Time optimal sampled-data controls for heat equations

TL;DR

This paper develops a framework for time optimal sampled-data controls for heat equations, providing error estimates and connections to distributed controls, advancing numerical methods for such control problems.

Contribution

It introduces a novel approach to approximate time optimal distributed controls using sampled-data controls, with proven error bounds and theoretical properties.

Findings

Error estimates for control and time between sampled-data and distributed controls

Connections established among sampled-data control problems, minimal norm, and minimization problems

Proven optimality of the error bounds in terms of sampling period

Abstract

In this paper, we first design a time optimal control problem for the heat equation with sampled-data controls, and then use it to approximate a time optimal control problem for the heat equation with distributed controls. Our design is reasonable from perspective of sampled-data controls. And it might provide a right way for the numerical approach of a time optimal distributed control problem, via the corresponding semi-discretized (in time variable) time optimal control problem. The study of such a time optimal sampled-data control problem is not easy, because it may have infinitely many optimal controls. We find connections among this problem, a minimal norm sampled-data control problem and a minimization problem. And obtain some properties on these problems. Based on these, we not only build up error estimates for optimal time and optimal controls between the time optimal sampled-data control problem and the time optimal distributed control problem, in terms of the sampling period, but also prove that such estimates are optimal in some sense.

Equations

$$(TP{)}^{M,y_0}:T(M,y_0)=\inf \left\{\hat{t}>0:\exists \hat{u}\in {\mathcal{U}}^M\text{mbox}s.t.y(\hat{t};y_0,\hat{u})\in B_r(0)\right\},$$

$${\mathcal{U}}^M\triangleq \left\{u\in L^2({\mathbb{R}}^{+};L^2(\Omega )):\Vert u{\Vert }_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}\le M\right\},$$

$$\displaystyle\left\{\begin{array}[]{lll}\partial_{t}y-\Delta y=\chi_{\omega}u&\mbox{in}&\Omega\times\mathbb{R}^{+},\\ y=0&\mbox{on}&\partial\Omega\times\mathbb{R}^{+},\\ y(0)=y_{0}&\mbox{in}&\Omega.\end{array}\right.$$

$$\displaystyle L_{\delta}^{2}(\mathbb{R}^{+};L^{2}(\Omega))\triangleq\Big{\{}u_{\delta}\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))\;:\;u_{\delta}\triangleq\sum_{i=1}^{\infty}\chi_{((i-1)\delta,i\delta]}u^{i},\;\{u^{i}\}_{i=1}^{\infty}\subset L^{2}(\Omega)\Big{\}},$$

$$\left\{\begin{array}[]{lll}\partial_{t}y-\Delta y=\chi_{\omega}u_{\delta}&\mbox{in}&\Omega\times\mathbb{R}^{+},\\ y=0&\mbox{on}&\partial\Omega\times\mathbb{R}^{+},\\ y(0)=y_{0}&\mbox{in}&\Omega.\end{array}\right.$$

$$(TP{)}_{\delta }^{M,y_0}:T_{\delta }(M,y_0)=\inf \left\{k\delta :\exists k\in {\mathbb{N}}^{+},\exists u_{\delta }\in {\mathcal{U}}_{\delta }^M\text{mbox}s.t.y(k\delta ;y_0,u_{\delta })\in B_r(0)\right\},$$

$${\mathcal{U}}_{\delta }^M\triangleq \left\{u_{\delta }\in L_{\delta }^2({\mathbb{R}}^{+};L^2(\Omega )):\Vert u_{\delta }{\Vert }_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}\le M\right\}.$$

$$0\le T_{\delta }(M,y_0)-T(M,y_0)\le 2\delta \text{mbox}forall\delta \in (0,{\delta }_0)\text{mbox}andM\in [M_1,M_2].$$

$$\delta >T_{\delta }(M,y_0)-T(M,y_0)>(1-\eta )\delta \text{mbox}foreach\delta \in {\mathcal{A}}_{M,\eta }.$$

$$\underset{M_1\le M\le M_2}{\sup }\Vert u_{M,\delta }^{\ast }-u_M^{\ast }{\Vert }_{L^2(0,T(M,y_0);L^2(\Omega ))}$$

$$\Vert u_{M,\delta }^{\ast }-u_M^{\ast }{\Vert }_{L^2(0,T(M,y_0);L^2(\Omega ))}$$

$$\underset{M_1\le M\le M_2}{\sup }\Vert u_{M,\delta }-u_M^{\ast }{\Vert }_{L^2(0,T(M,y_0);L^2(\Omega ))}$$

$$\Vert {\hat{u}}_{M,\delta }-u_M^{\ast }{\Vert }_{L^2(0,T(M,y_0);L^2(\Omega ))}\ge C_M\sqrt{(1-\eta )\delta },$$

$$(NP{)}^{T,y_0}:N(T,y_0)\triangleq \inf \{\Vert v{\Vert }_{L^2(0,T;L^2(\Omega ))}:y(T;y_0,v)\in B_r(0)\},$$

$$(NP{)}_{\delta }^{k\delta ,y_0}:N_{\delta }(k\delta ,y_0)\triangleq \inf \{\Vert v_{\delta }{\Vert }_{L_{\delta }^2(0,k\delta ;L^2(\Omega ))}:y(k\delta ;y_0,v_{\delta })\in B_r(0)\},$$

$$L_{\delta }^2(0,k\delta ;L^2(\Omega ))\triangleq \{f{|}_{(0,k\delta ]}:f\in L_{\delta }^2({\mathbb{R}}^{+};L^2(\Omega ))\},$$

$$(JP{)}^{T,y_0}:V(T,y_0)\triangleq \underset{z\in L^2(\Omega )}{\inf }{J}^{T,y_0}(z)$$

$$\left\{\begin{array}[]{lll}\partial_{t}\varphi+\Delta\varphi=0&\mbox{in}&\Omega\times(0,T),\\ \varphi=0&\mbox{on}&\partial\Omega\times(0,T),\\ \varphi(T)=z\in L^{2}(\Omega).\end{array}\right.$$

$$(JP{)}_{\delta }^{k\delta ,y_0}:V_{\delta }(k\delta ,y_0)\triangleq \underset{z\in L^2(\Omega )}{\inf }{J}_{\delta }^{k\delta ,y_0}(z)$$

$${\overline{\phi }}_{\delta }(t;k\delta ,z)\triangleq \sum _{i=1}^k{\chi }_{((i-1)\delta ,i\delta ]}(t)\frac{1}{\delta }{\int }_{(i-1)\delta }^{i\delta }\phi (s;k\delta ,z)\mathrm{d}s\text{mbox}foreacht\in (0,k\delta ].$$

$${\mathcal{O}}_{M,\delta }\triangleq \{u_{\delta }{|}_{(0,T(M,y_0))}:u_{\delta }\text{mbox}isanoptimalcontrolto(TP{)}_{\delta }^{M,y_0}\}.$$

$$\frac{1}{C(\epsilon ,\delta ,k)}\Vert u_{\delta }^{y_0}{\Vert }_{L^2(0,k\delta ;L^2(\Omega ))}^2+\frac{1}{\epsilon }\Vert y(k\delta ;y_0,u_{\delta }^{y_0}){\Vert }^2\le \Vert y_0{\Vert }^2.$$

$${\bar{f}}_{\delta }(t)\triangleq \sum _{i=1}^{\infty }{\chi }_{((i-1)\delta ,i\delta ]}(t)\frac{1}{\delta }{\int }_{(i-1)\delta }^{i\delta }f(s)\mathrm{d}s\text{mbox}foreacht\in {\mathbb{R}}^{+}.$$

$$⟨{\bar{f}}_{\delta },g{⟩}_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}=⟨f,{\bar{g}}_{\delta }{⟩}_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}=⟨{\bar{f}}_{\delta },{\bar{g}}_{\delta }{⟩}_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}.$$

$$⟨{\bar{f}}_{\delta },g{⟩}_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}=⟨{\bar{f}}_{\delta },{\bar{g}}_{\delta }{⟩}_{L^2({\mathbb{R}}^{+};L^2(\Omega ))}.$$

$$\displaystyle\langle\bar{f}_{\delta},g\rangle_{L^{2}(\mathbb{R}^{+};L^{2}(\Omega))}=\sum_{i=1}^{\infty}\langle\bar{f}_{\delta},g\rangle_{L^{2}((i-1)\delta,i\delta;L^{2}(\Omega))}=\sum_{i=1}^{\infty}\Big{\langle}\bar{f}_{\delta}(i\delta),\int_{(i-1)\delta}^{i\delta}g(t)\,\mathrm{d}t\Big{\rangle}$$

$$\displaystyle\|\varphi(0;T,z)\|\leq e^{C(1+\frac{1}{T-S})}\|z\|^{1/2}\Big{\|}\frac{1}{S}\int_{0}^{S}\varphi(t;T,z)\,\mathrm{d}t\Big{\|}_{\omega}^{1/2}\;\;\mbox{for all}\;\;z\in L^{2}(\Omega).$$

$$f^z\triangleq \frac{1}{S}{\int }_0^S{e}^{\Delta (S-t)}z\mathrm{d}t.$$

$$\Vert e^{\Delta (T-S)}{f}^z\Vert \le e^{C(1+\frac{1}{T-S})}\Vert f^z{\Vert }^{1/2}\Vert {\chi }_{\omega }{e}^{\Delta (T-S)}{f}^z{\Vert }^{1/2}.$$

$$\Vert f^z\Vert \le \Vert z\Vert \text{mbox}and{e}^{\Delta (T-S)}{f}^z=\frac{1}{S}{\int }_0^S\phi (t;T,z)\mathrm{d}t.$$

Full text

The study of such a time optimal sampled-data control problem is not easy, because it may have infinitely many optimal controls. We find connections among this problem, a minimal norm sampled-data control problem and a minimization problem. And obtain some properties on these problems. Based on these, we not only build up error estimates for optimal time and optimal controls between the time optimal sampled-data control problem and the time optimal distributed control problem, in terms of the sampling period, but also prove that such estimates are optimal in some sense.

Keywords. Sampled-data controls, time optimal control, the heat equation, error estimates

2010 Mathematics Subject Classifications. 35K05, 49J20, 93C57

1 Introduction

1.1 Motivation and problems

In most published literature on time optimal control problems, controls are distributed in time, i.e., they can vary at each instant of time. However, in practical application, it is more convenient to use controls which vary only finite times. Sampled-data controls are such kind of controls. In this paper, we will design and study a time optimal control problem for the heat equation with sampled-data controls. And then we use it to approximate a time optimal control problem for the heat equation with distributed controls, through building up several error estimates for optimal time and optimal controls between these two problems, in terms of the sampling period. Such errors estimates have laid foundation for us to replace distributed controls by sampled-data controls in time optimal control problems for heat equations.

Throughout this paper, $\mathbb{R}^{+}\triangleq(0,\infty)$; $\Omega\subset\mathbb{R}^{d}$ ($d\in\mathbb{N}^{+}\triangleq\{1,2,\dots\}$) is a bounded domain with a $C^{2}$ boundary $\partial\Omega$; $\omega\subset\Omega$ is an open and nonempty subset with its characteristic function $\chi_{\omega}$; $\lambda_{1}$ is the first eigenvalue of $-\Delta$ with the homogeneous Dirichlet boundary condition over $\Omega$; $B_{r}(0)$ denotes the closed ball in $L^{2}(\Omega)$, centered at [math] and of radius $r>0$; for each measurable set $\mathcal{A}$ in $\mathbb{R}$, $|\mathcal{A}|$ denotes its Lebesgue measure; $\langle\cdot,\cdot\rangle$ and $\|\cdot\|$ denote the usual inner product and norm of $L^{2}(\Omega)$, respectively; $\langle\cdot,\cdot\rangle_{\omega}$ and $\|\cdot\|_{\omega}$ stand for the usual inner product and norm in $L^{2}(\omega)$, respectively.

First, we introduce a time optimal distributed control problem for the heat equation. Take $B_{r}(0)$ (with $r>0$ arbitrarily fixed) as our target. For each $M\geq 0$ and $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, consider the following time optimal distributed control problem:

[TABLE]

where

[TABLE]

and where $y(\cdot;y_{0},u)$ is the solution to the following distributed controlled heat equation:

[TABLE]

Since $y(t;y_{0},0)\rightarrow 0$ as $t\rightarrow\infty$, we find that $T(M,y_{0})<\infty$, for all $M\geq 0$ and $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. About $(TP)^{M,y_{0}}$, we introduce some concepts in the following definition:

Definition 1.1**.**

(i) The number $T(M,y_{0})$ is called the optimal time; $\hat{u}\in\mathcal{U}^{M}$ is called an admissible control if $y(\hat{t};y_{0},\hat{u})\in B_{r}(0)$ for some $\hat{t}>0$; $u^{*}\in\mathcal{U}^{M}$ is called an optimal control if $y(T(M,y_{0});y_{0},u^{*})\in B_{r}(0)$. (ii) Two optimal controls are said to be different (or the same), if they are different (or the same) on their effective domain \big{(}0,T(M,y_{0})\big{)}.

Several notes on the problem $(TP)^{M,y_{0}}$ are given in order:

• •

It is shown in Theorem 3.1 that for each $M\geq 0$ and $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, $(TP)^{M,y_{0}}$ has a unique optimal control.

• •

In many time optimal distributed control problems for heat equations, controls are taken from $L^{\infty}(\mathbb{R}^{+};L^{2}(\Omega))$. However, the current setting is also significant (see, for instance, [16] and [38]).

Next, we are going to design a time optimal sampled-data control problem for the heat equation. For this purpose, we define the following space of sampled-data controls (where $\delta>0$ is arbitrarily fixed):

[TABLE]

endowed with the $L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$-norm. Here and in what follows, $\chi_{((i-1)\delta,i\delta]}$ denotes the characteristic function of the interval \big{(}(i-1)\delta,i\delta\big{]} for each $i\in\mathbb{N}^{+}$. The numbers $\delta$, $2\delta$, $\ldots$, $i\delta$, $\ldots$ are called the sampling instants, while $\delta$ is called the sampling period. Each $u_{\delta}$ in the space $L_{\delta}^{2}(\mathbb{R}^{+};L^{2}(\Omega))$ is called a sampled-data control. For each $u_{\delta}\in L_{\delta}^{2}(\mathbb{R}^{+};L^{2}(\Omega))$ and each $y_{0}\in L^{2}(\Omega)$, write $y(\cdot;y_{0},u_{\delta})$ for the solution to the following sampled-data controlled heat equation:

[TABLE]

For each $M\geq 0$, $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and each $\delta>0$, consider the following time optimal sampled-data control problem:

[TABLE]

where

[TABLE]

Since $y(t;y_{0},0)\rightarrow 0$ as $t\rightarrow\infty$, we see that $T_{\delta}(M,y_{0})<\infty$, for all $M\geq 0$, $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and $\delta>0$. With respect to $(TP)_{\delta}^{M,y_{0}}$, we introduce some concepts in the following definition:

Definition 1.2**.**

(i) The number $T_{\delta}(M,y_{0})$ is called the optimal time; $u_{\delta}\in\mathcal{U}^{M}_{\delta}$ is called an admissible control if $y(\hat{k}\delta;y_{0},u_{\delta})\in B_{r}(0)$ for some $\hat{k}\in\mathbb{N}^{+}$; $u_{\delta}^{*}\in\mathcal{U}^{M}_{\delta}$ is called an optimal control if $y(T_{\delta}(M,y_{0});y_{0},u_{\delta}^{*})\in B_{r}(0)$. (ii) A control $u^{*}_{\delta}$ is called the optimal control with the minimal norm, if $u^{*}_{\delta}$ is an optimal control and satisfies that $\|u^{*}_{\delta}\|_{L^{2}(0,T_{\delta}(M,y_{0});L^{2}(\Omega))}\leq\|v^{*}_{\delta}\|_{L^{2}(0,T_{\delta}(M,y_{0});L^{2}(\Omega))}$ for any optimal control $v^{*}_{\delta}$. (iii) Two optimal controls are said to be different (or the same), if they are different (or the same) over $(0,T_{\delta}(M,y_{0}))$.

Several notes on this problem are given in order:

• •

The optimal time $T_{\delta}(M,y_{0})$ is a multiple of $\delta$ (see (1.8)). For each $M\geq 0$, each $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and each $\delta>0$, $(TP)_{\delta}^{M,y_{0}}$ has a unique optimal control with the minimal norm (see (ii) in Theorem 3.1); Given $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, there are infinitely many pairs $(M,\delta)$ so that $(TP)_{\delta}^{M,y_{0}}$ has infinitely many different optimal controls (see Theorem 3.2).

