# Equivalence of three different kinds of optimal control problems for heat
equations and its applications

Gengsheng Wang Yashan Xu  School of Mathematics and Statistics, Wuhan
University, Wuhan, 430072, China<EMAIL_ADDRESS>The author was partially
supported by National Basis Research Program of China (973 Program) under
grant 2011CB808002 and the National Natural Science Foundation of China under
grant 11161130003 and 11171264.School of Mathematical Sciences, Fudan
University, KLMNS, Shanghai 200433, China<EMAIL_ADDRESS>This work
was partially supported by NNSF Grant 10801041, 10831007.

###### Abstract

This paper presents an equivalence theorem for three different kinds of
optimal control problems, which are optimal target control problems, optimal
norm control problems and optimal time control problems. Controlled systems in
this study are internally controlled heat equations. With the aid of this
theorem, we establish an optimal norm feedback law and build up two algorithms
for optimal norms (together with optimal norm controls) and optimal time
(along with optimal time controls), respectively.

AMS Subject Classifications. 35K05, 49N90

Keywords. optimal controls, optimal norm, optimal time, feedback law, heat
equations

##  1 Introduction

We begin with introducing the controlled system. Let $T$ be a positive number
and $\Omega\subseteq{\mathbb{R}}^{d}$ be a bounded domain with a smooth
boundary $\partial\Omega$. Let $\omega$ be an open and non-empty subset of
$\Omega$. Write $\chi_{\omega}$ for the characteristic function of $\omega$.
Consider the following controlled heat equation:

$\left\\{\begin{array}[]{lll}\partial_{t}y-\triangle
y=\chi_{\omega}\chi_{(\tau,T)}u&\mbox{in}&\Omega\times(0,T),\\\
y=0&\mbox{on}&\partial\Omega\times(0,T).\\\
y(0)=y_{0}&\mbox{in}&\Omega,\end{array}\right.$ (1.1)

Here $y_{0}\in L^{2}(\Omega)$, $u\in L^{\infty}(0,T;L^{2}(\Omega))$,
$\tau\in[0,T)$ and $\chi_{(\tau,T)}$ stands for the characteristic function of
$(\tau,T)$. In this equation, controls are restricted over
$\omega\times(\tau,T)$. It is well known that for each $u\in
L^{\infty}(0,T;L^{2}(\Omega))$ and each $y_{0}\in L^{2}(\Omega)$, Equation
(1.1) has a unique solution in $C([0,T];L^{2}(\Omega))$. We denote, by
$y(\cdot;\chi_{(\tau,T)}u,y_{0})$, the solution of Equation (1.1)
corresponding to the control $u$ and the initial state $y_{0}$.  Throughout
this paper, $\|\cdot\|$ and $<\cdot,\cdot>$ stand for the usual norm and inner
product of the space $L^{2}(\Omega)$, respectively.

Next, we will set up, for each $y_{0}\in L^{2}(\Omega)$, three kinds of
optimal control problems associated with Equation (1.1). For this purpose, we
take a target $z_{d}\in L^{2}(\Omega)$ such that

$\displaystyle z_{d}\notin\Bigr{\\{}y(T;\chi_{(0,T)}u,0):~{}u\in
L^{\infty}(0,T;L^{2}(\Omega))\Bigl{\\}}.$ (1.2)

The set on the right hand side of (1.2) is called the attainable set of
Equation (1.1). Then we introduce the following target sets:

$B(z_{d},r)=\\{\hat{y}\in L^{2}(\Omega):\|\hat{y}-z_{d}\|\leq r\\},\;\;r>0.$

For each $M\geq 0$, each $r>0$ and each $\tau\in[0,T)$, we define three sets
of controls as follows:

  * •

$\mathcal{U}_{\tau,M}=\\{v\in L^{\infty}(0,T;L^{2}(\Omega)):\|v(t)\|\leq
M\;\;\mbox{for a.e.}\;\;t\in(\tau,T)\\}$;

  * •

$\mathcal{U}_{M,r}=\\{v:\exists\;\tau\in[0,T)\;\mbox{s.t.}\;v\in\mathcal{U}_{\tau,M}\;\;\mbox{and}\;\;y(T;\chi_{(\tau,T)}v,y_{0})\in
B(z_{d},r)\\}$;

  * •

$\mathcal{U}_{r,\tau}=\\{v\in
L^{\infty}(0,T;L^{2}(\Omega)):y(T;\chi_{(\tau,T)}v,y_{0})\in B(z_{d},r)\\}.$

For each $u\in\mathcal{U}_{M,r}$, we set

$\displaystyle\widetilde{\tau}_{M,r}(u)=\sup\\{\tau\in[0,T):u\in\mathcal{U}_{\tau,M}\;\mbox{and}\;y(T;\chi_{(\tau,T)}u,y_{0})\in
B(z_{d},r)\\}.$ (1.3)

Three kinds of optimal control problems studied in this paper are as follows:

  * •

$(OP)^{\tau,M}$:
$\inf\\{\|y(T;\chi_{(\tau,T)}u,y_{0})-z_{d}\|^{2}:u\in\mathcal{U}_{\tau,M}\\}$;

  * •

$(TP)^{M,r}$: $\sup\\{\widetilde{\tau}_{M,r}(u):u\in\mathcal{U}_{M,r}\\}$;

  * •

$(NP)^{r,\tau}$:
$\inf\\{\|u\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}:u\in\mathcal{U}_{r,\tau}\\}$.

We call $(OP)^{\tau,M}$ as an optimal target control problem, which is a kind
of optimal control problem with the observation of the final state (see [9],
page 177). The problem $(NP)^{r,\tau}$ is an optimal norm control problem,
which is related to the approximate controllability (see [4]). The problem
$(TP)^{M,r}$ is an optimal time control problem. The aim of controls in
$(TP)^{M,r}$ is to delay initiation of active control as late as possible,
such that the corresponding solution reaches the target $B(z_{d},r)$ at the
ending time $T$ (see [11]).

The above three problems provide the following three values, respectively:

  * •

$r(\tau,M)\equiv\inf\\{\|y(T;\chi_{(\tau,T)}u,y_{0})-z_{d}\|:u\in\mathcal{U}_{\tau,M}\\};$

  * •

$\tau(M,r)\equiv\sup\\{\widetilde{\tau}_{M,r}(u):u\in\mathcal{U}_{M,r}\\}$;

  * •

$M(r,\tau)\equiv\inf\\{\|u\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}:u\in\mathcal{U}_{r,\tau}\\}.$

The value $r(\tau,M)$ is called the optimal distance to the target for
$(OP)^{\tau,M}$; while values $\tau(M,r)$ and $M(r,\tau)$ are called the
optimal time for $(TP)^{M,r}$ and the optimal norm for $(NP)^{r,\tau}$,
respectively. The optimal controls to these problems are defined as follows:

  * •

$u^{*}$ is called an optimal control to $(OP)^{\tau,M}$ if
$u^{*}=\chi_{(\tau,T)}v^{*}$ for some $v^{*}\in\mathcal{U}_{\tau,M}$ such that
$\|y(T;\chi_{(\tau,T)}v^{*},y_{0})-z_{d}\|=r(\tau,M)$;

  * •

$u^{*}$ is called an optimal control to $(TP)^{M,r}$ if
$u^{*}=\chi_{(\tau(M,r),T)}v^{*}$ for some $v^{*}\in\mathcal{U}_{\tau(M,r),M}$
such that $y(T;\chi_{(\tau(M,r),T)}v^{*},y_{0})\in B(z_{d},r).$

  * •

$u^{*}$ is called an optimal control to $(NP)^{r,\tau}$ if
$u^{*}=\chi_{(\tau,T)}v^{*}$ for some $v^{*}\in\mathcal{U}_{r,\tau}$ and
$\|u^{*}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M(r,\tau)$.

Throughout the paper, the following notation will be used frequently:

$\displaystyle r_{T}(y_{0})\equiv\|y(T;0,y_{0})-z_{d}\|.$ (1.4)

The main purpose of this study is to present an equivalence theorem for the
above-mentioned three kinds of optimal control problems and its applications.
This theorem can be stated, in plain language, as follows:

  * •

$(OP)^{\tau,M}\Leftrightarrow(TP)^{M,r(\tau,M)}\Leftrightarrow(NP)^{r(\tau,M),\tau}$
when $M>0$ and $\tau\in[0,T)$;

  * •

$(NP)^{r,\tau}\Leftrightarrow(OP)^{\tau,M(r,\tau)}\Leftrightarrow(TP)^{M(r,\tau),r}$
when $r\in(0,r_{T}(y_{0}))$ and $\tau\in[0,T)$;

  * •

$(TP)^{M,r}\Leftrightarrow(NP)^{r,\tau(M,r)}\Leftrightarrow(OP)^{\tau(M,r),M}$
when $M>0$ and $r\in[r(0,M),r_{T}(y_{0}))$.

Here, by $(P_{1})\Leftrightarrow(P_{2})$, we mean that problems $(P_{1})$ and
$(P_{2})$ have the same optimal controls. Based on the equivalence theorem,
the study of one kind of optimal control problem can be carried out by
investigating one of the other two kinds of optimal control problems. In
particular, one can use some existing fine properties for optimal target
controls to derive properties of optimal norm controls and optimal time
controls.

An important application of the equivalence theorem is to build up a feedback
law for norm optimal control problems. We will roughly present this result in
what follows. Notice that Problem $(NP)^{r,\tau}$ depends on $\tau\in[0,T)$
and $y_{0}\in L^{2}(\Omega)$, when $r$ and $z_{d}$ are fixed. To stress this
dependence, we denote, by $(NP)^{r,\tau}_{y_{0}}$, the problem $(NP)^{r,\tau}$
with the initial state $y_{0}$. Throughout this paper, we let $A$ be the
operator on $L^{2}(\Omega)$ with domain $D(A)=H^{1}_{0}(\Omega)\bigcap
H^{2}(\Omega)$ and defined by $Ay=\triangle y$ for each $y\in D(A)$. Write
$\\{e^{t\triangle}:\;t\geq 0\\}$ for the semigroup generated by $A$. By the
equivalence theorem and some characteristics of the target optimal control
problems, we construct a map $F:[0,T)\times L^{2}(\Omega)\rightarrow
L^{2}(\Omega)$ holding properties:

$(i)$ For each $y_{0}\in L^{2}(\Omega)$ and each $\tau\in[0,T)$, the evolution
equation

$\displaystyle\left\\{\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\dot{y}(t)-Ay(t)=\chi_{\omega}\chi_{(\tau,T)}(t)F(t,y(t)),&t\in(0,T)\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr y(0)=y_{0},\end{array}\right.$

has a unique mild solution, which will be denoted by
$y_{F,\tau,y_{0}}(\cdot)$. Here $\chi_{\omega}$ is treated as an operator on
$L^{2}(\Omega)$ in the usual way.

${(ii)}$ For each $y_{0}\in L^{2}(\Omega)$ and each $\tau\in[0,T)$,
$\chi_{(\tau,T)}(\cdot)F(\cdot,y_{F,\tau,y_{0}}(\cdot))$ is the optimal
control to Problem $(NP)^{r,\tau}_{y_{0}}$.

Consequently, the map $F$ is an optimal feedback law for the family of optimal
norm control problems as follows:
$\big{\\{}(NP)^{r,\tau}_{y_{0}}\;:\;\tau\in[0,T),\,y_{0}\in
L^{2}(\Omega)\big{\\}}.$

With the aid of the equivalence theorem, we also build up two algorithms for
the optimal norm, along with the optimal control, to $(NP)^{r,\tau}$ and the
optimal time, together with the optimal control, to $(TP)^{M,r}$,
respectively. These algorithms show that the optimal norm and the optimal
control to $(NP)^{r,\tau}$ can be approximated through solving a series of
two-point boundary value problems, and the same can be said about the optimal
time and the optimal control to $(TP)^{M,r}$.

It deserves to mention that all results obtained in this paper still stand
when Equation (1.1) is replaced by

$\left\\{\begin{array}[]{lll}\partial_{t}y-\triangle
y+ay=\chi_{\omega}\chi_{(\tau,T)}u&\mbox{in}&\Omega\times(0,T),\\\
y=0&\mbox{on}&\partial\Omega\times(0,T),\\\
y(0)=y_{0}&\mbox{in}&\Omega,\end{array}\right.$ (1.6)

where $a\in L^{\infty}(\Omega\times(0,T))$ and $\Omega$ is convex (see Remark
2.13).

The equivalence between optimal time and norm control problems have been
studied in [13], [7] and [5] and the references therein. The optimal time
control problem studied in these papers is to initiate control from the
beginning such that the corresponding solution (to a controlled system)
reaches a target set in the shortest time. Though problems studied in the
current paper differ from those in [13], our study is partially inspired by
[13]. To the best of our knowledge, the equivalence theorem of the above-
mentioned three kinds of optimal control problems has not been touched upon.
Moreover, the feedback law and the algorithms established in this paper seem
to be new.

The rest of the paper is organized as follows: Section 2 presents the
equivalence theorem and its proof. Section 3 provides the above-mentioned two
algorithms. In section 4, we build up an optimal norm feedback law.

##  2 Equivalence of three optimal control problems

Throughout this section, the initial state $y_{0}$ is fixed in
$L^{2}(\Omega)$. For simplicity, we write $y(\cdot;\chi_{(\tau,T)}u)$ and
$r_{T}$ for $y(\cdot;\chi_{(\tau,T)}u,y_{0})$ and $r_{T}(y_{0})$ (which is
defined by (1.4)), respectively. The purpose of this section is to prove the
following equivalence theorem:

######  Theorem 2.1.

When $M>0$ and $\tau\in[0,T)$, the problems $(OP)^{\tau,M}$,
$(TP)^{M,r(\tau,M)}$ and $(NP)^{r(\tau,M),\tau}$ have the same optimal
control; When $r\in(0,r_{T})$ and $\tau\in[0,T)$, the problems
$(NP)^{r,\tau}$, $(OP)^{\tau,M(r,\tau)}$ and $(TP)^{M(r,\tau),r}$ have the
same optimal control; When $M>0$ and $r\in[r(0,M),r_{T})$, the problems
$(TP)^{M,r}$, $(NP)^{r,\tau(M,r)}$ and $(OP)^{\tau(M,r),M}$ have the same
optimal control.

###  2.1 Some properties on optimal target control problems

######  Lemma 2.2.

Let $M\geq 0$ and $\tau\in[0,T)$. Then, $(i)$ $(OP)^{\tau,M}$ has optimal
controls; $(ii)$ $r(\tau,M)>0$; $(iii)$ $u^{*}$ is an optimal control to
$(OP)^{\tau,M}$ if and only if $u^{*}\in L^{\infty}(0,T;L^{2}(\Omega))$, with
$u^{*}=0$ over $(\tau,T)$, satisfies

$\displaystyle\int_{0}^{T}<\chi_{(\tau,T)}(t)\chi_{\omega}p^{*}(t),u^{*}(t)>dt=\max_{v(\cdot)\in\mathcal{U}_{\tau,M}}\int_{0}^{T}<\chi_{(\tau,T)}(t)\chi_{\omega}p^{*}(t),v(t)>dt,$
(2.1)

where $p^{*}$ is the solution to the equation:

$\left\\{\begin{array}[]{lll}\partial_{t}p^{*}+\triangle
p^{*}=0&\mbox{in}&\Omega\times(0,T),\\\
p^{*}=0&\mbox{on}&\partial\Omega\times(0,T),\\\
p^{*}(T)=-(y^{*}(T)-z_{d})&\mbox{in}&\Omega\end{array}\right.$ (2.2)

with $y^{*}(\cdot)$ solving the equation:

$\left\\{\begin{array}[]{lll}\partial_{t}y^{*}-\triangle
y^{*}=\chi_{\omega}\chi_{(\tau,T)}u^{*}&\mbox{in}&\Omega\times(0,T),\\\
y^{*}=0&\mbox{on}&\partial\Omega\times(0,T),\\\
y^{*}(0)=y_{0}&\mbox{in}&\Omega.\end{array}\right.$ (2.3)

###### Proof.

$(i)$ and $(iii)$ have been proved in [9] (see the proof of Theorem 7.2,
Chapter III in [9]). The remainder is to show $(ii)$. For this purpose, we let
$u^{*}$ be an optimal control to $(OP)^{\tau,M}$ and write $y^{*}(\cdot)$ for
$y(\cdot;\chi_{(\tau,T)}u^{*},y_{0})$. Then it holds that
$r(\tau,M)=\|y^{*}(T)-z_{d}\|$ and

$\displaystyle y^{*}(T)\in\big{\\{}y(T;\chi_{(0,T)}u,y_{0})\;:\;u\in
L^{\infty}(0,T;L^{2}(\Omega))\big{\\}}.$ (2.4)

On the other hand, by the null controllability for the heat equation (see, for
instance, [3] or [6]), one can easily check that

$\big{\\{}y(T;\chi_{(0,T)}u,y_{0})\;:\;u\in
L^{\infty}(0,T;L^{2}(\Omega))\big{\\}}=\big{\\{}y(T;\chi_{(0,T)}u,0)\;:\;u\in
L^{\infty}(0,T;L^{2}(\Omega))\big{\\}}.$

This, along with (2.4) and the assumption (1.2), indicates that that
$y^{*}(T)\neq z_{d},$ which implies that $r(\tau,M)>0.$ This completes the
proof. ∎

######  Lemma 2.3.

Let $M\geq 0$ and $\tau\in[0,T)$. Then, $(i)$ $u^{*}$ is an optimal control to
$(OP)^{\tau,M}$ if and only if $u^{*}\in L^{\infty}(0,T;L^{2}(\Omega))$, with
$u^{*}=0$ over $(0,\tau)$, satisfies the following equality:

$\displaystyle
u^{*}(t)=M\frac{\chi_{\omega}p^{*}(t)}{\|\chi_{\omega}p^{*}(t)\|},\;\;\mbox{for
a.e.}\;\;t\in(\tau,T),$ (2.5)

where $p^{*}$ is the solution to (2.2), with $y^{*}(\cdot)$ solving the
equation (2.3); $(ii)$ $(OP)^{\tau,M}$ holds the bang-bang property: any
optimal control $u^{*}$ satisfies that $\|u^{*}(t)\|=M$ for a.e.
$t\in(\tau,T)$; $(iii)$ the optimal control of $(OP)^{\tau,M}$ is unique.

