t,x))=\min\\{\max\\{-n,v(t,x)\\},+n\\},$

where $P_{[-n,n]}(\xi)$ denotes the metric projection of $\xi$ on $[-n,n]$ in
$\mathbb{R}$. It is clear that $v_{n}\in L^{\infty}(Q)$ and $v_{n}(t,x)\to
v(t,x)$ a.e. $t\in[0,1]$. By the nonexpansive property of metric projection,
we have

$|v_{n}(t,x)|=|P_{[-n,n]}(v(t,x))-P_{[-n,n]}(0)|\leq|v(t,x)|.$

The dominated convergence theorem implies that $v_{n}\to v$ in $L^{p}(Q)$ and
in $L^{2}(Q)$. Let us put $y_{n}=z_{v_{n}}$. We now claim that
$(y_{n},v_{n})\in\mathcal{C}_{p}[(\overline{x},\overline{u})]$. Indeed,
condition $(c_{2})$ is obvious. To verify condition $(c_{3})$, we note that if
$\overline{u}(t,x)=a$, then $v(t,x)\geq 0$. Hence
$v_{n}(t,x)=\min\\{v(t,x),n\\}\geq 0$. If $\overline{u}(t,x)=b$, then
$v(t,x)\leq 0$. Hence $v_{n}(t,x)=\max\\{-n,v(t)\\}\leq 0$. To check
$(c_{1})$, we have

$\displaystyle\int_{Q}(L_{y}[t,x]y_{n}(t,x)+L_{u}[t,x]v_{n}(t,x))dtdx=\int_{Q}e(t,x)v_{n}(t,x)dtdx$
$\displaystyle=\int_{Q_{a}}e(t,x)v_{n}(t,x)dt+\int_{Q_{b}}e(t,x)v_{n}(t,x)dtdx=\int_{Q_{a}}e(t,x)\min\\{v(t,x),n\\}dtdx+\int_{Q_{b}}e(t,x)\max\\{-n,v(t,x)\\}dtdx$
$\displaystyle\leq\int_{Q_{a}}e(t,x)v(t,x)dtdx+\int_{Q_{b}}e(t,x)v(t,x)dtdx\leq\int_{Q}e(t,x)v(t,x)dtdx\leq
0.$

Therefore, the claim is justified. Now we have

$\displaystyle y=k_{1-\alpha}\ast(\Delta y+v)$ $\displaystyle
y_{n}=k_{1-\alpha}\ast(\Delta y_{n}+v_{n})$

Hence $y-y_{n}=k_{1-\alpha}\ast(\Delta(y-y_{n})+(v-v_{n})).$ By Proposition
3.2, there exist positive constants $C$ and $C^{\prime}$ such that

$\displaystyle\|y_{n}-y\|_{L^{p}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}(y_{n}-y)\|_{L^{p}(J,H)}\leq
C\|v-v_{n}\|_{L^{p}(J,H)}\leq C^{\prime}\|v-v_{n}\|_{L^{p}(Q)}.$ (62)

This implies that $\|y_{n}-y\|_{L^{p}(J,D)}\to 0$ as $n\to\infty$. Since
$L^{p}(J,D)\hookrightarrow L^{p}(J,L^{\infty}(\Omega)$, we also have
$\|y_{n}-y\|_{L^{p}(J,L^{\infty}(\Omega))}\to 0$. By Step 1, we have

$\int_{Q}(L_{yy}[t,x]y_{n}^{2}+L_{yu}[t,x]y_{n}v_{n}+L_{uy}[t,x]y_{n}v_{n}+L_{uu}[t,x]v_{n}^{2})dtdx\geq
0.$

By letting $n\to\infty$ and using similar arguments as in Step 1 with noting
that $y_{n}\to y$, $u_{n}\to u$, we obtain

$\int_{Q}(L_{yy}[t,x]y^{2}+L_{yu}[t,x]yv+L_{uy}[t,x]yv+L_{uu}[t,x]v^{2})dtdx\geq
0.$

The proof of the theorem is completed. $\Box$

The remain of this section, we give second-order sufficient optimal conditions
for locally optimal solution to (1)-(5). To do this, we need some tools of
variational analysis.

