Let \({\mathcal {U}}\) be a Hilbert space of control variables, and let B be an operator,
$$ B \in{\mathcal {L}} \bigl({ \mathcal {U}},L^{2} \bigl(0,T;L^{2} \bigr) \bigr), $$
(4.1)
called a controller.
We consider the following nonlinear control system:
$$ \textstyle\begin{cases} y_{tt}(u) - \Delta y_{tt}(u) + \Delta^{2} y(u) + \int^{t}_{0} k(t-s) \Delta ^{2} y(u;s)\,ds = [y(u), v(u)] + Bu \quad\mbox{in } Q, \\ \Delta^{2} v(u) = -[y(u), y(u) ] \quad\mbox{in } Q, \\ y(u)= \frac{\partial y(u)}{\partial\nu} = v(u) = \frac{\partial v(u)}{\partial\nu} = 0 \quad\mbox{on } \Sigma, \\ y(u;0,x) = y_{0}(x),\qquad y_{t} (u;0,x) = y_{1} (x) \quad\mbox{in } \Omega, \end{cases} $$
(4.2)
where \(y_{0} \in H^{2}_{0}\), \(y_{1} \in H^{1}_{0}\), and \(u \in{ \mathcal {U}}\) is a control. By Theorem 3.1 and (4.1) we can define uniquely the solution map \(u\to y(u)\) of \({ \mathcal {U}}\) into \(S(0, T)\). The observation of the state is assumed to be given by
$$ Y(u)=Cy(u),\qquad C \in{ \mathcal {L}} \bigl({S}(0,T), M \bigr), $$
(4.3)
where C is an operator called the observer, and M is a Hilbert space of observation variables. The quadratic cost function associated with the control system (4.2) is given by
$$ J(u)= \bigl\Vert Cy(u)-Y_{d} \bigr\Vert ^{2}_{M}+(Ru,u)_{\mathcal {U}} \qquad\mbox{for } u \in { \mathcal {U}}, $$
(4.4)
where \(Y_{d} \in M\) is a desired value of \(y(u)\), and \(R \in{ \mathcal {L}}({ \mathcal {U}},{\mathcal {U}}) \) is symmetric and positive, that is,
$$ (Ru,u)_{{\mathcal {U}}}=(u,Ru)_{\mathcal {U}} \geq d \Vert u \Vert ^{2}_{{\mathcal {U}}} $$
(4.5)
for some \(d > 0\). Let \({\mathcal {U}}_{\mathrm{ad}}\) be a closed convex subset of \({\mathcal {U}}\), which is called the admissible set. An element \(u^{*} \in {\mathcal {U}}_{\mathrm{ad}}\) that attains the minimum of J over \({\mathcal {U}}_{\mathrm{ad}}\) is called an optimal control for the cost (4.4).
4.1 Existence of an optimal control
As indicated in Introduction, we need to show the existence of an optimal control and to give its characterization. The existence of an optimal control \(u^{*}\) for the cost (4.4) can be stated by the following theorem.
Theorem 4.1
Assume that the hypotheses of Theorem
3.1
are satisfied. Then there exists at least one optimal control
u
for the control problem (4.2) with the cost (4.4).