###### Proof.

First, the maximal condition (2.1) is equivalent to the following condition:

$\displaystyle<\chi_{\omega}p^{*}(t),u^{*}(t)>=\max_{v^{0}\in
B(0,M)}<\chi_{\omega}p^{*}(t),v^{0}>\;\;\mbox{for a.e.}\;\;t\in(\tau,T),$
(2.6)

where $B(0,M)$ is the closed ball (in $L^{2}(\Omega)$), centered at the origin
and of radius $M$. Since $p^{*}(T)=-(y^{*}(T)-z_{d})\neq 0$ (see $(ii)$ of
Lemma 2.2), it follows from the unique continuation property of the heat
equation (see [8]) that

$\displaystyle\|\chi_{\omega}p^{*}(t)\|\neq 0\;\;\mbox{for
each}\;\;t\in[0,T).$ (2.7)

Thus, the condition (2.6) is equivalent to the condition (2.5). This, along
with $(iii)$ of Lemma 2.2, yields $(i)$. Next, $(ii)$ follows at once from
(2.5). Finally, $(iii)$ follows from $(ii)$ (see [5] or [14]). This completes
the proof. ∎

######  Lemma 2.4.

Let $M\geq 0$ and $\tau\in[0,T)$. Then the two-point boundary value problem:

$\left\\{\begin{array}[]{ccll}\partial_{t}\varphi-\Delta\varphi=M\chi_{(\tau,T)}\displaystyle\frac{\chi_{\omega}\psi}{\|\chi_{\omega}\psi\|},&\partial_{t}\psi+\triangle\psi=0&\mbox{in}&\Omega\times(0,T),\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\varphi=0,&\psi=0&\mbox{on}&\partial\Omega\times(0,T),\\\ \vskip 3.0pt
plus 1.0pt minus
1.0pt\cr\varphi(0)=y_{0},&\psi(T)=-(\varphi(T)-z_{d})&\mbox{in}&\Omega\end{array}\right.$
(2.8)

admits a unique solution $(\varphi^{\tau,M},\psi^{\tau,M})$ in
$C([0,T];L^{2}(\Omega))\times C([0,T];L^{2}(\Omega))$. Furthermore, the
control, defined by

$\displaystyle
u^{\tau,M}(t)=M\chi_{(\tau,T)}(t)\frac{\chi_{\omega}\psi^{\tau,M}(t)}{\|\chi_{\omega}\psi^{\tau,M}(t)\|},\;\;t\in[0,T),$
(2.9)

is the optimal control to $(OP)^{\tau,M}$, while $\varphi^{\tau,M}$ is the
corresponding optimal state. Consequently, it holds that

$\|\varphi^{\tau,M}(T)-z_{d}\|=r(\tau,M).$ (2.10)

###### Proof.

By Lemma 2.2, $(OP)^{\tau,M}$ has an optimal control $u^{*}$. Let $y^{*}$ and
$p^{*}$ be the corresponding solutions to Equation (2.3) and Equation (2.2),
respectively. Clearly, they belong to $C([0,T];L^{2}(\Omega))$. It follows
from $(i)$ of Lemma 2.3 that $u^{*}$ satisfies (2.5). This, together with
(2.2) and (2.3), shows that $(y^{*},p^{*})$ solves Equation (2.8).

Next, we prove the uniqueness. Suppose that $(\varphi_{1},\psi_{1})$ and
$(\varphi_{2},\psi_{2})$ are two solutions of Equation (2.8). Define $u_{1}$
and $u_{2}$ by (2.5), where $p^{*}$ is replaced by $\psi_{1}$ and $\psi_{2}$,
respectively. It follows from $(i)$ of Lemma 2.3 that $u_{1}$ and $u_{2}$ are
the optimal control to $(OP)^{\tau,M}$ and
$\varphi_{i}(\cdot)=y(\cdot;\chi_{(\tau,T)}u_{i})$, $i=1,2$. Then by $(iii)$
of Lemma 2.3, $\varphi_{1}=\varphi_{2}$. Thus, it holds that
$\psi_{1}(T)=\psi_{2}(T)$, from which, it follows that $\psi_{1}=\psi_{2}$.

Finally, if $(\varphi^{\tau,M},\psi^{\tau,M})$ is the solution of Equation
(2.8), then it follows from $(i)$ of Lemma 2.3 that $u^{\tau,M}$ (defined by
(2.9)) and $\varphi^{\tau,M}$ are the optimal control and the optimal state to
$(OP)^{\tau,M}$. This completes the proof. ∎

######  Remark 2.5.

The unique continuation property (2.7) for the adjoint equation plays a very
important role in this paper. This property also holds for the adjoint
equation of Equation (1.6), where $\Omega$ is convex (see [12]). With the help
of this fact, one can easily check that all results in previous lemmas still
stand when the controlled system is Equation (1.6).

###  2.2 Equivalence of optimal target and norm control problems

######  Lemma 2.6.

Let $\tau\in[0,T)$. Then the map $M\rightarrow r(\tau,M)$ is strictly
monotonically decreasing and Lipschitz continuous from $[0,\infty)$ onto
$(0,r_{T}]$. Furthermore, it holds that

$\displaystyle r=r(\tau,M(r,\tau))\;\;\mbox{for each}\;\;r\in(0,r_{T}]$ (2.11)

and

$\displaystyle M=M(r(\tau,M),\tau)\;\;\mbox{for each}\;\;M\geq 0.$ (2.12)

Consequently, for each $\tau\in[0,T)$, the maps $M\rightarrow r(\tau,M)$ and
$r\rightarrow M(r,\tau)$ are the inverse of each other.

###### Proof.

The proof will be carried out by several steps as follows:

Step 1. It holds that $r(\tau,0)=r_{T}$ and
$\lim_{M\rightarrow\infty}r(\tau,M)=0$.

The first equality above follows directly from the definitions of $r_{T}$ and
$r(\tau,0)$. Now, we prove the second one. Let $\varepsilon>0$. By the
approximate controllability for the heat equation (see [4]), there is a
control $u_{\varepsilon}\in L^{\infty}(\tau,T;L^{2}(\Omega))$ such that
$y(T;\chi_{(\tau,T)}u_{\varepsilon})\in B(z_{d},\varepsilon)$. Clearly,
$u_{\varepsilon}\in\mathcal{U}_{\tau,M}$ for all
$M\geq\|u_{\varepsilon}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}$. Then, by the
optimality of $r(\tau,M)$, we deduce that
$r(\tau,M)\leq\|y(T;\chi_{(\tau,T)}u_{\varepsilon})-z_{d}\|\leq\varepsilon$
for each $M\geq\|u_{\varepsilon}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))},$ from
which, it follows that $\lim_{M\rightarrow\infty}r(\tau,M)=0$.

Step 2. The map $M\rightarrow r(\tau,M)$ is strictly monotonically decreasing.

Let $0\leq M_{1}<M_{2}$. We claim that $r(\tau,M_{2})<r(\tau,M_{1})$. Seeking
for a contradiction, suppose that $r(\tau,M_{2})\geq r(\tau,M_{1})$. Then
optimal control $u_{1}$ to $(OP)^{\tau,M_{1}}$ would satisfy that
$\|y(T;\chi_{(\tau,T)}u_{1})-z_{d}\|=r(\tau,M_{1})\leq r(\tau,M_{2})$ and
$u_{1}\in\mathcal{U}_{\tau,M_{1}}\subset\mathcal{U}_{\tau,M_{2}}.$ These yield
that $u_{1}$ is the optimal control to $(OP)^{\tau,M_{2}}$. By the bang-bang
property of $(OP)^{\tau,M_{2}}$ (see $(ii)$ of Lemma 2.3), it holds that
$\|u_{1}(t)\|=M_{2}$ for a.e. $t\in(\tau,T)$. This contradicts to that
$u_{1}\in\mathcal{U}_{\tau,M_{1}}$, since $M_{1}<M_{2}$.

Step 3. The map $M\rightarrow r(\tau,M)$ is Lipschitz continuous.

Let $M_{1},M_{2}\in[0,\infty)$. Without loss of generality, we can assume that
$0\leq M_{1}<M_{2}$. Let $u^{*}$ be optimal control to $(OP)^{\tau,M_{2}}$.
Then by the monotonicity of the map $M\rightarrow r(\tau,M)$ and the
optimality of $u^{*}$ to $(OP)^{\tau,M_{2}}$, we see that

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus 1.0pt\cr&\displaystyle
r(\tau,M_{1})>r(\tau,M_{2})=\left\|e^{T\triangle}y_{0}+\int^{T}_{\tau}e^{(T-s)\triangle}u^{*}(s)ds-
z_{d}\right\|\\\ \vskip 3.0pt plus 1.0pt minus
1.0pt\cr\geq&\displaystyle\left\|e^{T\triangle}y_{0}+\int^{T}_{\tau}e^{(T-s)\triangle}\frac{M_{1}}{M_{2}}u^{*}(s)ds-
z_{d}\right\|-\frac{(M_{2}-M_{1})}{M_{2}}\left\|\int^{T}_{\tau}e^{(T-s)\triangle}u^{*}(s)ds\right\|.\end{array}$

Since $\displaystyle\frac{M_{1}}{M_{2}}u^{*}\in\mathcal{U}_{\tau,M_{1}}$, it
follows from the definition of $r(\tau,M_{1})$ that

$\displaystyle\left\|e^{T\triangle}y_{0}+\int^{T}_{\tau}e^{(T-s)\triangle}\frac{M_{1}}{M_{2}}u^{*}(s)ds-
z_{d}\right\|\geq r(\tau,M_{1}).$

Because $\|u^{*}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}\leq M_{2}$, we find that

$\int_{\tau}^{T}\|e^{(T-s)\triangle}\|\|u^{*}(s)\|ds\leq M_{2}(T-\tau).$

Putting the above three estimates together leads to the estimate as follows:

$r(\tau,M_{1})>r(\tau,M_{2})\geq r(\tau,M_{1})-(M_{2}-M_{1}),$

from which, it follows that

$|r(\tau,M_{1})-r(\tau,M_{2})|\leq|M_{1}-M_{2}|(T-\tau)\;\;\mbox{for
all}\;\;M_{1},M_{2}\in[0,\infty).$

Step 4. The proof of (2.11)

First of all, by the definitions of $r_{T}$ , one can easily check that
$M(r_{T},\tau)=0$ and

$\displaystyle r_{T}=r(\tau,0)=r(\tau,M(r_{T},\tau)).$ (2.13)

Then, let $r\in(0,r_{T})$. By Step 2, $M(r,\tau)>0$ for this case. We are
going to prove the following two claims:

Claim one: $r\geq r(\tau,M(r,\tau))$ and Claim two: $r\leq r(\tau,M(r,\tau))$.

Clearly, these claims, together with (2.13), lead to (2.11). To prove the
first claim, we let $u$ be an optimal control to $(NP)^{r,\tau}$ (the
existence of such a control is provided in [4]). Then it holds that
$\|y(T;\chi_{(\tau,T)}u)-z_{d}\|\leq r$ and
$u\in\mathcal{U}_{\tau,M(r,\tau)}$. These, along with the definition of
$r(\tau,M)$, shows Claim one.

Now we show the second claim. Seeking a contradiction, suppose that
$r>r(\tau,M(r,\tau))$. Since the map $M\rightarrow r(\tau,M)$ is continuous
and strictly monotonically decreasing, there would be a
$M_{1}\in(0,M(r,\tau))$ such that $r(\tau,M_{1})=r$. Thus, the optimal control
$u_{1}$ to $(OP)^{\tau,M_{1}}$ satisfies that

$\displaystyle\|u_{1}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M_{1}<M(r,\tau)\;\;\mbox{and}\;\;\|y(T;\chi_{(\tau,T)}u_{1})-z_{d}\|=r(\tau,M_{1})=r.$
(2.14)

The second equality in (2.14) implies that $u_{1}\in\mathcal{U}_{r,\tau}$,
which, together with the optimality of $M(r,\tau)$, indicates that
$M(r,\tau)\leq\|u_{1}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}$. This contradicts
to the first inequality in (2.14).

Step 5. The proof of (2.12).

One can easily check that $r(\tau,M)\in(0,r_{T}]$ whenever $M\geq 0$ and
$\tau\in[0,T)$. Thus, we can make use of (2.11) to get that

$\displaystyle r(\tau,M)=r(\tau,M(r(\tau,M),\tau)),\;\;M\geq 0,\tau\in[0,T).$
(2.15)

Since the map $M\rightarrow r(\tau,M)$ is strictly monotone, (2.12) follows
from (2.15) at once.

In summary, we complete the proof. ∎

######  Proposition 2.7.

$(i)$ The optimal control to $(OP)^{\tau,M}$, where $M\geq 0$ and
$\tau\in[0,T)$, is an optimal control to $(NP)^{r(\tau,M),\tau}$. $(ii)$ Any
optimal control to $(NP)^{r,\tau}$, where $\tau\in[0,T)$ and $r\in(0,r_{T}]$,
is the optimal control to $(OP)^{\tau,M(r,\tau)}$. $(iii)$ For each
$\tau\in[0,T)$ and each $r\in(0,r_{T}]$, $(NP)^{r,\tau}$ holds the bang-bang
property (i.e., any optimal control $u^{*}$ satisfies that
$\|u^{*}(t)\|=M(r,\tau)$ for a.e. $t\in(\tau,T)$) and the optimal control to
$(NP)^{r,\tau}$ is unique.

###### Proof.

$(i)$ The optimal control $u$ to $(OP)^{\tau,M}$ satisfies that
$y(T;\chi_{(\tau,T)}u)\in B(z_{d},r(\tau,M))$,
$\|u\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M$ and $u=0$ over $(0,\tau)$. These,
together with (2.12), indicate that $u$ is an optimal control to
$(NP)^{r(\tau,M),\tau}$. $(ii)$ An optimal control $v$ to $(NP)^{r,\tau}$,
where $\tau\in[0,T)$ and $r\in(0,r_{T}]$, satisfies that
$\|v\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M(r,\tau)$,
$\|y(T;\chi_{(\tau,T)}v)-z_{d}\|\leq r$ and $v=0$ over $(0,\tau)$. These,
along with (2.11), yields that that $v$ is the optimal control to
$(OP)^{\tau,M(r,\tau)}$. $(iii)$ The bang-bang property and the uniqueness of
$(NP)^{r,\tau}$ follow from $(ii)$ and Lemma 2.3. This completes the proof.

∎

###  2.3 Equivalence of optimal norm and time control problems

######  Lemma 2.8.

Let $r\in(0,r_{T})$ and $M\geq M(r,0)$. Then, $(TP)^{M,r}$ has optimal
controls. Moreover, it holds that $\tau(M,r)<T$.

###### Proof.

We first claim that when $u\in\mathcal{U}_{M,r}$, the supremum in (1.3) can be
reached, i.e.

$\displaystyle y(T;\chi_{(\widetilde{\tau}(u),T)}u)\in
B(z_{d},r)\;\mbox{and}\;u\in\mathcal{U}_{\widetilde{\tau}(u),M}\;\;\mbox{when
}\;\;u\in\mathcal{U}_{M,r}.$ (2.16)

Here, we simply write $\widetilde{\tau}(u)$ for $\widetilde{\tau}_{M,r}(u)$,
which is defined by (1.3). To this end, we let $u\in\mathcal{U}_{M,r}$. Then
by the definition of $\widetilde{\tau}(u)$, there is a sequence
$\\{\tau_{n}\\}\subset[0,T)$ such that
$\tau_{n}\rightarrow\widetilde{\tau}(u)$, $y(T;\chi_{(\tau_{n},T)}u)\in
B(z_{d},r)$ and $u\in\mathcal{U}_{\tau_{n},M}$. From these, (2.16) follows at
once.

Next we notice that $(NP)^{r,0}$ has optimal controls (see [4]) and any
optimal control to $(NP)^{r,0}$ belongs to
$\mathcal{U}_{M(r,0),r}\subset\mathcal{U}_{M,r}$ (since $M\geq M(r,0)$). These
imply that $\mathcal{U}_{M,r}\neq\emptyset$. Thus, there is a sequence
$\\{u_{n}\\}\subset\mathcal{U}_{M,r}$ such that
$\widetilde{\tau}(u_{n})\rightarrow\tau(M,r)$. On the other hand, by (2.16),
$y(T;\chi_{(\widetilde{\tau}(u_{n}),T)}u_{n})\in B(z_{d},r)$ and
$u_{n}\in\mathcal{U}_{\widetilde{\tau}(u_{n}),M}$. Hence, there exist a
subsequence of $\\{u_{n}\\}$, still denoted in the same way, and a control
$v^{*}\in L^{\infty}(0,T;L^{2}(\Omega))$ such that

$\chi_{(\widetilde{\tau}(u_{n}),T)}u_{n}\rightarrow\chi_{(\tau(M,r),T)}v^{*}\;\;\mbox{weakly
star in}\;L^{\infty}(0,T;L^{2}(\Omega))$

and

$y(T;\chi_{(\widetilde{\tau}(u_{n}),T)}u_{n})\rightarrow
y(T;\chi_{(\tau(M,r),T)}v^{*}).$

From these, it follows that $y(T;\chi_{(\tau(M,r),T)}v^{*})\in B(z_{d},r)$ and
$v^{*}\in\mathcal{U}_{\tau(M,r),M}$. Hence, $\chi_{(\tau(M,r),T)}v^{*}$ is an
optimal control to $(TP)^{M,r}$.