• •

We may design a time optimal sampled-data control problem in another way: To find a control $u_{\delta}^{*}$ in $\mathcal{U}_{\delta}^{M}$ so that $y(\cdot;y_{0},u_{\delta}^{*})$ enters $B_{r}(0)$ in the shortest time $\hat{T}_{\delta}(M,y_{0})$ (which may not be a multiple of $\delta$). We denote this problem by $(\widehat{TP})_{\delta}^{M,y_{0}}$. Several reasons for us to design time optimal sampled-data control problem to be $(TP)_{\delta}^{M,y_{0}}$ are as follows: (i) Each sampled-data control $u_{\delta}$ has the form: $\sum_{i=1}^{\infty}\chi_{((i-1)\delta,i\delta]}u^{i}$ with some $\{u^{i}\}_{i=1}^{\infty}\subset L^{2}(\Omega)$. From the perspective of sampled-data controls, each $u^{i}$ should be active in the whole subinterval $((i-1)\delta,i\delta]$. Thus, our definition for $T(M,y_{0})$ is reasonable. (ii) In the definition $(\widehat{TP})_{\delta}^{M,y_{0}}$, in order to make sure if the control process should be finished, we need to observe the solution (of the controlled equation) at each time. However, in our definition $(TP)_{\delta}^{M,y_{0}}$, we only need to observe the solution at time points $i\delta$, with $i=1,2,\dots$. (iii) Our design on $(TP)_{\delta}^{M,y_{0}}$ might provide a right way to approach numerically $(TP)^{M,y_{0}}$ via a discretized time optimal control problem. For instance, if we semi-discretize $(TP)^{M,y_{0}}$ in time variable, then our design on $(TP)_{\delta}^{M,y_{0}}$ can be borrowed to define a semi-discretized (in the time variable) time optimal control problem.

1.2 Main results

The main results of this paper are presented in the following three theorems.

Theorem 1.3**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then the following conclusions are true:

(i) For each $(M_{1},M_{2})$ with $0<M_{1}\leq M_{2}$, there is $\delta_{0}\triangleq\delta_{0}(M_{1},M_{2},y_{0},r)>0$ so that

[TABLE]

Moreover, $\delta_{0}$ depends on $M_{1}$, $M_{2}$, $y_{0}$ and $r$ continuously.

(ii) For each $M>0$ and $\eta\in(0,1)$, there exists a measurable set $\mathcal{A}_{M,\eta}\subset(0,1)$ (depending also on $y_{0}$ and $r$) with $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{A}_{M,\eta}\cap(0,h)|=\eta$ so that

[TABLE]

Theorem 1.4**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. For each $M>0$ and $\delta>0$, let $u^{*}_{M}$ and $u^{*}_{M,\delta}$ be the optimal control to $(TP)^{M,y_{0}}$ and the optimal control with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$, respectively. Then the following conclusions are true:

(i) For each $(M_{1},M_{2})$ with $0<M_{1}\leq M_{2}$, there is $C\triangleq C(M_{1},M_{2},y_{0},r)>0$ so that

[TABLE]

Moreover, $C$ depends on $M_{1}$, $M_{2}$, $y_{0}$ and $r$ continuously.

(ii) For each $M>0$ and $\eta\in(0,1)$, there is a measurable set $\mathcal{A}_{M,\eta}\subset(0,1)$ (depending also on $y_{0}$ and $r$) with $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{A}_{M,\eta}\cap(0,h)|=\eta$ so that

[TABLE]

Theorem 1.5**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. For each $M>0$, let $u^{*}_{M}$ be the optimal control to $(TP)^{M,y_{0}}$. Then the following conclusions are true:

(i) For each $(M_{1},M_{2})$ with $0<M_{1}\leq M_{2}$, there is $C\triangleq C(M_{1},M_{2},y_{0},r)>0$ so that

[TABLE]

where $u_{M,\delta}$ is any optimal control to $(TP)_{\delta}^{M,y_{0}}$. Moreover, $C$ depends on $M_{1}$, $M_{2}$, $y_{0}$ and $r$ continuously.

(ii) For each $M>0$ and $\eta\in(0,1)$, there is a measurable set $\mathcal{A}_{M,\eta}\subset(0,1)$ (depending also on $y_{0}$ and $r$) with $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{A}_{M,\eta}\cap(0,h)|=\eta$ so that for each $\delta\in\mathcal{A}_{M,\eta}$, there is an optimal control $\hat{u}_{M,\delta}$ to $(TP)_{\delta}^{M,y_{0}}$ so that

[TABLE]

for some positive constant $C_{M}\triangleq C_{M}(y_{0},r)$.

Several remarks on the main results are given in order.

• •

Theorem 1.3 and Theorem 1.4 present two facts. First, the error between $T_{\delta}(M,y_{0})$ and $T(M,y_{0})$ and the error between $u_{M}^{*}$ and $u^{*}_{M,\delta}$ have the order $1$ with respect to the sampling period $\delta$. Second, this order is optimal, because of the lower bound estimates (1.11) and (1.13), and because of the property that $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{A}_{M,\eta}\cap(0,h)|=\eta$ with any $\eta\in(0,1)$. Notice that when $\delta\in(0,1)\setminus\mathcal{A}_{M,\eta}$, (1.11) may not be true (see Theorem 6.2, as well as Remark 6.3).

• •

Theorem 1.5, as well as Theorem 1.4, presents two facts. First, in $(TP)_{\delta}^{M,y_{0}}$, the optimal control with the minimal norm differs from some of other optimal controls, from perspective of the order of the errors. Second, the order of the error between any optimal control of $(TP)_{\delta}^{M,y_{0}}$ and the optimal control to $(TP)^{M,y_{0}}$ is $1/2$, with respect to $\delta$. Moreover, this order is optimal in the sense (ii) of Theorem 1.5.

• •

Since we aim to approximate $u^{*}_{M}$ by $u^{*}_{M,\delta}$ and because the efficient domain of $u^{*}_{M}$ is $(0,T(M,y_{0}))$, we take the $L^{2}(0,T(M,y_{0});L^{2}(\Omega))$-norm in the estimates in Theorem 1.4 and Theorem 1.5.

• •

There have been many publications on optimal sampled-data control problems (with fixed ending time point). In [6] (see also [5]), the authors built up the Pontryagin maximum principle for some optimal sampled-data control problems. In [7], the authors showed that for some LQ problem, the optimal sampled-data control converges to the optimal distributed control as the sampling period tends to zero. In [33], the authors built up some error estimates between the optimal distributed control and the optimal sampled-data control for some periodic heat equations. About more works on sampled-data controls, we would like to mention [1, 3, 4, 8, 13, 14, 19, 20, 23, 32] and the references therein.

• •

There have been some literatures on the approximations of time optimal control problems for the parabolic equations. We refer to [15, 41] for semi-discrete finite element approximations, and [34, 43] for perturbations of equations. About more works on time optimal control problems, we would like to mention [2, 10, 11, 16, 17, 18, 21, 22, 25, 27, 30, 31, 35, 37, 38, 39, 40, 42, 44] and the references therein.

• •

About approximations of time optimal sampled-data controls, we have not found any literature in the past publications.

1.3 The strategy to get the main results

The strategy to prove the main theorems is as follows: We first introduce two norm optimal control problems which correspond to time optimal control problems $(TP)^{M,y_{0}}$ and $(TP)_{\delta}^{M,y_{0}}$ respectively; then get error estimates between the above two norm optimal control problems (in terms of $\delta$); finally, obtain the desired error estimates between $(TP)^{M,y_{0}}$ and $(TP)_{\delta}^{M,y_{0}}$ (in terms of $\delta$), through using connections between the time optimal control problems and the corresponding norm optimal control problems (see (iii) of Theorem 3.1 and Theorem 4.1, respectively).

To explain our strategy more clearly, we will introduce two norm optimal control problems. The first one corresponds to $(TP)^{M,y_{0}}$ and is as:

[TABLE]

where $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, $T>0$ and $y(\cdot;y_{0},v)$ is the solution of (1.5) with $u$ being replaced by the zero extension of $v$ over $\mathbb{R}^{+}$. The second one corresponds to $(TP)_{\delta}^{M,y_{0}}$ and is defined by

[TABLE]

where $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, $T>0$, $(\delta,k)\in\mathbb{R}^{+}\times\mathbb{N}^{+}$,

[TABLE]

and $y(\cdot;y_{0},v_{\delta})$ is the solution of (1.5) with $u$ being replaced by the zero extension of $v_{\delta}$ over $\mathbb{R}^{+}$.

Some concepts about the above two norm optimal control problems are given in the following definition:

Definition 1.6**.**

(i) In the problem $(NP)^{T,y_{0}}$, $N(T,y_{0})$ is called the optimal norm; $v\in L^{2}(0,T;L^{2}(\Omega))$ is called an admissible control if $y(T;y_{0},v)\in B_{r}(0)$; $v^{*}$ is called an optimal control if $y(T;y_{0},v^{*})\in B_{r}(0)$ and $\|v^{*}\|_{L^{2}(0,T;L^{2}(\Omega))}=N(T,y_{0})$.

(ii) In the problem $(NP)_{\delta}^{k\delta,y_{0}}$, $N_{\delta}(k\delta,y_{0})$ is called the optimal norm; $v_{\delta}\in L_{\delta}^{2}(0,k\delta;L^{2}(\Omega))$ is called an admissible control if $y(k\delta;y_{0},v_{\delta})\in B_{r}(0)$; and $v_{\delta}^{*}$ is called an optimal control if $y(k\delta;y_{0},v_{\delta}^{*})\in B_{r}(0)$ and $\|v_{\delta}^{*}\|_{L_{\delta}^{2}(0,k\delta;L^{2}(\Omega))}=N_{\delta}(k\delta,y_{0})$.

We mention that both $(NP)^{T,y_{0}}$ and $(NP)_{\delta}^{k\delta,y_{0}}$ have unique nonzero minimizers in $L^{2}(\Omega)$ (see Theorems 4.2-4.3).

Inspired by [9], we study the above two minimal norm control problems by two minimization problems. The first one corresponds to $(NP)^{T,y_{0}}$ and reads

[TABLE]

where $\varphi(\cdot;T,z)$ is the solution to the adjoint heat equation:

[TABLE]

The second minimization problem corresponds to $(NP)^{T,y_{0}}_{\delta}$ and is as:

[TABLE]

where $\overline{\varphi}_{\delta}(\cdot;k\delta,z)$ is defined by

[TABLE]

We mention that both $(JP)^{T,y_{0}}$ and $(JP)_{\delta}^{k\delta,y_{0}}$ have unique nonzero minimizers in $L^{2}(\Omega)$ (see Theorems 4.2-4.3).

We prove Theorem 1.3 by the following steps:

• (a)

Build up connections between $(TP)^{M,y_{0}}_{\delta}$ and $(NP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$ (see (iii) of Theorem 3.1); and connections between $(TP)^{M,y_{0}}$ and $(NP)^{T(M,y_{0}),y_{0}}$ (see Theorem 4.1).

• (b)

Obtain the lower and upper bounds for the derivative of the map $T\rightarrow N(T,y_{0})$ (see Theorem 5.2).

• (c)

Compute the error estimate between the map $T\rightarrow N(T,y_{0})$ and the map $k\delta\rightarrow N_{\delta}(k\delta,y_{0})$ (see Theorem 5.3).

• (d)

Get (i) of Theorem 1.3, with the aid of the above (a)-(c).

• (e)

By using the above (a)-(c) again, we build up sets $\mathcal{A}_{M,\eta}$ and obtain related properties in Theorem 5.4, which leads to (ii) of Theorem 1.3.

The steps to prove Theorem 1.4 and Theorem 1.5 are as follows:

• (1)

Build up connections between $(TP)^{M,y_{0}}_{\delta}$ and $(NP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$ (see (iii) of Theorem 3.1), and connections between $(TP)^{M,y_{0}}$ and $(NP)^{T(M,y_{0}),y_{0}}$ (see Theorem 4.1).

• (2)

With the aid of the connections obtained in (1), we can transfer the estimate in (i) of Theorem 1.4 into an estimate between optimal controls of $(NP)^{T(M,y_{0}),y_{0}}$ and $(NP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$.

• (3)

Find connections between $(NP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$ (or $(NP)^{T(M,y_{0}),y_{0}}$) and $(JP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$ (or $(JP)^{T(M,y_{0}),y_{0}}$) (see Theorem 4.3 and Theorem 4.2, respectively).

• (4)

Obtain the error estimate between the minimizers of $(JP)^{T(M,y_{0}),y_{0}}$ and $(JP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$.

• (5)

Using the connections obtained in (3) and the estimate obtained in (4), we get an error estimate between optimal controls of $(NP)^{T(M,y_{0}),y_{0}}$ and $(NP)^{T_{\delta}(M,y_{0}),y_{0}}_{\delta}$. This, along with results in (2), leads to the estimate in (i) of Theorem 1.4.

• (6)

Using connections obtained in (1) and (3), and using Theorem 5.4, we prove the estimate in (ii) of Theorem 1.4.

• (7)

Obtain the least order of the diameter of the set $\mathcal{O}_{M,\delta}$ (in $L^{2}(0,T(M,y_{0});L^{2}(\Omega)))$, in terms of $\delta$, (see Lemma 6.1). Here,

[TABLE]

• (8)

Derive the estimates in Theorem 1.5, with the aid of Lemma 6.1 and the estimates in Theorem 1.4.

We would like to give the following note:

• •

The above introduced strategy was used to study other properties of time optimal distributed control problems (see, for instance, [34] and [42]). It could be used to study numerical approximations of time optimal distributed control problems, via discrete time optimal control problems.

The rest of the paper is organized as follows: Section 2 shows a kind of approximate null controllability for the equation (1.7). Section 3 concerns with the existence and uniqueness of time optimal control problems. Section 4 provides some connections among time optimal control problems, norm optimal control problems and some minimization problems. Section 5 presents several auxiliary estimates. Section 6 proves the main results. Section 7 (Appendix) gives two lemmas. The first one presents an equivalence between controllability and observability in an abstract setting, which was taken from [36]. The second one gives an inequality which was quoted from [29]. Since both [36] and [29] have not appeared, we put them and their proofs in Appendix.

2 $L^{2}$-approximate null controllability with a cost

In this section, we present a kind of approximate null controllability for the sampled-data controlled equation (1.7). Such controllability will be defined in the next Definition 2.1 and will play a key role in getting some estimates in Section 5.

Definition 2.1**.**

(i) Let $(\delta,k)\in\mathbb{R}^{+}\times\mathbb{N}^{+}$. Equation (1.7) is said to have the $L^{2}$-approximate null controllability with a cost over $[0,k\delta]$, if for any $\varepsilon>0$, there is $C(\varepsilon,\delta,k)>0$ so that for each $y_{0}\in L^{2}(\Omega)$, there is $u^{y_{0}}_{\delta}\in L_{\delta}^{2}(0,k\delta;L^{2}(\Omega))$ (see (1.18)) satisfying that

[TABLE]

(ii) Equation (1.7) is said to have the $L^{2}$-approximate null controllability with a cost, if it has the $L^{2}$-approximate null controllability with a cost over $[0,k\delta]$, for each $(\delta,k)\in\mathbb{R}^{+}\times\mathbb{N}^{+}$.

To prove the $L^{2}$-approximate null controllability with a cost for Equation (1.7), we need some preliminaries. For each $f\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$ and $\delta>0$, we let

[TABLE]

Lemma 2.2**.**

For each $f,\,g\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$ and each $\delta>0$,

[TABLE]

Proof.

Arbitrarily fix $\delta>0$ and $f,\,g\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$. To prove (2.3), it suffices to show

[TABLE]

By (2.2), one can directly check that

[TABLE]

which leads to (2.4). This ends the proof of this lemma. ∎

The following interpolation inequality plays an important role in the proof of the $L^{2}$-approximate null controllability with a cost.

Lemma 2.3**.**

There exists $C\triangleq C(\Omega,\omega)>0$ so that when $0<S<T$,

[TABLE]

Proof.

Let $0<S<T$. Arbitrarily fix $z\in L^{2}(\Omega)$. We define a function $f^{z}$ over $\Omega$ by

[TABLE]

By [29, (iii) of Theorem 2.1], (Since the paper [29] has not appeared, we put [29, (iii) of Theorem 2.1] and its proof in Appendix, see Lemma 7.2.), there is $C\triangleq C(\Omega,\omega)>0$ so that

[TABLE]

Two facts are given in order: First, it follows from (2.6) that

[TABLE]

Second, write $\{\lambda_{j}\}_{j=1}^{\infty}$ for the family of all eigenvalues of $-\Delta$ with the zero Dirichlet boundary condition so that $\lambda_{1}<\lambda_{2}\leq\cdots$. Let $\{e_{j}\}_{j=1}^{\infty}$ be the family of the corresponding normalized eigenvectors. Let $z=\sum_{j=1}^{\infty}z_{j}e_{j}$ for some $\{z_{j}\}_{j=1}^{\infty}\in l^{2}$. Then it follows that

[TABLE]

Since $\frac{1}{S}\int_{0}^{S}e^{\lambda_{j}t}\,\mathrm{d}t\geq 1$ for each $j\in\mathbb{N}^{+}$, it follows from (2.6) and the above equality that

[TABLE]

Finally, the facts (2.8) and (2.9), along with (2.7), lead to (2.5). This ends the proof. ∎

The next Theorem 2.4 contains the main results of this section. The conclusion (iii) in Theorem 2.4 will play an important role in our further studies.

