In this section, we first state two time optimal control problems, and then study the solvability of these problems, obtain the bang-bang property of the time optimal controls. Throughout this section, for all \(i\in\{1,2,\ldots ,K\}\), \(E_{i}=(0,T)\), and
$$ u_{i}\in\mathcal{U}_{\mathrm{ad}}^{i}\equiv \bigl\{ u\in L^{ \infty}\bigl(0,+\infty; L^{2}(\Omega)\bigr) \mid \bigl\Vert u(t)\bigr\Vert _{L^{2}(\Omega)} \leq M_{i} \mbox{ a.e. } t\in(0,+\infty)\bigr\} . $$
(2.1)
When \(K=1\), for simplicity, we write \(\mathcal{U}_{\mathrm{ad}}\) for \(\mathcal{U}_{\mathrm{ad}}^{i}\).
In the following, we consider the following two time optimal control problems subject to (1.1):
Problem (TP1)
$$T^{*}=\inf\bigl\{ T\mid y\bigl(\cdot, T; \{ \chi_{\omega_{i}}u_{i}\} , y _{0}\bigr)=0, u_{i} \in\mathcal{U}_{\mathrm{ad}}^{i} \mbox{ for all } i\in\{1,2,\ldots ,K\} \bigr\} . $$
Problem (TP2)
For \(K=1\),
$$T_{\varepsilon }^{*}=\inf\bigl\{ T\mid y(\cdot, T; \chi_{\omega}u, y_{0}) \in\bar{B}(0,\varepsilon ), u\in \mathcal{U}_{\mathrm{ad}} \bigr\} . $$
Here and in what follows, we denote by \(B(u, r)\) the open ball in \(L^{2}(\Omega)\) with center \(u\in L^{2}(\Omega)\) and radius \(r>0\), and by \(\bar{B}(u,r)\) the closed ball in \(L^{2}(\Omega)\) with center \(u\in L^{2}(\Omega)\) and radius \(r>0\).
In order to obtain the solvability of Problem (TP1), we assume that there exists a constant \(M>0\) such that
$$ M_{i}\leq M \quad\mbox{for all } i\in\{1,2,\ldots ,K\}. $$
(2.2)
Notice that the hypothesis (2.2) is reasonable: for a single control system
$$\textstyle\begin{cases} \partial_{t}y(x,t)-\Delta y(x,t)=\rho(x)\chi_{\omega}(x) u(x,t), &\mbox{in } \Omega\times(0,+\infty), \\ y(x,t)=0, &\mbox{on } \partial\Omega\times(0, +\infty), \\ y(x,0)=y_{0}(x), &\mbox{in } \Omega, \end{cases} $$
its optimal time
$$T^{*}\equiv\inf\bigl\{ T\mid y(\cdot, T; \chi_{\omega}u, y_{0})=0, u \in\mathcal{U}_{\mathrm{ad}} \bigr\} \rightarrow0 $$
as \(M\rightarrow\infty\).
It is obvious that Problem (TP1) is related to null controllable problem of (1.1), while Problem (TP2) is related to approximately controllable problem of (1.1). It is well known that, when \(K=1\) and \(\rho\equiv1\), the system (1.1) is null controllable for the measurable control domain ω (see [5]), even if the characteristic function \(\chi_{\omega}\) can be relaxed by a measurable function \(\beta\in L^{2}(\Omega)\) with \(0\leq\beta\leq1\) for a.e. \(x\in\Omega\) and \(\int_{\Omega}\beta^{2}(x)\,dx=\alpha\vert \Omega\vert\) (see [6]). Here \(\alpha\in(0,1)\) is a given constant and \(\vert\Omega\vert\) is the Lebesgue measure of Ω. It is natural that there exist a positive constant T and a control \(u\in L^{\infty}(0,T; L^{2}(\Omega))\) such that \(y(x, T;u,y_{0})=0\) (see [5, 6]).
The following result is related to the solvability of Problem (TP1).
Theorem 2.1
Let
\(\{M_{i}\}\)
be a given positive real number sequence satisfying (2.2). Then there exists
\(T^{*}>0\), such that
\(T^{*}\)
is the solution to Problem
(TP1). Moreover, for each
\(i=1,2,\ldots ,K\), there exists a unique
\(u_{i}^{*}\in L^{\infty}(0, T^{*}; L^{2}( \Omega))\), such that
$$ \bigl\Vert u_{i}^{*}\bigr\Vert _{L^{2}(\Omega)}=M_{i}\quad \textit{for a.e. } t\in\bigl(0,T^{*} \bigr) \textit{ with }M_{i}\leq M \textit{ for all } i\in\{1,2,\ldots ,K\}, $$
(2.3)
i.e., the time optimal controls sequence of Problem
(TP1)
has the bang-bang property.
The following lemma is needed in proving Theorem 2.1, which comes from [5, 9].
Lemma 2.2
Let
\(E\subset[0,T]\)
and
\(\omega\subset\Omega\)
be two positive measurable sets. Then, for each
\(y_{0}\in L^{2}(\Omega)\), there is a bounded control function
\(u(\cdot)\in L^{\infty}(0,T; L^{2}(\Omega ))\)
with
$$\Vert u\Vert _{L^{\infty}(0,T; L^{2}(\Omega))}\leq C\Vert y_{0}\Vert _{L^{2}(\Omega)}, $$
such that the solution to the equation
$$\textstyle\begin{cases} \partial_{t} y(x,t)-\Delta y(x,t)=\rho(x)\chi_{{E}}(t)\chi_{\omega }(x) u(x,t), &\textit{in } \Omega\times(0,T), \\ y(x,t)=0, &\textit{on } \partial\Omega\times(0,T), \\ y(x,0)=y_{0}(x), &\textit{in } \Omega, \end{cases} $$
satisfies
\(y(\cdot, T; u, y_{0})=0\). Here
\(C=C(\Omega, T, \vert E\vert, \vert \omega\vert)\)
is a constant.