Finally, since $r<r_{T}$ and $y(T;\chi_{(\tau(M,r),T)}v^{*})\in B(z_{d},r)$,
it follows that $\tau(M,r)<T$. This completes the proof. ∎

######  Lemma 2.9.

Let $r\in(0,r_{T})$. Then the map $\tau\rightarrow M(r,\tau)$ is strictly
monotonically increasing and continuous from $[0,T)$ onto
$[M(0,\tau),\infty)$. Furthermore, it holds that

$\displaystyle M=M(r,\tau(M,r))\;\;\mbox{for each}\;\;M\in[M(r,0),\infty)$
(2.17)

and

$\displaystyle\tau=\tau(M(r,\tau),r)\;\;\mbox{for each}\;\;\tau\in[0,T).$
(2.18)

Consequently, the maps $\tau\rightarrow M(r,\tau)$ and $M\rightarrow\tau(M,r)$
are the inverse of each other.

###### Proof.

We carry out the proof by several steps as follows:

Step 1. This map is strictly monotonically increasing over $[0,T)$.

Let $0\leq\tau_{1}<\tau_{2}<T$. We claim that $M(r,\tau_{1})<M(r,\tau_{2})$.
Seeking for a contradiction, suppose that $M(r,\tau_{2})\leq M(r,\tau_{1})$.
Then the optimal control $u_{2}$ to $(NP)^{r,\tau_{2}}$ would satisfy

$\displaystyle\|\chi_{(\tau_{2},T)}u_{2}\|_{L^{\infty}(0,T;L^{2}(\Omega))}=M(r,\tau_{2})\leq
M(r,\tau_{1})\;\;\mbox{and}\;\;y(T;\chi_{(\tau_{2},T)}u_{2})\in B(z_{d},r).$

These imply that $\chi_{(\tau_{2},T)}u_{2}$ is the optimal control to
$(NP)^{r,\tau_{1}}$. Then, it follows from the bang-bang property of
$(NP)^{r,\tau_{1}}$ (see Proposition 2.7) that
$\|\chi_{(\tau_{2},T)}u_{2}(t)\|=M(r,\tau_{1})$ over $(\tau_{1},\tau_{2})$.
This contradicts to the facts that $\tau_{1}<\tau_{2}$ and $M(r,\tau_{1})>0$
(which follows from $r<r_{T}$).

Step 2. $0\leq\tau_{1}<\tau_{2}<\cdots<\tau_{n}\rightarrow\tau<T\Rightarrow
M(r,\tau_{n})\rightarrow M(r,\tau)$.

If this did not hold, then by the monotonicity of the map $\tau\rightarrow
M(r,\tau)$, we would have

$\displaystyle\lim_{n\rightarrow\infty}M(r,\tau_{n})=M(r,\tau)-\delta\;\;\mbox{for
some}\;\;\delta>0.$ (2.19)

Let $u_{n}$ and $y_{n}$ be the optimal control and the optimal state to
$(OP)^{\tau_{n},M(r,\tau_{n})}$, respectively. Then, it follows from Lemma 2.2
that

$\displaystyle\int_{0}^{T}<\chi_{\omega}p_{n},\chi_{(\tau_{n},T)}u_{n}>dt\geq\int_{0}^{T}<\chi_{\omega}p_{n},\chi_{(\tau_{n},T)}v_{n}>dt\;\;\mbox{for
each}\;\;v_{n}\in\mathcal{U}_{\tau_{n},M(r,\tau_{n})},$ (2.20)
$\displaystyle\left\\{\begin{array}[]{ll}\partial_{t}y_{n}-\triangle
y_{n}=\chi_{\omega}\chi_{(\tau_{n},T)}u_{n}&\textrm{in }\Omega\times(0,T),\\\
y_{n}=0&\textrm{on }\partial\Omega\times(0,T),\\\ y_{n}(0)=y_{0}&\textrm{in
}\Omega,\end{array}\right.$
$\displaystyle\left\\{\begin{array}[]{ll}\partial_{t}p_{n}+\triangle
p_{n}=0&\textrm{in }\Omega\times(0,T),\\\ p_{n}=0&\textrm{on
}\partial\Omega\times(0,T),\\\ p_{n}(T)=-(y_{n}(T)-z_{d})&\textrm{in
}\Omega.\end{array}\right.$

Besides, by the optimality of $y_{n}$ and (2.11) (in Lemma 2.6), we see that

$\displaystyle\|y_{n}(T)-z_{d}\|=r(M(r,\tau_{n}),\tau_{n})=r\;\;\mbox{for
all}\;\;n\in\mathbb{N}.$ (2.23)

Since $\tau_{n}\rightarrow\tau$ and
$\|u_{n}\|_{L^{\infty}(\tau_{n},T;L^{2}(\Omega))}=M(r,\tau_{n})\leq
M(r,\tau)-\delta$, there exist a subsequence, still denoted in the same way,
and a control $\widetilde{u}\in L^{\infty}(0,T;L^{2}(\Omega))$ such that

$\displaystyle\chi_{(\tau_{n},T)}u_{n}\rightarrow\chi_{(\tau,T)}\widetilde{u}\;\;\mbox{weakly
star in}\;\;L^{\infty}(0,T;L^{2}(\Omega)).$ (2.24)

This, together with the equations satisfied by $y_{n}$ and $p_{n}$
respectively, indicates that

$\displaystyle
y_{n}\rightarrow\widetilde{y}\;\;\mbox{and}\;\;p_{n}\rightarrow\widetilde{p}\;\;\mbox{in}\;\;C([0,T];L^{2}(\Omega)),$
(2.25)
$\displaystyle\left\\{\begin{array}[]{ll}\partial_{t}\widetilde{y}-\triangle\widetilde{y}=\chi_{\omega}\chi_{(\tau,T)}\widetilde{u}&\textrm{in
}\Omega\times(0,T),\\\ \widetilde{y}=0&\textrm{on
}\partial\Omega\times(0,T),\\\ \widetilde{y}(0)=y_{0}&\textrm{in
}\Omega\end{array}\right.$ (2.29)

and

$\displaystyle\left\\{\begin{array}[]{ll}\partial_{t}\widetilde{p}+\triangle\widetilde{p}=0&\textrm{in
}\Omega\times(0,T),\\\ \widetilde{p}=0&\textrm{on
}\partial\Omega\times(0,T),\\\
\widetilde{p}(T)=-(\widetilde{y}(T)-z_{d})&\textrm{in
}\Omega.\end{array}\right.$ (2.33)

In addition, it follows from (2.23) and (2.25) that
$\|\widetilde{y}(T)-z_{d}\|=r$. By making use of (2.11) again, we deduce that

$\displaystyle\|\widetilde{y}(T)-z_{d}\|=r(\tau,M(r,\tau)).$ (2.34)

Now, we take a $v\in\mathcal{U}_{\tau,M(r,\tau)-\delta}$. Since
$M(r,\tau)-\delta>0$, it holds that

$\frac{M(r,\tau_{n})}{M(r,\tau)-\delta}\chi_{(\tau_{n},T)}v\in\mathcal{U}_{\tau_{n},M(r,\tau_{n})}.$

Then, it follows from (2.20) that

$\displaystyle\int_{0}^{T}<\chi_{\omega}p_{n},\chi_{(\tau_{n},T)}u_{n})>dt\geq\int_{0}^{T}<\chi_{\omega}p_{n},\frac{M(r,\tau_{n})}{M(r,\tau)-\delta}\chi_{(\tau_{n},T)}v>dt.$

By (2.19), (2.24) and (2.25), we can pass to the limit in the above to get
that

$\displaystyle\int_{0}^{T}<\chi_{\omega}\widetilde{p}\;,\chi_{(\tau,T)}\widetilde{u}>dt\geq\int_{0}^{T}<\chi_{\omega}\widetilde{p}\;,\chi_{(\tau,T)}v>dt\;\;\mbox{for
all}\;\;v\in\mathcal{U}_{\tau,M(r,\tau)-\delta,}.$

This, along with the fact that
$\widetilde{u}\in\mathcal{U}_{\tau,M(r,\tau)-\delta,}$ (which follows from
(2.24)), indicates that

$\displaystyle\int_{0}^{T}<\chi_{\omega}\widetilde{p}\;,\chi_{(\tau,T)}\widetilde{u}>dt=\displaystyle{\max_{v\in\mathcal{U}_{\tau,M(r,\tau)-\delta,}}}\int_{0}^{T}<\chi_{\omega}\widetilde{p}\;,\chi_{(\tau,T)}>dt.$

According to Lemma 2.2, the above equality, together with (2.29) and (2.33),
shows that $\chi_{(\tau,T)}\widetilde{u}$ and $\widetilde{y}$ are the optimal
control and the optimal state to $(OP)^{\tau,M(r,\tau)-\delta,}$. Therefore,
it stands that $\|\widetilde{y}(T)-z_{d}\|=r(\tau,M(r,\tau)-\delta),$ which,
combined with (2.34), indicates that
$r(\tau,M(r,\tau))=r(\tau,M(r,\tau)-\delta).$ This contradicts with the strict
monotonicity of the map $M\rightarrow r(\tau,M)$ (see Lemma 2.6).

Step 3. $T>\tau_{1}>\cdots>\tau_{n}\rightarrow\tau\geq 0\Rightarrow
M(r,\tau_{n})\rightarrow M(r,\tau)$.

If this did not hold, then by the monotonicity of the map $\tau\rightarrow
M(r,\tau)$, we would have that
$\lim_{n\rightarrow\infty}M(r,\tau_{n})=M(r,\tau)+\delta\;\;\mbox{for
some}\;\;\delta>0.$ Following the same argument as that in Step 2, we can
derive that $r(\tau,M(r,\tau))=r(\tau,M(r,\tau)+\delta).$ This contradicts to
the strict monotonicity of the map $M\rightarrow r(\tau,M)$.

Step 4. $\lim_{\tau\rightarrow T}M(r,\tau)=\infty.$

Seeking for a contradiction, we suppose that
$0<\tau_{1}<\cdots<\tau_{n}\rightarrow T$ and
$\lim_{n\rightarrow\infty}M(r,\tau_{n})=M<\infty$. Let $u_{n}$ and $y_{n}$ be
the optimal control and state for $(NP)^{r,\tau_{n}}.$ Then we would have that
$\chi_{(\tau_{n},T)}u_{n}\rightarrow 0$ weakly star in
$L^{\infty}(0,T;L^{2}(\Omega))$ and $y_{n}(\cdot)\rightarrow y(\cdot;0)$ in
$C([0,T];L^{2}(\Omega))$. Thus, it holds that
$r_{T}\equiv\|y(T;0)-z_{d}\|=\lim_{n\rightarrow\infty}\|y_{n}(T)-z_{d}\|\leq
r$, which contradicts to the assumption that $r<r_{T}$.

Step 5. The proof of (2.17)

By Lemma 2.8, the problem $(TP)^{M,r}$ has an optimal control $u$. It holds
that

$\displaystyle y(T;\chi_{(\tau(M,r),T)}u)\in B(z_{d},r)\;\;\mbox{
and}\;\;\|u\|_{L^{\infty}(\tau(M,r),T;L^{2}(\Omega))}\leq M.$ (2.35)

From the first fact in (2.35), we see that $u\in\mathcal{U}_{r,\tau(M,r)}$.
This, together with the optimality of $M(r,\tau)$ and the second fact in
(2.35), shows that

$\displaystyle M\geq M(r,\tau(M,r)).$ (2.36)

Seeking for a contradiction, suppose that $M>M(r,\tau(M,r))$. Since the map
$\tau\rightarrow M(r,\tau)$ is continuous and strictly monotonically
increasing, there would be a $\tau_{1}$, with $\tau_{1}\in(\tau(M,r),T)$, such
that $M(r,\tau_{1})=M$. Clearly, the optimal control $u_{1}$ to
$(NP)^{r,\tau_{1}}$ satisfies that

$\displaystyle\|u_{1}\|_{L^{\infty}(\tau_{1},T;L^{2}(\Omega))}=M(r,\tau_{1})=M\;\;\mbox{and}\;\;y(T;\chi_{(\tau_{1},T)}u_{1})\in
B(z_{d},r).$ (2.37)

From these, it follows that $u_{1}\in\mathcal{U}_{M,r}$. Then, by the
optimality of $\tau(M,r)$ , (1.3) and (2.37), we deduce that
$\tau(M,r)\geq\widetilde{\tau}(u_{1})\geq\tau_{1},$ which contradicts with
that $\tau_{1}\in(\tau(M,r),T)$.

Step 6. The proof of (2.18).

Let $\tau\in[0,T)$. By Step 1, it follows that $M(r,\tau)\geq M(r,0)$. Then we
can apply (2.17) to deduce that $M(r,\tau)=M(r,\tau(M(r,\tau),r))$. By making
use of Step 1 again, we obtain that $\tau=\tau(M(r,\tau),r)$.

In summary, we complete the proof. ∎

######  Proposition 2.10.

$(i)$ Any optimal control to $(TP)^{M,r}$, where $r\in(0,r_{T})$ and $M\geq
M(r,0)$, is the optimal control to $(NP)^{r,\tau(M,r)}$. $(ii)$ The optimal
optimal control to $(NP)^{r,\tau}$, with $\tau\in[0,T)$ and $r\in(0,r_{T})$,
is an optimal control to $(TP)^{M(r,\tau),r}$. $(iii)$ For each
$r\in(0,r_{T})$ and each $M\geq M(r,0)$, $(TP)^{M,r}$ holds the bang-bang
property (i.e., any optimal control $u^{*}$ satisfies that $\|u^{*}(t)\|=M$
for a.e. $t\in(\tau(M,r),T)$) and the optimal control to $(TP)^{M,r}$ is
unique.

###### Proof.

$(i)$ An optimal control $u$ to $(TP)^{M,r}$ satisfies that $u=0$ over
$(\tau(M,r),T)$,

$y(T;\chi_{(\tau(M,r),T)}u)\in B(z_{d},r)\;\mbox{ and
}\;\|u\|_{L^{\infty}(\tau(M,r),T;L^{2}(\Omega))}\leq M.$

These, together with (2.17), yields that $u$ is the optimal control to
$(NP)^{r,\tau(M,r)}$. $(ii)$ The optimal control $v$ to $(NP)^{r,\tau}$
satisfies that $v=0$ over $(\tau,T)$,
$\|u\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M(r,\tau)$ and
$y(T;\chi_{(\tau,T)}u)\in B(z_{d},r)$. These, together with (2.18), yields
that $u$ is an optimal control to $(TP)^{M(r,\tau),r}$. $(iii)$ The bang-bang
property and the uniqueness of $(TP)^{M,r}$ follow from $(iii)$ of Proposition
2.7 and $(i)$ above. This completes the proof. ∎

###  2.4 Equivalence of optimal target and time control problems

Though the equivalence between optimal target and time control problems can be
derived from Proposition 2.7 and Proposition 2.10, the properties of maps
$\tau\rightarrow r(\tau,M)$ and $r\rightarrow\tau(M,r)$ are independently
interesting and will be used in the next section. This is why we introduce
what follows.

######  Lemma 2.11.

Let $M>0$. Then the map $\tau\rightarrow r(\tau,M)$ is strictly monotonically
increasing and continuous from $[0,T)$ onto $[r(0,M),r_{T})$. Furthermore, it
holds that

$\displaystyle r=r(\tau(M,r),M)\;\;\mbox{for each}\;\;r\in[r(0,M),r_{T}),$
(2.38) $\displaystyle\tau=\tau(M,r(\tau,M))\;\;\mbox{for
each}\;\;\tau\in[0,T).$ (2.39)

Consequently, the maps $\tau\rightarrow r(\tau,M)$ and $r\rightarrow\tau(M,r)$
are the inverse of each other.

###### Proof.

We carry out the proof by several steps as follows:

Step 1. The map $\tau\rightarrow r(\tau,M)$ is strictly monotonically
increasing.

Let $0\leq\tau_{1}<\tau_{2}<T$. It follows from (2.12) that

$\displaystyle M(r(\tau_{1},M),\tau_{1})=M(r(\tau_{2},M),\tau_{2}).$ (2.40)

We first claim that

$\displaystyle r(\tau_{2},M)\in(0,r_{T})\;\;\mbox{when}\;\;M>0.$ (2.41)

In fact, on one hand, it is clear that $r(\tau_{2},M)>0$ (see Lemma 2.2). On
the other hand, since the map $M\rightarrow r(\tau_{2},M)$ is strictly
monotonically decreasing (see Lemma 2.6), it holds that
$r(\tau_{2},M)<r(\tau_{2},0)=\|y(T;0)-z_{d}\|=r_{T}$. Then by (2.41), we can
apply Lemma 2.9 to get that
$M(r(\tau_{2},M),\tau_{2})>M(r(\tau_{2},M),\tau_{1})$. This, together with
(2.40), yields that

$\displaystyle M(r(\tau_{1},M),\tau_{1})>M(r(\tau_{2},M),\tau_{1}).$ (2.42)

Since the map $r\rightarrow M(r,\tau_{1})$ is strictly monotonically
decreasing (see Lemma 2.6), it follows from (2.42) that
$r(\tau_{1},M)<r(\tau_{2},M)$.

Step 2. The map $\tau\rightarrow r(\tau,M)$ is continuous.