Let $K$ be a nonempty and closed subset in a Banach space $X$ and
$\overline{w}\in K$. The sets

$\displaystyle T^{\flat}(K;\overline{w})$ $\displaystyle:=\left\\{h\in
X\;|\;\forall t_{n}\to 0^{+},\exists h_{n}\to h,\overline{w}+t_{n}h_{n}\in K\
\ \forall n\in\mathbb{N}\right\\},$ $\displaystyle T(K,\overline{w}):$
$\displaystyle=\left\\{h\in X\;|\;\exists t_{n}\to 0^{+},\exists h_{n}\to
h,\overline{w}+t_{n}h_{n}\in K\ \ \forall n\in\mathbb{N}\right\\},$

are called the adjacent tangent cone and the contingent cone to $K$ at
$\overline{w}$, respectively. It is well-known that when $K$ is convex, then

$T^{\flat}(K;\overline{w})=T(K;\overline{w})=\overline{\rm
cone}(K-\overline{w}),$

where

${\rm
cone}(K-\overline{w}):=\\{\lambda(h-\overline{w})\;|\;h\in\Omega,\lambda>0\\}.$

Let $K_{p}$ be defined by (37). Then from [16, Lemma 2.4], we have

$\displaystyle T^{\flat}(K_{p},\overline{u})=\big{\\{}v\in L^{p}:v(t,x)\
\text{satisfies}\ (c_{3})\big{\\}}.$

Hence condition $(c_{3})$ in $\mathcal{C}_{p}[(\overline{y},\overline{u})]$ is
equivalent to saying that $v\in T^{\flat}(K_{p},\overline{u})$.

Below, we set $Y=L^{p}(J,D)$ and $U=L^{\infty}(Q)$. Given $\epsilon>0$, the
symbols $B(\overline{y},\epsilon)_{Y}$ is a ball in $Y$ with center at
$\overline{y}\in Y$ and radius $\epsilon>0$.

We now have

###### Theorem 4.3

Suppose $(A4)$, $p\geq 2$, $1>\alpha>1/2$ and $(\varphi,e)$ satisfies the
conclusion of Theorem 4.1. Furthermore, suppose that the following conditions
are satisfied:

$(i)$ $L_{yy}[\cdot,\cdot],L_{yu}[\cdot,\cdot]\in L^{\infty}(Q)$ and there
exists a number $\Lambda>0$ such that

$\displaystyle L_{uu}(t,x,y,u)\geq\Lambda\quad\forall(t,x,y,u)\in
Q\times\mathbb{R}\times\mathbb{R}.$ (63)

$(ii)$ (the strictly second-order condition)

$\displaystyle\nabla^{2}_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)[(z,v),(z,v)]>0\quad\forall(z,v)\in\mathcal{C}_{2}[(\overline{y},\overline{u})]\setminus\\{(0,0)\\}.$
(64)

Then there exist positive constants $\varrho$ and $\epsilon$ such that

$\displaystyle\psi(y,u)\geq\psi(\overline{y},\overline{u})+\varrho\|u-\overline{u}\|_{2}^{2}\quad\forall(y,u)\in
B_{Y}(\overline{y},\epsilon)\times B_{U}(\overline{u},\epsilon).$

Proof. On the contrary, we suppose the conclusion were false. Then, we could
find sequences $\\{(y_{n},u_{n})\\}\subset\Phi$ and
$\\{\varrho_{n}\\}\subset\mathrm{int}\,\mathbb{R}_{+}$ such that
$y_{n}\to\overline{y}$ in $Y$, $u_{n}\to\overline{u}$ in $U$, $\varrho_{n}\to
0$ and

$\psi(y_{n},u_{n})<\psi(\overline{y},\overline{u})+\varrho_{k}\|u_{n}-\overline{u}\|_{2}^{2}.$
(65)

If $u_{n}=\overline{u}$ then we can show that $y_{n}=\overline{y}$. It follows
that $\psi(\overline{x},\overline{u})<\psi(\overline{x},\overline{u})+0$ which
is absurd. Therefore, we can assume that $u_{n}\neq\overline{u}$ for all
$n\geq 1$.