Proof
Set \(J_{0}= \inf_{u \in{ \mathcal {U}}_{\mathrm{ad}}}J(u)\). Since \({\mathcal {U}}_{\mathrm{ad}}\) is nonempty, there is a sequence \(\{ u_{n} \}\) in \({\mathcal {U}}\) such that
$$\inf_{u \in{\mathcal {U}}_{\mathrm{ad}}}J(u)=\lim _{n \rightarrow \infty}J(u_{n})=J_{0}. $$
Obviously, \(\{J(u_{n})\}\) is bounded in \({\mathbf {R}}^{+}\). Then by (4.5) there exists a constant \(K_{0}>0\) such that
$$ d\Vert u_{n}\Vert _{\mathcal {U}}^{2} \leq(Ru_{n},u_{n})_{\mathcal {U}} \leq J(u_{n}) \leq K_{0}. $$
(4.6)
This shows that \(\{u_{n}\}\) is bounded in \({\mathcal {U}}\). Since \({\mathcal {U}}_{\mathrm{ad}}\) is closed and convex, we can choose a subsequence (denoted again by \(\{ u_{n} \}\)) of \(\{u_{n}\}\) and find \(u \in{\mathcal {U}}_{\mathrm{ad}} \) such that
$$ u_{n} \rightarrow u^{*} \quad\mbox{weakly in } {\mathcal {U}} $$
(4.7)
as \(n \rightarrow\infty\). From now on, each state \(y_{n}=y(u_{n}) \in{ S}(0,T)\) corresponding to \(u_{n}\) is a solution of
$$ \textstyle\begin{cases} y_{n,tt} - \Delta y_{n,tt} + \Delta^{2} y_{n} + \int^{t}_{0} k(t-s) \Delta^{2} y_{n}(s)\,ds = [y_{n}, v_{n}] + Bu_{n} \quad\mbox{in } Q, \\ \Delta^{2} v_{n} = -[y_{n}, y_{n}] \quad\mbox{in } Q, \\ y_{n} = \frac{\partial y_{n} }{\partial\nu} = v_{n} = \frac{\partial v_{n}}{\partial\nu} = 0 \quad\mbox{on } \Sigma, \\ y_{n} (0) = y_{0},\qquad y_{n, t} (0) = y_{1} \quad\mbox{in } \Omega. \end{cases} $$
(4.8)
By (4.6) the term \(Bu_{n}\) is estimated as
$$\begin{aligned} \Vert Bu_{n}\Vert _{L^{2}(0,T;L^{2} )} \leq& \Vert B \Vert _{ { \mathcal {L}}({\mathcal {U}}, L^{2}(0,T;L^{2})) } \Vert u_{n}\Vert _{\mathcal {U}} \\ \leq& \Vert B\Vert _{ {\mathcal {L}}({ \mathcal {U}}, L^{2}(0,T;L^{2} )) } \sqrt{K_{0} d^{-1}} \equiv K_{1}. \end{aligned}$$
(4.9)
Hence, noting that \(y(0,0,0,t) =0\) and \(v(0,0,0,t)=0\), it follows from Theorem 3.1 that
$$\begin{aligned}& \Vert y_{n} \Vert _{{ W}(0, T)} + \bigl\Vert y_{n} (t) \bigr\Vert _{ H^{2}_{0}}+ \bigl\Vert y'_{n} (t) \bigr\Vert _{H^{1}_{0}}+ \bigl\Vert v_{n} (t) \bigr\Vert _{W^{2, \infty}} \\& \quad \le C \bigl( \Vert y_{0} \Vert _{H^{2}_{0} } + \Vert y_{1} \Vert _{H^{1}_{0}} + K_{1} \bigr). \end{aligned}$$
(4.10)
By (4.10) we easily verify that \([y_{n}, v_{n}]\) is bounded in \(L^{2}(0, T;L^{2})\). Therefore, by the extraction theorem of Rellich we can find a subsequence of \(\{ y_{n}\}\), say again \(\{ y_{n} \}\), and find \(y\in { W}(0,T) \cap L^{\infty}(0, T; H^{2}_{0}) \) with \(y' \in L^{\infty} (0, T; H^{1}_{0})\) and \(F \in L^{2}(0, T; L^{2})\) such that
$$\begin{aligned}& y_{n} \to y \quad\mbox{weakly in } W(0,T), \end{aligned}$$
(4.11)