We are now in the position to prove Theorem 2.1.
Proof of Theorem 2.1
Since the proof is long, we separate it into two steps.
Step 1. For fixed \(i_{0}\in\{1,2,\ldots ,K\}\), consider the following system:
$$\textstyle\begin{cases} \partial_{t}y(x,t)-\Delta y(x,t)=\rho(x)\chi_{\omega_{i_{0}}}(x) u _{i_{0}}(x,t), &\mbox{in } \Omega\times(0,+\infty), \\ y(x,t)=0, &\mbox{on } \partial\Omega\times(0, +\infty), \\ y(x,0)=y_{0}(x), &\mbox{in } \Omega. \end{cases} $$
By Lemma 2.1 of [6], we know that there exist a control \(u_{i_{0}}\in\mathcal{U}_{\mathrm{ad}}^{i_{0}}\) and T, such that \(y(\cdot ,T,\chi_{\omega_{i_{0}}}u_{i_{0}},y_{0})=0\). we also know that \(0\neq y_{0}\in L^{2}(\Omega)\) is a given function. Therefore,
$$\begin{aligned} 0&< T^{*} \equiv\inf\bigl\{ T\mid y \bigl(\cdot, T; \{\chi_{\omega_{i}}u _{i}\}, y_{0}\bigr)=0, u^{i}\in\mathcal{U}_{\mathrm{ad}}^{i} \mbox{ for all } i\in \{1,2,\ldots ,K\} \bigr\} \\ &\leq\inf\bigl\{ T\mid y(\cdot, T; \chi_{\omega_{i_{0}}}u_{i_{0}}, y_{0})=0, u_{i_{0}}\in\mathcal{U}_{\mathrm{ad}}^{i_{0}} \bigr\} < \infty. \end{aligned} $$
Hence, there exists a sequence \(\{T_{n}\}\), such that \(\{T_{n}\}\) is a monotone decreasing sequence with \(y(\cdot, T_{n}; \{\chi_{\omega_{i}}u _{i}^{n}\}, y_{0})=0\) and
$$T_{n}\rightarrow T^{*}\equiv\inf\bigl\{ T\mid y \bigl(\cdot, T; \{ \chi_{\omega_{i}}u_{i}\}, y_{0}\bigr)=0, u_{i}\in\mathcal{U}_{\mathrm{ad}}^{i} \mbox{ for all } i\in \{1,2,\ldots ,K\} \bigr\} . $$
Without loss of generality, we assume that \(T_{n}\leq T^{*}+1\) for all \(n\in\mathbb{N}\). Then \(y^{n}\equiv y(\cdot, \cdot; \{\chi_{\omega _{i}}u_{i}^{n}\}, y_{0})\) is a solution to the following equation:
$$\textstyle\begin{cases} \partial_{t}y^{n}(x,t)-\Delta y^{n}(x,t)=\rho(x)\sum_{i=1}^{K} \chi_{\omega_{i}} (x) u_{i}^{n}(x,t), &\mbox{in } \Omega\times(0,T _{n}), \\ y^{n}(x,t)=0, &\mbox{on } \partial\Omega\times(0, T_{n}), \\ y^{n}(x,0)=y_{0}(x), y^{n}(x,T_{n})=0, &\mbox{in } \Omega. \end{cases} $$
Now, denote
$$\tilde{u}_{i}^{n}(x,t)= \textstyle\begin{cases} u_{i}^{n}(x,t), & (x,t)\in\Omega\times(0,T_{n}), \\ 0, & (x,t)\in\Omega\times[T_{n}, +\infty). \end{cases} $$
Then
$$\tilde{y}^{n}(x,t)= \textstyle\begin{cases} y^{n}(x,t), & (x,t)\in\Omega\times(0,T_{n}), \\ 0, & (x,t)\in\Omega\times[T_{n}, +\infty), \end{cases} $$
solves the following system:
$$\textstyle\begin{cases} \partial_{t}\tilde{y}^{n}(x,t)-\Delta\tilde{y}^{n}(x,t)=\rho(x) \sum_{i=1}^{K} \chi_{\omega_{i}} (x) \tilde{u}_{i}^{n}(x,t), &\mbox{in } \Omega\times(0, +\infty), \\ \tilde{y}^{n}(x,t)=0, &\mbox{on } \partial\Omega\times(0, + \infty), \\ \tilde{y}^{n}(x,0)=y_{0}(x), &\mbox{in } \Omega, \\ \tilde{y}_{n}(x,t)=0, &\mbox{in } \Omega\times[T_{n}, +\infty). \end{cases} $$
Moreover, by the definition of \(\tilde{y}^{n}\), it is easy to see that \(\tilde{y}^{n}\) solves the following system:
$$\textstyle\begin{cases} \partial_{t}\tilde{y}^{n}(x,t)-\Delta\tilde{y}^{n}(x,t)=\rho(x) \sum_{i=1}^{K} \chi_{\omega_{i}} (x) \tilde{u}_{i}^{n}(x,t), &\mbox{in } \Omega\times(0, T^{*}+1), \\ \tilde{y}^{n}(x,t)=0, &\mbox{on } \partial\Omega\times(0, T^{*}+1), \\ \tilde{y}^{n}(x,0)=y_{0}(x), &\mbox{in } \Omega, \\ \tilde{y}_{n}(x,t)=0, &\mbox{in } \Omega\times[T_{n}, T^{*}+1). \end{cases} $$
Note that \(\Vert \tilde{u}_{1}^{n}\Vert _{L^{2}(\Omega)}\leq M_{1}\) for all \(n\in\mathbb{N}\). Then there exist a subsequence \(\{\tilde{u}_{1} ^{n_{1}}\}\subset L^{\infty}(0,T^{*}+1; L^{2}(\Omega))\) of \(\{\tilde{u}_{1}^{n}\}\) and \(\tilde{u}_{1}^{0}\in L^{\infty}(0,T^{*}+1; L^{2}(\Omega))\) such that
$$\tilde{u}_{1}^{n_{1}}\rightarrow\tilde{u}_{1}^{0} \quad\mbox{weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T^{*}+1; L^{2}(\Omega)\bigr) \mbox{ as } n_{1} \rightarrow\infty. $$