Theorem 2.4**.**

The following conclusions are true:

(i) Equation (1.7) has the $L^{2}$-approximate null controllability with a cost if and only if given $\varepsilon>0$, $\delta>0$ and $k\in\mathbb{N}^{+}$, there is $C(\varepsilon,\delta,k)>0$ (which also depends on $\Omega$ and $\omega$) so that

[TABLE]

where $\overline{\varphi}_{\delta}(\cdot;k\delta,z)$ is given by (1.22).

(ii) Given $\delta>0$ and $k\geq 2$, Equation (1.7) has the $L^{2}$-approximate null controllability with a cost over $[0,k\delta]$.

(iii) Given $\varepsilon>0$, $\delta>0$ and $k\geq 2$, the constants $C(\varepsilon,\delta,k)$ in (2.10) and (2.1) can be taken as

[TABLE]

Proof.

We first prove the conclusion (i). Arbitrarily fix $\delta>0$, $k\in\mathbb{N}^{+}$ and $\varepsilon>0$. We will put our problems under the framework of [36, Lemma 5.1] in the following manner: Let $X\triangleq L^{2}(\Omega)$, $Y\triangleq L^{2}_{\delta}(0,k\delta;L^{2}(\Omega))$ and $Z\triangleq L^{2}(\Omega)$. Define operators $\mathcal{R}:Z\rightarrow X$ and $\mathcal{O}:Z\rightarrow Y$ by

[TABLE]

One can directly check that $\mathcal{R}^{*}:X^{*}\rightarrow Z^{*}$ and $\mathcal{O}^{*}:Y^{*}\rightarrow Z^{*}$ are given respectively by

[TABLE]

From these, Definition 2.1 and (2.10), we can apply [36, Lemma 5.1] to get the conclusion (i) of Theorem 2.4. (Since the paper [36] has not appeared, we put [36, Lemma 5.1] and its proof in Appendix, see Lemma 7.1.)

We next prove the conclusions (ii) and (iii). Arbitrarily fix $\varepsilon>0$, $\delta>0$ and $k\geq 2$. By the conclusion (i), we find that it suffices to show (2.10) with the triplet $(\varepsilon,\delta,k)$. To this end, we use (2.5) (where $T=k\delta$ and $S=[k/2]\delta$, with $[k/2]$ the integer so that $k/2-1<[k/2]\leq k/2$) to get that for each $z\in L^{2}(\Omega)$,

[TABLE]

where $C$ is given by (2.5). Then by Young’s inequality, we find that for each $z\in L^{2}(\Omega)$,

[TABLE]

Two observations are given in order: First, it follows from (1.22) that for each $z\in L^{2}(\Omega)$,

[TABLE]

Second, since

[TABLE]

one can directly check that

[TABLE]

Finally, from (2.12), (2.13) and (2.14), we get (2.10), with $C(\varepsilon,\delta,k)$ given by (2.11), where $C(\Omega,\omega)$ may differ from that in (2.14). This proves (ii), as well as (iii).

In summary, we end the proof of Theorem 2.4. ∎

3 Existence and uniqueness of optimal controls

In this section, we will prove that for each $(M,y_{0})$, $(TP)^{M,y_{0}}$ has the unique optimal control, while for some $(M,y_{0},\delta)$, $(TP)^{M,y_{0}}_{\delta}$ has infinitely many optimal controls. The later may cause difficulties in our studies. Fortunately, we observe that the optimal control with the minimal norm to $(TP)^{M,y_{0}}_{\delta}$ (see Definition 1.2) is unique. The first main theorem in this section is stated in the next Theorem 3.1. It deserves mentioning what follows: The conclusion (iii) of Theorem 3.1 should belong to the materials in the next section. The reason that we put it here is that we will use it in the proof of the non-uniqueness of optimal controls to $(TP)_{\delta}^{M,y_{0}}$. More precisely, in the proof of Lemma 3.4, we will use it.

Theorem 3.1**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Let $M\geq 0$. The following conclusions are true:

(i) The problem $(TP)^{M,y_{0}}$ has a unique optimal control.

(ii) For each $\delta>0$, $(TP)_{\delta}^{M,y_{0}}$ has a unique optimal control with the minimal norm.

(iii) Let $u^{*}_{\delta}$ (with $\delta>0$) be the optimal control with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$. Then $u^{*}_{\delta}|_{(0,T_{\delta}(M,y_{0}))}$ (the restriction of $u^{*}_{\delta}$ over $(0,T_{\delta}(M,y_{0}))$) is an optimal control to $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$ and the $L^{2}(0,T_{\delta}(M,y_{0});L^{2}(\Omega))$-norm of $u^{*}_{\delta}$ is $N_{\delta}(T_{\delta}(M,y_{0}),y_{0})$.

Proof.

Arbitrarily fix $M\geq 0$ and $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. We will prove conclusions (i), (ii) and (iii) one by one.

(i) Because $y(t;y_{0},0)\rightarrow 0$ as $t\rightarrow\infty$, the null control is an admissible control to $(TP)^{M,y_{0}}$, which implies that $(TP)^{M,y_{0}}$ has an admissible control. Then by the standard way as that used in the proof of [10, Lemma 1.1], one can show that $(TP)^{M,y_{0}}$ has an optimal control.

To show the uniqueness of the optimal control to $(TP)^{M,y_{0}}$, we first notice that each optimal control $u^{*}$ to $(TP)^{M,y_{0}}$ has the property that

[TABLE]

(The property (3.1) can be proved by the same way as that used to show [16, Lemma 4.3].) Next, we notice that if $u^{*}_{1}$ and $u^{*}_{2}$ are optimal controls to $(TP)^{M,y_{0}}$, then $(u^{*}_{1}+u^{*}_{2})/2$ is also an optimal control to $(TP)^{M,y_{0}}$. From this, (3.1) and the parallelogram law in $L^{2}(0,T(M,y_{0});L^{2}(\Omega))$, we can easily use the contradiction argument to get the uniqueness. This ends the proof of the conclusion (i).

(ii) Arbitrarily fix $\delta>0$. We first show that $(TP)_{\delta}^{M,y_{0}}$ has an optimal control. Indeed, since the null control is clearly an admissible control to $(TP)^{M,y_{0}}_{\delta}$, it follows by the definition of $T_{\delta}(M,y_{0})$ (see (1.8)) that there exists $\hat{k}\in\mathbb{N}^{+}$ so that

[TABLE]

Meanwhile, since $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, by the definition of the infimum in (1.8), we see that there is $k_{0}\in\mathbb{N}^{+}$ and $u_{\delta}^{0}\in L^{2}_{\delta}(\mathbb{R}^{+};L^{2}(\Omega))$ so that

[TABLE]

[TABLE]

From (3.2) and (3.3), we find that $\hat{k}\delta\leq k_{0}\delta\leq\hat{k}\delta+\delta/2$, which leads to that $k_{0}=\hat{k}$. This, along with (3.2) and (3.4), yields that $T_{\delta}(M,y_{0})=k_{0}\delta\in(0,\infty)$, which, together with (3.4), implies that $u_{\delta}^{0}$ is an optimal control to $(TP)_{\delta}^{M,y_{0}}$.

Next, we will prove that $(TP)_{\delta}^{M,y_{0}}$ has a unique optimal control with the minimal norm. Indeed, since $L^{2}_{\delta}(\mathbb{R}^{+};L^{2}(\Omega))$ is a closed subspace of $L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$, by Definition 1.2, one can use a standard way (i.e., taking a minimization sequence) to show the existence of the optimal control with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$. To show the uniqueness, we let $u_{1}$ and $u_{2}$ be two optimal controls with the minimal norm. By Definition 1.2, one can easily check that $(u_{1}+u_{2})/2$ is also an optimal control with the minimal norm to $(TP)^{M,y_{0}}_{\delta}$. By making use of Definition 1.2 again, we find that

[TABLE]

These, along with the parallelogram law for $L^{2}(0,T_{\delta}(M,y_{0});L^{2}(\Omega))$, yields that

[TABLE]

So $(TP)_{\delta}^{M,y_{0}}$ has a unique optimal control with the minimal norm.

(iii) Let $u^{*}_{\delta}$ be the optimal control with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$. We will show that $u^{*}_{\delta}|_{(0,T_{\delta}(M,y_{0})}$ is an optimal control to $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$. Indeed, we have that

[TABLE]

from which, one can easily check that $u^{*}_{\delta}|_{(0,T_{\delta}(M,y_{0}))}$ is an admissible control to $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$. Then by the optimality of $N_{\delta}(T_{\delta}(M,y_{0}),y_{0})$ and the second inequality in (3.5), we see that

[TABLE]

Meanwhile, since $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$ has an admissible control, we can use a standard argument (see for instance the proof of [10, Lemma 1.1]) to show that $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$ has an optimal control $v^{*}_{\delta}$. Write $\widetilde{v}^{*}_{\delta}$ for the zero extension of $v^{*}_{\delta}$ over $\mathbb{R}^{+}$. Then we have that

[TABLE]

From (3.7) and (3.6), it follows that $\widetilde{v}^{*}_{\delta}$ is an optimal control to $(TP)_{\delta}^{M,y_{0}}$. Since $u^{*}_{\delta}$ is the optimal control with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$, we see from (3.6), (ii) of Definition 1.2 and the second equality in (3.7) that

[TABLE]

The above, together with the first conclusion in (3.5), implies that $u^{*}_{\delta}|_{(0,T_{\delta}(M,y_{0}))}$ is an optimal control to $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$ and that

[TABLE]

In summary, we end the proof of Theorem 1.5. ∎

The next theorem concerns with the non-uniqueness of optimal controls to $(TP)_{\delta}^{M,y_{0}}$.

Theorem 3.2**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then there are sequences $\{M_{n}\}$ dense in $\mathbb{R}^{+}$ and $\{\delta_{n}\}\subset\mathbb{R}^{+}$, with $\lim_{n\rightarrow\infty}\delta_{n}=0$, so that for each $n$, the problem $(TP)_{\delta_{n}}^{M_{n},y_{0}}$ has infinitely many different optimal controls.

To prove Theorem 3.2, we need two lemmas.

Lemma 3.3**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then for each $(M,\delta)\in\mathbb{R}^{+}\times\mathbb{R}^{+}$, with $2\delta\leq T_{\delta}(M,y_{0})<\infty$, it stands that

[TABLE]

Proof.

Let $(M,\delta)\in\mathbb{R}^{+}\times\mathbb{R}^{+}$ so that $2\delta\leq T_{\delta}(M,y_{0})<\infty$. Then by (1.8), we see that

[TABLE]

Thus, (3.8) is equivalent to the following inequality:

[TABLE]

To prove (3.10), we let $u_{\delta}^{1}$ be an optimal control to $(TP)_{\delta}^{M,y_{0}}$. Then we have that

[TABLE]

According to (3.11) and (3.9), $u_{\delta}^{1}|_{(0,\hat{k}\delta)}$ is an admissible control to $(NP)_{\delta}^{\hat{k}\delta,y_{0}}$. Then by the optimality of $N_{\delta}(\hat{k}\delta,y_{0})$ and the first inequality in (3.11), we get that

[TABLE]

which leads to the first inequality in (3.10).

We now show the second inequality in (3.10). By contradiction, we suppose that

[TABLE]

Then we would obtain from (3.12) that $(NP)_{\delta}^{(\hat{k}-1)\delta,y_{0}}$ has an admissible control, since $M<\infty$. Thus, by a standard way (see for instance the proof of [10, Lemma 1.1]), one can prove that $(NP)_{\delta}^{(\hat{k}-1)\delta,y_{0}}$ has an optimal control $v_{\delta}^{1}$. Hence,

[TABLE]

Write $\widetilde{v}_{\delta}^{1}$ for the zero extension of $v_{\delta}^{1}$ over $\mathbb{R}^{+}$. From (3.13) and (3.12), we find that $\widetilde{v}_{\delta}^{1}$ is an admissible control to $(TP)_{\delta}^{M,y_{0}}$. Then by the optimality of $T_{\delta}(M,y_{0})$, we get that $T_{\delta}(M,y_{0})\leq(\hat{k}-1)\delta$, which contradicts (3.9). Thus, the second inequality in (3.10) is true. We end the proof of this lemma. ∎

Lemma 3.4**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then for each $M>0$ and $N>0$, there exists an integer $n\geq N$ so that $2/2^{n}\leq T_{1/2^{n}}(M,y_{0})<\infty$.

Proof.

Arbitrarily fix $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. It is clear that $T_{1/2^{n}}(M,y_{0})<\infty$ for all $M>0$ and $n\in\mathbb{N}^{+}$. Thus, we only need to show that for any $M>0$ and $N>0$, $2/2^{n}\leq T_{1/2^{n}}(M,y_{0})$, when $n\geq N$. By contradiction, suppose that it were not true. Then there would be $M>0$ and $N>0$ so that

[TABLE]

Let $u^{*}_{n}$, with $n\geq N$, be an optimal control to $(TP)_{1/2^{n}}^{M,y_{0}}$ (see (ii) of Theorem 3.1). Then we have that

[TABLE]

By the last inequality in (3.15), Hölder’s inequality and (3.14), we can easily check that

[TABLE]

This, along with (3.14) and the first conclusion in (3.15), yields that

[TABLE]

which contradicts the assumption that $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. This ends the proof. ∎

We are now on the position to prove Theorem 3.2.

Proof of Theorem 3.2.

Choose a sequence $\{M_{n}\}_{n=1}^{\infty}$ dense in $\mathbb{R}^{+}$ so that

[TABLE]

By Lemma 3.4, there exists an increasing subsequence $\{k_{n}\}_{n=1}^{\infty}$ (in $\mathbb{N}^{+}$), with $\lim_{n\rightarrow\infty}k_{n}=\infty$, so that

[TABLE]

Write $\delta_{n}\triangleq 1/2^{k_{n}}$, $n\in\mathbb{N}^{+}$. Then, by (3.17), we can apply Lemma 3.3 to get that

[TABLE]

This, along with (3.16), yields that

[TABLE]

The key to show Theorem 3.2 is to claim that for each $n\in\mathbb{N}^{+}$, $(TP)^{M_{n},y_{0}}_{\delta_{n}}$ has at least two different optimal controls. By contradiction, we suppose that for some $n_{0}\in\mathbb{N}^{+}$, $(TP)^{M_{n_{0}},y_{0}}_{\delta_{n_{0}}}$ had a unique optimal control. To get a contradiction, we define two convex subsets in $L^{2}(\Omega)$ as follows:

[TABLE]

We first show that

[TABLE]

i.e., $A_{n_{0}}\cap B_{r}(0)$ contains only one element. In fact, by (ii) and (iii) of Theorem 3.1, the optimal control with the minimal norm $u^{*}_{\delta_{n_{0}}}$ to $(TP)^{M_{n_{0}},y_{0}}_{\delta_{n_{0}}}$ satisfies that

[TABLE]

These imply that $A_{n_{0}}\cap B_{r}(0)\neq\emptyset$. We next show that $A_{n_{0}}\cap B_{r}(0)$ contains only one element. Suppose, by contradiction, that it contained two different elements $y_{1}$ and $y_{2}$. Then by the definition of $A_{n_{0}}$, there would be two different controls $u_{1}$ and $u_{2}$ so that

[TABLE]

[TABLE]

Since $y_{1},y_{2}\in B_{r}(0)$, we have that ${(y_{1}+y_{2})}/{2}\in B_{r}(0)$. From this (3.20) and (3.21), one can easily check that

[TABLE]

Meanwhile, since $u_{1}\neq u_{2}$, by the second inequality in (3.20) and the second inequality in (3.21), using the parallelogram law, we find that

[TABLE]

which, together with (3.18), indicates that

[TABLE]

From this and (3.22), we see that $(u_{1}+u_{2})/2$ is an optimal control to $(TP)^{M_{n_{0}},y_{0}}_{\delta_{n_{0}}}$. This, along with Definition 1.2 and the conclusions (ii) and (iii) of Theorem 3.1, yields that

[TABLE]

which contradicts (3.23). Hence, (3.19) is true.

Next, by the definitions of $A_{n_{0}}$ and $B_{n_{0}}$, one can easily check that each element of $(A_{n_{0}}+B_{n_{0}})\cap B_{r}(0)$ can be expressed as:

[TABLE]

Since it was assumed that $(TP)^{M_{n_{0}},y_{0}}_{\delta_{n_{0}}}$ had a unique optimal control, it follows from (3.24) that $(A_{n_{0}}+B_{n_{0}})\cap B_{r}(0)$ contains only one element. This, along with (3.19), yields that

[TABLE]

By (3.25), we can apply the Hahn-Banach separation theorem to find $\eta^{*}\in L^{2}(\Omega)$, with $\|\eta^{*}\|=r>0$, so that

[TABLE]

This, along with (3.25), yields that

[TABLE]

From now on and throughout the proof of Theorem 3.2, we simply write $T_{\delta_{n_{0}}}$ for $T_{\delta_{n_{0}}}(M_{n_{0}},y_{0})$; simply write $\varphi(\cdot)$ and $\overline{\varphi}_{\delta_{n_{0}}}(\cdot)$ for $\varphi(\cdot;T_{\delta_{n_{0}}},\eta^{*})$ (see (1.20)) and $\overline{\varphi}_{\delta_{n_{0}}}(t;T_{\delta_{n_{0}}},\eta^{*})$ (see (1.22)), respectively.