Define $t_{n}=\|u_{n}-\overline{u}\|_{2}$,
$z_{n}=\frac{y_{n}-\overline{y}}{t_{n}}$ and
$v_{n}=\frac{u_{n}-\overline{u}}{t_{n}}$. Then $t_{n}\to 0^{+}$ and
$\|v_{n}\|_{2}=1$. By the reflexivity of $L^{2}(Q)$, we may assume that
$v_{n}\rightharpoonup v$. From the above, we have

$\psi(y_{n},u_{n})-\psi(\overline{y},\overline{u})\leq
t_{n}^{2}\varrho_{n}\leq o(t_{n}^{2}).$ (66)

We claim that $y_{n}$ converges uniformly to some $y$ in $L^{2}(J,D)$. In
fact, since $(y_{n},u_{n})\in\Phi$, we have

$\displaystyle y_{n}(t)$ $\displaystyle=y_{0}+k_{1-\alpha}\ast(\Delta
y_{n}+u_{n}),$ $\displaystyle\overline{y}$
$\displaystyle=y_{0}+k_{1-\alpha}\ast(\Delta\overline{y}+\overline{u}).$

This implies that

$\displaystyle z_{n}=k_{1-\alpha}\ast(\Delta z_{n}+v_{n}),\quad z_{n}(0)=0.$

By Proposition 3.2, there exist constant $C>0$ such that

$\displaystyle\|z_{n}\|_{L^{2}(J,D)}+\|\frac{d^{\alpha}}{dt^{\alpha}}z_{n}\|_{L^{2}(J,H)}\leq
C\|v_{n}\|_{L^{2}(J,H)}=C\|v_{n}\|_{L^{2}(Q)}.$

Hence $\\{z_{n}\\}$ is bounded in $L^{2}(J,D)$. Without loss of generality, we
may assume that $z_{n}\rightharpoonup z$ in $L^{2}(J,D)$.

Let us claim that if $2>\frac{1}{\alpha}$, then $z_{n}\to z$ strongly in
$L^{2}(J,V)$. In fact, it is known that the embeddings

$D\hookrightarrow V\hookrightarrow H$

are compact. For $\alpha>1/2$, we have

$\displaystyle\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\|z_{k}(s)\|_{D}ds$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}(\int_{0}^{t}(t-s)^{(\alpha-1)2}ds)^{1/2}\|z_{k}\|_{L^{2}(J,D)}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\frac{1}{(\alpha-1/2)^{1/2}}t^{\alpha-1/2}\|z_{k}\|_{L^{p}(J,D)}$
$\displaystyle\leq\frac{1}{\Gamma(\alpha)}\frac{1}{(\alpha-1/2)^{1/2}}T^{\alpha-1/2}M:=M^{\prime}.$

Hence

$\sup_{t\in(0,T)}\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\|z_{k}(s)\|_{D}ds\leq
M^{\prime}.$

Besides, we have $2>\frac{2}{1+2\alpha}$, and
$\\{D_{0+}^{\alpha}z_{k}\\}=\\{\frac{d^{\alpha}}{dt^{\alpha}}z_{k}\\}$ is
bounded in $L^{2}(J,H)$. By Theorem 4.1 in [19], we see that $\\{z_{k}\\}$ is
a relatively compact set in $L^{2}(J,V)$. Therefore, we may assume that
$z_{k}\to z$ strongly in $L^{2}(J,V)$. The claim is justified.

By repeating the procedure in the the proof of Theorem 3.1 we can show that

$\displaystyle z=k_{1-\alpha}\ast(\Delta z+v).$ (67)

The remain of the proof will be divided into two steps.

Step 1. Showing that $(y,v)\in\mathcal{C}_{2}[(\overline{y},\overline{u})]$.

By a Taylor expansion, we have from (66) that

$\psi_{y}(\overline{y},\overline{u})z_{k}+\psi_{u}(\overline{y},\overline{u})v_{k}+\frac{o(t_{k})}{t_{k}}\leq\frac{o(t^{2}_{k})}{t_{k}}.$
(68)

Note that $L_{u}[\cdot,\cdot]\in L^{\infty}(Q)$ and
$\psi_{u}(\overline{x},\overline{u})\colon L^{2}(Q)\to\mathbb{R}$ is a
continuous linear mapping, where

$\langle\psi_{u}(\overline{x},\overline{u}),u\rangle:=\int_{Q}L_{u}[t,x]u(t,x)dtdx\
\quad\forall u\in L^{2}(Q).$

By [6, Theorem 3.10], $\psi_{u}(\overline{y},\overline{u})$ is weakly
continuous on $L^{2}(Q)$. On the other hand, $L_{y}[\cdot,\cdot]\in
L^{q}(J,H)\subset(L^{p}(J,D))^{*}$. Hence
$\lim_{k\to\infty}\int_{Q}L_{y}[t,x]z_{k}(t,x)dtdx=\int_{Q}L_{y}[t,x]z(t,x)dtdx$.
Letting $k\to\infty$ in (68), we get