$$\begin{aligned}& y_{n} \to y\quad\mbox{weakly * in } L^{\infty} \bigl(0, T; H^{2}_{0} \bigr), \end{aligned}$$
(4.12)
$$\begin{aligned}& y'_{n} \to y'\quad \mbox{weakly * in } L^{\infty } \bigl(0, T; H^{1}_{0} \bigr), \end{aligned}$$
(4.13)
$$\begin{aligned}& [y_{n}, v_{n} ] \to F\quad\mbox{weakly in } L^{2} \bigl(0,T; L^{2} \bigr). \end{aligned}$$
(4.14)
To prove \(F = [y, -G[y,y]]\), we employ the idea given in Dautray and Lions [7]. By similar manipulations given in Dautray and Lions [7], pp.564-566, we can deduce that the weak limit y in (4.11) is a weak solution of the linear problem
$$ \textstyle\begin{cases} y_{tt} - \Delta y_{tt} + \Delta^{2} y + \int^{t}_{0} k(t-s) \Delta^{2} y(s)\,ds = F + Bu^{*} \quad \mbox{in } Q, \\ y = \frac{\partial y }{\partial\nu} = 0 \quad \mbox{on } \Sigma, \\ y (0) = y_{0},\qquad y_{ t} (0) = y_{1} \quad \mbox{in } \Omega. \end{cases} $$
(4.15)
As in (3.5), the weak solution y of Eq. (4.15) satisfies the following energy equality:
$$\begin{aligned}& \bigl\Vert y'(t) \bigr\Vert ^{2} + \bigl\Vert \nabla y'(t) \bigr\Vert ^{2} + \bigl\Vert \Delta y (t) \bigr\Vert ^{2} \\& \qquad{} + 2 \bigl( k * \Delta y(t), \Delta y(t) \bigr)_{2} \\& \quad = 2 \int^{t}_{0} \bigl(k' * \Delta y, \Delta y \bigr)_{2} \,ds + 2 \int^{t}_{0} k(0) \Vert \Delta y \Vert ^{2} \,ds \\& \qquad{} + 2 \int^{t}_{0} \bigl(F + Bu^{*}, y' \bigr)_{2} \,ds + \Vert y_{1} \Vert ^{2} + \Vert \nabla y_{1} \Vert ^{2} + \Vert \Delta y_{0} \Vert ^{2}. \end{aligned}$$
(4.16)
We can also deduce, as in (3.5), that the weak solution \(y_{n}\) of Eq. (4.8) satisfies the following energy equality:
$$\begin{aligned}& \bigl\Vert y'_{n} (t) \bigr\Vert ^{2} + \bigl\Vert \nabla y'_{n}(t) \bigr\Vert ^{2} + \bigl\Vert \Delta y_{n} (t) \bigr\Vert ^{2} \\& \qquad{} + 2 \bigl( k * \Delta y_{n} (t), \Delta y_{n} (t) \bigr)_{2} \\& \quad = 2 \int^{t}_{0} \bigl(k' * \Delta y_{n}, \Delta y_{n} \bigr)_{2} \,ds + 2 \int^{t}_{0} k(0) \Vert \Delta y_{n} \Vert ^{2} \,ds \\& \qquad{} + 2 \int^{t}_{0} \bigl([y_{n}, v_{n}] + Bu_{n}, y'_{n} \bigr)_{2} \,ds + \Vert y_{1} \Vert ^{2} + \Vert \nabla y_{1} \Vert ^{2} + \Vert \Delta y_{0} \Vert ^{2}. \end{aligned}$$
(4.17)
We note the following simple equalities:
$$\begin{aligned}& \Vert a \Vert ^{2} + \Vert b \Vert ^{2} = \Vert a- b \Vert ^{2} + 2 (a,b)_{2},\quad \forall a, b \in L^{2}; \\& (a_{1}, a_{2} )_{2} + (b_{1}, b_{2} )_{2} = (a_{1} - b_{1}, a_{2} - b_{2} )_{2} + (b_{1}, a_{2} )_{2} + ( a_{1}, b_{2})_{2}, \quad \forall a_{i}, b_{i} ( i=1,2) \in L^{2}. \end{aligned}$$
Adding (4.16) to (4.17), denoting \(y_{n} - y\) by \(\phi_{n}\), and using the above equalities, we have