Arbitrarily fix $u_{\delta_{n_{0}}}\in L^{2}_{\delta_{n_{0}}}(\mathbb{R}^{+};L^{2}(\Omega))$. Three facts are given in order. Fact one: Since $M_{n_{0}}>N_{\delta_{n_{0}}}(T_{\delta_{n_{0}}},y_{0})$ (see (3.18)), it follows from the definition of $B_{n_{0}}$ that

[TABLE]

This, along with (3.26), yields that

[TABLE]

Fact two: One can directly check that

[TABLE]

Fact three: we have that

[TABLE]

The proof of (3.29) is as follows: Let $f=u_{\delta_{n_{0}}}$ and let $g$ be the zero extension of $\chi_{\omega}\varphi(\cdot)$ over $\mathbb{R}^{+}$. Since $u_{\delta_{n_{0}}}\in L^{2}_{\delta_{n_{0}}}(\mathbb{R}^{+};L^{2}(\Omega))$, it follows by (1.6), (2.2) and (1.22) that $\bar{f}_{\delta_{n_{0}}}=u_{\delta_{n_{0}}}$ and $\bar{g}_{\delta_{n_{0}}}(\cdot)=\overline{\varphi}_{\delta_{n_{0}}}(\cdot)$ (where $\overline{\varphi}_{\delta_{n_{0}}}(\cdot)$ is treated as its zero extension over $\mathbb{R}^{+}$). Then, by Lemma 2.2, we obtain (3.29).

Now, from facts (3.27), (3.28) and (3.29), we see that

[TABLE]

Since $u_{\delta_{n_{0}}}$ was arbitrarily taken from $L^{2}_{\delta_{n_{0}}}(\mathbb{R}^{+};L^{2}(\Omega))$, the above inequality implies that

[TABLE]

Since $T_{\delta_{n_{0}}}\geq 2\delta_{n_{0}}$ (see (3.17)), we apply (3.30) and Lemma 2.3 (where $T=T_{\delta_{n_{0}}}$ and $S=\delta_{n_{0}}$) to get that $\varphi(0)=0$ in $L^{2}(\Omega)$. Then from the backward uniqueness property for the heat equation (see, for instance, [26]), we deduce that $\eta^{*}=0$. This leads to a contradiction. Hence, we ends the proof of the key claim: For each $n\in\mathbb{N}^{+}$, $(TP)^{M_{n},y_{0}}_{\delta_{n}}$ has at least two different optimal controls.

Finally, we observe that any convex combination of optimal controls to $(TP)^{M_{n},y_{0}}_{\delta_{n}}$ (with $n\in\mathbb{N}^{+}$) is still an optimal control to $(TP)^{M_{n},y_{0}}_{\delta_{n}}$. Therefore, for each $n\in\mathbb{N}^{+}$, $(TP)^{M_{n},y_{0}}_{\delta_{n}}$ has infinitely many different optimal controls. This ends the proof of Theorem 3.2. ∎

4 Connections among different problems

This section presents connections among $(TP)^{M,y_{0}}_{\delta}$, $(NP)^{k\delta,y_{0}}_{\delta}$ and $(JP)_{\delta}^{k\delta,y_{0}}$ (and among $(TP)^{M,y_{0}}$, $(NP)^{T,y_{0}}$ and $(JP)^{T,y_{0}}$). We define, for each $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$,

[TABLE]

[TABLE]

We mention that $0<T^{*}_{y_{0}}<\infty$ for each $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ (because the semigroup $\{e^{\Delta t}\}_{t\geq 0}$ has the exponential decay).

4.1 Connections between time optimal control problems and norm optimal control problems

We first present the following equivalence theorem. We will omit its proof, because it can be proved by the same way as one of proofs of [34, Proposition 4.1], [43, Proposition 3.1] and [42, Theorem 1.1 and Theorem 2.1].

Theorem 4.1**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Let $T_{y_{0}}^{*}$ be given by (4.1). Then the following conclusions are true:

(i) The function $N(\cdot,y_{0})$ is strictly decreasing and continuous from $(0,T^{*}_{y_{0}})$ onto $(0,+\infty)$. Moreover, $\lim_{T\rightarrow{T^{*}_{y_{0}}}^{-}}N(T,y_{0})=0$.

(ii) When $M>0$ and $T\in(0,T^{*}_{y_{0}})$, $N(T(M,y_{0}),y_{0})=M$ and $T(N(T,y_{0}),y_{0})=T$.

(iii) The function $T(\cdot,y_{0})$ is strictly decreasing and continuous from $(0,+\infty)$ onto $(0,T^{*}_{y_{0}})$.

(iv) For each $M>0$, the optimal control to $(TP)^{M,y_{0}}$, when restricted on $(0,T(M,y_{0}))$, is the optimal control to $(NP)^{T(M,y_{0}),y_{0}}$. For each $T\in(0,T^{*}_{y_{0}})$, the zero extension of the optimal control to $(NP)^{T(M,y_{0}),y_{0}}$ is the optimal control to $(TP)^{M,y_{0}}$.

We next recall (iii) of Theorem 3.1 for the connections between $(TP)_{\delta}^{M,y_{0}}$ and $(NP)_{\delta}^{T_{\delta}(M,y_{0}),y_{0}}$.

4.2 Connections between norm optimal control problems and the minimization problems

The first theorem of this subsection concerns with connections between problems $(NP)^{T,y_{0}}$ and $(JP)^{T,y_{0}}$ (given by (1.19)). Its proof can be done by the same methods as those in the proofs of of Lemma 3.5 and Proposition 3.6 in [34]. We omit it here.

Theorem 4.2**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and let $T\in(0,T^{*}_{y_{0}})$, with $T_{y_{0}}^{*}$ given by (4.1). Then the following conclusions are true:

(i) The functional $J^{T,y_{0}}$ has a unique nonzero minimizer $z^{*}$ in $L^{2}(\Omega)$.

(ii) Problem $(NP)^{T,y_{0}}$ has a unique optimal control $v^{*}$, which satisfies that

[TABLE]

and that

[TABLE]

(iii) It holds that $V(T,y_{0})=-\frac{1}{2}N(T,y_{0})^{2}=-\frac{1}{2}\|\chi_{\omega}\varphi(\cdot;T,z^{*})\|_{L^{2}(0,T;L^{2}(\Omega))}^{2}$.

The next theorem deals with connections between $(NP)^{k\delta,y_{0}}_{\delta}$ (given by (1.17)) and $(JP)^{k\delta,y_{0}}_{\delta}$ (given by (1.21)). Recall (1.22) for the definition of $\overline{\varphi}(\cdot;k\delta,z)$.

Theorem 4.3**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Let $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$ (given (4.2)). Then the following conclusions are true:

(i) The functional $J_{\delta}^{k\delta,y_{0}}$ has a unique minimizer $z_{\delta}^{*}$ in $L^{2}(\Omega)$. Moreover, $z_{\delta}^{*}\neq 0$ and

[TABLE]

(ii) Problem $(NP)_{\delta}^{k\delta,y_{0}}$ has a unique optimal control $v_{\delta}^{*}$, which verifies that

[TABLE]

(where $z_{\delta}^{*}$ is the minimizer of $J_{\delta}^{k\delta,y_{0}}(\cdot)$) and that

[TABLE]

(iii) $V_{\delta}(k\delta,y_{0})=-\frac{1}{2}N_{\delta}(k\delta,y_{0})^{2}=-\frac{1}{2}\|\chi_{\omega}\overline{\varphi}_{\delta}(\cdot;k\delta,z_{\delta}^{*})\|_{L^{2}(0,k\delta;L^{2}(\Omega))}^{2}$.

Proof.

(i) First of all, we show the existence of minimizers of $(JP)_{\delta}^{k\delta,y_{0}}$. Indeed, by (1.21), one can easily see that $J_{\delta}^{k\delta,y_{0}}$ is continuous and convex over $L^{2}(\Omega)$. We now show its coercivity. Since $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$, we have that $k\geq 2$ (see (4.2)). Thus, we can apply Theorem 2.4 to see that both (2.10) and (2.11) are true. By taking \varepsilon=\big{(}\frac{r}{2\|y_{0}\|}\big{)}^{2} in (2.10), we find that for each $z\in L^{2}(\Omega)$,

[TABLE]

where $C\triangleq C(\Omega,\omega)$ is given by (2.11). The above yields that for each $z\in L^{2}(\Omega)$,

[TABLE]

From this and (1.21), one can easily check that

[TABLE]

which leads to the coercivity of $J_{\delta}^{k\delta,y_{0}}$ over $L^{2}(\Omega)$. Hence, $J_{\delta}^{k\delta,y_{0}}$ has at least one minimizer in $L^{2}(\Omega)$.

Next, we claim that [math] is not a minimizer of $J_{\delta}^{k\delta,y_{0}}$. By contradiction, suppose that it were not true. Then we would find from (1.21) that for all $z\in L^{2}(\Omega)$ and $\varepsilon>0$,

[TABLE]

Sending $\varepsilon$ to [math] in the above leads to that

[TABLE]

This yields that

[TABLE]

Since $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, the above, along with (4.1), indicates that $k\delta\geq T^{*}_{y_{0}}$, which contradicts the assumption that $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$ (given by (4.2)). Thus, [math] is not a minimizer of $J_{\delta}^{k\delta,y_{0}}$.

We now show the uniqueness of the minimizer of $J_{\delta}^{k\delta,y_{0}}$. To this end, we claim that the first term on the right hand side of (1.21) is strictly convex. When this claim is proved, it follows from (1.21) that $J_{\delta}^{k\delta,y_{0}}$ is strictly convex over $L^{2}(\Omega)$. So its minimizer is unique.

To show the above claim, we first observe from (1.22) that

[TABLE]

By this, we see that the first term on the right hand side of (1.21) is convex. Next, we suppose, by contradiction, that this term were not strictly convex. Then, by the convexity of this term, there would be $\hat{\lambda}\in(0,1)$ and $z_{1},\,z_{2}\in L^{2}(\Omega)$, with $z_{1}\neq z_{2}$, so that

[TABLE]

which, along with (4.9), yields that for each $t\in(0,k\delta)$,

[TABLE]

From this and the strict convexity of $\|\cdot\|^{2}$, we see that for each $t\in(0,k\delta)$,

[TABLE]

Notice that $k\geq 2$. Thus, we can apply Lemma 2.3 (where $S=(k-1)\delta$, $T=k\delta$ and $z=z_{1}-z_{2}$), and use (4.10) to obtain that $\varphi(0;k\delta,z_{1}-z_{2})=0$. This, together with the backward uniqueness of the heat equation, yields that $z_{1}=z_{2}$ in $L^{2}(\Omega)$, which leads to a contradiction. Hence, the first term on the right hand side of (1.21) is strictly convex.

In summary, conclude that $J_{\delta}^{k\delta,y_{0}}$ has a unique minimizer $z_{\delta}^{*}\neq 0$.

Finally, we prove that the minimizer $z_{\delta}^{*}$ satisfies (4.5). By contradiction, suppose that it were not true. Then we would have that

[TABLE]

Since $\overline{\varphi}_{\delta}(\cdot;k\delta,z_{\delta}^{*})$ is a piece-wise constant function from $(0,k\delta]$ to $L^{2}(\Omega)$ (see (1.22)), it follows from (4.11) that

[TABLE]

By (4.12), we can apply Lemma 2.3 (where $T=(k+1-i_{0})\delta$, $S=\delta$ and $z=e^{\Delta(i_{0}-1)\delta)}z_{\delta}^{*}$) to get that

[TABLE]

This, along with the backward uniqueness for the heat equation, yields that $z_{\delta}^{*}=0$ in $L^{2}(\Omega)$, which leads to a contradiction. Therefore, (4.5) holds. This ends the proof of the conclusion (i) of Theorem 4.3.

(ii) Let $z_{\delta}^{*}$ be the minimizer of $J_{\delta}^{k\delta,y_{0}}$. Let $v_{\delta}^{*}$ be given by (4.6). It suffices to show that $v_{\delta}^{*}$ is the unique optimal control to $(NP)_{\delta}^{k\delta,y_{0}}$ and satisfies (4.7). From now on and throughout the proof of Theorem 4.3, we simply write $\varphi(\cdot)$ and $\overline{\varphi}_{\delta}(\cdot)$ for $\varphi(\cdot;k\delta,z_{\delta}^{*})$ and $\overline{\varphi}_{\delta}(\cdot;k\delta,z_{\delta}^{*})$; and simply write $L^{2}(0,k\delta)$ for $L^{2}(0,k\delta;L^{2}(\Omega))$.

We first show that $v_{\delta}^{*}$ is an admissible control to $(NP)_{\delta}^{k\delta,y_{0}}$ and satisfies (4.7). By (1.21), one can easily check that the Euler-Lagrange equation associated with the minimizer $z_{\delta}^{*}$ is as follows:

[TABLE]

We claim that for each $z\in L^{2}(\Omega)$,

[TABLE]

To this end, we arbitrarily fix $z\in L^{2}(\Omega)$. Let $f(\cdot)$ and $g(\cdot)$ be the zero extensions of $\chi_{\omega}\varphi(\cdot)$ and $\chi_{\omega}\varphi(\cdot;k\delta,z)$ over $\mathbb{R}^{+}$. Then by (2.2) and (1.22), we see that

[TABLE]

where $\overline{\varphi}_{\delta}(\cdot)$ and $\overline{\varphi}_{\delta}(\cdot;k\delta,z)$ are treated as their zero extensions over $\mathbb{R}^{+}$. Then by Lemma 2.2, we have that

[TABLE]

which leads to (4.14). Now, from (4.13) and (4.14), it follows that for each $z\in L^{2}(\Omega)$,

[TABLE]

This, along with (4.13), yields that

[TABLE]

From (4.15), $v_{\delta}^{*}$ is an admissible control to $(NP)_{\delta}^{k\delta,y_{0}}$, and satisfies (4.7).

We next prove that $v_{\delta}^{*}$ is an optimal control to $(NP)_{\delta}^{k\delta,y_{0}}$. To this end, we arbitrarily fix an admissible control $v_{\delta}$ to $(NP)_{\delta}^{k\delta,y_{0}}$. Then we have that $\|y(k\delta;y_{0},v_{\delta})\|\leq r$. This, together with (4.15), implies that

[TABLE]

Meanwhile, by Lemma 2.2 (where $(f,g)$ are taken as the zero extensions of $(v^{*}_{\delta},\chi_{\omega}\varphi)$ and $(v_{\delta},\chi_{\omega}\varphi)$ over $\mathbb{R}^{+}$ respectively), and by (2.2) and (1.22), one can easily verify that

[TABLE]

Since $v_{\delta}$ and $v^{*}_{\delta}$ are piece-wise constant functions (see (1.18) and (1.6)), it follows from (4.6), (4.17) and (4.16) that

[TABLE]

This, along with (4.5), yields that $\|v_{\delta}^{*}\|_{L^{2}(0,k\delta)}\leq\|v_{\delta}\|_{L^{2}(0,k\delta)}$. Because $v_{\delta}$ is an arbitrarily fixed admissible control to $(NP)_{\delta}^{k\delta,y_{0}}$, we see that $v_{\delta}^{*}$ is an optimal control to $(NP)_{\delta}^{k\delta,y_{0}}$.

Finally, we prove the uniqueness of the optimal control to $(NP)_{\delta}^{k\delta,y_{0}}$. By contradiction, we suppose that $(NP)_{\delta}^{k\delta,y_{0}}$ had two different optimal controls $v^{*}_{\delta,1}$ and $v^{*}_{\delta,2}$. Then one could easily check that $(v^{*}_{\delta,1}+v^{*}_{\delta,2})/2$ is still an optimal control. Since $v^{*}_{\delta,1}\neq v^{*}_{\delta,2}$, we can use the parallelogram law to get that

[TABLE]

which contradicts the optimality of $N_{\delta}(k\delta,y_{0})$ to $(NP)_{\delta}^{k\delta,y_{0}}$. This proves the conclusion (ii) of Theorem 4.3.

(iii) Taking $z=z_{\delta}^{*}$ in (4.13) leads to that

[TABLE]

Since $z_{\delta}^{*}$ is the minimizer of $J_{\delta}^{k\delta,y_{0}}$, the above equality, along with (1.21), indicates that

[TABLE]

Meanwhile, from (ii) of Theorem 4.3, we see that

[TABLE]

This, along with (4.18) leads to the conclusion (iii) of Theorem 4.3.

In summary, we end the proof of Theorem 4.3. ∎

5 Several auxiliary estimates

This section presents several estimates, as well as properties, on minimizers (of $J_{\delta}^{k\delta,y_{0}}$ and $J^{T,y_{0}}$), the minimal norm functions and the minimal time functions. These estimate will play important roles in the proofs of the main theorems.

5.1 Some estimates on minimizers

The following theorem concerns with the $H_{0}^{1}(\Omega)$-estimates on the minimizers of the functionals $J_{\delta}^{k\delta,y_{0}}$ and $J^{T,y_{0}}$.