$\displaystyle\psi_{y}(\overline{y},\overline{u})z+\psi_{u}(\overline{y},\overline{u})v=\int_{Q}(L_{y}[t,x]z(t,x)+L_{u}[t,x]v(t,x))dtdx\leq
0.$ (69)

Hence condition $(c_{1})$ in $\mathcal{C}_{2}[(\overline{y},\overline{u})]$ is
valid. By (67), condition $(c_{2})$ is fulfilled. For $(c_{3})$, we note that
$u_{k}-\overline{u}\in K_{\infty}-\overline{u}$. This implies that

$v_{k}\in\frac{1}{t_{k}}(K_{\infty}-\overline{u})\subset
T^{\flat}(K_{\infty};\overline{u})),$

where $T^{\flat}(K_{\infty};\overline{u}))$ is the tangent cone to
$K_{\infty}$ at $\overline{u}$ in $L^{2}([0,1],\mathbb{R})$. Note that
$K_{\infty}=K_{2}$. Since $T^{\flat}(K_{\infty};\overline{u}))$ is convex and
closed in $L^{2}$, it is weakly closed. Passing to the limit, we obtain $v\in
T^{\flat}(K_{\infty};\overline{u}).$ Hence condition $(c_{3})$ of
$\mathcal{C}_{2}[(\overline{y},\overline{u})]$ is fulfilled.

Step 2. Showing that $(z,v)=0$.

Note that $\langle e,u_{n}-\overline{u}\rangle\geq 0$ and

$\displaystyle y_{n}$ $\displaystyle=y_{0}+k_{1-\alpha}\ast(\Delta
y_{n}+u_{n})$ $\displaystyle\overline{y}$
$\displaystyle=y_{0}+k_{1-\alpha}\ast(\Delta\overline{y}+\overline{u}).$

This implies that

$(y_{n}-\overline{y})=k_{1-\alpha}\ast(\Delta(y_{n}-\overline{y})+u_{n}-\overline{u})=0.$

Hence

$\frac{d^{\alpha}}{dt^{\alpha}}(y_{n}-\overline{y})-\Delta(y_{n}-\overline{y})-(u_{n}-\overline{u})=0.$

This and Lemma 2.1 yield

$\displaystyle\int_{Q}[(y_{n}-\overline{y})D_{*T-}^{\alpha}\varphi-\varphi(\Delta(y_{n}-\overline{y})-\varphi(u_{n}-\overline{u})]dtdx$
$\displaystyle=\int_{Q}\varphi(t)[\frac{d^{\alpha}}{dt^{\alpha}}(y_{n}-\overline{y})-\Delta(y_{n}-\overline{y})-(u_{n}-\overline{u})]dtdx=0.$

Combining these with (66) and definition of Lagrange function (57), we have

$\displaystyle\mathcal{L}(y_{n},u_{n},\varphi,e)-\mathcal{L}(\overline{y},\overline{u},\varphi,e)$
$\displaystyle=\psi(y_{n},u_{n})-\psi(\overline{y},\overline{u})-\langle
e,u_{n}-\overline{u}\rangle$
$\displaystyle\leq\psi(y_{n},u_{n})-\psi(\overline{y},\overline{u})\leq
o(t_{n}^{2}).$

From this, (58) and a second-order Taylor expansion, we get

$\frac{t_{n}^{2}}{2}\nabla^{2}_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)[(z_{n},v_{n}),(z_{n},v_{n})]+o(t_{n}^{2})\leq
o(t_{n}^{2}),$

or, equivalently,

$\nabla^{2}_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)[(z_{n},v_{n}),(z_{n},v_{n})]\leq\frac{o(t_{n}^{2})}{t_{n}^{2}}.$
(70)

This means

$\displaystyle\int_{Q}(L_{yy}[t,x]z_{n}^{2}+2L_{yu}[t,x]z_{n}v_{n}+L_{uu}[t,x]v_{n}^{2})dtdx\leq\frac{o(t_{n}^{2})}{t_{n}^{2}}.$
(71)

Let us claim that

$\displaystyle\lim_{n\to\infty}\int_{Q}(L_{yy}[t,x]z_{n}^{2}+2L_{yu}[t,x]z_{n}v_{n}+L_{uu}[t,x]v_{n}^{2})dtdx\geq\int_{Q}(L_{yy}[t,x]z^{2}+2L_{yu}[t,x]zv+L_{uu}[t,x]v^{2})dtdx.$
(72)