$$\begin{aligned}& \bigl\Vert \phi'_{n} (t) \bigr\Vert ^{2} + \bigl\Vert \nabla\phi'_{n}(t) \bigr\Vert ^{2} + \bigl\Vert \Delta\phi _{n} (t) \bigr\Vert ^{2} \\& \qquad{} + 2 \bigl( k * \Delta\phi_{n} (t), \Delta \phi_{n} (t) \bigr)_{2} \\& \quad = 2 \int^{t}_{0} \bigl(k' * \Delta \phi_{n}, \Delta\phi_{n} \bigr)_{2} \,ds + 2 \int^{t}_{0} k(0) \Vert \Delta\phi_{n} \Vert ^{2} \,ds + \Phi^{0} + \sum ^{5}_{i=1} \Phi^{i}_{n}, \end{aligned}$$
(4.18)
where
$$\begin{aligned}& \Phi^{0} = 2 \bigl(\Vert y_{1} \Vert ^{2} + \Vert \nabla y_{1} \Vert ^{2} + \Vert \Delta y_{0} \Vert ^{2} \bigr), \end{aligned}$$
(4.19)
$$\begin{aligned}& \Phi^{1}_{n} = -2 \bigl( \bigl(y'_{n} (t), y' (t) \bigr)_{2} + \bigl(\nabla y'_{n} (t), \nabla y'(t) \bigr)_{2} + \bigl(\Delta y_{n} (t), \Delta y (t) \bigr)_{2} \bigr), \end{aligned}$$
(4.20)
$$\begin{aligned}& \Phi^{2}_{n} = -2 \bigl( \bigl(k* \Delta y_{n} (t), \Delta y(t) \bigr)_{2} + \bigl(k* \Delta y(t), \Delta y_{n} (t) \bigr)_{2} \bigr), \end{aligned}$$
(4.21)
$$\begin{aligned}& \Phi^{3}_{n} = 2 \biggl( \int^{t}_{0} \bigl(k' * \Delta y_{n}, \Delta y \bigr)_{2} \,ds + \int ^{t}_{0} \bigl(k' * \Delta y, \Delta y_{n} \bigr)_{2} \,ds \biggr), \end{aligned}$$
(4.22)
$$\begin{aligned}& \Phi^{4}_{n} = 4 \int^{t}_{0} k(0) (\Delta y_{n}, \Delta y)_{2} \,ds, \end{aligned}$$
(4.23)
$$\begin{aligned}& \Phi^{5}_{n} = 2 \biggl( \int^{t}_{0} \bigl([y_{n}, v_{n}] + Bu_{n}, y'_{n} \bigr)_{2} \,ds + \int^{t}_{0} \bigl(F + Bu^{*}, y' \bigr)_{2} \,ds \biggr). \end{aligned}$$
(4.24)
Then by routine calculations in (4.18), as in the proof of Theorem 3.1, we derive the inequality
$$ \bigl\Vert \phi'_{n} (t) \bigr\Vert ^{2} + \bigl\Vert \nabla\phi'_{n}(t) \bigr\Vert ^{2} + \bigl\Vert \Delta\phi_{n} (t) \bigr\Vert ^{2} \le C(k, T) \Biggl\vert \Phi^{0} + \sum ^{5}_{i=1} \Phi^{i}_{n} \Biggr\vert . $$
(4.25)
By virtue of (4.11)-(4.13) together with [7], pp.518-520, we can extract a subsequence \(\{ y_{n_{k} } \}\) of \(\{ y_{n} \}\) such that, as \(k \to \infty\),
$$\begin{aligned}& \Phi^{1}_{n_{k}} \to -2 \bigl( \bigl\Vert y' (t) \bigr\Vert ^{2} + \bigl\Vert \nabla y'(t) \bigr\Vert ^{2} + \bigl\Vert \Delta y (t) \bigr\Vert ^{2} \bigr), \end{aligned}$$
(4.26)
$$\begin{aligned}& \Phi^{2}_{n_{k}} \to -4 \bigl(k* \Delta y(t), \Delta y(t) \bigr)_{2}, \end{aligned}$$
(4.27)
$$\begin{aligned}& \Phi^{3}_{n_{k}} \to 4 \int^{t}_{0} \bigl(k' * \Delta y, \Delta y \bigr)_{2} \,ds, \end{aligned}$$
(4.28)
$$\begin{aligned}& \Phi^{4}_{n_{k}} \to 4 \int^{t}_{0} k(0) \Vert \Delta y \Vert ^{2} \,ds. \end{aligned}$$
(4.29)