Similarly, since \(\Vert u_{2}^{n_{1}}\Vert _{L^{2}(\Omega)}\leq M_{2}\) for all \(n_{1}\in\mathbb{N}\), there exist a subsequence \(\{\tilde{u}_{2}^{n _{2}}\}\) of \(\{\tilde{u}_{2}^{n_{1}}\}\) and \(\tilde{u}_{2}^{0}\in L ^{\infty}(0,T^{*}+1; L^{2}(\Omega))\) such that
$$\tilde{u}_{2}^{n_{2}}\rightarrow\tilde{u}_{2}^{0} \quad\mbox{weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T^{*}+1; L^{2}(\Omega)\bigr)\mbox{ as } n_{2} \rightarrow\infty. $$
By inductive argument, for each \(i\in\{1,2,\ldots ,K\}\), there exist a subsequence \(\{\tilde{u}_{i}^{n_{i}}\}\) of \(\{\tilde{u}_{i}^{n_{i-1}} \}\) and \(\tilde{u}_{i}^{0}\in L^{\infty}(0,T^{*}+1; L^{2}(\Omega))\) such that
$$\tilde{u}_{i}^{n_{i}}\rightarrow\tilde{u}_{i}^{0} \quad\mbox{weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T^{*}+1; L^{2}(\Omega)\bigr)\mbox{ as } n_{i} \rightarrow\infty. $$
By the diagram argument, for all \(i\in\{1,2,\ldots ,K\}\), we can abstract a subsequence \(\{\tilde{u}_{i}^{n_{n}}\}\) of \(\{\tilde{u}_{i}^{n}\}\) such that
$$\tilde{u}_{i}^{n_{n}}\rightarrow\tilde{u}_{i}^{0} \quad\mbox{weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T^{*}+1; L^{2}(\Omega)\bigr)\mbox{ as } n\rightarrow\infty. $$
Since \(\omega_{i}\cap\omega_{j}=\emptyset\) for all \(i, j\in\{ 1,2,\ldots ,K \}\) with \(i\neq j\), one can get
$$\rho(x)\sum_{i=1}^{K} \chi_{\omega_{i}} \tilde{u}_{i}^{n_{n}} \rightarrow\rho(x)\sum _{i=1}^{K} \chi_{\omega_{i}}\tilde{u}_{i}^{0} \quad\mbox{weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T^{*}+1; L^{2}(\Omega)\bigr) \mbox{ as } n\rightarrow\infty. $$
On the other hand, \(\tilde{y}^{n_{n}}\) is the solution to the following system:
$$\textstyle\begin{cases} \partial_{t}\tilde{y}^{n_{n}}(x,t)-\Delta\tilde{y}^{n_{n}}(x,t)= \rho(x)\sum_{i=1}^{K} \chi_{\omega_{i}} (x) \tilde{u}_{i}^{n_{n}}(x,t), &\mbox{in } \Omega\times(0, T^{*}+1), \\ \tilde{y}^{n_{n}}(x,t)=0, &\mbox{on } \partial\Omega\times(0, T ^{*}+1), \\ \tilde{y}^{n_{n}}(x,0)=y_{0}(x), &\mbox{in } \Omega, \\ \tilde{y}_{n_{n}}(x,t)=0, &\mbox{in } \Omega\times[T_{n_{n}}, T ^{*}+1). \end{cases} $$
Then there exist a subsequence of \(\{\tilde{y}^{n_{n}}\}\), still so denoted, and \(\tilde{y}^{0}\) such that
$$\tilde{y}^{n_{n}}\rightarrow\tilde{y}^{0}\quad \mbox{weakly in } L^{2}\bigl( \Omega\times\bigl(0,T^{*}+1\bigr)\bigr) \mbox{ as } n\rightarrow\infty $$
and
$$\textstyle\begin{cases} \partial_{t}\tilde{y}^{0}(x,t)-\Delta\tilde{y}^{0}(x,t)=\rho(x) \sum_{i=1}^{K} \chi_{\omega_{i}} (x) \tilde{u}_{i}^{0}(x,t), &\mbox{in } \Omega\times(0, T^{*}+1), \\ \tilde{y}^{0}(x,t)=0, &\mbox{on } \partial\Omega\times(0, T^{*}+1), \\ \tilde{y}^{0}(x,0)=y_{0}(x), &\mbox{in } \Omega. \end{cases} $$
Note that \(\tilde{y}^{n_{n}}=0\) in \(\Omega\times[T_{n_{n}}, T^{*}+1)\). Since for \(n\in\mathbb{N}\), one has \(\tilde{y}^{n_{n}}=0\) in \(\Omega\times[T_{n_{n}}, T^{*}+1)\),
$$\tilde{y}^{n_{n}}\rightarrow\tilde{y}^{0}\quad \mbox{in } L^{2}\bigl(\Omega\times\bigl(0,T^{*}+1\bigr)\bigr) \mbox{ as } n\rightarrow\infty. $$
We get \(\tilde{y}^{0}=0\) in \(\Omega\times[T_{n_{n}}, T^{*}+1)\). We get \(\tilde{y}^{0}=0\) in \([T^{*}, T^{*}+1)\) since \(T_{n_{n}}\rightarrow T^{*}\) as \(n\rightarrow\infty\). Take
$$u_{i}^{0}=\tilde{u}_{i}^{0}\vert _{\Omega\times(0,T^{*})},\quad \mbox{and} \quad y ^{0}=\tilde {y}^{0}\vert _{\Omega\times(0,T^{*})}. $$
By the fact that \(\tilde{y}^{0} \in C((0,T]; L^{2}(\Omega))\), \(y^{0}\) is the solution to the following system:
$$\textstyle\begin{cases} \partial_{t} y^{0}(x,t)-\Delta y^{0}(x,t)=\rho(x)\sum_{i=1}^{K} \chi_{\omega_{i}} (x) u_{i}^{0}(x,t), &\mbox{in } \Omega\times(0, T ^{*}), \\ y^{0}(x,t)=0, &\mbox{on } \partial\Omega\times(0, T^{*}), \\ y^{0}(x,0)=y_{0}(x),\qquad y^{0}(x,T^{*})=0, &\mbox{in } \Omega, \end{cases} $$
which implies that \(\{u_{i}^{0}\}\) are the desired controls respect to the optimal time \(T^{*}\).