Theorem 5.1**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Let $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$ (given by (4.2)) and $0<T<T^{*}_{y_{0}}$ (given by (4.1)). Write $z_{\delta}^{*}$ and $z^{*}$ for the minimizers of $J_{\delta}^{k\delta,y_{0}}$ and $J^{T,y_{0}}$, respectively. Then the following conclusions are true:

(i) There is a positive constant $C_{1}\triangleq C_{1}(\Omega,\omega)$ so that

[TABLE]

(ii) There is a positive constant $C_{2}\triangleq C_{2}(\Omega,\omega)$ so that

[TABLE]

Proof.

Throughout the proof, $C(\Omega,\omega)$ stands for a positive constant depending only on $\Omega$ and $\omega$. It may vary in different contexts.

(i) We begin with proving (5.1). From (4.8), we find that

[TABLE]

Since $z_{\delta}^{*}$ is the minimizer of $J_{\delta}^{k\delta,y_{0}}$, the above inequality, along with (1.21), implies that

[TABLE]

which leads to (5.1).

To show (5.2), we need two estimates related to the optimal control $u_{\delta}^{*}$ of $(NP)_{\delta}^{k\delta,y_{0}}$. We first claim that

[TABLE]

Indeed, since $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$, we have that $k\geq 2$. Thus, by (ii) of Theorem 2.4, Equation (1.7) has the $L^{2}$-approximate null controllability with a cost. From this, Definition 2.1 (see (2.1)), and (iii) of Theorem 2.4 (see (2.11)), we find that for $\varepsilon_{0}=(r/\|y_{0}\|)^{2}$, there is $u_{\delta}\in L_{\delta}^{2}(0,k\delta;L^{2}(\Omega))$ so that

[TABLE]

Since $\varepsilon_{0}=(r/\|y_{0}\|)^{2}$, it follows from (5.6) that $u_{\delta}$ is an admissible control to $(NP)_{\delta}^{k\delta,y_{0}}$. Then by the optimality of $u^{*}_{\delta}$ and $N_{\delta}(k\delta,y_{0})$, and by (5.6), we find that

[TABLE]

which leads to (5.5).

Next, we claim that

[TABLE]

For this purpose, we consider the following equation:

[TABLE]

where $z\in C_{0}^{\infty}(\Omega)$ and $f\in C_{0}^{\infty}(\Omega\times\mathbb{R}^{+})$. Multiplying $y$ on both sides of Equation (5.11) and then integrating it over $\Omega$, after some computations, we obtain that for each $S>0$,

[TABLE]

Meanwhile, multiplying $-t\Delta y(t)$ on both sides of Equation (5.11) and then integrating it over $\Omega$, after some computations, we obtain that for each $S>0$,

[TABLE]

From (5.13) and (5.12), we deduce that for each $S>0$, $z\in C_{0}^{\infty}(\Omega)$ and $f\in C_{0}^{\infty}(\Omega\times\mathbb{R}^{+})$,

[TABLE]

Then by a standard density argument, we can easily derive from the above inequality that

[TABLE]

Since $\|y_{0}\|>r$, the above, along with (5.5), leads to (5.7).

We now show the second inequality in (5.2). From (4.7), we see that

[TABLE]

which, together with (5.1) and (5.7), leads to the second inequality in (5.2).

Then, we show the first inequality in (5.2). Simply write $\varphi(\cdot)$ for $\varphi(\cdot;k\delta,z^{*}_{\delta})$. Multiplying by $\Delta\varphi$ on both sides of the equation satisfied by $\varphi(\cdot;k\delta,z^{*}_{\delta})$, and then integrating it over $\Omega$, after some computations, we obtain that

[TABLE]

From this, it follows that

[TABLE]

which leads to the first inequality in (5.2). This ends the proof of the conclusion (i).

(ii) Arbitrarily fix $k_{0}\in\mathbb{N}^{+}$ so that $k_{0}\geq\max\{2,2/T\}$. For each integer $k\geq k_{0}$, let $n_{k}$ be the integer so that

[TABLE]

We first claim

[TABLE]

In fact, for each $k\geq k_{0}$, $(NP)_{1/k}^{n_{k}/k,y_{0}}$ has a unique optimal control $v^{*}_{k}$ (see (ii) of Theorem 4.3). Then, by (5.14), one can easily check that the zero extension of $v^{*}_{k}$ over $(0,T)$ is an admissible control to $(NP)^{T,y_{0}}$. From this and the optimality of $N(T,y_{0})$, one can easily check that

[TABLE]

Since $0<T<T^{*}_{y_{0}}$ and because $(1/k,n_{k})\in\mathcal{P}_{T^{*}_{y_{0}}}$ for all $k\geq k_{0}$ (which follows from (5.14) and (4.2)), we can apply (iii) of Theorem 4.2 and (iii) of Theorem 4.3 (with $(\delta,k)$=$(1/k,n_{k})$), and use (5.16) to obtain (5.15).

For each integer $k\geq k_{0}$, write $z_{1/k}^{*}$ for the minimizer of $(JP)_{1/k}^{n_{k}/k,y_{0}}$. The key is to show that on a subsequence of $\{z^{*}_{1/k}\}_{k\geq k_{0}}$, still denoted in the same manner,

[TABLE]

(Here, $z^{*}$ is the minimizer of $J^{T,y_{0}}$.) To this end, we notice that $(1/k,n_{k})\in\mathcal{P}_{T^{*}_{y_{0}}}$ for all $k\geq k_{0}$ (which follows from (5.14) and (4.2)). Thus, we can use the second inequality in (5.2) (where $\delta=1/k$; $k=n_{k}$) to find that $\{z^{*}_{1/k}\}_{k\geq k_{0}}$ is bounded in $H_{0}^{1}(\Omega)$. So there exists a subsequence of $\{z^{*}_{1/k}\}_{k\geq k_{0}}$, still denoted in the same manner, and some $\hat{z}\in H_{0}^{1}(\Omega)$ so that

[TABLE]

From the above, we see that in order to show (5.17), it suffices to prove that $z^{*}=\hat{z}$. For this purpose, we first claim that for each $k\geq k_{0}$,

[TABLE]

[TABLE]

To show (5.19), we arbitrarily fix $k\geq k_{0}$. By (5.14), we see that $0\leq T-n_{k}/k\leq 1/k$. This, along with the time-invariance of Equation (1.20), yields

[TABLE]

Meanwhile, since $\{e^{t\Delta}:t\geq 0\}$ is contractive, we have that

[TABLE]

Using the triangle inequality, by (5.21) and (5.22), we obtain (5.19).

To show (5.20), we arbitrarily fix $k\geq k_{0}$ and $t\in(0,n_{k}/k)$. Three facts are given in order. Fact one: Since $0\leq T-n_{k}/k\leq 1/k$, we can use the time-invariance of Equation (1.20) to get that

[TABLE]

Fact two: Since $0\leq T-n_{k}/k\leq 1/k$, by (1.22) and the time-invariance of Equation (1.20), we can easily check that

[TABLE]

Fact three: Since $\{e^{t\Delta}:t\geq 0\}$ is contractive, by (1.22), we see that

[TABLE]

The above three facts (5.23), (5.1) and (5.25), together with the triangle inequality, leads to (5.20).

Two observations are given in order: First, since $\varphi(\cdot;T,\hat{z})$ is uniformly continuous on $[0,T]$, we see that two supremums in (5.19) and (5.20) tend to zero as $k\rightarrow\infty$. Second, it follows by (5.14) that $\lim_{k\rightarrow\infty}n_{k}/k=T$. From these two observations, (5.18), (5.19) and (5.20), one can easily check that

[TABLE]

These, together with (1.19), (1.21) and (5.18), indicate that

[TABLE]

This, along with (5.15) and (1.19), yields that

[TABLE]

Hence, $\hat{z}$ is a minimizer of $J^{T,y_{0}}$. Then, by the uniqueness of the minimizer, we see that $\hat{z}=z^{*}$. Hence, (5.17) is true.

Finally, since $0<T<T^{*}_{y_{0}}$ and because $(1/k,n_{k})\in\mathcal{P}_{T^{*}_{y_{0}}}$ for all $k\geq k_{0}$ (which follows from (5.14) and (4.2)), the conclusion (i) in Theorem 5.1 is available for $(\delta,k)$=$(1/k,n_{k})$. Thus, by (5.1), the second inequality in (5.2) (with $(\delta,k)$=$(1/k,n_{k})$) and (5.17), using the fact that $n_{k}/k\rightarrow T$ (see (5.14)), we can easily obtain (5.3) and the second inequality in (5.4). Besides, by the same way as that used to prove the first inequality in (5.2), we get the first inequality in (5.4).

In summary, we end the proof of Theorem 5.1. ∎

5.2 Some estimates related to minimal norm functions

Several inequalities related to the minimal norm functions $T\rightarrow N(T,y_{0})$ and $k\delta\rightarrow N_{\delta}(k\delta,y_{0})$ will be presented in the following two theorems.

Theorem 5.2**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then there is $C_{3}\triangleq C_{3}(\Omega,\omega)>0$ so that for each pair $(T_{1},T_{2})$, with $0<T_{1}\leq T_{2}<T^{*}_{y_{0}}$ (given by (4.1)),

[TABLE]

Proof.

Arbitrarily fix a pair $(T_{1},T_{2})$, with $0<T_{1}<T_{2}<T^{*}_{y_{0}}$. The proof is organized by the following two steps:

Step 1. To show the first inequality in (5.26)

By (i) of Theorem 4.1, we have that

[TABLE]

Then by (iii) in Theorem 4.1, we see that

[TABLE]

Let $u^{*}_{1}$ be an optimal control to $(TP)^{M_{1},y_{0}}$. Then we find that

[TABLE]

It follows from the first inequality in (5.29) that

[TABLE]

Since

[TABLE]

the above, along with Hölder’s inequality and the second inequality in (5.29), yields that

[TABLE]

Next, we define a control $u_{2}$ over $\mathbb{R}^{+}$ as follows:

[TABLE]

From (5.33) and the second inequality in (5.29), it follows that

[TABLE]

Meanwhile, we let

[TABLE]

Since $u_{2}=0$ over \big{(}T(M_{1},y_{0}),\infty\big{)}, by (5.33), (5.30) and (5.35), one can easily check that

[TABLE]

Now, it follows from (5.34) and (5.36) that $u_{2}$ is an admissible control to $(TP)^{M_{2},y_{0}}$, which drives the solution to $B_{r}(y_{0})$ at time $T(M_{1},y_{0})+\widehat{T}$. This, along with the optimality of $T(M_{2},y_{0})$, yields that

[TABLE]

From this, (5.28) and (5.35), we find that

[TABLE]

Since $M_{1}\triangleq(N(T_{1},y_{0})$ and $M_{2}\triangleq N_{2}(T_{2},y_{0}))$ (see (5.27)), the above leads to the first inequality in (5.26). This ends the proof of Step 1.

Step 2. To show the second inequality in (5.26)

Let $z^{*}_{1}$ be the minimizer of $J^{T_{1},y_{0}}$. Throughout this step, we simply write $\varphi_{1}(\cdot)$ and $\varphi_{2}(\cdot)$ for $\varphi(\cdot;T_{1},z_{1}^{*})$ and $\varphi(\cdot;T_{2},z_{1}^{*})$ respectively. First, we claim that

[TABLE]

Indeed, according to [28, Theorem 6.13 in Chapter 2], there is $C_{22}\triangleq C_{22}(\Omega)>0$ so that

[TABLE]

From this, we see that

[TABLE]

This, along with [12, Proposition 3.1], yields that for some $C_{23}\triangleq C_{23}(\Omega,\omega)>0$,

[TABLE]

Meanwhile, by (ii) of Theorem 4.3 and the time-invariance of Equation (1.20), we see that

[TABLE]

This, along with (5.38), yields that

[TABLE]

which leads to (5.37).

Next, since $0<T_{1}\leq T_{2}<T^{*}_{y_{0}}$, it follows by (i) and (ii) of Theorem 4.1 that $N(T_{1},y_{0})\geq N(T_{2},y_{0})$. From this and (iii) of Theorem 4.2, it follows that

[TABLE]

This, along with (1.19), yields that

[TABLE]

At the same time, by the time-invariance of Equation (1.20), we have that

[TABLE]

Since the semigroup $\{e^{\Delta t}\}_{t\geq 0}$ is contractive, from (5.41), we see that

[TABLE]

From (5.41), we also have that

[TABLE]

Now, by (5.40), (5.42) and (5.2), we obtain that

[TABLE]

By this, (5.39) and (5.37), we get that

[TABLE]

Finally, by [12, Proposition 3.1], we can find $u_{T_{1}}\in L^{2}(0,T_{1};L^{2}(\Omega))$ so that

[TABLE]

From the first equality in the above, we see that $u_{T_{1}}$ is an admissible to $(NP)^{T_{1},y_{0}}$. This, along with the second inequality in the above and the optimality of $N(T_{1},y_{0})$, indicates

[TABLE]

which, along with (5.44), leads to the second inequality in (5.26) for some $C_{3}\triangleq C_{3}(\Omega,\omega)$.

In summary, we finish the proof of Theorem 5.2. ∎

Theorem 5.3**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Let $\mathcal{P}_{T^{*}_{y_{0}}}$ and $T^{*}_{y_{0}}$ be given by (4.2) and (4.1) respectively. Then there is $C_{4}\triangleq C_{4}(\Omega,\omega)>0$ so that for each $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$,

[TABLE]

Proof.

Arbitrarily fix $(\delta,k)\in\mathcal{P}_{T^{*}_{y_{0}}}$. Let $z^{*}_{\delta}$ be the minimizer of $J^{k\delta,y_{0}}_{\delta}$. Throughout the proof of Theorem 5.3, we simply write respectively $\varphi(\cdot)$ and $\overline{\varphi}_{\delta}(\cdot)$ for $\varphi(\cdot;k\delta,z^{*}_{\delta})$ (see (1.20)) and $\overline{\varphi}(\cdot;k\delta,z^{*}_{\delta})$ (see (1.22)). We organize the proof by several steps as follows:

Step 1. To prove that

[TABLE]

Since $L^{2}_{\delta}(0,k\delta;L^{2}(\Omega))\subset L^{2}(0,k\delta;L^{2}(\Omega))$ (see (1.6)), we find that each admissible control to $(NP)_{\delta}^{k\delta,y_{0}}$ is also an admissible control to $(NP)^{k\delta,y_{0}}$. This, along with (1.16) and (1.17), yields that $N(k\delta,y_{0})\leq N_{\delta}(k\delta,y_{0})$, from which, as well as (iii) of Theorems 4.2 and (iii) of Theorem 4.3, it follows that

[TABLE]

This, along with (1.19) and (1.21), yields that

[TABLE]

which leads to (5.46).

Step 2. To show that

[TABLE]

First, we claim that for each $f\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$,

[TABLE]

where $\bar{f}_{\delta}$ is given by (2.2). Indeed, for an arbitrarily fixed $f\in L^{2}(\mathbb{R}^{+};L^{2}(\Omega))$, one can directly check that

[TABLE]

Meanwhile, it follows by (2.2) that $\bar{g}_{\delta}=0$, where $g\triangleq f-\bar{f}_{\delta}$. Then by Lemma 2.2, we obtain that

[TABLE]

This, along with (5.2), leads to (5.49).

Next, by taking $f$ to be the zero extension of $\varphi$ over $\mathbb{R}^{+}$ in (5.49), we obtain (5.48). Here, we used the fact that in this case, $\bar{f}_{\delta}(\cdot)$ is the zero extension of $\chi_{\omega}\overline{\varphi}_{\delta}(\cdot)$ over $\mathbb{R}^{+}$, which follows from (2.2) and (1.22).

Step 3. To verify that there exists $C_{41}\triangleq C_{41}(\Omega,\omega)>0$ so that

[TABLE]

From (1.22), it follows that

[TABLE]

Applying the Hölder inequality to the above leads to that

[TABLE]

This, along with (5.2), implies (5.50) for some $C_{41}\triangleq C_{41}(\Omega,\omega)$.

Step 4. To show (5.45)

We first claim that

[TABLE]

In fact, by (i) of Theorem 4.1, we have that

[TABLE]

This, along with the first inequality in (5.26) (where $T_{1}=k\delta$), yields that

[TABLE]

Since we clearly have that

[TABLE]

(5.51) follows from (5.52) at once.

Meanwhile, from (5.46), (5.48) and (5.50), we obtain that

[TABLE]

From this, (5.47) and (5.51), we find that

[TABLE]

Since $k\delta<T^{*}_{y_{0}}$, the above leads to (5.45) for some $C_{4}\triangleq C_{4}(\Omega,\omega)$.

In summary, we end the proof of Theorem 5.3. ∎

5.3 Some properties on minimal time functions

Some inequalities, as well as properties, related to the minimal time functions $M\rightarrow T_{\delta}(M,y_{0})$ and $M\rightarrow T(M,y_{0})$ will be given in this subsection.

Theorem 5.4**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. For each $M>0$ and $\eta\in(0,1)$, there is a measurable subset $\mathcal{A}_{M,\eta}\subset(0,1)$ (depending also on $y_{0}$ and $r$), with $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{A}_{M,\eta}\cap(0,h)|=\eta$, so that for each $\delta\in\mathcal{A}_{M,\eta}$, there is $a_{\delta}\in(0,\eta)$ so that

[TABLE]

Proof.