Since for each $\tau\in[0,T)$, the map $r\rightarrow M(r,\tau)$ is continuous
and monotonic over $(0,r_{T})$ (see Lemma 2.6), and for each $r\in(0,r_{T})$,
the map $\tau\rightarrow M(r,\tau)$ is continuous (and monotonic) over $[0,T)$
(see Lemma 2.9), it follows that

$\displaystyle\mbox{the map}\;\;(r,\tau)\rightarrow M(r,\tau)\;\;\mbox{is
continuous over}\;\;(0,r_{T})\times[0,T).$ (2.43)

Now we prove that the map $\tau\rightarrow r(\tau,M)$ is continuous from left.
For this purpose, we let
$0\leq\tau_{1}<\tau_{2}<\cdots<\tau_{n}\rightarrow\tau<T$. Then by the
monotonicity of $\\{\tau_{n}\\}$, $\lim_{n\rightarrow\infty}r(\tau_{n},M)$
exists. Thus, it follows from (2.43) that

$\lim_{n\rightarrow\infty}M(r(\tau_{n},M),\tau_{n})=M(\lim_{n\rightarrow\infty}r(\tau_{n},M),\tau).$

On the other hand, by (2.12), it stands that

$M(r(\tau_{n},M),\tau_{n})=M=M(r(\tau,M),\tau)\;\;\mbox{for all}\;\;n.$

These yield that
$M(\lim_{n\rightarrow\infty}r(\tau_{n},M),\tau)=M(r(\tau,M),\tau)$. This,
together with the strict monotonicity of the map $r\rightarrow M(r,\tau)$ (see
Lemma 2.6), indicates that $\lim_{n\rightarrow\infty}r(\tau_{n},M)=r(\tau,M)$.
Thus, the map $\tau\rightarrow r(\tau,M)$ is continuous from left. Similarly,
we can prove that it is continuous from right.

Step 3. It holds that $\lim_{\tau\rightarrow T}r(\tau,M)=r_{T}$.

Clearly, the optimal control $u_{\tau}$ to $(OP)^{\tau,M}$ satisfies that
$\|y(T;\chi_{(\tau,T)}u_{\tau})-z_{d}\|=r(\tau,M)$ and
$\|u_{\tau}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}\leq M.$ One can easily see
that $\chi_{(\tau,T)}u_{\tau}\rightarrow
0\;\;\mbox{in}\;\;L^{\infty}(0,T;L^{2}(\Omega))$ as $\tau$ tends to $T$, from
which, it follows that $y(T;\chi_{(0,T)}u_{\tau})\rightarrow y(T;0)$ as $\tau$
tends to $T$. Therefore, it holds that
$r_{T}\equiv\|y(T;0)-z_{d}\|=\lim_{\tau\rightarrow
T}\|y(T;\chi_{(0,T)}u_{\tau})-z_{d}\|=\lim_{\tau\rightarrow T}r(\tau,M).$

Step 4. The proof of (2.38) and (2.39).

We start with proving the following:

$\displaystyle\mathcal{A}_{1}=\mathcal{A}_{2},$ (2.44)

where $\mathcal{A}_{1}=\\{(M,r):r\in(0,r_{T}),M\geq M(r,0)\\}$ and
$\mathcal{A}_{2}=\\{(M,r):M>0,r\in[r(0,M),r_{T})\\}.$ In fact, if
$(M,r)\in\mathcal{A}_{1}$, since $r>0$, it follows that $M>0$. On the other
hand, because $M\geq M(r,0)$, we can apply Lemma 2.6 to get that $r(0,M)\geq
r(0,M(r,0))=r.$ Thus, it stands that $(M,r)\in\mathcal{A}_{2}$. Similarly, we
can prove that $\mathcal{A}_{2}\subset\mathcal{A}_{1}$.

Next, it follows from (2.44) and (2.17) that $M=M(r,\tau(M,r))$ when $M>0$ and
$r\in[r(0,M),r_{T})$. This, together with (2.11), indicates that

$r(\tau(M,r),M)=r(\tau(M,r),M(r,\tau(M,r)))=r\;\;\mbox{for
each}\;r\in[r(0,M),r_{T}),$

which leads to (2.38).

Finally, because $r(\tau,M)\in(0,r_{T})$ (see (2.41)), we can make use of
(2.18) to get that $\tau(M(r(\tau,M),\tau),r(\tau,M))=\tau,$ which, along with
(2.12), gives (2.39).

In summary, we complete the proof. ∎

######  Proposition 2.12.

The optimal control to $(TP)^{M,r}$, where $M>0$ and $r\in[r(0,M),r_{T})$, is
the optimal control to $(OP)^{\tau(M,r),M}$. Conversely, the optimal control
to $(OP)^{\tau,M}$, where $M>0$ and $\tau\in[0,T)$, is the optimal control to
$(TP)^{M,r(\tau,M)}$.

This proposition can be directly derived from Lemma 2.11. Also it is a
consequence of Proposition 2.7, Proposition 2.10 and (2.44). We omit its
proof.

###  2.5 Proof of Theorem 2.1

Let $(P_{1})$ and $(P_{2})$ be two optimal control problems. By
$(P_{1})\Rightarrow(P_{2})$, we mean that the optimal control to $(P_{1})$ is
the optimal control to $(P_{2})$. The proof will be carried out by several
steps as follows:

Step 1.
$(OP)^{\tau,M}\Rightarrow(TP)^{M,r(\tau,M)}\Rightarrow(NP)^{r(\tau,M),\tau}\Rightarrow(OP)^{\tau,M}$,
$M>0$, $\tau\in[0,T)$.

$(OP)^{\tau,M}\Rightarrow(TP)^{M,r(\tau,M)}$: It follows from Proposition 2.7.

$(TP)^{M,r(\tau,M)}\Rightarrow(NP)^{r(\tau,M),\tau}$: We first claim that

$\displaystyle
r(\tau,M)\in(0,r_{T})\;\;\mbox{when}\;\;M>0\;\;\mbox{and}\;\;\tau\in[0,T).$
(2.45)

In fact, it follows from Lemma 2.2 that $r(\tau,M)>0$. On the other hand,
since $M>0$ and the map $M\rightarrow r(\tau,M)$ is strictly monotonically
decreasing (see Lemma 2.6), it holds that $r(\tau,M)<r(0,\tau)=r_{T}$. These
lead to (2.45).

We next claim that

$\displaystyle M\geq
M(r(\tau,M),0)\;\;\mbox{when}\;\;M>0\;\;\mbox{and}\;\;\tau\in[0,T).$ (2.46)

Indeed, since the map $\tau\rightarrow r(\tau,M)$ is monotonically increasing
(see Lemma 2.9), it holds that $r(0,M)\leq r(\tau,M)$. Because the map
$r\rightarrow M(r,0)$ is monotonically decreasing (see Lemma 2.6), it stands
that $M(r(0,M),0)\geq M(r(\tau,M),0)$. This, combined with (2.12), shows
(2.46). Now, by (2.45) and (2.46), we can apply Proposition 2.10, together
with (2.18), to get $(TP)^{M,r(\tau,M)}\Rightarrow(NP)^{r(\tau,M),\tau}$.

$(NP)^{r(\tau,M),\tau}\Rightarrow(OP)^{\tau,M}$: By (2.45), we can make use of
Proposition 2.7, together with (2.12), to get
$(NP)^{r(\tau,M),\tau}\Rightarrow(OP)^{\tau,M}$.

Step 2.
$(NP)^{r,\tau}\Rightarrow(OP)^{\tau,M(r,\tau)}\Rightarrow(TP)^{M(r,\tau),r}\Rightarrow(NP)^{r,\tau}$,
$r\in(0,r_{T})$, $\tau\in[0,T)$.

$(NP)^{r,\tau}\Rightarrow(OP)^{\tau,M(r,\tau)}$: It follows from Proposition
2.7.

$(OP)^{\tau,M(r,\tau)}\Rightarrow(TP)^{M(r,\tau),r}$: We first claim that

$\displaystyle
M(r,\tau)>0\;\;\mbox{when}\;\;r\in(0,r_{T})\;\;\mbox{and}\;\;\tau\in[0,T).$
(2.47)

In fact, since $r_{T}=\|y(T;0)-z_{d}\|$, it holds that $M(r_{T},\tau)=0$. On
the other hand, since $r<r_{T}$ and the map $r\rightarrow M(r,\tau)$ is
strictly monotonically decreasing (see Lemma 2.6), we see that
$M(r,\tau)>M(r_{T},\tau)$. Thus, (2.47) follows immediately. Now, by (2.47),
we can apply Proposition 2.12, along with (2.11), to derive
$(OP)^{\tau,M(r,\tau)}\Rightarrow(TP)^{M(r,\tau),r}$.

$(TP)^{M(r,\tau),r}\Rightarrow(NP)^{r,\tau}$: Since $r\in(0,r_{T})$, the map
$\tau\rightarrow M(r,\tau)$ is monotonically increasing (see Lemma 2.9). Thus,
it holds that $M(r,\tau)\geq M(r,0)$. Then we can make use of Proposition
2.10, together with (2.18), to yield
$(TP)^{M(r,\tau),r}\Rightarrow(NP)^{r,\tau}$.

Step 3.
$(TP)^{M,r}\Rightarrow(NP)^{r,\tau(M,r)}\Rightarrow(OP)^{\tau(M,r),M}\Rightarrow(TP)^{M,r}$,
$M>0$, $r\in[r(0,M),r_{T})$.

$(TP)^{M,r}\Rightarrow(NP)^{r,\tau(M,r)}$: It follows from (2.44) and
Proposition 2.10.

$(NP)^{r,\tau(M,r)}\Rightarrow(OP)^{\tau(M,r),M}$: Since $r>0$ in this case
(see (2.44)), we can apply Proposition 2.7, together with (2.12), to get
$(NP)^{r,\tau(M,r)}\Rightarrow(OP)^{\tau(M,r),M}$.

$(OP)^{\tau(M,r),M}\Rightarrow(TP)^{M,r}$: It follows from Proposition 2.12,
together with (2.38).

In summary, we complete the proof of Theorem 2.1.

######  Remark 2.13.

All results in this section hold for the case where the controlled system is
Equation (1.6). In fact, these results hold for the three kinds of optimal
control problems studied in this paper, when the adjoint equation of the
controlled heat equation has the unique continuation property (2.7).

##  3 Applications I: Algorithms for $M(r,\tau)$ and $\tau(M,r)$

Throughout this section, we fix an initial state $y_{0}\in L^{2}(\Omega)$ and
write $r_{T}$ for $r_{T}(y_{0})$. For each $M>0$ and $\tau\in[0,T)$,
$(\varphi^{\tau,M},\psi^{\tau,M})$ denotes the unique solution to the two-
point boundary value problem (2.8) and $\varphi^{\tau,M}$ (or
$\psi^{\tau,M})$) stands for the first (or second) component of this solution
when it appears alone.

######  Proposition 3.1.

Let $\tau\in[0,T)$ and $r\in(0,r_{T})$. Then $M^{*}$, $u^{*}$ and $y^{*}$ are
the optimal norm, the optimal control and the optimal state to $(NP)^{r,\tau}$
if and only if $M^{*}$, $u^{*}$ and $y^{*}$ satisfy that $M^{*}>0$,

$\displaystyle\|y^{*}(T)-z_{d}\|=r,$ (3.1) $\displaystyle
u^{*}(t)=M^{*}\chi_{(\tau,T)}(t)\displaystyle\frac{\chi_{\omega}\psi^{\tau,M^{*}}(t)}{\|\chi_{\omega}\psi^{\tau,M^{*}}(t)\|},\;\;t\in[\tau,T)$
(3.2)

and

$\displaystyle y^{*}(t)=\varphi^{\tau,M^{*}}(t),\;\;t\in[0,T].$ (3.3)

###### Proof.

Suppose that $M^{*}$, $u^{*}$ and $y^{*}$ are the optimal norm, the optimal
control and the optimal state to $(NP)^{r,\tau}$. Clearly, $M^{*}=M(r,\tau)$.
It follows from Lemma 2.6 that $M(r,\tau)>M(r_{T},\tau)$. Hence, $M^{*}>0$.
Then, by Theorem 2.1, $u^{*}$ and $y^{*}$ are the optimal control and the
optimal state to $(OP)^{\tau,M(r,\tau)}=(OP)^{\tau,M^{*}}$, respectively. On
the other hand, it follows from Lemma 2.4 that
$M^{*}\chi_{(\tau,T)}\displaystyle\frac{\chi_{\omega}\psi^{\tau,M^{*}}}{\|\chi_{\omega}\psi^{\tau,M^{*}}\|}$
and $y^{\tau,M^{*}}$ are also the optimal control and the optimal state to
$(OP)^{\tau,M^{*}}$. Then, by the uniqueness of the optimal control to this
problem, (3.2) and (3.3) follow at once. Besides, by the optimality of $y^{*}$
to $(OP)^{\tau,M(r,\tau)}$, we see that
$\|y^{*}(T)-z_{d}\|=r(\tau,M(r,\tau))$. This, together with (2.11), gives
(3.1).

Conversely, suppose that a triplet $(M^{*},u^{*},y^{*})$, with $M^{*}>0$,
enjoys (3.1), (3.2) and (3.3). According to Lemma 2.4, it follows from (3.2)
and (3.3) that $u^{*}$ and $y^{*}$ are the optimal control and the optimal
state to $(OP)^{\tau,M^{*}}$ and that $\|y^{*}(T)-z_{d}\|=r(M^{*},\tau)$,
which, together with (3.1), shows that $r=r(M^{*},\tau)$. Then, by Theorem
2.1, $u^{*}$ and $y^{*}$ are the optimal control and the optimal state to
$(NP)^{r(M^{*},\tau),\tau}=(NP)^{r,\tau}$. Hence,
$\|u^{*}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=M(r,\tau)$, which, along with
(3.2), indicates that $M^{*}=M(r,\tau)$, i.e., $M^{*}$ is the optimal norm to
$(NP)^{r,\tau}$. This completes the proof. ∎

By Theorem 2.1, Lemma 2.4 and Lemma 2.11, following a very similar argument
used to prove Proposition 3.1, we can verify the following property for
$(TP)^{M,r}$.

######  Proposition 3.2.

Let $r\in(0,r_{T})$ and $M\geq M(r,0)$. Then $\tau^{*}$, $u^{*}$ and $y^{*}$
are the optimal time, the optimal control and the optimal state to
$(TP)^{M,r}$ if and only if $\tau^{*}$, $u^{*}$ and $y^{*}$ satisfy that
$\tau^{*}\in[0,T)$,

$\displaystyle\|y^{*}(T)-z_{d}\|=r,$ $\displaystyle
u^{*}(t)=M\chi_{(\tau^{*},T)}(t)\displaystyle\frac{\chi_{\omega}\psi^{\tau^{*},M}(t)}{\|\chi_{\omega}\psi^{\tau^{*},M}(t)\|},\;\;t\in(\tau^{*},T)$

and

$\displaystyle y^{*}(t)=\varphi^{\tau^{*},M}(t),\;\;t\in[0,T].$

The above two propositions not only are independently interesting, but also
hint us to find two algorithms for the optimal norm, together with the optimal
control, to $(NP)^{r,\tau}$ and the optimal time, along with the optimal
control, to $(TP)^{M,r}$, respectively. First of all, we build up,
corresponding to each $r\in(0,r_{T})$ and each $\tau\in(0,T)$, a sequence of
numbers as follows:

  * •

Structure of $\\{M_{n}\\}_{n=0}^{\infty}$: Let $M_{0}>0$ be arbitrarily taken.
Let $K\in\mathbb{N}$ be such that

$K=\min\\{k:r(\tau,kM_{0})<r,k=1,2,\cdots\\}.$

( The existence of such a $K$ is guaranteed by Lemma 2.6.) Set $a_{0}=0$ and
$b_{0}=KM_{0}$. Write $M_{1}=\displaystyle\frac{a_{0}+b_{0}}{2}$. In general,
when $M_{n}=\displaystyle\frac{a_{n-1}+b_{n-1}}{2}$ with $a_{n-1}$ and
$b_{n-1}$ being given, it is defined that

$\displaystyle\\{a_{n},b_{n}\\}=\left\\{\begin{array}[]{ll}\\{M_{n},b_{n-1}\\}&\;\mbox{if}\;\;r(\tau,M_{n})>r,\\\
\\{a_{n-1},M_{n}\\}&\;\mbox{if}\;\;r(\tau,M_{n})\leq r\end{array}\right.$

and $M_{n+1}=\displaystyle\frac{a_{n}+b_{n}}{2}$.

######  Remark 3.3.

Let $\tau\in[0,T)$ and $r\in(0,r_{T})$ be given. For each $M\geq 0$, we can
determine the value $r(\tau,M)$ by solving the two-point boundary value
problem (2.8) corresponding to $(\tau,M)$, since
$r(\tau,M)=\|\varphi^{\tau,M}(T)-z_{d}\|$ (see Lemma 2.4). Clearly, $M_{1}$ is
determined by $K$. Since the map $M\rightarrow r(\tau,M)$ is strictly
monotonically decreasing and $r(\tau,M)$ tends to $0$ as $M$ goes to $\infty$
(see Lemma 2.6), $K$ can be determined by solving limited number of two-point
boundary value problems (2.8) corresponding to $(\tau,M)$ with $M=kM_{0}$,
$k=1,2,\cdots,K$. On the other hand, when $n\geq 1$ $M_{n+1}$ is determined by
$\varphi^{M_{n},\tau}$, which can be solved from (2.8) corresponding to
$(\tau,M_{n})$. In summary, we conclude that the sequence
$\\{M_{n}\\}_{n=0}^{\infty}$ can be solved from a series of two-point boundary
value problems (2.8) corresponding to $(\tau,M)$, with $M=kM_{0}$,
$k=1,2,\cdots,K$ and with $M=M_{n}$, $n=1,2,\cdots$.