Step 2. In the following, we shall show that the time optimal control of Problem (TP1) has the bang-bang property. Otherwise, we suppose that there exist \(i_{0}\in\{1,2,\ldots ,K\}\) and a subset \(E^{0}\subset[\alpha, T^{*}-\alpha]\) with positive measure for some \(\alpha>0\) and a positive number \(\varepsilon _{0}\), such that
$$u_{i_{0}}^{*}\in\mathcal{U}_{\mathrm{ad}}^{i_{0}}\quad\mbox{and}\quad M_{i_{0}}-\bigl\Vert u_{i_{0}}^{*}\bigr\Vert _{L^{2}(\Omega)}\geq \varepsilon _{0},\quad \mbox{for each } t \mbox{ in the set } E^{0}, $$
where \(u_{i_{0}}^{*}\) is the time optimal control respect to \(T^{*}\). It is obvious that the solution to (1.1) satisfies
$$y\bigl(\cdot, T^{*}; \bigl\{ \chi_{\omega_{i}}u_{i}^{*} \bigr\} , y_{0}\bigr)=0, $$
and, for each \(t\in E^{0}\), \(B(u_{i_{0}}^{*}(t), \frac{\varepsilon _{0}}{2}) \subset B(0, M_{i_{0}})\).
We denote by \(e^{\Delta t}\) the semigroup generated by Δ with the Dirichlet boundary condition. Set
$$ h_{\delta}= \int_{0}^{\delta}e^{\Delta(\delta-\sigma)}\rho\sum _{i=1}^{K} \chi_{\omega_{i}} u_{i}^{*}( \sigma)\,d\sigma+\bigl(e^{\Delta \delta}-I\bigr)y_{0}. $$
(2.4)
Considering the following system:
$$ \textstyle\begin{cases} z_{t}^{\delta}(x,t)-\Delta z^{\delta}(x,t)=\rho(x) \chi_{E_{\delta}^{0}}(t)\chi_{\omega_{i_{0}}}(x) w_{\delta}(x,t), &\mbox{in } \Omega\times(0,T^{*}-\delta), \\ z(x,t)=0, &\mbox{on } \partial\Omega\times(0,T^{*}-\delta), \\ z(x,0)=-h_{\delta}(x), &\mbox{in } \Omega, \end{cases} $$
(2.5)
where \(E_{\delta}^{0}\) is the set \(\{t\mid t+\delta\in E^{0}\}\). By Lemma 2.2, there exist positive constants \(\delta_{1}\) and \(L=L(\Omega, T, \vert E_{0}\vert, \vert\omega_{i_{0}}\vert)\), such that, for each δ with \(0<\delta\leq\delta_{1}\), there is a control \(w_{\delta}\) in the space \(L^{\infty}(0,T^{*}-\delta; L^{2}(\Omega ))\) with the estimate
$$\Vert w_{\delta} \Vert _{L^{\infty}(0,T^{*}-\delta; L^{2}(\Omega ))}\leq L \Vert h_{\delta} \Vert _{L^{2}(\Omega)}, $$
and the solution to (2.5) satisfies
$$ z^{\delta}\bigl(\cdot ,T^{*}-\delta;\omega _{\delta },-h_{\delta }\bigr)=0. $$
(2.6)
On the other hand, by (2.4), there exists a positive number \(\delta_{2}\) such that, for each positive number δ with \(\delta\leq\delta_{2}\), one has
$$\Vert h_{\delta} \Vert _{L^{2}(\Omega)}\leq\frac{\varepsilon _{0}}{2L}. $$
Therefore, for each \(\delta\leq\delta_{0}\equiv\min\{\delta_{1}, \delta_{2}\}\), there exists a control \(w_{\delta}\) satisfying
$$ \Vert w_{\delta} \Vert _{L^{\infty}(0,T^{*}-\delta ; L^{2}(\Omega))}\leq \frac{\varepsilon _{0}}{2}, $$
(2.7)
and the corresponding solution to (2.5) satisfies (2.6).