Arbitrarily fix $M>0$ and $\eta\in(0,1)$. Throughout the proof of Theorem 5.4, we simply write $T^{*}\triangleq T(M,y_{0})$. For each $k\in\mathbb{N}^{+}$ and $a\in(0,\eta)$, we define a subset of $\mathbb{R}^{+}$ in the following manner:

[TABLE]

We then define another subset of $\mathbb{R}^{+}$ as follows:

[TABLE]

The rest proof is divided into the following two steps:

Step 1. To prove that $\lim_{h\rightarrow 0^{+}}\frac{1}{h}|\mathcal{B}_{M,\eta}\cap(0,h)|=\eta$

From (5.54), we see that

[TABLE]

From this and (5.55), it follows that

[TABLE]

For each $j\in\mathbb{N}^{+}$, we let $h_{j}\triangleq T^{*}/j$. For each $h\in(0,T^{*})$, we let $j(h)$ be the integer so that $h_{j(h)+1}\leq h<h_{j(h)}$. Then, by (5.56), one can easily verify that

[TABLE]

[TABLE]

From (5.57) and (5.58), we can easily obtain the conclusion in Step 1.

Step 2. To show (5.53)

We first claim that for each $\delta\in\mathcal{B}_{M,\eta}\cap(0,1)$, there is a unique pair $(k_{\delta},a_{\delta})$ so that

[TABLE]

Indeed, the existence of such a pair follows from (5.55) and (5.54) at once, while the uniqueness of such pairs can be directly checked.

Thus, for each $\delta\in\mathcal{B}_{M,\eta}\cap(0,1)$, we can define $k_{\delta}$ to be the first component of the unique pair satisfying (5.59). We next claim that there exists $\delta^{1}_{M,\eta}\in(0,1)$ so that

[TABLE]

To this end, we notice that $T^{*}<T^{*}_{y_{0}}$ (see (iii) of Theorem 4.1). Arbitrarily fix $\delta\in\mathcal{B}_{M,\eta}\cap(0,1)$ so that

[TABLE]

(The existence of such $\delta$ is ensured by (5.56).) Then it follows from (5.61) and (5.59) that

[TABLE]

This, along with (4.2), yields that

[TABLE]

By these, we can apply Theorem 5.3 (with $(\delta,k)=(\delta,k_{\delta})$) and Theorem 5.2 (with $T_{1}=T^{*}$ and $T_{2}=(k_{\delta}+1)\delta$) to get that

[TABLE]

where $C_{4}$ is given by (5.45). Meanwhile, by (5.59) and (5.61), we find that

[TABLE]

These, along with (5.62) and (ii) of Theorem 4.1, yield that

[TABLE]

By this and (5.61), we obtain (5.60).

Define a set $\mathcal{A}_{M,\eta}$ in the following manner:

[TABLE]

We now show that the second conclusion in (5.53) holds for each $\delta$ in $\mathcal{A}_{M,\eta}$ defined by (5.63). To this end, we arbitrarily fix $\delta\in\mathcal{A}_{M,\eta}$. We claim that

[TABLE]

To show the first inequality in (5.64), we let $u_{\delta}$ be an admissible control to $(NP)^{(k_{\delta}+1)\delta,y_{0}}_{\delta}$ and let $\tilde{u}_{\delta}$ be the zero extension of $u_{\delta}$ over $\mathbb{R}^{+}$. Since $N_{\delta}((k_{\delta}+1)\delta,y_{0})\leq M$ (see (5.60)), one can easily check that $\tilde{u}_{\delta}$ is an admissible control (to $(TP)^{M,y_{0}}_{\delta}$), which drives the solution to $B_{r}(0)$ at time $(k_{\delta}+1)\delta$. This, along with the optimality of $T_{\delta}(M,y_{0})$, leads to the first inequality in (5.64). To prove the second inequality in (5.64), we notice that $\mathcal{U}^{M}_{\delta}\subset\mathcal{U}^{M}$. This, along with (1.1) and (1.8), yields that $T^{*}\leq T_{\delta}(M,y_{0})$. From this and (5.59), we obtain the second inequality in (5.64).

Since $T_{\delta}(M,y_{0})$ is a multiple of $\delta$ (see (1.9)), it follows from (5.64) that

[TABLE]

This, along with (5.60), implies that the second conclusion in (5.53).

Finally, from (5.65) and (5.59), we see that the first conclusion in (5.53) is true for each $\delta\in\mathcal{A}_{M,\eta}$ defined by (5.63).

In summary, we end the proof of Theorem 5.4. ∎

Theorem 5.5**.**

For each $M\geq 0$,

[TABLE]

Besides, $T(0,y_{0})=T^{*}_{y_{0}}$ for each $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Here, $T^{*}_{y_{0}}$ is given by (4.1).

Proof.

It follows from (4.1) and (1.1) that $T(0,y_{0})=T^{*}_{y_{0}}$. To prove (5.66), we arbitrarily fix $y_{1},\,y_{2}\in L^{2}(\Omega)\setminus B_{r}(0)$. We first claim that

[TABLE]

Let $\hat{u}_{2}\triangleq\chi_{(0,T(M,y_{2}))}u_{2}$, where $u_{2}$ is the optimal control to $(TP)^{M,y_{2}}$. Then we have that

[TABLE]

Define a number $\widehat{T}_{1}$ by

[TABLE]

Since $\hat{u}_{2}=0$ over \big{(}T(M,y_{2}),\infty\big{)}, by the first conclusion in (5.68) and (5.69), we see that

[TABLE]

From this and the second inequality in (5.68), we see that $\hat{u}_{2}$ is an admissible control (to $(TP)^{M,y_{1}}$) which drives the solution to $B_{r}(0)$ at time $T(M,y_{2})+\widehat{T}_{1}$. This, along with the optimality of $T(M,y_{1})$, yields that $T(M,y_{1})\leq T(M,y_{2})+\widehat{T}_{1}$. From this and (5.69), we deduce (5.67).

Next, by a very similar way as that to show (5.67), we can easily check that

[TABLE]

Finally, (5.66) follows from (5.67) and (5.70) immediately. This completes the proof of Theorem 5.5. ∎

6 The proofs of the main theorems

In this section, we will prove Theorems 1.3-1.5. The strategies to show them have been introduced in Subsection 1.3.

6.1 The proof of Theorem 1.3

Proof of Theorem 1.3.

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. We will prove the conclusions (i) and (ii) in Theorem 1.3 one by one.

(i) Arbitrarily fix $0<M_{1}<M_{2}$. Then arbitrarily fix $M\in[M_{1},M_{2}]$. From the conclusion (iii) in Theorem 4.1, it follows

[TABLE]

We take $\delta$ so that

[TABLE]

Let $\hat{k}_{\delta}\in\mathbb{N}$ satisfy that

[TABLE]

We first claim that

[TABLE]

Indeed, from (4.2) and (6.1)-(6.3), one can easily check that

[TABLE]

Three facts are given in order: (a) By the second conclusion in (6.5), we can apply Theorem 5.3, with $(\delta,k)=(\delta,\hat{k}_{\delta}+1)$, to obtain that

[TABLE]

where $C_{4}\triangleq C_{4}(\Omega,\omega)$ is given by (5.45). (b) By the first conclusion in (6.5), we can use the first inequality in (5.26) in Theorem 5.2 (where $T_{1}=T(M,y_{0})$ and $T_{2}=(\hat{k}_{\delta}+1)\delta$) to get that

[TABLE]

(c) By (ii) of Theorem 4.1, we have that $N(T(M,y_{0}),y_{0})=M$.

From above three facts (a)-(c), we find that

[TABLE]

Meanwhile, from (6.3), (6.2) and (6.1), one can easily check that

[TABLE]

This, along with (6.6), leads to (6.4).

We next claim that

[TABLE]

where $\delta_{1}$ is given by (6.2) and $\delta_{2}$ is defined by

[TABLE]

In fact, for an arbitrarily fixed $\delta\in(0,\delta_{0})$, by (6.4) and (6.8), after some computations, we find that

[TABLE]

Let $u_{\delta}$ be the zero extension of an admissible control (to $(NP)^{(\hat{k}_{\delta}+1)\delta,y_{0}}_{\delta}$) over $\mathbb{R}^{+}$. Then by (6.9), one can easily check that $u_{\delta}$ is an admissible control (to $(TP)^{M,y_{0}}_{\delta}$), which drives the solution to $B_{r}(0)$ at time $(\hat{k}_{\delta}+1)\delta$. This, along with the optimality of $T_{\delta}(M,y_{0})$, leads to (6.7).

We now show (1.10) with $\delta_{0}$ given by (6.7). For this purpose, we arbitrarily fix $\delta\in(0,\delta_{0})$. Since $\mathcal{U}^{M}_{\delta}\subset\mathcal{U}^{M}$, it follows by (1.1) and (1.8) that $T(M,y_{0})\leq T_{\delta}(M,y_{0})$. This, along with (6.3) and (6.7), leads to (1.10).

Finally, we will prove that the above $\delta_{0}$ is continuous in $(M_{1},M_{2},y_{0},r)$. Indeed, according to the definitions of $\delta_{0}$, $\delta_{1}$ and $\delta_{2}$ (see (6.7), (6.2) and (6.8), respectively), it suffices to show that the maps $y_{0}\rightarrow T^{*}_{y_{0}}$ and $(\hat{M},y_{0})\rightarrow T(\hat{M},y_{0})$ are continuous. To show the first one, we notice from Theorem 5.5 that for each $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, $T(0,y_{0})=T^{*}_{y_{0}}$ and that the map $y_{0}\rightarrow T(0,y_{0})$ is continuous outside $B_{r}(0)$. Hence, the map $y_{0}\rightarrow T^{*}_{y_{0}}$ is continuous outside $B_{r}(0)$. To show the continuity of the map $(\hat{M},y_{0})\rightarrow T(\hat{M},y_{0})$, two observations are given in order: First, it follows from Theorem 5.5 that $T(\hat{M},y_{0})$ continuously depends on $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$; Second, according to the conclusion (iii) of Theorem 4.1, for an fixed $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, the map $\hat{M}\rightarrow T(\hat{M},y_{0})$ is continuous and decreasing. From these two observations, the desired continuity follows at once. This ends the proof of the conclusion (i).

(ii) Let $\mathcal{A}_{M,\eta}$, with $M>0$ and $\eta\in(0,1)$, be given by Theorem 5.4. Then the conclusion (ii) of Theorem 1.3 follows from the first conclusion in (5.53) at once.

In summary, we end the proof of Theorem 1.3. ∎

6.2 The proof of Theorem 1.4

Proof of Theorem 1.4.

Arbitrarily fix $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. For each $M>0$ and $\delta>0$, we let $u_{M}^{*}$ and $u^{*}_{\delta,M}$ be the optimal control and the optimal control with the minimal norm to $(TP)^{M,y_{0}}$ and $(TP)^{M,y_{0}}_{\delta}$ respectively (see Theorem 3.1). We will prove the conclusions (i)-(ii) one by one.

(i) Let $0<M_{1}<M_{2}$. Let $\delta_{0}=\delta_{0}(M_{1},M_{2},y_{0},r)$ and $C_{3}=C_{3}(\Omega,\omega)$ be given by Theorem 1.3 and Theorem 5.2, respectively. Arbitrarily fix $M\in[M_{1},M_{2}]$ and $\delta>0$. In the proof of (i) of Theorem 1.4, we simply write $T^{*}$ and $T^{*}_{\delta}$ for $T(M,y_{0})$ and $T_{\delta}(M,y_{0})$, respectively; simply write $u^{*}$ and $u^{*}_{\delta}$ for $u_{M}^{*}$ and $u^{*}_{\delta,M}$, respectively.

Since $T^{*}_{\delta}$ is a multiple of $\delta$ (see (1.8)), we can write

[TABLE]

In the case that

[TABLE]

one can easily show (1.12) and the desired continuity of $C(M_{1},M_{2},y_{0},r)$. In fact, it follows from (6.11) that

[TABLE]

Since $(M_{2},y_{0})\rightarrow T(M_{2},y_{0})$ is continuous (see the proof of the continuity of $\delta_{0}$ with respect to $(M_{1},M_{2},y_{0},r)$ in Theorem 1.3), and because $T(M_{2},y_{0})>0$ for all $M_{2}>0$ and $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$, we see from (6.12) that $\hat{C}(M_{1},M_{2},y_{0},r)$ is continuous in $M_{1}$, $M_{2}$, $y_{0}$ and $r$.

Thus, we only need to show (1.12), as well as the continuous dependence of $C$ in (1.12), for the case that

[TABLE]

For this purpose, some preliminaries are needed. First we claim that

[TABLE]

Indeed, (6.14) follows from the next three facts at once. Fact one: From Theorem 4.1, we have that $0<T(M_{2},y_{0})\leq T^{*}\leq T(M_{1},y_{0})<T^{*}_{y_{0}}$; Fact two: Since $\mathcal{U}^{M}_{\delta}\subset\mathcal{U}^{M}$, we find from (1.1) and (1.8) that $T^{*}\leq T_{\delta}^{*}$; Fact three: By Theorem 1.3 and (6.13), we see that

[TABLE]

Then it follows from (6.14), (6.13) and (4.2) that

[TABLE]

Next, we let $z^{*}\neq 0$ and $z^{*}_{\delta}\neq 0$ be the minimizers of $J^{T^{*},y_{0}}$ and $J_{\delta}^{T^{*}_{\delta},y_{0}}$, respectively (see Theorem 4.2 and Theorem 4.3). Write

[TABLE]

By (iii) of Theorem 4.1, $u^{*}|_{(0,T^{*})}$ is the optimal control to $(NP)^{T^{*},y_{0}}$. Then by (ii) of Theorem 4.2 (with $T=T^{*}\in(0,T^{*}_{y_{0}})$), we see that $u^{*}(\cdot)=\chi_{\omega}\varphi(\cdot;T^{*},z^{*})$ over $(0,T^{*})$. Meanwhile, by (ii) of Theorem 4.1, we find that $N(T^{*},y_{0})=M$. These, along with (6.16), yield that

[TABLE]

Finally, it follows from (iii) of Theorem 3.1 that $u_{\delta}^{*}|_{(0,T^{*}_{\delta})}$ is an optimal control to $(NP)_{\delta}^{T^{*}_{\delta},y_{0}}$. This, along with the fact that $u^{*}_{\delta}$ is an optimal control to $(TP)_{\delta}^{M,y_{0}}$, yields

[TABLE]

Meanwhile, by (6.15), we can apply Theorem 4.3 (with $(\delta,k)=(\delta,T^{*}_{\delta}/\delta)\triangleq(\delta,k_{\delta})$), as well as (6.16), to obtain that

[TABLE]

Here, $\overline{\varphi}_{\delta}(\cdot;T^{*}_{\delta},z_{\delta}^{*})$ and $\overline{\varphi}_{\delta}(\cdot;T^{*}_{\delta},\hat{z}_{\delta}^{*})$ are given by (1.22) with $(\delta,k)=(\delta,k_{\delta})$ and $M_{\delta}$ is defined by

[TABLE]

We now prove (1.12) for an arbitrarily fixed $\delta$ (satisfying (6.13)) by several steps.

Step 1. To show that

[TABLE]

One can easily check the following two estimates:

[TABLE]

[TABLE]

From these, we see that to prove (6.21), it suffices to show that

[TABLE]

For this purpose, let $\bar{k}$ be the integer so that $\bar{k}\delta<T^{*}\leq(\bar{k}+1)\delta$. Because $T^{*}\leq T^{*}_{\delta}$ (see (6.14)) and $T^{*}_{\delta}$ is a multiple of $\delta$ (see (1.8)), we find that $T^{*}_{\delta}\geq(\bar{k}+1)\delta$. Since $u^{*}_{\delta}$ is a piece-wise constant function over $(0,T^{*}_{\delta})$, and because

[TABLE]

one can easily check that

[TABLE]

This yields that

[TABLE]

Meanwhile, from the first equality in (6.19), one can easily verify that when $j=1,\dots,\bar{k}$,

[TABLE]

This, along with (6.23), (6.19), (6.20) and the contractivity of $\{e^{\Delta t}\}_{t\geq 0}$, yields that

[TABLE]

Since $\|\partial_{\tau}\varphi(\cdot;T^{*}_{\delta},z^{*}_{\delta})\|_{L^{2}(0,T^{*}_{\delta};L^{2}(\Omega))}\leq\|z^{*}_{\delta}\|_{H_{0}^{1}(\Omega)}$ (see (5.2)), by applying the Hölder inequality to (6.24), we obtain (6.22). This ends the proof of (6.21).