######  Theorem 3.4.

Suppose that $r\in(0,r_{T})$ and $\tau\in[0,T)$. Let
$\\{M_{n}\\}_{n=0}^{\infty}$ be the sequence built up above. Let
$u_{n}=M_{n}\chi_{(\tau,T)}\displaystyle\frac{\chi_{\omega}\psi^{\tau,M_{n}}}{\|\chi_{\omega}\psi^{\tau,M_{n}}\|}$
and $u^{*}$ be the optimal control to $(NP)^{r,\tau}$. Then it holds that

$\displaystyle M_{n}\rightarrow M(r,\tau)$ (3.5)

and

$\displaystyle u_{n}\rightarrow
u^{*}\;\;\mbox{in}\;\;L^{2}(\tau,T;L^{2}(\Omega))\;\;\mbox{and
in}\;\;C([\tau,T-\delta];L^{2}(\Omega))\;\;\mbox{for
each}\;\;\delta\in(0,T-\tau).$ (3.6)

###### Proof.

For simplicity, we write $(\varphi_{n},\psi_{n})$ for the solution
$(\varphi^{\tau,M_{n}},\psi^{\tau,M_{n}})$ with $n=1,2,\cdots$. We start with
proving (3.5). From the structure of $\\{M_{n}\\}$, it follows that
$M_{n}\in[a_{n},b_{n}]\subset[a_{n-1},b_{n-1}]$ and
$b_{n}-a_{n}=\displaystyle\frac{b_{n-1}-a_{n-1}}{2}$. Thus, it holds that
$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}b_{n}=\lim_{n\rightarrow\infty}M_{n}$.
Since the map $M\rightarrow r(\tau,M)$ is continuous (see Lemma 2.6) and
$r(\tau,a_{n})>r\geq r(\tau,b_{n})$ (which follows also from the structure of
$\\{M_{n}\\}$), we find that $r(\tau,\lim_{n\rightarrow\infty}M_{n})=r$. This,
along with (2.11), indicates that

$\displaystyle r(\tau,\lim_{n\rightarrow\infty}M_{n})=r(\tau,M(r,\tau)).$
(3.7)

Then, (3.5) follows from (3.7) and the strict monotonicity of the map
$M\rightarrow r(\tau,M)$ (see Lemma 2.6).

Next, write $y^{*}(\cdot)$ and $y_{n}(\cdot)$ for the solutions
$y(\cdot;\chi_{(\tau,T)}u^{*})$ and $y(\cdot;\chi_{(\tau,T)}u_{n})$,
respectively. We claim that

$\displaystyle u_{n}\rightarrow u^{*}\;\;\mbox{weakly star
in}\;\;L^{\infty}(\tau,T;L^{2}(\Omega))\;\;\mbox{and}\;\;y_{n}\rightarrow
y^{*}\;\;\mbox{in}\;\;C([0,T];L^{2}(\Omega)).$ (3.8)

In fact, by the definitions of $u_{n}$ and $y_{n}$, it follows from Lemma 2.4
that they are the optimal control and the optimal state to
$(OP)^{\tau,M_{n}}$, respectively. We arbitrarily take subsequences of
$\\{u_{n}\\}$ and $\\{y_{n}\\}$, denoted by $\\{u_{n_{k}}^{\prime}\\}$ and
$\\{y_{n_{k}}^{\prime}\\}$, respectively. Clearly, there are subsequences
$\\{u_{n_{k}}\\}$ of $\\{u_{n_{k}}^{\prime}\\}$ and $\\{y_{n_{k}}\\}$ of
$\\{y_{n_{k}}^{\prime}\\}$ such that

$\displaystyle u_{n_{k}}\rightarrow\widetilde{u}\;\;\mbox{weakly star in
}\;L^{\infty}(\tau,T;L^{2}(\Omega))\;\;\mbox{and}\;\;y_{n_{k}}\rightarrow\widetilde{y}\;\;\mbox{in}\;\;C([0,T];L^{2}(\Omega)),$
(3.9)

where $\widetilde{y}(\cdot)=y(\cdot;\chi_{(\tau,T)}\widetilde{u})$. These,
along with (3.5) and (3.7), indicate that

$\|\widetilde{u}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}\leq\mathop{\underline{\rm
lim}}_{k\rightarrow\infty}\|u_{n_{k}}\|_{L^{\infty}(\tau,T;L^{2}(\Omega))}=\mathop{\underline{\rm
lim}}_{k\rightarrow\infty}M_{n_{k}}=M(r,\tau)$

and

$\|\widetilde{y}(T)-z_{d}\|=\lim_{k\rightarrow\infty}\|y_{n_{k}}(T)-z_{d}\|=\lim_{n\rightarrow\infty}r(\tau,M_{n_{k}})=r(\tau,\lim_{k\rightarrow\infty}M_{n_{k}})=r(\tau,M(r,\tau)).$

From these, we see that $\widetilde{u}$ and $\widetilde{y}$ are the optimal
control and the optimal state to $(OP)^{\tau,M(r,\tau)}$. Then, according to
Theorem 2.1, they are the optimal control and the optimal state to
$(NP)^{r,\tau}$. Since the optimal control to $(NP)^{r,\tau}$ is unique, (3.8)
follows from (3.9).

Now we verify the first convergence in (3.6). By the first convergence in
(3.8), we see that

$\displaystyle u_{n}\rightarrow u^{*}\;\;\mbox{weakly
in}\;\;L^{2}(\tau,T;L^{2}(\Omega)).$ (3.10)

On the other hand, according to Proposition 3.1, it stands that

$\displaystyle
u^{*}(t)=M(r,\tau)\chi_{(\tau,T)}(t)\frac{\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)}{\|\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)\|},\;\;t\in[0,T)$
(3.11) $\displaystyle
y^{*}=\varphi^{\tau,M(r,\tau)}\;\;\mbox{and}\;\;\|y^{*}(T)-z_{d}\|=r.$ (3.12)

By the definition of $u_{n}$, (3.11) and (3.5), we see that

$\|u_{n}\|_{L^{2}(\tau,T;L^{2}(\Omega))}\rightarrow\|u^{*}\|_{L^{2}(\tau,T;L^{2}(\Omega))}.$

This, along with (3.10), yields the first convergence in (3.6).

Finally, we show the second convergence in (3.6). By the first equality of
(3.12) and the second convergence in (3.9), we see that
$y_{n}(T)\rightarrow\varphi^{\tau,M(r,\tau)}(T)$ strongly in $L^{2}(\Omega)$.
This, together with the equations satisfied by $\psi_{n}$ and
$\psi^{\tau,M(r,\tau)}$, respectively, indicates that

$\displaystyle\psi_{n}\rightarrow\psi^{\tau,M(r,\tau)}\;\;\mbox{in}\;\;C([0,T];L^{2}(\Omega)).$
(3.13)

Then we arbitrarily fix a $\delta\in(0,T-\tau)$. By (3.11) and by the
definition of $u_{n}$, after some simple computation, we deduce that for each
$t\in[0,T-\delta]$,

$\displaystyle\begin{array}[]{ll}\|u_{n}(t)-u^{*}(t)\|&\leq|M_{n}-M(r,\tau)|+\displaystyle\frac{2M_{n}}{\|\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)\|}\|\chi_{\omega}\psi_{n}(t)-\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)\|.\end{array}$
(3.15)

On the other hand, by the second equality of (3.12) and the unique
continuation property (see [8]), it follows that
$\|\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)\|\neq 0$ for all $t\in[0,T)$. This,
together with the continuity of $\psi^{\tau,M(r,\tau)}(\cdot)$ over
$[0,T-\delta]$, yields that

$\displaystyle\max_{t\in[0,T-\delta]}\displaystyle\frac{1}{\|\chi_{\omega}\psi^{\tau,M(r,\tau)}(t)\|}\leq
C_{\delta}\;\;\mbox{for some positive}\;\;C_{\delta}.$ (3.16)

Now, the second convergence in (3.6) follows immediately from (3.15), (3.5),
(3.13), and (3.16). This completes the proof.

∎

We end this section by introducing an algorithm for the optimal time and the
optimal control to $(TP)^{M,r}$. For each pair $(M,r)$ with $r\in(0,r_{T})$
and $M\geq M(r,0)$, we construct a sequence
$\\{\tau_{n}\\}_{n=0}^{\infty}\subset[0,T)$ as follows.

  * •

Structure of $\\{\tau_{n}\\}_{n=1}^{\infty}$: Let $a_{0}=0$ and $b_{0}=T$. Set
$\tau_{1}=\displaystyle\frac{a_{0}+b_{0}}{2}$. In general, when
$\tau_{n}=\displaystyle\frac{a_{n-1}+b_{n-1}}{2}$ with $a_{n-1}$ and $b_{n-1}$
being given, it is defined that

$\displaystyle\\{a_{n},b_{n}\\}=\left\\{\begin{array}[]{ll}\\{a_{n-1},\tau_{n}\\}&\;\mbox{if}\;\;r(\tau_{n},M)>r,\\\
\\{\tau_{n},b_{n-1}\\}&\;\mbox{if}\;\;r(\tau_{n},M)\leq r\end{array}\right.$

and $\tau_{n+1}=\displaystyle\frac{a_{n}+b_{n}}{2}$.

######  Remark 3.5.

Since $r(\tau_{n},M)=\|\varphi^{\tau_{n},M}(T)-z_{d}\|$, $\tau_{n+1}$ is
determined by $\varphi^{\tau_{n},M}$, which can be solved from (2.8)
corresponding to $\tau=\tau_{n}$.

By Theorem 2.1, Lemma 2.4, Lemma 2.11 and Proposition 3.2, following a very
similar argument to prove Theorem 3.4, we can verify the next approximation
result.

######  Theorem 3.6.

Suppose that $r\in(0,r_{T})$ and $M\geq M(r,0)$. Let
$\\{\tau_{n}\\}_{n=1}^{\infty}$ be the sequence built up above. Let
$u_{n}=M\chi_{(\tau_{n},T)}\displaystyle\frac{\chi_{\omega}\psi^{\tau_{n},M}}{\|\chi_{\omega}\psi^{\tau_{n},M}\|}$
and $u^{*}$ be the optimal control to $(TP)^{M,r}$. Then it holds that

$\displaystyle\tau_{n}\rightarrow\tau(M,r)\;\;\mbox{as}\;\;n\rightarrow\infty$

and

$\displaystyle u_{n}\rightarrow
u^{*}\;\;\mbox{in}\;\;L^{2}(\tau(M,r),T;L^{2}(\Omega))\;\;\mbox{and
in}\;\;C([\tau(M,r),T-\delta];L^{2}(\Omega))$

for each $\delta\in(0,T-\tau(M,r))$.

######  Remark 3.7.

$(i)$ From the above-mentioned two algorithms, we observe that the optimal
norm and the optimal control to $(NP)^{r,\tau}$ and the optimal time and the
optimal control to $(TP)^{M,r}$ can be numerically solved, through numerically
solving the two-point boundary value problems (2.8) with parameters $M$ and
$\tau$ suitably chosen.
$(ii)$ All results obtained in this section hold for the case where the
controlled system is Equation (1.6) (see Remark 2.13).

##  4  Application II: Optimal Normal Feedback Law

Throughout this section, we arbitrarily fix a $r>0$. We aim to build up a
feedback law for norm optimal control problems.

###  4.1 Main results

We first introduce the following controlled equation:

$\left\\{\begin{array}[]{cl}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\partial_{t}y-\triangle y=\chi_{\omega}u{}{}&{\rm
in}~{}\Omega\times(t_{0},T),\\\ \vskip 3.0pt plus 1.0pt minus 1.0pt\cr
y=0{}&{\rm on}~{}\partial\Omega\times(t_{0},T),\\\ \vskip 3.0pt plus 1.0pt
minus 1.0pt\cr y(t_{0})=y_{0},{}&{\rm
in}~{}\Omega\times(t_{0},T).\end{array}\right.$ (4.1)

where $(t_{0},y_{0})\in[0,T)\times L^{2}(\Omega)$. Denote by
$y(\cdot;u,t_{0},y_{0})$ the solution to Equation (4.1) corresponding to the
control $u$ and the initial data $(t_{0},y_{0})$. Then, we define the
following optimal target control and optimal norm control problems.

  * •

$(OP)^{M}_{t_{0},y_{0}}$: $\inf\\{\|y(T;u,t_{0},y_{0})-z_{d}\|^{2}:u\in
L^{\infty}(t_{0},T;B(0,M))\\}$;

  * •

$(NP)_{t_{0},y_{0}}$: $\inf\\{\|u\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}:u\in
L^{\infty}(t_{0},T;L^{2}(\Omega)),y(T;u,t_{0},y_{0})\in B(z_{d},r)\\}$.

Throughout this section,

  * •

$\bar{u}^{M}_{t_{0},y_{0}}$ stands for the optimal control to
$(OP)^{M}_{t_{0},y_{0}}$;

  * •

$N(t_{0},y_{0})$ denotes the optimal norm to $(NP)_{t_{0},y_{0}}$.

Thus $N(\cdot,\cdot)$ defines an optimal norm functional over $[0,T)\times
L^{2}(\Omega)$.

The only difference between optimal target control problems $(OP)^{0,M}$
(which was introduced in Section 1) and $(OP)^{M}_{t_{0},y_{0}}$ is that the
initial data for the first one is $(0,y_{0})$ while the initial data for the
second one is $(t_{0},y_{0})$. The same can be said about the norm optimal
control problems. Therefore, corresponding to each result about $(OP)^{0,M}$
or $(NP)^{r,0}$, obtained in Section 2 or Section 3, there is an analogous
version for $(OP)^{M}_{t_{0},y_{0}}$ or $(NP)_{t_{0},y_{0}}$.

A feedback law for the norm optimal control problems will be established, with
the aid of the equivalence between norm and target optimal controls and some
properties of $(OP)^{M}_{t_{0},y_{0}}$. Those properties are related to the
following two-point boundary value problem associated with $M\geq 0$,
$t_{0}\in[0,T)$ and $y_{0}\in L^{2}(\Omega)$:

$\left\\{\begin{array}[]{ccll}\partial_{t}y-\Delta
y=M\displaystyle\frac{\chi_{\omega}\psi}{\|\chi_{\omega}\psi\|},&\partial_{t}\psi+\triangle\psi=0&\mbox{in}&\Omega\times(t_{0},T),\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr
y=0,&\psi=0&\mbox{on}&\partial\Omega\times(t_{0},T),\\\ \vskip 3.0pt plus
1.0pt minus 1.0pt\cr
y(t_{0})=y_{0},&\psi(T)=-(y(T)-z_{d})&\mbox{in}&\Omega.\end{array}\right.$
(4.2)

Similar to Lemma 2.5, for each triplet
$(M,t_{0},y_{0})\in[0,\infty)\times[0,T)\times L^{2}(\Omega)$, Equation (4.2)
has a unique solution in $C([0,T];L^{2}(\Omega))$. Throughout this section,

  * •

$(\bar{y}^{M}_{t_{0},y_{0}},\bar{\psi}^{M}_{t_{0},y_{0}})$ denotes the
solution of (4.2) corresponding to $M$, $t_{0}$ and $y_{0}$;

  * •

$\bar{y}^{M}_{t_{0},y_{0}}$ and $\bar{\psi}^{M}_{t_{0},y_{0}}$ denote the
first and the second component of the above solution, respectively, when one
of them appears alone.

Because of the assumption (1.2),
$\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(T)=-(\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(T)-z_{d})\neq
0$ (see the proof of Lemma 2.2). Thus, it follows from the unique continuation
property of the heat equation (see [8]) that

$\displaystyle{\chi}_{\omega}\bar{\psi}^{M(t_{0},y_{0})}_{t_{0},y_{0}}(t_{0})\neq
0\;\;\mbox{for all}\;\;(t_{0},y_{0})\in[0,T)\times L^{2}(\Omega).$ (4.3)

Now we define a feedback law $F:[0,T)\times L^{2}({\Omega})\mapsto
L^{2}({\Omega})$ by setting

$\displaystyle
F(t_{0},y_{0})=N(t_{0},y_{0})\frac{{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t_{0})}{\|{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t_{0})\|},\;(t_{0},y_{0})\in[0,T)\times
L^{2}({\Omega}).$ (4.4)

Because of the existence and uniqueness of the solution to (4.2), as well as
(4.3), the map $F$ is well defined. For each $(t_{0},y_{0})\in[0,T)\times
L^{2}(\Omega)$, consider the evolution equation:

$\left\\{\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\dot{y}(t)-Ay(t)=\chi_{\omega}F(t,y(t)),&t\in(t_{0},T),\\\ \vskip
6.0pt plus 2.0pt minus 2.0pt\cr y(t_{0})=y_{0},\end{array}\right.$ (4.5)

where the operator $A$ was defined in Section 1. Two main results in this
section are as follows:

######  Theorem 4.1.

For each pair $(t_{0},y_{0})\in[0,T)\times L^{2}({\Omega})$, Equation (4.5)
has a unique (mild) solution. Furthermore, this solution is exactly
$\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$.

######  Theorem 4.2.

For each pair $(t_{0},y_{0})\in[0,T)\times L^{2}(\Omega)$,
$F(\cdot,y_{F}(\cdot;t_{0},y_{0}))$ is the optimal control to
$(NP)_{t_{0},y_{0}}$, where $y_{F}(\cdot;t_{0},y_{0}))$ is the unique solution
to Equation (4.5) corresponding to the initial data $(t_{0},y_{0})$.

It follows directly from Theorem 4.1 that for each $y_{0}\in L^{2}(\Omega)$
and each $\tau\in[0,T)$, the evolution equation

$\left\\{\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\dot{y}(t)-Ay(t)=\chi_{\omega}\chi_{(\tau,T)}F(t,y(t)),&t\in(0,T),\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt\cr y(0)=y_{0}\end{array}\right.$ (4.6)

admits a unique (mild) solution, denoted by $y_{F,\tau,y_{0}}(\cdot)$. Thus,
the following result is a direct consequence of Theorem 4.2:

######  Corollary 4.3.

For each $y_{0}\in L^{2}(\Omega)$ and each $\tau\in[0,T)$,
$\chi_{(\tau,T)}(\cdot)F(\cdot,y_{F,\tau,y_{0}}(\cdot))$ is the optimal
control to Problem $(NP)^{r,\tau}$ with the initial state $y_{0}$.