Set
$$v_{i}(x,t)= \textstyle\begin{cases} u_{i}^{*}(x,\delta+t)+w_{\delta}(x,t), &\mbox{for } i=i_{0} \mbox{ and } (x,t)\in\Omega\times E_{\delta }^{0}, \\ u_{i}^{*}(x,\delta+t), &\mbox{for } i= i_{0} \mbox{ and } (x,t) \in\Omega\times((0,T^{*}-\delta)-E_{\delta }^{0}), \\ u_{i}^{*}(x,\delta+t), &\mbox{for } i\neq i_{0} \mbox{ and } (x,t) \in\Omega\times(0, T^{*}-\delta), \\ 0, &\mbox{otherwise}. \end{cases} $$
It is obvious that \(v_{i}\in\mathcal{U}_{\mathrm{ad}}^{i}\) for all \(i\in\{1,2,\ldots ,K\}\). Consider the following system:
$$ \textstyle\begin{cases} y_{t}(x,t)-\Delta y(x,t)=\rho(x)\sum_{i=1}^{K} \chi_{\omega_{i}} (x) v_{i}(x,t), &\mbox{in } \Omega\times(0,+\infty), \\ y(x,t)=0, &\mbox{on } \partial\Omega\times(0, +\infty), \\ y(x,0)=y_{0}(x), &\mbox{in } \Omega. \end{cases} $$
(2.8)
For any \(0<\delta<\min\{\frac{T^{*}}{2}, \delta_{0}\}\), it is easy to check that the solution to above equation satisfies
$$\begin{aligned} y\bigl(\cdot, T^{*}-\delta; \{v_{i}\}, y_{0}\bigr) =&e^{\Delta(T^{*}-\delta)}y_{0}+ \rho(x)\sum_{i=1}^{K} \int_{0}^{T^{*}-\delta} e^{\Delta(T ^{*}-\delta-\sigma)}\chi_{\omega_{i}} v_{i}(\sigma)\,d\sigma \\ =&e^{\Delta(T^{*}-\delta)}y_{0}+\sum_{i=1}^{K} \int_{0}^{T^{*}- \delta} e^{\Delta(T^{*}-\delta-\sigma)}\rho(x) \chi_{\omega_{i}} u _{i}^{*}(\delta+\sigma)\,d\sigma \\ &{} + \int_{0}^{T^{*}-\delta} e^{\Delta(T^{*}-\delta- \sigma)}\rho(x) \chi_{E_{\delta }^{0}}\chi_{\omega_{i_{0}}}w_{\delta}( \sigma)\,d\sigma \\ =&e^{\Delta(T^{*}-\delta)}y_{0}+\sum_{i=1}^{K} \int_{\delta}^{T ^{*}} e^{\Delta(T^{*}-\sigma)}\rho(x) \chi_{\omega_{i}} u_{i}^{*}( \sigma)\,d\sigma+e^{\Delta (T^{*}-\delta)}h_{\delta} \\ =&e^{\Delta(T^{*}-\delta)}y_{0}+\sum_{i=1}^{K} \int_{\delta}^{T ^{*}} e^{\Delta(T^{*}-\sigma)}\rho(x) \chi_{\omega_{i}} u_{i}^{*}( \sigma)\,d\sigma \\ &{} + \int_{0}^{\delta}e^{\Delta(T^{*}-\sigma)}\rho(x) \sum _{i=1}^{K} \chi_{\omega_{i}} u_{i}^{*}( \sigma)\,d\sigma+e^{ \Delta T^{*}}y_{0}-e^{\Delta(T^{*}-\delta)}y_{0} \\ =&e^{\Delta T^{*}}y_{0}+ \int_{0}^{T^{*}} e^{\Delta(T^{*}-\sigma)} \rho(x)\sum _{i=1}^{K} \chi_{\omega_{i}} u_{i}^{*}( \sigma)\,d\sigma \\ =&0, \end{aligned}$$
(2.9)
which shows that \(\{v_{i}\}\) are the desired controls such that \(y(\cdot, T^{*}-\delta; \{\chi_{\omega_{i}}v_{i}\}, y_{0})=0\). It contradicts the definition of \(T^{*}\) and we have proved the theorem. □
Remark 2.3
There are some relations between the optimal control problem of (1.1) and shape design problem. For more about shape design problem see [8, 14–16].
Before stating the results on Problem (TP2), we define the following reachable set:
$$ \mathcal{R}(T)=\bigl\{ y(\cdot, T; \chi_{\omega}u, y_{0})\mid u\in\mathcal{U}_{\mathrm{ad}} \bigr\} $$
(2.10)
for each \(T\in(0,+\infty)\).
Theorem 2.4
For any given positive constant
ε, Problem
(TP2)
has a solution
\(T_{\varepsilon }^{*}\), and
\(\mathcal{R}(T_{\varepsilon }^{*})\cap\bar{B}(0,\varepsilon )\)
has only one point belonging to the boundary of
\(B(0,\varepsilon )\). Moreover, the corresponding time optimal control
\(u_{\varepsilon }^{*}\)
is unique and has the bang-bang property.
Proof
Since the proof is long, we separate it into the following several steps.
Step 1. We shall show that there exists at least one time optimal control, i.e., there exists at least one \(u^{*}\in \mathcal{U}_{\mathrm{ad}}\) such that \(y(\cdot, T_{\varepsilon }^{*}; u^{*}, y_{0})\in\bar{B}(0,\varepsilon )\).