Step 2. To show that

[TABLE]

Observe that

[TABLE]

We first claim that

[TABLE]

Write $\{\lambda_{j}\}_{j=1}^{\infty}$ for the family of all eigenvalues of $-\Delta$ with the zero Dirichlet boundary condition so that $\lambda_{1}<\lambda_{2}\leq\cdots$. Let $\{e_{j}\}_{j=1}^{\infty}$ be the family of the corresponding normalized eigenvectors. Arbitrarily fix $f\in L^{2}(0,T^{*};L^{2}(\Omega))$. Write $f(t)=\sum_{j=1}^{\infty}f_{j}(t)e_{j}$, $t\in(0,T^{*})$. Then we have that

[TABLE]

By some computations, we obtain that

[TABLE]

Two facts are given in order: (a) It is clear that

[TABLE]

(b) By the Hölder inequality, we obtain that

[TABLE]

The above facts (6.29) and (6.30), along with (6.28) and (6.27), yield that

[TABLE]

This implies that

[TABLE]

Since $f$ was arbitrarily taken from $L^{2}(0,T^{*};L^{2}(\Omega))$, the above indicates that

[TABLE]

which leads to (6.26).

We next estimate $I_{2}$. Since $T^{*}_{\delta}=k_{\delta}\delta$ (see (6.10)) and because $T^{*}_{\delta}\geq T^{*}$, we see from (1.22) that

[TABLE]

By using the Hölder inequality in the above and by (5.2), we see that

[TABLE]

Finally, (6.25) follows from the above estimates on $I_{1}$ and $I_{2}$.

Step 3. To prove that

[TABLE]

*where $C_{1}(M_{2},y_{0})$ is continuous in $M_{2}$ and $y_{0}$ *

By (6.15), we can use Theorem 5.3, with $(\delta,k)=(\delta,k_{\delta})$ (where $k_{\delta}$ is given by (6.10)), to see that $N_{\delta}(T^{*}_{\delta},y_{0})\geq N(T^{*}_{\delta},y_{0})$. By (6.18) and (6.20), we find that $M\leq M_{\delta}$. These, along with (ii) of Theorem 4.1, yield that

[TABLE]

Meanwhile, by (6.14), we can use Theorem 5.2 (with $T_{1}=T^{*}$ and $T_{2}=T^{*}_{\delta}$) to see that

[TABLE]

where $C_{3}(\Omega,\omega)$ is given by (5.26). Since $T(M_{2},y_{0})\leq T^{*}$ (see (6.14)), the above, along with (6.32), yields that

[TABLE]

Since $\delta\in(0,\delta_{0})$, the above, along with Theorem 1.3, leads to (6.31). Finally, since $T(M_{2},y_{0})$ is continuous in $M_{2}$ and $y_{0}$ (see the proof of the continuity of $\delta_{0}$ with respect to $(M_{1},M_{2},y_{0},r)$ in Theorem 1.3), we obtain from (6.31) that $C_{1}(M_{2},y_{0})$ is continuous in $M_{2}$ and $y_{0}$.

Step 4. To show that

[TABLE]

*where $C_{2}(M_{1},M_{2},y_{0},r)$ is continuous in $M_{1}$, $M_{2}$, $y_{0}$ and $r$ *

Define an affiliated control $\hat{u}_{\delta}$ over $\mathbb{R}^{+}$ by

[TABLE]

We divide the rest of the proof of Step 4 by several parts.

Part 4.1. To prove that

[TABLE]

By (6.17) and (6.34), one can directly check that

[TABLE]

Hence, we have that

[TABLE]

This, along with (4.4), (4.7) and (6.16), yields that

[TABLE]

which leads to (6.35).

Part 4.2. To show that there exists $C_{21}\triangleq C_{21}(\Omega)>0$ so that

[TABLE]

Three facts are given in order. Fact one: By the Hölder inequality, we find that for some $C_{22}\triangleq C_{22}(\Omega)>0$,

[TABLE]

Fact two: Since $\|u^{*}_{\delta}\|_{L^{\infty}(0,T^{*}_{\delta};L^{2}(\Omega))}\leq\|z^{*}_{\delta}\|$ (which follows from (6.19) and (1.22)), and because $T^{*}\leq T^{*}_{\delta}$ (see (6.14)), we find that

[TABLE]

Fact three: By (6.14), we see that

[TABLE]

Now, by the triangle inequality, (6.37), (6.21), (6.38), (6.39) and the Poincaré inequality, we obtain (6.36) for some $C_{21}=C_{21}(\Omega)$.

Part 4.3. To show that there exists $C_{23}\triangleq C_{23}(\Omega)>1$ so that

[TABLE]

Recall (6.16) for the definition of $\hat{z}^{*}_{\delta}$. In Part 4.3, we simply write respectively $\varphi(\cdot)$ and $\overline{\varphi}_{\delta}(\cdot)$ for $\varphi(\cdot;T^{*},\hat{z}^{*}_{\delta})$ and $\overline{\varphi}_{\delta}(\cdot;T^{*}_{\delta},\hat{z}^{*}_{\delta})$; simply write $L^{2}(0,T^{*})$, $L^{2}(0,T^{*}_{\delta})$ and $L^{2}(T^{*},T^{*}_{\delta})$ for $L^{2}(0,T^{*};L^{2}(\Omega))$, $L^{2}(0,T^{*}_{\delta};L^{2}(\Omega))$ and $L^{2}(T^{*},T^{*}_{\delta};L^{2}(\Omega))$ respectively. From (6.34) and (6.19), using the triangle inequality, we obtain that

[TABLE]

By direct computations, we find that

[TABLE]

[TABLE]

From (6.41), (6.2) and (6.43), we deduce that

[TABLE]

Meanwhile, by (6.19) and (6.16), we see that $M_{\delta}=\|z_{\delta}^{*}\|\|\chi_{\omega}\overline{\varphi}_{\delta}(\cdot)\|_{L^{2}(0,T^{*}_{\delta})}$. This, together with (6.44) and (6.16), yields that

[TABLE]

Since $\|\overline{\varphi}_{\delta}(t;T^{*}_{\delta},z_{\delta}^{*})\|\leq\|z_{\delta}^{*}\|$ for each $t\in(0,T^{*}_{\delta})$ (which follows from (1.22)), we find from (6.45) and (6.25) that

[TABLE]

At the same time, since $M\geq M_{1}$, it follows from (6.31) and (6.13) that

[TABLE]

The above two inequalities, along with the Poincaré inequality, yield (6.40).

Part 4.4. To show (6.33) and the continuity of $C_{2}(M_{1},M_{2},y_{0},r)$

By (6.36) and (6.40), we can easily check that

[TABLE]

Meanwhile, by (6.15), we can use Theorem 5.1 (with $(\delta,k)=(\delta,k_{\delta})$, where $k_{\delta}$ is given by (6.10)), as well as (6.14), to get that

[TABLE]

where $C_{1}=C_{1}(\Omega,\omega)$ is given by (5.2). This, along with (6.46), leads to that

[TABLE]

where

[TABLE]

Notice that $\hat{C}_{3}(M_{1},M_{2},y_{0},r)$ is continuous in $M_{1}$, $M_{2}$, $y_{0}$ and $r$. This can be showed by the same way as that used to prove of the continuity of $\delta_{0}$ with respect to $(M_{1},M_{2},y_{0},r)$ in Theorem 1.3.

Now it follows from (6.35) that

[TABLE]

This, along with (6.48) and (6.31), yields (6.33), with

[TABLE]

Finally, along with the continuity of $C_{3}(M_{1},M_{2},y_{0},r)$ and $C_{1}(M_{2},y_{0})$, the above leads to the desired continuity of $C_{2}(M_{1},M_{2},y_{0},r)$. This ends the proof of Step 4.

Step 5. To show that

[TABLE]

where $C_{4}(M_{1},M_{2},y_{0},r)$ is continuous in $M_{1}$, $M_{2}$, $r$ and $y_{0}$

Recall (6.16) for the definitions of $\hat{z}^{*}_{\delta}$ and $\hat{z}^{*}$. In Step 5, we simply write $\varphi_{1}(\cdot)$ and $\varphi_{2}(\cdot)$ for $\varphi(\cdot;T^{*},\hat{z}^{*})$ and $\varphi(\cdot;T^{*},\hat{z}^{*}_{\delta})$, respectively; simply write $L^{2}(0,T^{*})$ for $L^{2}(0,T^{*};L^{2}(\Omega))$. By (6.17) and (6.34), we see that

[TABLE]

Meanwhile, from (6.17) and (6.16), we find that

[TABLE]

This, along with (6.50), yields that

[TABLE]

By (6.51), (6.40) and the triangle inequality, then using (6.33), we see that

[TABLE]

From this, (6.47) and (6.31), we obtain (6.49), with

[TABLE]

where $C_{1}=C_{1}(M_{2},y_{0})$ is given by (6.31), $C_{2}=C_{2}(M_{1},M_{2},y_{0},r)$ is given by (6.33) and $C_{23}=C_{23}(\Omega)$ is given by (6.40). Notice that $(a)$ $C_{1}(M_{2},y_{0})$ and $C_{2}(M_{1},M_{2},y_{0},r)$ are continuous; $(b)$ the maps $y_{0}\rightarrow T^{*}_{y_{0}}$ and $(M_{2},y_{0})\rightarrow T(M_{2},y_{0})$ are continuous (which were showed in the proof of the continuity of $\delta_{0}$ with respect to $(M_{1},M_{2},y_{0},r)$ in Theorem 1.3); $(c)$ $T(M_{2},y_{0})>0$. From these, we can easily show that the above $C_{4}(M_{1},M_{2},y_{0},r)$ is continuous in $M_{1}$, $M_{2}$, $y_{0}$ and $r$. This ends the proof of Step 5.

In summary, we end the proof of the conclusion (i) in Theorem 1.4.

(ii) Arbitrarily fix $M>0$ and $\eta>0$. Let $\mathcal{A}_{M,\eta}$ be given by Theorem 5.4. Then, by Theorem 5.4, we see that

[TABLE]

and that for each $\delta\in\mathcal{A}_{M,\eta}$,

[TABLE]

Arbitrarily fix $\delta\in\mathcal{A}_{M,\eta}$. Let $u^{*}_{M}$ and $u^{*}_{M,\delta}$ be the optimal control to $(TP)^{M,y_{0}}$ and the optimal control optimal with the minimal norm to $(TP)_{\delta}^{M,y_{0}}$, respectively. (see Theorem 3.1.) Three facts are given in order: (a) By Theorem 4.1, one can easily check

[TABLE]

(b) From (iii) of Theorem 3.1, we see that

[TABLE]

(c) Since $\mathcal{U}^{M}_{\delta}\subset\mathcal{U}^{M}$, by (1.1) and (1.8), we find that $T(M,y_{0})\leq T_{\delta}(M,y_{0})$. Combining the above facts (a)-(c) with (6.52), we find that

[TABLE]

which leads to (1.13). Thus, the conclusion (ii) in Theorem 1.4 is true.

In summary, we end the proof of Theorem 1.4. ∎

6.3 The proof of Theorem 1.5

This subsection devotes to the proof of Theorem 1.5. To show (1.15) in Theorem 1.5, we need the next lemma which gives a lower bound for the diameter of the subset $\mathcal{O}_{M,\delta}$ (in $L^{2}(0,T(M,y_{0});L^{2}(\Omega))$), which is defined by (1.23).

Lemma 6.1**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and $M>0$. Then there is $\delta_{M}\triangleq\delta_{M}(y_{0},r)>0$ so that for each $\eta\in(0,1)$ and $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{M})$ (where $\mathcal{A}_{M,\eta}$ is given by Theorem 5.4),

[TABLE]

Proof.

Arbitrarily fix $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$ and $M>0$. Let $\delta_{0}>0$ be given by (i) of Theorem 1.3 with $M_{1}=M/2$ and $M_{2}=M$. From (i) and (ii) of Theorem 4.1, we see that $0<T(M,y_{0})<T^{*}_{y_{0}}$. Thus we can take a positive number $\delta_{1}$ in the following manner:

[TABLE]

Arbitrarily fix $\eta\in(0,1)$ and $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{1})$. From now on and throughout the proof of Lemma 6.1, We simply write $T^{*}$ and $T^{*}_{\delta}$ for $T(M,y_{0})$ and $T_{\delta}(M,y_{0})$ respectively; simply write $L^{2}(0,T^{*})$ and $L^{2}(0,T^{*}_{\delta})$ for $L^{2}(0,T^{*};L^{2}(\Omega))$ and $L^{2}(0,T^{*}_{\delta};L^{2}(\Omega))$ respectively.

From (6.54) and Theorem 1.3, we see that

[TABLE]

Meanwhile, it follows from the second conclusion in (5.53) in Theorem 5.4 that

[TABLE]

To show (6.53), it suffices to find a subset $\mathcal{O}^{2}_{M,\delta}\subset\mathcal{O}_{M,\delta}$ so that for some $\hat{C}_{M}\triangleq C_{M}(y_{0},r)$, $\hat{C}_{M}\sqrt{(1-\eta)\delta}$ is a lower bound for the ”diam $\mathcal{O}^{2}_{M,\delta}$”. To this end, we first introduce an affiliated subset $\mathcal{O}_{M,\delta}^{1}\subset\mathcal{O}_{M,\delta}$ in the following manner: Let $u^{*}_{\delta}$ be the optimal control with the minimal norm to $(TP)^{M,y_{0}}_{\delta}$ (see (iii) of Theorem 3.1). Arbitrarily fix $\hat{v}_{\delta}\in L^{2}_{\delta}(0,T^{*}_{\delta})$ so that

[TABLE]

(The existence of such $\hat{v}_{\delta}$ can be easily verified.) Define $\mathcal{O}_{M,\delta}^{1}$ to be the set of all solutions $u_{\delta}$ to the following problem:

[TABLE]

From (6.58), we see that $\mathcal{O}_{M,\delta}^{1}\subset\mathcal{O}_{M,\delta}$.

We next characterize elements in $\mathcal{O}_{M,\delta}^{1}$ via studying the problem (6.58). To this end, we first claim

[TABLE]

where $z^{*}_{\delta}$ denotes the minimizer of $(JP)_{\delta}^{T^{*}_{\delta},y_{0}}$. Indeed, the first equality in (6.59) follows from (iii) of Theorem 3.1; To show the second one, two facts are given in order. Fact one: From (iii) of Theorem 3.1, we see that the restriction of $u^{*}_{\delta}$ over $(0,T^{*}_{\delta})$, denoted in the same manner, is an optimal control to $(NP)^{T^{*}_{\delta},y_{0}}_{\delta}$. Fact two: By (6.55) and (4.2), we find that $(\delta,T^{*}_{\delta}/\delta)\in\mathcal{P}_{T^{*}_{y_{0}}}$. By these two facts, we can use (4.7) in Theorem 4.3 (with $(\delta,k)=(\delta,T^{*}_{\delta}/\delta)$) to obtain the second equality in (6.59); To show the third equality in (6.59), we recall the above two facts. Then we can apply (ii) in Theorem 4.3 (with $(\delta,k)=(\delta,T^{*}_{\delta}/\delta)$) to get that

[TABLE]

(The first equality on the last line of (6.60) is obtained by the same way as that used to show (4.17).) Then the third equality in (6.59) follows from (6.60) and the first equality in (6.59) at once; To show the last equality in (6.59), we still recall the above two facts (given in the proof of the second equality in (6.59)). Then we can apply (ii) in Theorem 4.3 (with $(\delta,k)=(\delta,T^{*}_{\delta}/\delta)$) to see that

[TABLE]

(The first equality on the last line in (6.61) is obtained by the same way as that used to show (4.17).) From (6.61) and (6.57), we are led to the last equality in (6.59). Hence, (6.59) has been proved.

With the aid of (6.59), we can characterize elements $u_{\delta}$ of $\mathcal{O}_{M,\delta}^{1}$ as follows:

[TABLE]

where the pair $(a_{\delta},b_{\delta},c_{\delta})$ is given by

[TABLE]

Indeed, for each $u_{\delta}=\alpha u^{*}_{\delta}+\beta\hat{v}_{\delta}$, with $\alpha,\beta\in\mathbb{R}$, we have that

[TABLE]

Thus, from (6.59), (6.57) and (6.58), we can easily verify that $u_{\delta}$ is a solution to the problem (6.58) if and only if $u_{\delta}$ is a solution to the problem (6.65).