###  4.2  Proof of Theorem 4.1 (Part 1): The existence of solutions

By a very similar argument to prove Lemma 2.4, we can obtain that

$\displaystyle\bar{u}^{M}_{t_{0},y_{0}}(t)=M\frac{\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t)}{||\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t)||},\;t\in[t_{0},T).$
(4.7)

By the uniqueness and existence of the solution to (4.2), we can easily derive
the following consequence, which, in some sense, is a dynamic programming
principle.

######  Lemma 4.4.

Let $(t_{0},y_{0})\in[0,T)\times L^{2}({\Omega})$ and $M\geq 0$. Then, for
each $s\in(t_{0},T)$,

$\displaystyle(\bar{y}^{M}_{t_{0},y_{0}},\bar{\psi}^{M}_{t_{0},y_{0}}){\bigg{|}}_{[s,T]}=\bigr{(}\bar{y}^{M}_{s,\bar{y}^{M}_{t_{0},y_{0}}(s)},\bar{\psi}^{M}_{s,\bar{y}^{M}_{t_{0},y_{0}}(s)}\bigl{)}.$
(4.8)

######  Lemma 4.5.

Let $(t_{0},y_{0})\in[0,T)\times L^{2}({\Omega})$. Then

$M=N(t_{0},y_{0})\;\;\mbox{if and only
if}\;\;\displaystyle\left\|\bar{y}^{M}_{t_{0},y_{0}}(T)-z_{d}\right\|=r\wedge\left\|e^{(T-t_{0})\triangle}y_{0}-z_{d}\right\|,$
(4.9)

where ”$\wedge$” is the symbol taking the smaller. Moreover, the control,
defined by

$\displaystyle\bar{u}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)=N(t_{0},y_{0})\frac{\chi_{\omega}\psi^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)}{||\chi_{\omega}\psi^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)||},\;t\in(t_{0},T),$
(4.10)

is the unique optimal control of Problem $(NP)_{t_{0},y_{0}}$.

###### Proof.

First, we show (4.9) for the case where
$\|e^{(T-t_{0})\Delta}y_{0}-z_{d}\|>r$. In this case, we can apply the
analogous version of Proposition 3.1 for Problem $(NP)_{t_{0},y_{0}}$ to get
that $M=N(t_{0},y_{0})$ if and only if
$\|\bar{y}^{M}_{t_{0},y_{0}}(T)-z_{d}\|=r$. This leads to (4.9) for this case.

Next, we prove (4.9) for the case where
$\|e^{(T-t_{0})\Delta}y_{0}-z_{d}\|\leq r$. In this case, one can easily check
that $N(t_{0},y_{0})=0$, the null control is the optimal control to
$(OP)^{0}_{t_{0},y_{0}}$, and
$\bar{y}^{0}_{t_{0},y_{0}}(\cdot)=y(\cdot;0,t_{0},y_{0})=e^{(\cdot-
t_{0})\Delta}y_{0}$ over $[t_{0},T]$. Suppose that $M=N(t_{0},y_{0})$. Then it
holds that $M=0$ and
$\|\bar{y}^{0}_{t_{0},y_{0}}(T)-z_{d}\|=\|e^{(T-t_{0})\Delta}y_{0}-z_{d}\|$.
These lead to the statement on the right hand side of (4.9). Conversely,
suppose that there is an $M_{0}\geq 0$ such that

$\displaystyle\|\bar{y}^{M_{0}}_{t_{0},y_{0}}(T)-z_{d}\|=\|e^{(T-t_{0})\Delta}y_{0}-z_{d}\|.$
(4.11)

To show the statement on the left side of (4.9), it suffices to prove that
$M_{0}=0$. By the analogous version of Lemma 2.4 for
$(OP)^{M_{0}}_{t_{0},y_{0}}$ (see (2.10)), it holds that

$\displaystyle\|\bar{y}^{M_{0}}_{t_{0},y_{0}}(T)-z_{d}\|=r_{t_{0},y_{0}}(M_{0}),$
(4.12)

where $r_{t_{0},y_{0}}(\cdot)$ corresponds to the map $M\rightarrow r(0,M)$
given in Section 1, namely,

$r_{t_{0},y_{0}}(M)=\inf\\{\|y(T;u,t_{0},y_{0})-z_{d}\|:u\in
B(0,M))\\},\;M\geq 0.$

Since the null control is the optimal control to $(OP)^{0}_{t_{0},y_{0}}$, we
find that

$r_{t_{0},y_{0}}(0)=\|y(T;0,t_{0},y_{0})-z_{d}\|=\|e^{(T-t_{0})\Delta}y_{0}-z_{d}\|.$

Along with (4.11) and (4.12), this indicates that

$\displaystyle r_{t_{0},y_{0}}(0)=r_{t_{0},y_{0}}(M_{0}).$ (4.13)

By the analogous version of Lemma 2.6 for $(OP)^{M}_{t_{0},y_{0}}$, the map
$M\rightarrow r_{t_{0},y_{0}}(M)$ is strictly monotonically decreasing. This,
together with (4.13), yields that $M_{0}=0$.

In summary, we conclude that (4.9) stands.

Finally, we prove (4.10). In the case that $\|e^{(T-t_{0})}\Delta
y_{0}-z_{d}\|>r$, according to the analogous version of Proposition 3.1 for
Problem $(NP)_{t_{0},y_{0}}$, the control defined by (4.10) is the unique
optimal control of Problem $(NP)_{t_{0},y_{0}}$. In the case where
$\|e^{(T-t_{0})}\Delta y_{0}-z_{d}\|\leq r$, it is clear that
$N(t_{0},y_{0})=0$ and the null control is the optimal control to
$(NP)_{t_{0},y_{0}}$. Hence, (4.10) holds for this case. This completes the
proof.

∎

The following result shows that the functional $N(\cdot,\cdot)$ holds the
dynamic programming principle.

######  Lemma 4.6.

Let $(t_{0},y_{0})\in[0,T)\times L^{2}({\Omega})$. Then it stands that

$\displaystyle
N(t_{0},y_{0})=N\biggr{(}s,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)\biggl{)}\;\mbox{for
each}\;s\in(t_{0},T).$ (4.14)

Proof. In the case where $e^{(T-t_{0})\triangle}y_{0}\in B(z_{d},r)$, it is
clear that

$N(t_{0},y_{0})=0\;\;\mbox{and}\;\;\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)=e^{(\cdot-
t_{0})\triangle}y_{0}.$

Because

$e^{(T-s)\triangle}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)=e^{(T-s)\triangle}e^{(s-t_{0})\triangle}y_{0}=e^{(T-t_{0})\triangle}y_{0}\in
B(z_{d},r)\;\;\mbox{for each}\;\;s\in(t_{0},T),$

we see that
$N\bigr{(}s,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)\bigl{)}=0$. Therefore
the equality (4.14) holds for this case.

In the case where $e^{(T-t_{0})\triangle}y_{0}\notin B(z_{d},r)$, it is clear
that $r\wedge\left\|e^{(T-t_{0})\triangle}y_{0}-z_{d}\right\|=r$. By the
analogous version of Proposition 3.1 for Problem $(NP)_{t_{0},y_{0}}$, it
holds that $\|\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(T)-z_{d}\|=r$. Thus, it
follows from (4.9) that

$\displaystyle\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(T)\in\partial
B(z_{d},r).$ (4.15)

We claim that

$e^{(T-s)\triangle}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)\notin
B(z_{d},r)\;\;\mbox{for all}\;\;s\in(t_{0},T).$ (4.16)

If (4.16) did not hold, then there would exist a $\hat{s}\in(t_{0},T)$ such
that

$\displaystyle
e^{(T-\hat{s})\triangle}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\hat{s})\notin
B(z_{d},r).$ (4.17)

We construct a control $\hat{u}$ by setting

$\displaystyle\hat{u}(s)=\displaystyle\chi_{(t_{0},\hat{s})}(s)N(t_{0},y_{0})\frac{\chi_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)}{\|\chi_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)\|},\;\;s\in[t_{0},T).$
(4.18)

Clearly, the solution $y(\cdot;\hat{u},t_{0},y_{0})$ to (4.1), where
$u=\hat{u}$, coincides with $\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$
over $[t_{0},\hat{s}]$. This, along with (4.18) and (4.17), indicates that

$\displaystyle
y(T;\hat{u},t_{0},y_{0})=e^{(T-\hat{s})\triangle}y(\hat{s};\hat{u},t_{0},y_{0})\in
B(z_{d},r).$ (4.19)

On the other hand, it follows from (4.18) that
$\displaystyle\|\hat{u}||_{L^{\infty}(t_{0},T;L^{2}(\Omega))}=N(t_{0},y_{0}).$
This, together with (4.19), yields that $\hat{u}$ is the optimal control to
$(NP)_{t_{0},y_{0}}$. However, the problem $(NP)_{t_{0},y_{0}}$ holds the
bang-bang property (it follows from the analogous version of Proposition 2.7
for $(NP)_{t_{0},y_{0}}$). This implies that $\|\hat{u}(s)\|=N(t_{0},y_{0})$
for a.e. $s\in(t_{0},T)$, which contradicts to the structure of $\hat{u}$.
Hence, (4.16) stands.

Next, by (4.8) and (4.15), we see that

$\displaystyle\bar{y}^{N(t_{0},y_{0})}_{s,\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)}(T)=\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(T)\in\partial
B(z_{d},r)\;\mbox{for each}\;s\in(t_{0},T).$

This, together with (4.16), implies that

$\displaystyle\|\displaystyle\bar{y}^{N(t_{0},y_{0})}_{s,\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)}(T)-z_{d}\|=r\wedge\|e^{(T-s)\triangle}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)-z_{d}\|\;\mbox{for
each}\;s\in(t_{0},T).$ (4.20)

Now, we arbitrarily fix a $s\in(t_{0},T)$. By (4.20), we can apply (4.9), with
$t_{0}=s$ and $y_{0}=\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)$, to get that

$N(t_{0},y_{0})=N(s,\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(s)),$

which gives the equality (4.14) for the second case. In summary, we finish the
proof.

Proof of Theorem 4.1 (Part 1): The existence. It follows from (4.4) (the
definition of $F$) that

$\displaystyle\displaystyle
F(t,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t))=N(t,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t))\frac{\phi(t)}{\|\phi(t\|},\;t\in(t_{0},T),$

where
$\phi(t)={\chi}_{\omega}\bar{\psi}^{N(t,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t))}_{t,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)}(t)$.
This, together with (4.14) and (4.8), yields that

$\displaystyle
F(t,~{}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t))=N(t_{0},y_{0})\frac{{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)}{\|{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)\|}=\frac{d}{dt}\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)+A\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t),\;t\in(t_{0},T).$

Here, we used that $\chi_{\omega}\circ\chi_{\omega}=\chi_{\omega}$. From the
above equality and the fact that
$\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t_{0})=y_{0}$ , it follows that
$\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$ is a solution to (4.5). This
completes the proof.

###  4.3  Proof of Theorem 4.1 (Part 2): The uniqueness

The key to prove the uniqueness is showing the following properties of the
feedback law $F(\cdot,\cdot)$.

######  Proposition 4.7.

$(i)$ For each pair $(\bar{t}_{0},\bar{y}_{0})\in[0,T)\times L^{2}(\Omega)$,
there is a $\bar{\rho}>0$ such that $F(t_{0},\cdot)$ is Lipschitz continuous
in $B(\bar{y}_{0},\bar{\rho})$ uniformly with respect to
$t_{0}\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]$. $(ii)$ For
each $\bar{y}_{0}\in L^{2}(\Omega)$, $F(\cdot\,,\bar{y}_{0})$ is continuous
over $[0,T)$.

When it is proved, the uniqueness of the solution to Equation (4.5) follows
immediately from the generalized Picard-Lindelof Theorem (see [15]) and
Proposition 4.7. Consequently, the proof of Theorem 4.1 is completed.

The remainder is showing Proposition 4.7. To serve such purpose, we first
study some continuity properties of $N(\cdot,\cdot)$. These properties will be
concluded in Lemma 4.10. Two lemmas before it will play important roles in its
proof.

######  Lemma 4.8.

For each $t_{0}\in[0,T)$, the functional $N(t_{0},\cdot)$ is convex over
$L^{2}(\Omega)$.

Proof. Let $y_{0}^{1}$ and $y_{0}^{2}$ belong to $L^{2}(\Omega)$. The optimal
controls $\bar{u}^{i}$ to $(NP)_{t_{0},y_{0}^{i}}$, $i=1,2$, satisfy that
$N(t_{0},y_{0}^{i})=\|\bar{u}^{i}\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}$ and
$y(T;\bar{u}^{i},t_{0},y_{0}^{i})\in B(z_{d},r)$, $i=1,2.$ Since for each
$\lambda\in(0,1)$,

$y(T;\lambda\bar{u}^{1}+(1-\lambda)\bar{u}^{2},t_{0},\lambda
y_{0}^{1}+(1-\lambda)y_{0}^{2})=\lambda
y(T;\bar{u}^{1},t_{0},y_{0}^{1})+(1-\lambda)y(T;\bar{u}^{2},t_{0},y_{0}^{2})\in
B(z_{d},r),$

we obtain that

$\begin{array}[]{l }\vskip 3.0pt plus 1.0pt minus 1.0pt\cr N(t_{0},\lambda
y_{0}^{1}+(1-\lambda)y_{0}^{2})\leq\|\lambda\bar{u}^{1}+(1-\lambda)\bar{u}^{2}\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq\lambda\|\bar{u}^{1}\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}+(1-\lambda)\|\bar{u}^{2}\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr=\lambda
N(t_{0},y_{0}^{1})+(1-\lambda)N(t_{0},y_{0}^{2}).\end{array}.$

This completes the proof.

######  Lemma 4.9.

For each $\bar{t}_{0}\in[0,T)$ and each bounded subset $E$ of $L^{2}(\Omega)$,
the functional $N(\cdot,\cdot)$ is bounded on
$\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}\times
E$, where $\bar{\delta}=(T-\bar{t}_{0})/2$.

Proof. Write $C_{E}=\sup\\{\|y_{0}\|:y_{0}\in E\\}$. Let
$(t_{0},y_{0})\in\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}\times
E$. By the null controllability of the heat equation over
$(t_{0},\bar{t}_{0}+3\bar{\delta}/2)$ (see, for instance, [6]), there is a
control $u_{1}$ with

$\|u_{1}\|_{L^{\infty}(t_{0},\bar{t}_{0}+3\bar{\delta}/2;L^{2}(\Omega))}\leq
C_{1}\|y_{0}\|\leq C_{1}C_{E},$ (4.21)

where $C_{1}>0$ is independent of $t_{0}$ and $y_{0}$, such that

$\bar{y}\equiv y(\bar{t}_{0}+3\bar{\delta}/2;u_{1},t_{0},y_{0})=0.$

Here we used that $t_{0}\leq\bar{t}_{0}+\bar{\delta}$. Then, by the
approximate controllability of the heat equation over
$(\bar{t}_{0}+3\bar{\delta}/2,T)$ (see, for instance, [4]), there is another
control $u_{2}$ with

$\|u_{2}\|_{L^{\infty}(\bar{t}_{0}+3\bar{\delta}/2,T;L^{2}(\Omega))}\leq
C_{2},$

where $C_{2}>0$ is independent of $t_{0}$ and $y_{0}$, such that

$y(T;u_{2},\bar{t}_{0}+3\bar{\delta}/2,\bar{y})\in B(z_{d},r).$

Clearly, the control
$v\equiv\chi_{(t_{0},\bar{t}_{0}+3\bar{\delta}/2)}u_{1}+\chi_{(\bar{t}_{0}+3\bar{\delta}/2,T)}u_{2}$
satisfies that $y(T;v,t_{0},y_{0})\in B(z_{d},r)$. Therefore, it holds that

$N(t_{0},y_{0})\leq\|v\|_{L^{\infty}(t_{0},T;L^{2}(\Omega))}\leq\max\\{C_{1}C_{E},C_{2}\\}.$

This completes the proof.

######  Lemma 4.10.

$(i)$ For each $(\bar{t}_{0},\bar{y}_{0})\in[0,T)\times L^{2}(\Omega)$ and
each $\rho\in(0,1/2)$, $N(t_{0},\cdot)$ is Lipschitz continuous over
$B(\bar{y}_{0},\rho)$ uniformly w.r.t.
$t_{0}\in\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}$,
where $\bar{\delta}=(T-\bar{t}_{0})/2$; $(ii)$ For each $\bar{y}_{0}\in
L^{2}(\Omega)$, $N(\cdot,\bar{y}_{0})$ is continuous over $[0,T)$.