Let \(\{T_{n}\}\) be a monotone decreasing sequence such that \(T_{n}\rightarrow T_{\varepsilon }^{*}\) as \(n\rightarrow+\infty\), then there exists a sequence \(\{u_{n}\}\subset\mathcal{U}_{\mathrm{ad}}\) such that \(y(\cdot, T_{n}; \chi_{\omega}u_{n}, y_{0})\in \bar{B}(0,\varepsilon )\). Set
$$\tilde{u}_{n}(t)= \textstyle\begin{cases} u_{n}(t), & t\in(0, T_{n}), \\ 0, & t\in[T_{n}, T_{1}). \end{cases} $$
Since \(M\in L^{\infty}(0,T_{1}; L^{2}(\Omega))\), \(\{\tilde{u}_{n}\}\) is a bounded sequence in \(L^{\infty}(0,T_{1}; L^{2}(\Omega))\). Then there exist \(\tilde{u}^{*}\in L^{\infty}(0,T_{1}; L^{2}(\Omega))\) and a subsequence of \(\{u_{n}\}\), still so denoted, such that \(u_{n} \rightarrow u^{*}\) weakly∗ in \(L^{\infty }(0,T_{1};L^{2}(\Omega))\). Moreover,
$$\rho\chi_{\omega}\tilde{u}_{n}\rightarrow\rho \chi_{\omega} \tilde{u}^{*}\quad \mbox{ weakly}^{*} \mbox{ in } L^{\infty}\bigl(0,T_{1}; L ^{2}(\Omega)\bigr) \mbox{ as } n\rightarrow\infty. $$
Therefore the solution \(y_{n}(\cdot, \cdot ; \chi_{\omega}\tilde {u}_{n}, y _{0})\) to the following system:
$$\textstyle\begin{cases} \partial_{t} y_{n}(x,t)-\Delta y_{n}(x,t)=\rho(x)\chi_{\omega}(x) \tilde{u}_{n}(x,t), &\mbox{in }\Omega\times(0,T_{1}), \\ y_{n}(x,t)=0, &\mbox{on } \partial\Omega\times(0,T_{1}), \\ y_{n}(x,0)=y_{0}(x), &\mbox{in }\Omega, \end{cases} $$
satisfies \(y_{n}(\cdot, t; \chi_{\omega} \tilde{u}_{n}, y_{0})\in \bar{B}(0,\varepsilon )\) for \(t\geq T_{n}\). Denote by \(y^{*}\) the solution to the following system:
$$\textstyle\begin{cases} \partial_{t} y^{*}(x,t)-\Delta y^{*}(x,t)=\rho(x)\chi_{\omega}(x) \tilde{u}^{*}(x,t), &\mbox{in }\Omega\times(0,T_{1}), \\ y^{*}(x,t)=0, &\mbox{on }\partial\Omega\times(0,T_{1}), \\ y^{*}(x,0)=y_{0}(x), &\mbox{in } \Omega. \end{cases} $$
Then
$$\begin{aligned} y_{n}\rightarrow y^{*} & \quad\mbox{weakly in } L^{2}\bigl(0,T; H_{0}^{1}( \Omega) \bigr)\cap H^{1}\bigl(0,T; L^{2}(\Omega)\bigr), \\ &\quad \mbox{strongly in } C\bigl([0, T_{1}]; L^{2}(\Omega)\bigr) \mbox{ as } n\rightarrow\infty, \end{aligned} $$
for any \(\delta>0\). Since \(y_{n}(\cdot, t; \chi_{\omega} \tilde{u}^{n}, y_{0})\in\bar{B}(0,\varepsilon )\) for \(t\geq T_{n}\), \(y^{*}(\cdot, t; \chi _{\omega} \tilde{u}^{*}, y_{0})\in\bar{B}(0,\varepsilon )\) for all \(t\geq T_{n}\) and \(n\in\mathbb{N}\). Hence \(y^{*}(\cdot, T_{\varepsilon }^{*}; \chi _{\omega} \tilde{u}^{*}, y_{0})\in\bar{B}(0,\varepsilon )\). Set
$$u^{*}(t)=\tilde{u}^{*}(t),\quad t\in\bigl[0,T_{\varepsilon }^{*} \bigr]. $$
Then \(y^{*}\) satisfies the following system:
$$\textstyle\begin{cases} \partial_{t} y^{*}(x,t)-\Delta y^{*}(x,t)=\rho(x)\chi_{\omega}(x) u ^{*}(x,t), &\mbox{in }\Omega\times(0,T_{1}), \\ y^{*}(x,t)=0, &\mbox{on }\partial\Omega\times(0,T_{1}), \\ y^{*}(x,0)=y_{0}(x), &\mbox{in } \Omega, \end{cases} $$
and \(y^{*}(\cdot, T_{\varepsilon }^{*}; \chi_{\omega} u^{*}, y_{0})\in\bar{B}(0,\varepsilon )\).
Claim: \(\Vert u^{*}(t)\Vert _{L^{2}(\Omega)}\leq M(t)\) for a.e. \(t\in[0,T _{\varepsilon }^{*}]\). Indeed, let \(\{\zeta_{k}\}_{k\in\mathbb{N}}\) be the countable density subset of \(L^{2}(\Omega)\). Denote by \(\mathcal{L}\) the Lebesgue point of \(\langle u_{n}(t), \zeta_{k}\rangle\), \(t\in[0, T_{\varepsilon }^{*}]\), where \(\langle\cdot, \cdot\rangle\) is the inner product of \(u_{n}(t)\) and \(\zeta_{k}\) in \(L^{2}(\Omega)\). Since \(\langle u_{n}(t), \zeta_{k}\rangle, \langle u^{*}(t), \zeta_{k} \rangle\in L^{\infty}(0,T_{\varepsilon }^{*})\), for each
$$t_{0}\in E_{0}\equiv\bigcap_{n,k=1}^{\infty} \mathcal{L}\bigl(\langle u _{n}, \zeta_{k}\rangle\bigr)\cap \bigcap_{k=1}^{\infty}\mathcal{L}\bigl( \bigl\langle u^{*},\zeta_{k}\bigr\rangle \bigr), $$
we have
$$\begin{aligned}& \lim_{\delta\rightarrow0} \frac{1}{2\delta} \int_{t_{0}-\delta} ^{t_{0}+\delta}\bigl\langle u_{n}(t), \zeta_{k}\bigr\rangle \,dt =\bigl\langle u _{n}(t_{0}), \zeta_{k}\bigr\rangle , \\& \lim_{\delta\rightarrow0} \frac{1}{2\delta} \int_{t_{0}-\delta} ^{t_{0}+\delta}\bigl\langle u^{*}(t), \zeta_{k}\bigr\rangle \,dt =\bigl\langle u ^{*}(t_{0}), \zeta_{k}\bigr\rangle . \end{aligned}$$