We now on the position to introduce the desired subset $(\mathcal{O})_{M,\delta}^{2}$. Define a number $\hat{\lambda}$ by

[TABLE]

(Notice that since $\delta\in\mathcal{A}_{M,\eta}$, it follows from (5.53) in Theorem 5.4 that $M>M_{\delta}$.) Let $(\mathcal{O})_{M,\delta}^{2}$ be the set of solutions $u_{\delta}$ to the following problem:

[TABLE]

We claim that

[TABLE]

Since the second conclusion in (6.75) has been proved, we only need to show the first one. Arbitrarily fix $\hat{u}_{\delta}\triangleq\hat{\alpha}u^{*}_{\delta}+\hat{\beta}\hat{v}_{\delta}\in(\mathcal{O})_{M,\delta}^{2}$. We will show that $(\hat{u}_{\delta},\hat{\alpha},\hat{\beta})$ satisfies (6.65). Since $\hat{\lambda}\in(0,1)$ and $M>M_{\delta}$ (see (6.70) and (6.56)), it follows from (6.74) that

[TABLE]

Meanwhile, since $b_{\delta}\leq 0$ (see (6.69) and (6.57)), we find from (6.74) and (6.70) that

[TABLE]

From these, we see that $(\hat{u}_{\delta},\hat{\alpha},\hat{\beta})$ verifies (6.65). Hence, (6.75) is true. By (6.75), (6.74) and (6.57), we find that

[TABLE]

when $\beta$ satisfies that

[TABLE]

(Here, we agree that $\frac{1}{0}\triangleq\infty$.) Then by (6.76) and (6.70), we get that

[TABLE]

where $C_{M,\delta}$ is defined by

[TABLE]

To get a lower bound of $C_{M,\delta}$ w.r.t. $\delta$, we first present the following inequalities (their proofs will be given at the end of the proof of this lemma):

[TABLE]

where $C_{3}$ and $C_{1}$ are given by Theorem 5.2 and (i) of Theorem 5.1, respectively. We next define

[TABLE]

where $\delta_{1}$ is given by (6.54). From (6.79) and (6.80), we have that

[TABLE]

This, along with (6.78) and (6.79), yields that for each $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{M})$,

[TABLE]

By this and the last inequality in (6.79), we can find $C_{M}^{\prime}\triangleq C_{M}^{\prime}(y_{0},r)>0$ so that $C_{M,\delta}\geq C_{M}^{\prime}$, when $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{M})$. (Hence, $C_{M}^{\prime}$ is a lower bound for $C_{M,\delta}$ w.r.t. $\delta$.) This, along with (6.77), (6.56) and (6.80), yields that for each $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{M})$,

[TABLE]

By the above and (6.80), we obtain (6.53), with $\hat{C}_{M}=\frac{1}{\sqrt{2}}C_{M}^{\prime}\lambda_{1}^{3/4}\sqrt{r}$.

Finally, we show (6.79). By the Hölder inequality, (6.69) and (6.57), we find that

[TABLE]

Similarly, from (6.69) and (6.59), we can obtain the estimate for $c_{\delta}$ in (6.79). We now show the third inequality in (6.79). By (6.55), we have that

[TABLE]

From the first two conclusions in (6.81), we can apply (ii) of Theorem 4.1 and the first inequality in (5.45) in Theorem 5.3 (with $(\delta,k)=(\delta,T^{*}_{\delta}/\delta)$), as well as (6.56), to get that

[TABLE]

From (6.82), the second inequality in (5.26) in Theorem 5.2, with $T_{1}=T^{*}$ and $T_{2}=T^{*}_{\delta}$, (Notice that $T^{*}_{\delta}>T^{*}$.) and the last inequality in (6.81), we can easily derive the last inequality in (6.79). Hence, (6.79) is true. This ends the proof of Lemma 6.1. ∎

We are now on the position to prove Theorem 1.5.

Proof of Theorem 1.5.

Arbitrarily fix $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. For each $M>0$ and $\delta>0$, we let $u_{M}^{*}$ and $u^{*}_{\delta,M}$ be the optimal control and the optimal control with the minimal norm to $(TP)^{M,y_{0}}$ and $(TP)^{M,y_{0}}_{\delta}$ respectively (see Theorem 3.1). Throughout the proof of Theorem 1.5, we simply write respectively $T^{*}$ and $T^{*}_{\delta}$ for $T(M,y_{0})$ and $T_{\delta}(M,y_{0})$; simply write $L^{2}(0,T^{*})$ and $L^{2}(0,T^{*}_{\delta})$ for $L^{2}(0,T^{*};L^{2}(\Omega))$ and $L^{2}(0,T^{*}_{\delta};L^{2}(\Omega))$ respectively. We will prove the conclusions (i)-(ii) of Theorem 1.5 one by one.

(i) Arbitrarily fix $M_{1}$ and $M_{2}$, with $0<M_{1}<M_{2}$. Then arbitrarily fix $M\in[M_{1},M_{2}]$. For each $\delta>0$, there are only two possibilities: either (6.11) or (6.13) holds. In the case when $\delta$ verifies (6.11), we can obtain (1.14) by the similar way to that used to show (1.12). We next consider the case that $\delta$ satisfies (6.13). Recall (1.23) for the subset $\mathcal{O}_{M,\delta}$ (which consists of all optimal controls to $(TP)_{\delta}^{M,y_{0}}$). Then it follows from Definition 1.2 that

[TABLE]

Arbitrarily fix $v_{\delta}\in\mathcal{O}_{M,\delta}$. One can directly check that

[TABLE]

From this and (6.83), we find that for each $\lambda\in(0,1)$,

[TABLE]

Dividing the above by $\lambda$ and then sending $\lambda\rightarrow\infty$, we obtain that

[TABLE]

From this, (6.83) and (6.19) (as well as (6.20)), one can directly check that

[TABLE]

(Here, we used the fact that $M\geq N_{\delta}(T^{*}_{\delta},y_{0})$, which follows from (iii) of Theorem 3.1.) Hence, from (6.49), (6.84) and (6.31), we find that

[TABLE]

where $C_{1}(M_{2},y_{0})$ and $C_{4}(M_{1},M_{2},y_{0},r)$ are respectively given by (6.31) and (6.49). The continuity of $C_{5}(M_{1},M_{2},y_{0},r)$ follows from the continuity of $C_{1}(M_{2},y_{0})$ and $C_{4}(M_{1},M_{2},y_{0},r)$. This ends the proof of the conclusion (i) of Theorem 1.5.

(ii) We mainly use Lemma 6.1 to prove (1.15). Arbitrarily fix $M>0$ and $\eta\in(0,1)$. Let $\mathcal{A}_{M,\eta}$ be given by Theorem 5.4. Let $\hat{C}_{M}$ and $\delta_{M}$ be given by Lemma 6.1. Arbitrarily fix $\delta\in\mathcal{A}_{M,\eta}\cap(0,\delta_{M})$. We claim that there is $\hat{u}_{M,\delta}\in\mathcal{O}_{M,\delta}$ so that

[TABLE]

By contradiction, we suppose that it were not true. Then we would find that

[TABLE]

This, along with the definition of $\mathcal{O}_{M,\delta}$ (see (6.53)), implies that

[TABLE]

which contradicts Lemma 6.1. Thus, (6.85) is true.

Now, we arbitrarily fix $\hat{u}_{M,\delta}\in\mathcal{O}_{M,\delta}$ satisfying (6.85). Then by (6.85) and by (i) of Theorem 1.4 (with $M_{1}=M/2$ and $M_{2}=M$), there is $C(M,y_{0},r)>0$ so that

[TABLE]

Write

[TABLE]

Then, one can easily check that

[TABLE]

From (6.86), and (6.87), one can easily verify that

[TABLE]

which leads to (1.15), with $C_{M}\triangleq\hat{C}_{M}/6$. This ends the proof of Theorem 1.5. ∎

6.4 Further discussions on the main results

From (ii) of Theorem 1.3, we see that when $\delta\in\mathcal{A}_{M,\eta}$, $T_{\delta}(M,y_{0})-T(M,y_{0})$ has a lower bound $(1-\eta)\delta$. The next Theorem 6.2 tells us that when $\delta\notin\mathcal{A}_{M,\eta}$, $(1-\eta)\delta$ will not be a lower bound for $T_{\delta}(M,y_{0})-T(M,y_{0})$.

Theorem 6.2**.**

Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. Then for each $M>0$, there is $k_{0}\in\mathbb{N}^{+}$ and $\{\delta_{k}\}_{k=k_{0}}^{\infty}\subset\mathbb{R}^{+}$, with $\lim_{k\rightarrow\infty}\delta_{k}=0$, so that when $k\geq k_{0}$,

[TABLE]

Proof.

Arbitrarily fix $M>0$ and $\eta\in(0,1)$. Throughout the proof of Theorem 6.2, we simply write $T^{*}$ for $T(M,y_{0})$. Let

[TABLE]

We define a sequence $\{\delta_{k}\}_{k=k_{0}}^{\infty}$ of $\mathbb{R}^{+}$ in the following manner:

[TABLE]

One can easily check that

[TABLE]

We now claim that there exists $k_{0}\geq\hat{k}_{0}$ so that

[TABLE]

In fact, by (6.90), we can choose $\hat{k}_{1}\geq k_{0}$ large enough so that

[TABLE]

Arbitrarily fix $k\geq\hat{k}_{1}$. Since $T^{*}<T^{*}_{y_{0}}$ (see (iii) in Theorem 4.1), from (6.93) and (6.91), we can easily check that

[TABLE]

These, along with (4.2), yield that

[TABLE]

By (6.94), we can apply Theorem 5.3 (see the second inequality in (5.45), where $(\delta,k)$ is replaced by $(\delta,k+1)$) and Theorem 5.2 (see the first inequality in (5.26), with $T_{1}=T^{*}$ and $T_{2}=(k+1)\delta_{k}$) to obtain that

[TABLE]

where $C_{4}\triangleq C_{4}(\Omega,\omega)$ is given by (5.45). Meanwhile, by (6.91) and (6.93), we find

[TABLE]

These, along with (6.95) and (ii) of Theorem 4.1, yield that

[TABLE]

This, together with (6.89), leads to (6.92), with $k_{0}=\hat{k}_{1}$.

Next, we arbitrarily fix $k\geq k_{0}\triangleq\hat{k}_{1}$. Let $u_{\delta_{k}}$ be an admissible control to $(NP)^{(k+1)\delta_{k},y_{0}}_{\delta_{k}}$. Let $\tilde{u}_{\delta_{k}}$ be the zero extension of $u_{\delta_{k}}$ over $\mathbb{R}^{+}$. Then by (6.92), one can easily check that $\tilde{u}_{\delta_{k}}$ is an admissible control (to $(TP)^{M,y_{0}}_{\delta_{k}}$), which drives the solution to $B_{r}(0)$ at time $(k+1)\delta_{k}$. This, along with the optimality of $T_{\delta}(M,y_{0})$, yields that

[TABLE]

Meanwhile, Since $\mathcal{U}^{M}_{\delta_{k}}\subset\mathcal{U}^{M}$, we find from (1.1) and (1.8) that $T^{*}\leq T_{\delta_{k}}(M,y_{0})$. From this and (6.91), we get that

[TABLE]

Since $T_{\delta_{k}}(M,y_{0})$ is a multiple of $\delta_{k}$ (see (1.9)), from (6.96) and (6.97), we obtain that

[TABLE]

This, along with (6.91) and (6.89), yields (6.88), with $C_{M}=a(M,y_{0},r)$ and with $k_{0}$ given by (6.92). Thus, we end the proof of Theorem 6.2. ∎

Remark 6.3**.**

(i) Let $y_{0}\in L^{2}(\Omega)\setminus B_{r}(0)$. The above theorem implies that the following conclusion is not true: For each $M>0$, there exists $\delta_{1}>0$ and $C>0$ so that

[TABLE]

(ii) We think of that the similar result to that in Theorem 1.3 can be obtained for optimal controls. But it seems for us that the corresponding proof will be more complicated.

7 Appendix

The next Lemma 7.1 is the copy of [36, Lemma 5.1].

Lemma 7.1**.**

Let $\mathbb{K}$ be either $\mathbb{R}$ or $\mathbb{C}$. Let $X$, $Y$ and $Z$ be three Banach spaces over $\mathbb{K}$, with their dual spaces $X^{*}$, $Y^{*}$ and $Z^{*}$. Let $R\in\mathcal{L}(Z,X)$ and $O\in\mathcal{L}(Z,Y)$. Then the following two propositions are equivalent:

(i) There exists $\widehat{C}_{0}>0$ and $\hat{\varepsilon}_{0}>0$ so that for each $z\in Z$,

[TABLE]

(ii) There is $C_{0}>0$ and $\varepsilon_{0}>0$ so that for each $x^{*}\in X^{*}$, there is $y^{*}\in Y^{*}$ satisfying that

[TABLE]

Furthermore, when one of the above two propositions holds, the pairs $(C_{0},\varepsilon_{0})$ and $(\widehat{C}_{0},\hat{\varepsilon}_{0})$ can be chosen to be the same.

Proof of Lemma 7.1.

The proof is divided into the following several steps:

Step 1. To show that (ii)$\Rightarrow$(i)

Suppose that (ii) is true. Then, for each $x^{*}\in X^{*}$, there is $y^{*}_{x^{*}}\in Y^{*}$ so that (7.2), with $y^{*}=y^{*}_{x^{*}}$, is true. From this, one can easily check that for any $x^{*}\in X^{*}$ and $z\in Z$,

[TABLE]

By this and the Cauchy-Schwarz inequality, we deduce that for each $x^{*}\in X^{*}$ and $z\in Z$,

[TABLE]

Hence, (7.1), with (\widehat{C}_{0},\hat{\varepsilon}_{0})$$=(C_{0},\varepsilon_{0}), is true.

Step 2. To show that (i)$\Rightarrow$(ii)

Suppose that (i) is true. Define a subspace $E$ of $Y\times Z$ in the following manner:

[TABLE]

The norm of $E$ is inherited form the following usual norm of $Y\times Z$:

[TABLE]

Arbitrarily fix $x^{*}\in X^{*}$. Define an operator $\mathcal{T}_{x^{*}}:E\rightarrow\mathbb{K}$ in the following manner:

[TABLE]

By (7.1) and (7.4), we can easily check that $\mathcal{T}_{x^{*}}$ is well defined and linear. We now claim

[TABLE]

Indeed, by the definition of $E$, we see that given $(f,g)\in E$, there is $z\in Z$ so that

[TABLE]

Then by (7.4), we find that |\mathcal{T}_{x^{*}}\big{(}(f,g)\big{)}|=|\langle x^{*},Rz\rangle_{X^{*},X}|\leq\|x^{*}\|_{X^{*}}\|Rz\|_{X}. This, along with (7.1), shows (7.5).

Since $\mathcal{T}_{x^{*}}$ is a linear and bounded functional, we can apply the Hahn-Banach extension theorem to find $\widetilde{\mathcal{T}}_{x^{*}}$ in $(Y\times Z)^{*}$ so that

[TABLE]

and so that

[TABLE]

These, together with (7.3) and (7.5), yield that for all $f\in T$ and $g\in Z$,

[TABLE]

Thus, there exists $(y^{*}_{x^{*}},z^{*}_{x^{*}})\in Y^{*}\times Z^{*}$ so that for all $(f,g)\in Y\times Z$,

[TABLE]

from which, it follows that

[TABLE]

We now claim that

[TABLE]

When (7.9) is proved, the conclusion (ii) (with (C_{0},\varepsilon_{0})$$=(\widehat{C}_{0},\hat{\varepsilon}_{0})) follows at once.

To prove the first inequality in (7.9), we see from (7.8), (7.7) and (7.3) that

[TABLE]

Meanwhile, for each $\delta\in(0,1)$, we can choose $(f_{\delta},g_{\delta})\in Y\times Z$ so that

[TABLE]

where $o_{1}(1)$ and $o_{2}(1)$ are so that $\lim_{\delta\rightarrow 0^{+}}o_{1}(1)=\lim_{\delta\rightarrow 0^{+}}o_{2}(1)=0$. From these, it follows that

[TABLE]

Sending $\delta\rightarrow 0^{+}$ in the above inequality leads to the first inequality in (7.9).

To prove the second equality in (7.9), we find from (7.4), (7.6) and (7.8) that

[TABLE]

which yields that for all $z\in Z$,

[TABLE]

This leads to the second equality in (7.9).

Step 3. On the pairs $(C_{0},\varepsilon_{0})$ and $(\widehat{C}_{0},\hat{\varepsilon}_{0})$

From the proofs in Step 1 and Step 2, we see that when one of the propositions (i) and (ii) holds, $(C_{0},\varepsilon_{0})$ and $(\widehat{C}_{0},\hat{\varepsilon}_{0})$ can be chosen to be the same. This ends the proof of Lemma 7.1. ∎

The next Lemma 7.1 is a part copy of the proof of [29, Theorem 2.1].

Lemma 7.2**.**

There exists $C\triangleq C(\Omega,\omega)>0$ so that for each $t>0$ and $z\in L^{2}(\Omega)$,

[TABLE]

Proof.

Write $\{\lambda_{j}\}_{j=1}^{\infty}$ and $\{e_{j}\}_{j=1}^{\infty}$ for the eigenvalues and the normalized orthogonal eigenfunctions for $-\Delta$ with the zero Dirichlet boundary condition. Arbitrarily fix $z\in L^{2}(\Omega)$ and $0\leq t<T$. We have that

[TABLE]

Then it follows that

[TABLE]

Recall that the Lebeau-Robbiano inequality says that there exists $C\triangleq C(\Omega,\omega)>0$ so that for each $\lambda>0$,

[TABLE]

We apply the above inequality (where $a_{j}=e^{-\lambda_{j}t}z_{j}$), as well as (7.11), to get that for each $\lambda>0$,

[TABLE]

This, as well as the following facts

[TABLE]

indicates that for each $\lambda>0$,

[TABLE]

Since the function $\lambda\rightarrow e^{-\lambda t/2}$, $\lambda\in\mathbb{R}^{+}$ can take any value in $(0,1)$, it follows by (7.12) that for each $\varepsilon\in(0,1)$,

[TABLE]

Meanwhile, one can easily check that

[TABLE]

From this and (7.13), we deduce that for each $\varepsilon>0$,

[TABLE]

Take the infimum in (7.14) for $\varepsilon$ over $\mathbb{R}^{+}$ to get (7.10). This ends the proof.

∎