Proof. $(i)$ Let $(\bar{t}_{0},y_{0})\in[0,T)\times L^{2}(\Omega)$ and let
$\rho\in(0,1/2)$. We arbitrarily take two different points $y_{0}^{1}$ and
$y_{0}^{2}$ from $B(\bar{y}_{0},\rho)\subset B(\bar{y}_{0},1)$. Denote by
$\mathcal{L}$ the straight line passing through $y_{0}^{1}$ and $y_{0}^{2}$,
namely,
$\mathcal{L}\equiv\biggr{\\{}y_{0}^{\lambda}\mathop{\buildrel\Delta\over{=}}(1-\lambda)y_{0}^{1}+\lambda
y_{0}^{2}\bigm{|}\lambda\in(-\infty,+\infty)\biggl{\\}}$. Clearly,
$\mathcal{L}$ intersects with $B(\bar{y}_{0},1)$ at two different points,
denoted by $y_{0}^{\lambda_{1}}$ and $y_{0}^{\lambda_{2}}$, with
$\lambda_{1}<\lambda_{2}$. Since the segment
$\\{\,y_{0}^{\lambda}|\lambda\in[0,1]\,\\}\subseteq B(\bar{y}_{0},\rho)$ and
$B(\bar{y},\rho)\bigcap\partial B(\bar{y},1)=\emptyset$, it holds that
$\lambda_{1}<0<1<\lambda_{2}.$ Moreover, one can easily check that

$\displaystyle\frac{\|y_{0}^{1}-y_{0}^{\lambda_{1}}\|}{0-\lambda_{1}}=\frac{\|y_{0}^{2}-y_{0}^{1}\|}{1-0}=\frac{\|y_{0}^{\lambda_{2}}-y_{0}^{2}\|}{\lambda_{2}-1}.$
(4.22)

For each
$t_{0}\in\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}$,
we define a function $g_{t_{0}}(\cdot;y_{0}^{1},y_{0}^{2})$ by setting

$g_{t_{0}}(\lambda;y_{0}^{1},y_{0}^{2})=N(t_{0},(1-\lambda)y_{0}^{1}+\lambda
y_{0}^{2}),\qquad\lambda\in(-\infty,+\infty).$

Obviously, the convexity of $N(t_{0},\cdot)$ (see Lemma 4.8) implies the
convexity of $g_{t_{0}}(\cdot;y_{0}^{1},y_{0}^{2})$. By the property of convex
functions, one has that

$\displaystyle\frac{g_{t_{0}}(0;y_{0}^{1},y_{0}^{2})-g_{t_{0}}(\lambda_{1};y_{0}^{1},y_{0}^{2})}{0-\lambda_{1}}\leq\frac{g_{t_{0}}(1;y_{0}^{1},y_{0}^{2})-g_{t_{0}}(0;y_{0}^{1},y_{0}^{2})}{1-0}\leq\frac{g_{t_{0}}(\lambda_{2};y_{0}^{1},y_{0}^{2})-g_{t_{0}}(1;y_{0}^{1},y_{0}^{2})}{\lambda_{2}-1}.$

This, along with (4.22) and the nonnegativity of
$g_{t_{0}}(1;y_{0}^{1},y_{0}^{2})$, indicates that

$\displaystyle\begin{array}[]{l}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle~{}~{}~{}\frac{N(t_{0},y_{0}^{2})-N(t_{0},y_{0}^{1})}{\|y_{0}^{2}-y_{0}^{1}\|}=\frac{g_{t_{0}}(1;y_{0}^{1},y_{0}^{2})-g_{t_{0}}(0;y_{0}^{1},y_{0}^{2})}{\|y_{0}^{2}-y_{0}^{1}\|}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\displaystyle\leq\frac{g_{t_{0}}(\lambda_{2};y_{0}^{1},y_{0}^{2})-g_{t_{0}}(1;y_{0}^{1},y_{0}^{2})}{(\lambda_{2}-1)\|y_{0}^{2}-y_{0}^{1}\|}\leq\frac{g_{t_{0}}(\lambda_{2};y_{0}^{1},y_{0}^{2})}{(\lambda_{2}-1)\|y_{0}^{2}-y_{0}^{1}\|}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle=\frac{g_{t_{0}}(\lambda_{2};y_{0}^{1},y_{0}^{2})}{\|y_{0}^{\lambda_{2}}-y_{0}^{2}\|}.\end{array}$
(4.26)

Two observations are as follows. The triangle inequality implies that

$\|y_{0}^{\lambda_{2}}-y_{0}^{2}\|\geq\|y_{0}^{\lambda_{2}}-\bar{y}_{0}\|-\|y_{0}^{2}-\bar{y}_{0}\|\geq
1-\rho;$

The boundedness of $N(\cdot,\cdot)$ (see Lemma 4.9) gives that for each
$t_{0}\in\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}$,
$y_{0}^{1},y_{0}^{2}\in B(\bar{y}_{0},\rho)$,

$g_{t_{0}}(\lambda_{2};y_{0}^{1},y_{0}^{2})=N(t_{0},y_{0}^{\lambda_{2}})\leq\sup\biggr{\\{}N(s_{0},z_{0})\Bigm{|}(s_{0},z_{0})\in\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}\times
B(\bar{y}_{0},1)\biggl{\\}}\equiv C.$

Along with these two observations, (4.26) yields that

${N(t_{0},y_{0}^{1})-N(t_{0},y_{0}^{2})}\leq\frac{C}{1-\rho}{\|y_{0}^{2}-y_{0}^{1}\|}\equiv
C(\rho){\|y_{0}^{2}-y_{0}^{1}\|}.$

Similarly, we can obtain ${N(t_{0},y_{0}^{2})-N(t_{0},y_{0}^{1})}\leq
C(\rho){\|y_{0}^{2}-y_{0}^{1}\|}.$ These lead to the desired Lipschitz
continuity.

$(ii)$ Let $\bar{y}_{0}\in L^{2}(\Omega)$. Arbitrarily take $\bar{t}_{0}$ from
$[0,T)$ and write $\bar{\delta}=(T-\bar{t}_{0})/2$. It suffices to show that
$N(\cdot,\bar{y}_{0})$ is continuous over
$\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}$. For
this purpose, we arbitrarily take two different $t_{0}^{1}$ and $t_{0}^{2}$
from this interval. Without lose of generality, we can assume that
$t_{0}^{1}<t_{0}^{2}$. Then by (4.14) (see Lemma 4.6), the part $(i)$ of the
current lemma and Lemma 4.9, we can easily deduce that

$\begin{array}[]{l}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr~{}~{}~{}|N(t_{0}^{1},\bar{y}_{0})-N(t_{0}^{2},\bar{y}_{0})|\\\ \vskip
3.0pt plus 1.0pt minus
1.0pt\cr=\left|N\Bigr{(}t_{0}^{2},\,\bar{y}^{N(t_{0}^{1},\bar{y}_{0})}_{t_{0}^{1},\bar{y}_{0}}(t_{0}^{2})\Bigr{)}-N(t_{0}^{2},\bar{y}_{0})\right|\leq
C\left\|\bar{y}^{N(t_{0}^{1},\bar{y}_{0})}_{t_{0}^{1},\bar{y}_{0}}(t_{0}^{2})-\bar{y}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle=C\left\|\Bigr{[}e^{(t_{0}^{2}-t_{0}^{1})\triangle}-I\Bigl{]}\bar{y}_{0}+\int^{t_{0}^{2}}_{t_{0}^{1}}e^{(t_{0}^{2}-s)\triangle}\bar{u}^{N(t_{0}^{1},\bar{y}_{0})}_{t_{0}^{1},\bar{y}_{0}}(s)ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\displaystyle\leq
C\left\|e^{(t_{0}^{2}-t_{0}^{1})\triangle}-I\right\|\|\bar{y}_{0}\|+C|t_{0}^{2}-t_{0}^{1}|\end{array}$

where $C$ stands for a positive constant independent of $t_{0}^{1}$ and
$t_{0}^{2}$. It varies in different contexts. Clearly, the continuity of
$N(\cdot,\bar{y}_{0})$ over
$\Bigr{[}(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}\Bigl{]}$
follows from the above inequality at once.

In summary, we finish the proof.

Next, we study some properties for the map ${\cal N}:[0,T)\times
L^{2}(\Omega)\times[0,+\infty)\mapsto L^{2}(\Omega)$ defined by

$\displaystyle{\cal
N}(t_{0},y_{0},M)=\bar{u}^{M}_{t_{0},y_{0}}(t_{0})=M\frac{\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})\|}.$
(4.27)

######  Lemma 4.11.

$(i)$ For each $(\bar{t}_{0},\bar{y}_{0},\bar{M})\in[0,T)\times
L^{2}(\Omega)\times[0,\infty)$, there is a $\bar{\rho}>0$ such that ${\cal
N}(t_{0},\cdot,\cdot)$ is Lipschitz continuous over
$B(\bar{y}_{0},\bar{\rho})\times[(\bar{M}-\bar{\rho})^{+},\bar{M}+\bar{\rho}]$
uniformly with respect to $t_{0}\in B(\bar{t}_{0},\bar{\rho})\bigcap[0,T)$.
$(ii)$ ${\cal N}(\cdot,\bar{y}_{0},\cdot)$ is continuous over
$[0,T)\times[0,\infty)$.

Proof. $(i)$ Let $(\bar{t}_{0},\bar{y}_{0},\bar{M})\in[0,T)\times
L^{2}(\Omega)\times[0,\infty)$. The proof of the first continuity will be
carried by several steps as follows:

Step 1. For all $t_{0}\in[0,T)$, $0\leq M_{1}\leq M_{2}$ and
$y_{0}^{1},~{}y_{0}^{2}\in L^{2}(\Omega)$, it holds that

$\begin{array}[]{rl}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\|{\cal N}(t_{0},y_{0}^{1},M_{1})-{\cal
N}(t_{0},y_{0}^{2},M_{2})\|\leq|M_{1}-M_{2}|+\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr&\displaystyle\frac{4M_{1}}{\|\chi_{\omega}\bar{\psi}^{M_{1}}_{t_{0},y^{1}_{0}}\|}\left[M_{1}\|y_{0}^{1}-y_{0}^{2}\|+(\|y_{0}^{2}\|+\|z_{d}\|){|M_{1}-M_{2}|}\right]\;\;\mbox{when}\;M_{1}>0;\end{array}$
(4.28) $\|{\cal N}(t_{0},y_{0}^{1},M_{1})-{\cal
N}(t_{0},y_{0}^{2},M_{2})\|=|M_{1}-M_{2}|\;\;\mbox{when}\;M_{1}=0,$ (4.29)

The equality (4.29) follows directly from the definition of ${\cal N}$. Now we
prove (4.28). For simplification of notation, we write

$\bar{y}^{i}\mathop{\buildrel\Delta\over{=}}\bar{y}^{M_{i}}_{t_{0},y^{i}_{0}},\quad\bar{\psi}^{i}\mathop{\buildrel\Delta\over{=}}\bar{\psi}^{M_{i}}_{t_{0},y^{i}_{0}},\quad\bar{u}^{i}\mathop{\buildrel\Delta\over{=}}\bar{u}^{M_{i}}_{t_{0},y^{i}_{0}},\qquad
i=1,2.$

It is clear that that
$(1-\varepsilon)\bar{u}^{1}+\varepsilon\frac{M_{1}}{M_{2}}\bar{u}^{2}\in
L^{\infty}(t_{0},T;B(0,M_{1}))$ for any $\varepsilon\in[0,1],$ which, together
with the optimality of $\bar{u}^{1}$ to $(OP)^{M_{1}}_{t_{0},y_{0}^{1}}$,
shows that

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus 1.0pt\cr
0&\displaystyle\leq\mathop{\underline{\rm lim}}\limits_{\varepsilon\rightarrow
0+}\frac{1}{2\varepsilon}\left\\{\biggr{\|}e^{(T-t_{0})\triangle}y_{0}^{1}+\int^{T}_{t_{0}}e^{(T-s)\triangle}[(1-\varepsilon)\bar{u}^{1}+\varepsilon\frac{M_{1}}{M_{2}}\bar{u}^{2}]ds-
z_{d}\biggl{\|}^{2}\right.\\\ \vskip 3.0pt plus 1.0pt minus
1.0pt\cr&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\left.\displaystyle-\biggr{\|}e^{(T-t_{0})\triangle}y_{0}^{1}+\int^{T}_{t_{0}}e^{(T-s)\triangle}\bar{u}^{1}ds-
z_{d}\biggl{\|}^{2}\right\\}\\\ \vskip 6.0pt plus 2.0pt minus
2.0pt\cr&\displaystyle=\left\langle
e^{(T-t_{0})\triangle}y_{0}^{1}+\int^{T}_{t_{0}}e^{(T-s)\triangle}\bar{u}^{1}ds-
z_{d}\,,~{}\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{M_{1}}{M_{2}}\bar{u}^{2}-\bar{u}^{1}\right]ds\right\rangle.\end{array}$

Similarly, we can prove that

$\displaystyle 0\leq\left\langle
e^{(T-t_{0})\triangle}y_{0}^{2}+\int^{T}_{t_{0}}e^{(T-s)\triangle}\bar{u}^{2}ds-
z_{d}\,,~{}\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{M_{2}}{M_{1}}\bar{u}^{1}-\bar{u}^{2}\right]ds\right\rangle.$

Dividing the first inequality above by $M_{1}^{2}$ and the second one by
$M_{2}^{2}$, then adding them together, we obtain that

$\begin{array}[]{l}\displaystyle~{}~{}~{}\left\langle\frac{e^{(T-t_{0})\triangle}y_{0}^{1}-z_{d}}{M_{1}}-\frac{e^{(T-t_{0})\triangle}y_{0}^{2}-z_{d}}{M_{2}}\,,~{}\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{\bar{u}^{2}}{M_{2}}-\frac{\bar{u}^{1}}{M_{1}}\right]ds\right\rangle\\\
\geq\displaystyle\biggr{\|}\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{\bar{u}^{2}}{M_{2}}-\frac{\bar{u}^{1}}{M_{1}}\right]ds\biggl{\|}^{2},\end{array}$

which implies that

$\biggr{\|}\frac{e^{(T-t_{0})\triangle}y_{0}^{1}-z_{d}}{M_{1}}-\frac{e^{(T-t_{0})\triangle}y_{0}^{2}-z_{d}}{M_{2}}\biggl{\|}\geq\biggr{\|}\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{\bar{u}^{2}}{M_{2}}-\frac{\bar{u}^{1}}{M_{1}}\right]ds\biggl{\|}.$
(4.30)

Since $(\bar{y}^{i},\bar{\psi}^{i})$, $i=1,2$, solve (4.2) and the semigroup
$\\{e^{t\triangle}:t\geq 0\\}$ is contractive, we can use (4.30) to derive
that

$\begin{array}[]{rl}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\biggr{\|}\frac{\bar{\psi}^{1}(t_{0})}{M_{1}}-\frac{\bar{\psi}^{2}(t_{0})}{M_{2}}\biggl{\|}=\biggr{\|}e^{(T-t_{0})\triangle}\left(\frac{\bar{\psi}^{1}(T)}{M_{1}}-\frac{\bar{\psi}^{2}(T)}{M_{2}}\right)\biggl{\|}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\biggr{\|}\frac{\bar{\psi}^{1}(T)}{M_{1}}-\frac{\bar{\psi}^{2}(T)}{M_{2}}\biggl{\|}=\biggr{\|}\frac{\bar{y}^{1}(T)-z_{d}}{M_{1}}-\frac{\bar{y}^{2}(T)-z_{d}}{M_{2}}\biggl{\|}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\left\|\frac{e^{(T-t_{0})\triangle}y^{1}_{0}-z_{d}}{M_{1}}-\frac{e^{(T-t_{0})\triangle}y^{2}_{0}-z_{d}}{M_{2}}\right\|+\left\|\int^{T}_{t_{0}}e^{(T-s)\triangle}\left[\frac{\bar{u}^{1}}{M_{1}}-\frac{\bar{u}^{2}}{M_{2}}\right]ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\frac{2}{M_{1}}\|y_{0}^{1}-y_{0}^{2}\|+2(\|y_{0}^{2}\|+\|z_{d}\|)\frac{|M_{1}-M_{2}|}{M_{1}M_{2}}.\end{array}$
(4.31)

By direct computation, we obtain that

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\|{\cal N}(t_{0},y_{0}^{1},M_{1})-{\cal
N}(t_{0},y_{0}^{2},M_{2})\|=\biggr{\|}M_{1}\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{1}(t_{0})\|}-M_{2}\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{2}(t_{0})\|}\biggl{\|}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle~{}M_{1}\biggr{\|}\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{1}(t_{0})\|}-\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{2}(t_{0})\|}\biggl{\|}+|M_{1}-M_{2}|\biggr{\|}\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{\|\chi_{\omega}\bar{\psi}^{2}(t_{0})\|}\biggl{\|}\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr=&\displaystyle\frac{M_{1}}{\|{\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{M_{1}}}\|\|{\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}}\|}{\biggr{\|}}\left[\|\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}\|\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{M_{1}}-\|\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{M_{1}}\|\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}\right]\biggl{\|}+|M_{1}-M_{2}|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\frac{2M_{1}}{\|{\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{M_{1}}}\|\|{\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}}\|}\left\|\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}\right\|\left\|\frac{\chi_{\omega}\bar{\psi}^{1}(t_{0})}{M_{1}}-\frac{\chi_{\omega}\bar{\psi}^{2}(t_{0})}{M_{2}}\right\|+|M_{1}-M_{2}|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\frac{2M_{1}^{2}}{\|\chi_{\omega}\bar{\psi}^{1}(t_{0})\|}\left\|\frac{\bar{\psi}^{1}(t_{0})}{M_{1}}-\frac{\bar{\psi}^{2}(t_{0})}{M_{2}}\right\|+|M_{1}-M_{2}|.\end{array}$

This, together with (4.31), shows (4.28).

Step 2. When $\bar{M}>0$, there is $\bar{\rho}>0$ such that for each
$(t_{0},y_{0},M)\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]\times
B(\bar{y}_{0},\bar{\rho})\times[\bar{M}-\bar{\rho},\bar{M},\bar{\rho}]$,

$\displaystyle\left\|\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})\right\|\geq\frac{1}{2}\left\|\chi_{\omega}\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|>0.$
(4.32)

The second inequality in (4.32) follows from (4.3). The first one will be
proved by the following two cases:

Case 1: $t_{0}\leq\bar{t}_{0}$. In this case, the following three estimates
hold for all $y_{0}\in L^{2}(\Omega)$ and $M\in[\bar{M}/2,(3\bar{M})/2]$:

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|=\displaystyle\left\|e^{(\bar{t}_{0}-t_{0})\triangle}\bar{\psi}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\leq&\left\|e^{(\bar{t}_{0}-t_{0})\triangle}\right\|\cdot\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\|;\end{array}$

(Here $I$ denotes the identity operator on $L^{2}(\Omega)$.)