By virtue of
$$\frac{1}{2\delta} \int_{t_{0}-\delta}^{t_{0}+\delta}\bigl\langle u_{n}(t), \zeta_{k}\bigr\rangle \,dt\rightarrow\frac{1}{2\delta} \int_{t_{0}- \delta}^{t_{0}+\delta}\bigl\langle u^{*}(t), \zeta_{k}\bigr\rangle \,dt \quad\mbox{as }n\rightarrow\infty, $$
\(u_{n}\rightarrow u^{*}\) weakly∗ in \(L^{\infty}(0,T_{\varepsilon }^{*}; L ^{2}(\Omega))\), and the arbitrary of \(\delta>0\), we get
$$\begin{aligned}& \bigl\langle u_{n}(t_{0}), \zeta_{k}\bigr\rangle =\lim_{\delta\rightarrow0} \frac{1}{2 \delta} \int_{t_{0}-\delta}^{t_{0}+\delta}\bigl\langle u_{n}(t),\zeta _{k}\bigr\rangle \,dt\quad\rightarrow \\& \quad \lim_{\delta\rightarrow0} \frac{1}{2 \delta} \int_{t_{0}-\delta}^{t_{0}+\delta}\bigl\langle u^{*}(t),\zeta _{k}\bigr\rangle \,dt=\bigl\langle u^{*}(t_{0}), \zeta_{k}\bigr\rangle \quad \mbox{as }n\rightarrow\infty. \end{aligned}$$
Since \(\{\zeta_{k}\}\) is dense in \(L^{2}(\Omega)\), we have
$$\bigl\langle u_{n}(t_{0}), \zeta\bigr\rangle \rightarrow \bigl\langle u^{*}(t_{0}), \zeta\bigr\rangle \quad\mbox{as }n \rightarrow\infty $$
for each \(\zeta\in L^{2}(\Omega)\). That implies \(u_{n}(t_{0})\rightarrow u^{*}(t_{0})\) weakly in \(L^{2}(\Omega)\), and hence \(\Vert u^{*}(t _{0})\Vert _{L^{2}(\Omega)}\leq\liminf_{n\rightarrow\infty }\Vert u_{n}(t _{0})\Vert _{L^{2}(\Omega)}\leq M \). Applying
$$\vert E_{0}\vert=T_{\varepsilon }^{*}, $$
we obtain \(\Vert u^{*}(t)\Vert _{L^{2}(\Omega)} \leq M \) for a.e. \(t\in[0,T_{\varepsilon }^{*}]\). That proves the claim.
Step 2. We show that \(\mathcal{R}(T_{\varepsilon }^{*})\cap\bar {B}(0,\varepsilon )\) has only one point. If so, it is obvious that this point belongs to the boundary of \(B(0,\varepsilon )\).
In Step 1 we get \(\mathcal{R}(T_{\varepsilon }^{*})\cap\bar{B}(0,\varepsilon )\neq \emptyset\). By contradiction, we assume that \(\mathcal{R}(T_{\varepsilon } ^{*})\cap\bar{B}(0,\varepsilon )\) has at least two points, i.e., there exist \(y_{1}\equiv y(\cdot, T_{\varepsilon }^{*}; \chi_{\omega} u_{1}^{*}, y_{0})\), \(y_{2}\equiv y(\cdot, T_{\varepsilon }^{*};\chi_{\omega} u_{2}^{*}, y_{0})\in \mathcal{R}(T_{\varepsilon }^{*})\cap\bar{B}(0,\varepsilon )\) with \(y_{1}\neq y_{2}\). It is obvious that \(u_{1}^{*}\neq u_{2}^{*}\) in \(\mathcal{U}_{\mathrm{ad}}\). Define
$$\hat{u}\equiv\frac{u_{1}^{*}+u_{2}^{*}}{2}. $$
Since \(\hat{u}\in\mathcal{U}_{\mathrm{ad}}\), and \(B(0,\varepsilon )\) is strongly convex in \(L^{2}(\Omega)\), we get \(\hat{y}\equiv y(\cdot, T_{\varepsilon }^{*}; \chi_{\omega} \hat{u}, y_{0})= \frac{1}{2}y_{1}+\frac{1}{2}y_{2}\) is an inner point of \(B(0,\varepsilon )\), i.e., there exists \(\gamma>0\) such that \(B(\hat{y}, \gamma )\subset B(0,\varepsilon )\).
For any \(\xi>0\), define
$$h_{\xi}\equiv\hat{y}-y\bigl(\cdot, T_{\varepsilon }^{*}-\xi; \chi_{\omega} \hat{u}, y_{0}\bigr). $$
It is easy to check that
$$\begin{aligned} h_{\xi} =& e^{\Delta(T_{\varepsilon }^{*}-\xi)} \bigl[ e^{\Delta\xi}-I \bigr] y_{0}+ \int_{0}^{T_{\varepsilon }^{*}-\xi} \bigl[ e^{\Delta\xi}-I \bigr] e^{ \Delta(T_{\varepsilon }^{*}-\xi-\sigma)} \rho\chi_{\omega}\hat{u}(\sigma )\,d\sigma \\ &{}+ \int_{T_{\varepsilon }^{*}-\xi}^{T_{\varepsilon }^{*}} e^{\Delta(T_{\varepsilon } ^{*}-\sigma)}\rho \chi_{\omega}\hat{u}(\sigma )\,d\sigma . \end{aligned}$$
Hence, ξ can be chosen small enough such that \(\Vert h_{\xi }\Vert _{L ^{2}(\Omega)}\leq\gamma\). Therefore, we can get \(y(\cdot, T_{\varepsilon } ^{*}-\xi; \chi_{\omega} \hat{u}, y_{0})\in\bar{B}(0,\varepsilon )\). This contradicts the optimal time \(T_{\varepsilon }^{*}\). Subsequently, the set \(\mathcal{R}(T_{\varepsilon }^{*})\cap\bar{B}(0,\varepsilon )\) has only one point, and this point belongs to the boundary of \(B(0,\varepsilon )\).
Step 3. The time optimal control \(u^{*}\) has the bang-bang property.