Indeed, since $L_{uu}[t,x]\geq\Lambda>0$, the functional
$u\mapsto\int_{Q}L_{uu}[t,x]u^{2}dtdx$ is convex and so it is sequentially
lower semicontinous. Hence

$\lim_{n\to\infty}\int_{Q}L_{uu}[t,x]v_{n}^{2}dtdx\geq\int_{Q}L_{uu}[t,x]v^{2}dtdx.$

Since $z_{n}\to z$ in $L^{2}(J,V)$, $\|z_{n}+z\|_{L^{2}(J,V)}\leq M$ for some
constant $M>0$. Using assumption $L_{yy}[\cdot,\cdot]\in L^{\infty}(Q)$ in
$(i)$, we have

$\displaystyle|\int_{Q}L_{yy}[t,x](z_{n}^{2}-z^{2})dtdx|$
$\displaystyle\leq\int_{0}^{T}\int_{\Omega}|L_{yy}[t,x]||z_{n}(t,x)-z(t,x)||z_{n}(t,x)+z(t,x)|dxdt$
$\displaystyle\leq\int_{0}^{T}\|L_{yy}[t,\cdot]\|_{L^{\infty}(\Omega)}\|z_{n}(t,\cdot)-z(t,\cdot)\|_{L^{2}(\Omega)}\|z_{n}(t,\cdot)+z(t,\cdot)\|_{L^{2}(\Omega)}$
$\displaystyle\leq\|L_{yy}[\cdot,\cdot]\|_{L^{\infty}(Q)}\|\|z_{n}(\cdot,\cdot)-z(\cdot,\cdot)\|_{L^{2}(Q)}\|z_{n}(\cdot,\cdot)+z(\cdot,\cdot)\|_{L^{2}(Q)}$
$\displaystyle\leq\|L_{yy}[\cdot,\cdot]\|_{L^{\infty}(Q)}\|\|z_{n}(\cdot,\cdot)-z(\cdot,\cdot)\|_{L^{2}(Q)}M.$

This implies that

$\lim_{n\to\infty}\int_{Q}L_{yy}[t,x]z_{n}^{2}(t,x)dtdx=\int_{Q}L_{yy}[t,x]z^{2}(t,x)dtdx.$

Similarly, we can show that

$\lim_{n\to\infty}\int_{Q}L_{yu}[t,x]z_{n}(t,x)v_{n}(t,x)dtdx=\int_{Q}L_{yu}[t,x]z(t,x)v(t,x)dtdx.$

Hence the claim is justified. By letting $n\to\infty$, we obtain from (71)
that

$\nabla^{2}_{(y,u)}\mathcal{L}(\overline{y},\overline{u},\varphi,e)[(z,v),(z,v)]\leq
0.$

Combining this with (64), we conclude that $(z,v)=(0,0)$.

Finally, from (63), (71) and $\|v_{n}\|_{L^{2}(Q)}=1$, we have

$\displaystyle\int_{Q}(L_{yy}[t,x]z_{n}^{2}+2L_{yu}[t,x]z_{n}v_{n})dtdx+\Lambda$
$\displaystyle\leq\int_{Q}(L_{yy}[t,x]z_{n}^{2}+2L_{yu}[t,x]z_{n}v_{n}+L_{uu}[t,x]v_{n}^{2})dtdx\leq\frac{o(t_{n}^{2})}{t_{n}^{2}}.$

By letting $n\to\infty$ and using $(y,v)=(0,0)$, we obtain $\Lambda\leq 0$
which is absurd. Therefore, the theorem is proved. $\Box$

###### Remark 4.1

When $0<\alpha\leq 1/2$, the embebded theorem is not applicable and the proof
of Theorem 4.2 is collapsed. Therefore, establishing second-order sufficient
conditions for this case is an open problem.

When $1/2<\alpha<1$ and $p=2$, Theorem 4.1 and Theorem 4.2 give no-gap second-
order conditions on the common critical cone
$\mathcal{C}_{2}[(\overline{y},\overline{u})]$.

## Acknowledgments

The research of the first author was partially supported by Vietnam Academy of
Science and Technology under grant number QTRU 01-01/21-22. The second author
was supported by Taiwan MOST grant 111-2811-M-110-017.

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