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|=\left\|\bar{\psi}^{M}_{\bar{t}_{0},\bar{y}^{M}_{t_{0},y_{0}}(\bar{t}_{0})}(\bar{t}_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\leq&\displaystyle
M\left\|\frac{\bar{\psi}^{M}_{\bar{t}_{0},\bar{y}^{M}_{t_{0},y_{0}}(\bar{t}_{0})}(\bar{t}_{0})}{M}-\frac{\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})}{\bar{M}}\right\|+\frac{|M-\bar{M}|}{\bar{M}}\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\leq&\displaystyle
2\left\|\bar{y}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{y}_{0}\right\|+\left(2\|\bar{y}_{0}\|+2\|z_{d}\|+\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\right)\frac{|M-\bar{M}|}{\bar{M}};\end{array}$

(Here, (4.8) and (4.31) have been used.) and

$\begin{array}[]{ll}{}{}{}{}&\displaystyle\left\|\bar{y}^{M}_{t_{0},y_{0}}(\bar{t}_{0})-\bar{y}_{0}\right\|=\left\|e^{(\bar{t}_{0}-t_{0})\triangle}y_{0}-\bar{y}_{0}+\int^{\bar{t}_{0}}_{t_{0}}e^{(\bar{t}_{0}-s)\triangle}\bar{u}^{M}_{t_{0},y_{0}}(s)ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\left\|e^{(\bar{t}_{0}-t_{0})\triangle}\right\|\cdot\|y_{0}-\bar{y}_{0}\|+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{y}_{0}\|+\int^{\bar{t}_{0}}_{t_{0}}\left\|e^{(\bar{t}_{0}-s)\triangle}\right\|\cdot\left\|\bar{u}^{M}_{t_{0},y_{0}}(s)\right\|ds\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\|y_{0}-\bar{y}_{0}\|+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{y}_{0}\|+M(\bar{t}_{0}-{t_{0}}).\end{array}$

Combining the above-mentioned three inequalities together leads to

$\begin{array}[]{l}~{}~{}\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\leq
2\|y_{0}-\bar{y}_{0}\|+2M(\bar{t}_{0}-{t_{0}})\\\ \vskip 3.0pt plus 1.0pt
minus
1.0pt\cr\displaystyle\par+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\left(2\|\bar{y}_{0}\|+\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\|\right)\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\displaystyle+\left(2\|\bar{y}_{0}\|+2\|z_{d}\|+\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\right)\frac{|M-\bar{M}|}{\bar{M}}.\end{array}$
(4.33)

Clearly, the right hand side of (4.33) is continuous with respect to
$(t_{0},y_{0},M)$. This, along with the second inequality in (4.32), indicates
that there exists a $\rho_{1}$ with

$\displaystyle 0<\rho_{1}<\frac{T-\bar{t}_{0}}{2}\wedge\frac{\bar{M}}{2},$

such that for each
$(t_{0},y_{0},M)\in[(\bar{t}_{0}-\rho_{1})^{+},\bar{t}_{0}]\times
B(\bar{y}_{0},\rho_{1})\times[\bar{M}-\rho_{1},\bar{M}+\rho_{1}]$,

$\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\leq\displaystyle\frac{1}{2}\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|,$
(4.34)

Case 2: $t_{0}\geq\bar{t}_{0}$. In this case, the following two estimates hold
for all $y_{0}\in L^{2}(\Omega)$ and $M\in[\bar{M}/2,(3\bar{M})/2]$:

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\leq&\displaystyle
M\left\|\frac{\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})}{M}-\frac{\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})}{\bar{M}}\right\|+\frac{|M-\bar{M}|}{\bar{M}}\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr=&\displaystyle
M\left\|\frac{\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})}{M}-\frac{\bar{\psi}^{\bar{M}}_{t_{0},\bar{y}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})}(t_{0})}{\bar{M}}\right\|+\frac{|M-\bar{M}|}{\bar{M}}\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\leq&\displaystyle
2\left\|y_{0}-\bar{y}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|+\left(2\left\|\bar{y}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|+2\|z_{d}\|+\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|\right)\frac{|M-\bar{M}|}{\bar{M}}\end{array}$

(Here, (4.8) and (4.31) have been used.) and

$\begin{array}[]{ll}{}{}{}{}&\displaystyle\left\|\bar{y}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})-y_{0}\right\|=\left\|e^{(t_{0}-\bar{t}_{0})\triangle}\bar{y}_{0}-y_{0}+\int^{t_{0}}_{\bar{t}_{0}}e^{(t_{0}-s)\triangle}\bar{u}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(s)ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr=&\displaystyle\left\|e^{(t_{0}-\bar{t}_{0})\triangle}\bar{y}_{0}-y_{0}+\int^{t_{0}}_{\bar{t}_{0}}e^{(t_{0}-s)\triangle}\bar{u}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(s)ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{y}_{0}\|+\|y_{0}-\bar{y}_{0}\|+\int^{t_{0}}_{\bar{t}_{0}}\left\|e^{(t_{0}-s)\triangle}\right\|\cdot\left\|\bar{u}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(s)\right\|ds\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\|y_{0}-\bar{y}_{0}\|+\left\|e^{(\bar{t}_{0}-t_{0})\triangle}-I\right\|\cdot\|\bar{y}_{0}\|+\bar{M}(t_{0}-\bar{t_{0}}).\end{array}$

From the above-mentioned two estimates, we derive that

$\begin{array}[]{c}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr~{}~{}~{}~{}\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\leq
2\|y_{0}-\bar{y}_{0}\|+2\bar{M}(\bar{t}_{0}-{t_{0}})+2\left\|e^{(t_{0}-\bar{t}_{0})\triangle}-I\right\|\cdot\|\bar{y}_{0}\|\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\cr\displaystyle\par+\left(2\left\|\bar{y}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})\right\|+\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|+2\|z_{d}\|\right)\frac{|M-\bar{M}|}{\overline{M}}+\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\end{array}$
(4.35)

By the same argument used to get (4.34) (notice the continuity of
$\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\cdot)$), we can find a
$\rho_{2}$ with $\displaystyle
0<\rho_{2}<\frac{T-\bar{t}_{0}}{2}\wedge\frac{\bar{M}}{2}$, such that for each
triplet $(t_{0},y_{0},M)\in[\bar{t}_{0},\bar{t}_{0}+\rho_{2}]\times
B(\bar{y}_{0}),\rho_{2})\times[\bar{M}-\rho_{2},\bar{M}+\rho_{2}]$,

$\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\leq\displaystyle\frac{1}{2}\left\|\bar{\psi}^{\bar{M}}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|.$
(4.36)

Now we set $\bar{\rho}=\rho_{1}\wedge\rho_{2}$. Then the first inequality of
(4.32) follows from (4.34) and (4.36).

Step 3. When $\bar{M}=0$, there is $\bar{\rho}>0$ such that for each
$t_{0}\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]$, each
$y_{0}\in B(\bar{y}_{0},\bar{\rho})$ and each $M\in[0,\bar{\rho}]$,

$\displaystyle\left\|\chi_{\omega}\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})\right\|\geq\frac{1}{2}\left\|\chi_{\omega}\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|>0.$
(4.37)

The second inequality in (4.37) follows from (4.3). The remainder is to show
the first one. The following two inequalities can be checked by direct
computation:

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{0}_{t_{0},\bar{y}_{0}}(t_{0})\right\|=\left\|e^{(T-\bar{t}_{0})\triangle}\left[\bar{\psi}^{M}_{t_{0},y_{0}}(T)-\bar{\psi}^{0}_{t_{0},\bar{y}_{0}}(T)\right]\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr=&\displaystyle\left\|e^{(T-\bar{t}_{0})\triangle}\left[\bar{y}^{M}_{t_{0},y_{0}}(T)-\bar{y}^{0}_{t_{0},\bar{y}_{0}}(T)\right]\right\|\leq\left\|\bar{y}^{M}_{t_{0},y_{0}}(T)-\bar{y}^{0}_{t_{0},\bar{y}_{0}}(T)\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr=&\displaystyle\left\|e^{(T-t_{0})\triangle}(y_{0}-\bar{y}_{0})+\int^{T}_{t_{0}}e^{(T-s)\triangle}\bar{u}^{M}_{t_{0},y_{0}}(s)ds\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\left\|y_{0}-\bar{y}_{0}\right\|+M(T-t_{0})\end{array}$

and

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\left\|\bar{\psi}^{0}_{t_{0},\bar{y}_{0}}(t_{0})-\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr=&\left\|\left[e^{2(T-t_{0})\triangle}-e^{2(T-\bar{t}_{0})\triangle}\right]\bar{y}_{0}+\left[e^{(T-t_{0})\triangle}-e^{(T-\bar{t}_{0})\triangle}\right]z_{d}\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\left\|e^{2(|t_{0}-\bar{t}_{0}|)\triangle}-I\right\|\|\bar{y}_{0}\|+\left[e^{(|t_{0}-\bar{t}_{0}|)\triangle}-I\right]\|z_{d}\|.\end{array}$

From these, we deduce that

$\begin{array}[]{ll}\vskip 3.0pt plus 1.0pt minus
1.0pt\cr&\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{0}_{t_{0},\bar{y}_{0}}(t_{0})\right\|+\left\|\bar{\psi}^{0}_{t_{0},\bar{y}_{0}}(t_{0})-\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\\\
\vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq&\displaystyle\|y_{0}-\bar{y}_{0}\|+M(T-t_{0})+\left\|e^{2(|t_{0}-\bar{t}_{0}|)\triangle}-I\right\|\|\bar{y}_{0}\|+\left[e^{(|t_{0}-\bar{t}_{0}|)\triangle}-I\right]\|z_{d}\|.\end{array}$
(4.38)

Clearly, the right hand side of (4.38) is continuous with respect to
$t_{0},y_{0}$ and $M$. This, together with the second inequality of (4.37),
yields that there exists
$\bar{\rho}\in(0,\displaystyle\frac{T-\bar{t}_{0}}{2})$ such that for each
$(t_{0},y_{0},M)\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]\times
B(\bar{y}_{0},\bar{\rho})\times[0,\bar{\rho}]$,

$\displaystyle\left\|\bar{\psi}^{M}_{t_{0},y_{0}}(t_{0})-\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|\leq\displaystyle\frac{1}{2}\left\|\bar{\psi}^{0}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\right\|,$

from which, the first inequality in (4.37) follows at once.

Step 4. The required Lipschitz continuity of the map ${\cal N}$

Clearly, we can take the same constant $\bar{\rho}$ in Step 2 and Step 3 such
that (4.32) and (4.37) stand. When $\bar{M}=0$, it follows from (4.28), (4.29)
and (4.37) that the map ${\cal N}(t_{0},\cdot,\cdot)$ is Lipschitz continuous
over $B(y_{0},\bar{\rho})\times[(\bar{M}-\bar{\rho})^{+},\bar{M}+\bar{\rho}]$
uniformly with respect to
$t_{0}\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]$; When
$\bar{M}>0$, the same conclusion follows from (4.28), (4.29) and (4.32).

$(ii)$ Fix a $\bar{y}_{0}\in L^{2}(\Omega)$. Let
$(\bar{M},\bar{t}_{0})\in[0,\infty)\times[0,T)$. Since
$\|\chi_{\omega}\bar{\psi}^{M}_{\bar{t}_{0},\bar{y}_{0}}(\bar{t}_{0})\|\neq 0$
(see (4.3)), the continuity of the map $(t_{0},M)\rightarrow{\cal
N}(t_{0},\bar{y}_{0},M)$ at $(\bar{t}_{0},\bar{M})$ follows from the
continuity of the map
$(t_{0},M)\rightarrow\bar{\psi}^{M}_{t_{0},\bar{y}_{0}}(t_{0})$ at
$(\bar{t}_{0},\bar{M})$. When $\bar{M}>0$, the continuity of the map
$(t_{0},M)\rightarrow\bar{\psi}^{M}_{t_{0},\bar{y}_{0}}(t_{0})$ at
$(\bar{t}_{0},\bar{M})$ follows from (4.33) and (4.35). When $\bar{M}=0$, the
continuity of this map at $(\bar{t}_{0},\bar{M})$ follows from (4.38). Thus,
${\cal N}(\cdot,\bar{y}_{0},\cdot)$ is continuous over
$[0,T)\times[0,\infty)$.

In summary, we complete the proof.

Proof of Proposition 4.7. $(i)$ Let $(\bar{t}_{0},\bar{y}_{0})\in[0,T)\times
L^{2}(\Omega)$. By the definition of maps ${\cal N}$ and $F$ (see (4.4) and
(4.27)), we see that

$\displaystyle F(t_{0},y_{0})={\cal N}(t_{0},y_{0},N(t_{0},y_{0}))\;\;{\rm
for~{}all}~{}(t_{0},y_{0})\in[0,T)\times L^{2}(\Omega).$ (4.39)

According to Lemma 4.11, there are $\bar{\rho}_{1}>0$ and $C_{1}>0$ such that
when $(t_{0},y_{0}^{1},M_{1})$ and $(t_{0},y_{0}^{2},M_{2})$ belong to
$[(\bar{t}_{0}-\bar{\rho}_{1})^{+},\bar{t}_{0}+\bar{\rho}_{1}]\times
B(\bar{y}_{0},\bar{\rho}_{1})\times[(\bar{M}-\bar{\rho}_{1})^{+},\bar{M}+\bar{\rho}_{1}]$,

$\|{\cal N}(t_{0},y_{0}^{1},M_{1})-{\cal N}(t_{0},y_{0}^{2},M_{2})\|\leq
C_{1}\Bigr{(}\|y_{0}^{1}-y_{0}^{2}\|+|M_{1}-M_{2}|\Big{)}.$ (4.40)

According to Lemma 4.10, there are $\bar{\rho}_{2}>0$ and $C_{2}>0$ such that

$|N(t_{0},y_{0}^{1})-N(t_{0},y_{0}^{2})|\leq C_{2}\|y_{0}^{1}-y_{0}^{2}\|,$

for all $(t_{0},y_{0}^{1})$ and $(t_{0},y_{0}^{2})$ belong to
$[(\bar{t}_{0}-\bar{\delta})^{+},\bar{t}_{0}+\bar{\delta}]\times
B(\bar{y}_{0},\bar{\rho}_{2})$, where $\bar{\delta}=(T-\bar{t}_{0})/2$.

Let
$\bar{\rho}=\min\\{\bar{\rho}_{1},\bar{\rho}_{2},\displaystyle\frac{\bar{\rho}_{1}}{2C_{2}},\bar{\delta}\\}$.
Then it follows from the above inequality that

$|N(t_{0},y_{0}^{1})-N(t_{0},\bar{y}_{0}^{2})|\leq
2C_{2}\bar{\rho}\leq\bar{\rho}_{1}\;\;\mbox{for all}\;y_{0}^{1},y_{0}^{2}\in
B(\bar{y}_{0},\bar{\rho})\;\mbox{and}\;t_{0}\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}].$

This, along with (4.40), indicates that

$\begin{array}[]{l}~{}~{}~{}\|F(t_{0},y_{0}^{1})-F(t_{0},y_{0}^{2})\|\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr=\|{\cal
N}(t_{0},y_{0}^{1},N(t_{0},y_{0}^{1}))-{\cal
N}(t_{0},y_{0}^{2},N(t_{0},y_{0}^{2}))\|\\\ \vskip 3.0pt plus 1.0pt minus
1.0pt\cr\leq
C_{1}\Bigr{(}\|y_{0}^{1}-y_{0}^{2}\|+|N(t_{0},y_{0}^{1})-N(t_{0},y_{0}^{2})|\Bigr{)}\\\
\vskip 3.0pt plus 1.0pt minus 1.0pt\cr\leq
C_{1}\Bigr{(}\|y_{0}^{1}-y_{0}^{2}\|+C_{2}\|y_{0}^{1}-y_{0}^{2}\|\Bigr{)}=C_{1}(1+C_{2})\|y_{0}^{1}-y_{0}^{2}\|\end{array}.$

for all $y_{0}^{1},y_{0}^{2}\in B(\bar{y}_{0},\bar{\rho})$ and
$t_{0}\in[(\bar{t}_{0}-\bar{\rho})^{+},\bar{t}_{0}+\bar{\rho}]$. The desired
Lipschitz continuity follows from the above inequality at once.

$(ii)$ Let $\bar{y}_{0}\in L^{2}(\Omega)$. Since $F(t_{0},\bar{y}_{0})={\cal
N}(t_{0},\bar{y}_{0},N(t_{0},\bar{y}_{0}))$ for all $t_{0}\in[0,T)$ (see
(4.39)), the desired continuity of $F(\cdot,\bar{y}_{0})$ follows directly
from the continuity of $N(\cdot,\bar{y}_{0})$ and ${\cal
N}(\cdot,\bar{y}_{0},\cdot)$.

In summary, we complete the proof.

###  4.4 Proof of Theorem 4.2

Proof of Theorem 4.2. Let $(t_{0},y_{0})\in[0,T)\times L^{2}({\Omega})$. By
theorem 4.1, $\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$ is the unique
solution to Equation (4.5), i.e.,
$y_{F}(\cdot;t_{0},y_{0})=\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$ over
$[t_{0},T)$. Then, by (4.4), (4.14), (4.8) and (4.10), we see that

$F\Bigr{(}t,y_{F}(t;t_{0},y_{0})\Bigl{)}=F\left(t,\bar{y}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)\right)=N(t_{0},y_{0})\frac{{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)}{\|{\chi}_{\omega}\bar{\psi}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t)\|}=\bar{u}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(t),\quad\forall~{}t\in(t_{0},T).$

Since $\bar{u}^{N(t_{0},y_{0})}_{t_{0},y_{0}}(\cdot)$ is optimal norm control
for Problem $(NP)_{t_{0},y_{0}}$ (see Lemma4.5), the above equality implies
that $F(\cdot;y_{F}(\cdot;t_{0},y_{0})$ is the optimal control to
$(NP)_{t_{0},y_{0}}$. This completes the proof.

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