Since the imbedding \(H^{1}_{0} \hookrightarrow L^{2}\) is compact, by virtue of (4.11), we can refer to the result of the Aubin-Lions-Temam compact imbedding theorem (see Temam [16]; p.271) to verify that \(\{ y'_{n} \}\) is precompact in \(L^{2}(0,T; L^{2})\). Hence, there also exists a subsequence \(\{y'_{n_{k}} \} \subset\{ y'_{n} \}\) such that
$$ y'_{n_{k}} \longrightarrow y' \quad \mbox{strongly in } L^{2} \bigl(0,T;L^{2} \bigr) \mbox{ as } k \rightarrow\infty. $$
(4.30)
From (4.7), (4.14), and (4.30) we have
$$ \Phi^{5}_{n_{k}} \to 4 \int^{t}_{0} \bigl(F + Bu^{*}, y' \bigr)_{2} \,ds \quad\mbox{as } k \to\infty. $$
(4.31)
In view of (4.16), the sum of (4.19) and all the limits from (4.26) to (4.29) and (4.31) are 0, so that
$$ \Phi^{0} + \sum^{5}_{i=1} \Phi^{i}_{n_{k}} \to 0\quad\mbox{as } k \to\infty. $$
(4.32)
Therefore, from (4.25) and (4.32) we get that
$$ y_{n_{k}} (t) \to y(t)\quad\mbox{strongly in } H^{2}_{0} \mbox{ as } k \to \infty, \forall t \in[0, T]. $$
(4.33)
Thus, by Lemma 2.2, Theorem 3.1, and (4.33) it follows that
$$\begin{aligned}& \bigl\Vert [y_{n_{k}}, v_{n_{k}}] - [y, v] \bigr\Vert _{L^{2}(0, T; L^{2})} \\& \quad = \bigl\Vert [y_{n_{k}} - y, v_{n_{k}}] + [y, v_{n_{k}} - v ] \bigr\Vert _{L^{2}(0, T; L^{2})} \\& \quad \le \bigl\Vert [y_{n_{k}} - y, v_{n_{k}}] \bigr\Vert _{L^{2}(0, T; L^{2})} + \bigl\Vert \bigl[y, G[y,y] - G [y_{n_{k}}, y_{n_{k}}] \bigr] \bigr\Vert _{L^{2}(0, T; L^{2})} \\& \quad = \bigl\Vert [y_{n_{k}} - y, v_{n_{k}}] \bigr\Vert _{L^{2}(0, T; L^{2})} + \bigl\Vert \bigl[y, G[y- y_{n_{k}}, y_{n_{k}} + y] \bigr] \bigr\Vert _{L^{2}(0, T; L^{2})} \\& \quad \le C \bigl( \Vert v_{n_{k}} \Vert _{L^{\infty}(0, T;W^{2, \infty})} + \Vert y \Vert _{L^{\infty} (0, T; H^{2}_{0} )} \bigl( \Vert y_{n_{k}} \Vert _{L^{\infty} (0, T; H^{2}_{0} )} \\& \qquad{} + \Vert y \Vert _{L^{\infty} (0, T; H^{2}_{0} )} \bigr) \bigr) \Vert y_{n_{k}} - y \Vert _{L^{2}(0, T; H^{2}_{0})} \\& \quad \le C \bigl( \Vert y \Vert ^{2}_{L^{\infty} (0, T; H^{2}_{0} )} + \Vert y_{n_{k}} \Vert ^{2}_{L^{\infty} (0, T; H^{2}_{0} )} \bigr) \Vert y_{n_{k}} - y \Vert _{L^{2}(0, T; H^{2}_{0})} \\& \quad \le C \bigl( \bigl\Vert p^{*} \bigr\Vert ^{2}_{{\mathcal {P}}} + \Vert p_{n_{k}} \Vert ^{2}_{{\mathcal {P}}} \bigr) \Vert y_{n_{k}} - y \Vert _{L^{2}(0, T; H^{2}_{0})} \to 0 \end{aligned}$$
(4.34)
as \(k \to \infty\), where \(p^{*} = (y_{0}, y_{1}, Bu^{*})\) and \(p_{n_{k}} = (y_{0}, y_{1}, Bu_{n_{k}})\). Hence, by the uniqueness of the weak limits, from (4.14) and (4.34) it follows that
$$ F = [y, v] \equiv \bigl[y, -G[y,y] \bigr]. $$
(4.35)
We replace \(y_{n}\) by \(y_{n_{k}}\) and take \(k\to\infty\) in (4.8). Then by the standard argument in Dautray and Lions ([7], pp.561-565) we conclude that the limit y is a weak solution of