Since \(\mathcal{R}(T_{\varepsilon }^{*})\cap\bar{B}(0,\varepsilon )\) has only one point (denote this point by \(y^{*}=y(\cdot, T_{\varepsilon }^{*}; \chi_{\omega} u^{*}, y_{0})\)), and \(\mathcal{R}(T_{\varepsilon }^{*})\) and \(\bar{B}(0,\varepsilon )\) are two convex sets, by hyperplane separation theorem, there exists \(\eta^{*}\in L^{2}(\Omega)\) such that
$$ \sup_{y\in\mathcal{R}(T_{\varepsilon }^{*})}\bigl\langle y, \eta^{*} \bigr\rangle \leq\inf_{z\in\bar{B}(0,\varepsilon )} \bigl\langle z, \eta^{*} \bigr\rangle \leq\bigl\langle y ^{*}, \eta^{*}\bigr\rangle . $$
(2.11)
Notice that \(y\in\mathcal{R}(T_{\varepsilon }^{*})\) can be written by
$$y\bigl(\cdot, T_{\varepsilon }^{*}; \chi_{\omega} u, y_{0}\bigr)= e^{\Delta T_{\varepsilon }^{*}}y _{0}+ \int_{0}^{T_{\varepsilon }^{*}} e^{\Delta(T_{\varepsilon }^{*}-\sigma)} \rho \chi_{\omega}u(\sigma)\,d\sigma. $$
Then (2.11) can be written as
$$\sup_{\bar{u}\in\mathcal{U}_{1}} \int_{0}^{T_{\varepsilon }^{*}} \bigl\langle e ^{\Delta(T_{\varepsilon }^{*}-\sigma )}\rho \chi_{\omega} M \bar{u}(\sigma), \eta^{*} \bigr\rangle \,d\sigma\leq \int_{0}^{T _{\varepsilon }^{*}} \bigl\langle e^{\Delta(T_{\varepsilon }^{*}-\sigma)}\rho \chi_{\omega} M \bar{u}^{*}(\sigma), \eta^{*} \bigr\rangle \,d\sigma. $$
Here
$$\bar{u}^{*}\in\mathcal{U}_{1}\equiv\bigl\{ \bar{u}\in L^{\infty}\bigl(0,T _{\varepsilon }^{*}; L^{2}(\Omega) \bigr)\mid \bigl\Vert \bar{u}(t)\bigr\Vert _{L^{2}(\Omega)} \leq1 \mbox{ for a.e. } t\in\bigl[0, T_{\varepsilon }^{*}\bigr] \bigr\} , $$
and
$$ u^{*}(t)= M \bar{u}^{*}(t)\quad \mbox{for all } t\in \bigl[0,T_{\varepsilon }^{*}\bigr]. $$
(2.12)
Hence, we have
$$\sup_{\bar{u}\in\mathcal{U}_{1}} \int_{0}^{T_{\varepsilon }^{*}} \bigl\langle \bar{u}( \sigma), e^{\Delta(T_{\varepsilon }^{*}-\sigma)}\rho\chi_{\omega} M \eta^{*} \bigr\rangle \,d\sigma\leq \int_{0}^{T_{\varepsilon }^{*}} \bigl\langle \bar{u} ^{*}( \sigma), e^{\Delta(T_{\varepsilon }^{*}-\sigma)}\rho\chi_{\omega} M \eta^{*} \bigr\rangle \,d\sigma. $$
For given \(t_{0}\in E_{0}\), choosing
$$\bar{u}(t)= \textstyle\begin{cases} \bar{u}^{*}(t), &\mbox{for } t\in(0,T_{\varepsilon }^{*})\setminus(t_{0}- \lambda, t_{0}+\lambda), \\ \zeta, &\mbox{for } t\in(t_{0}-\lambda, t_{0}+\lambda)\subset(0,T _{\varepsilon }^{*}), \end{cases} $$
where \(\zeta\in L^{2}(\Omega )\), we get
$$\sup_{\zeta\in L^{2}(\Omega)} \bigl\langle \zeta, e^{\Delta (T_{\varepsilon } ^{*}-t_{0})}\rho \chi_{\omega} M \eta^{*} \bigr\rangle \leq\bigl\langle \bar{u}^{*}(t_{0}), e^{\Delta (T_{\varepsilon }^{*}-t_{0})}\rho\chi_{\omega} M \eta^{*} \bigr\rangle , $$
i.e.,
$$\begin{aligned} \bigl\Vert e^{\Delta(T_{\varepsilon }^{*}-t_{0})}\rho\chi_{\omega} \eta^{*}\bigr\Vert _{L^{2}(\Omega)} =&\sup_{\zeta\in L^{2}(\Omega)} \bigl\langle \zeta, e^{\Delta(T_{\varepsilon }^{*}-t_{0})}\rho\chi_{\omega }\eta^{*} \bigr\rangle \\ \leq& \bigl\langle \bar{u}^{*}(t), e^{\Delta(T_{\varepsilon }^{*}-t_{0})} \rho \chi_{\omega}\eta^{*} \bigr\rangle \\ \leq&\bigl\Vert \bar{u}^{*}(t)\bigr\Vert _{L^{2}(\Omega )}\bigl\Vert e^{\Delta(T_{\varepsilon } ^{*}-t_{0})}\rho\chi_{\omega}\eta^{*}\bigr\Vert _{L^{2}(\Omega)}. \end{aligned}$$
This implies that
$$ \bigl\Vert \bar{u}^{*}(t_{0})\bigr\Vert _{L^{2}(\Omega)}=1 $$
(2.13)
by \(\bar{u}^{*}\in\mathcal{U}_{1}\). Equation (2.13), together with (2.12) and \(\vert E_{0}\vert=T_{\varepsilon }^{*}\), yields
$$\bigl\Vert u^{*}(t)\bigr\Vert _{L^{2}(\Omega)}= M \quad\mbox{for a.e. } t\in \bigl[0,T_{\varepsilon }^{*}\bigr]. $$
From the above, we get the time optimal control’s bang-bang property. That completes the proof. □