$$ \textstyle\begin{cases} y_{tt} - \Delta y_{tt} + \Delta^{2} y + \int^{t}_{0} k(t-s) \Delta^{2} y(s)\,ds = [y, v] + Bu^{*} \quad \mbox{in } Q, \\ \Delta^{2} v = - [y,y] \quad \mbox{in } Q, \\ y = \frac{\partial y }{\partial\nu} = v = \frac{\partial v }{\partial \nu} = 0 \quad \mbox{on } \Sigma, \\ y (0) = y_{0},\qquad y_{ t}(0) = y_{1} \quad \mbox{in } \Omega. \end{cases} $$
(4.36)
Also, since Eq. (4.36) has a unique weak solution \(y \in{S}(0,T)\) by Theorem 3.1, we conclude that \(y=y(u^{*})\) in \({ S}(0,T)\) by the uniqueness of solutions, which implies that \(y(u_{n})\to y(u^{*})\) weakly in \({ W}(0,T)\). Since C is continuous on \(S(0, T) \subset{ W}(0,T)\) and \(\Vert \cdot \Vert _{M}\) is lower semicontinuous, it follows that
$$\bigl\Vert Cy \bigl(u^{*} \bigr)-Y_{d} \bigr\Vert _{M} \leq \liminf_{n\rightarrow\infty} \bigl\Vert Cy(u_{n})-Y_{d} \bigr\Vert _{M}. $$
It is also clear from \(\liminf_{k\rightarrow\infty} \Vert R^{\frac{1}{2}}u_{n}\Vert _{ \mathcal {U}} \geq \Vert R^{\frac{1}{2}}u^{*} \Vert _{\mathcal {U}}\) that \(\liminf_{k\rightarrow\infty} (Ru_{n},u_{n})_{\mathcal {U}}\geq (Ru^{*},u^{*})_{\mathcal {U}}\). Hence,
$$J_{0}=\liminf_{n\rightarrow\infty}J(u_{n})\geq J \bigl(u^{*} \bigr). $$
But since \(J(u^{*})\geq J_{0}\) by definition, we conclude that \(J(u^{*})=\inf_{u\in{\mathcal {U}}_{\mathrm{ad}}}J(u)\). This completes the proof. □
In this section, we shall characterize the optimal controls by giving necessary conditions for optimality. For this, it is necessary to write down the necessary optimality condition
$$ DJ \bigl(u^{*} \bigr) \bigl(u-u^{*} \bigr) \ge0\quad\mbox{for all } u \in{ \mathcal {U}}_{\mathrm{ad}} $$
(4.37)
and to analyze (4.37) in view of the proper adjoint state system, where \(DJ(u^{*})\) denotes the Gâteaux derivative of \(J(u)\) at \(u=u^{*}\). That is, we have to prove that the mapping \(u \to y(u)\) of \({\mathcal {U}} \to{ S}(0,T)\) is Gâteaux differentiable at \(u=u^{*}\). First, we can see the continuity of the mapping.
Lemma 4.1
Let
\(w\in{\mathcal {U}}\)
be arbitrarily fixed. Then
$$ \lim_{\lambda\to0}y(u+\lambda w)=y(u)\quad \textit{strongly in } S(0,T). $$
(4.38)
Proof
The proof is an immediate consequence of Theorem 3.1. □
The solution map \(u \to y(u)\) of \({\mathcal {U}}\) into \({ S}(0,T)\) is said to be Gâteaux differentiable at \(u=u^{*}\) if for any \(w\in {\mathcal {U}}\), there exists a \(Dy(u^{*})\in{\mathcal {L}}({\mathcal {U}}, {S}(0,T))\) such that
$$\biggl\Vert \frac{1}{\lambda} \bigl(y \bigl(u^{*} +\lambda w \bigr)-y \bigl(u^{*} \bigr) \bigr)-Dy \bigl(u^{*} \bigr)w \biggr\Vert _{{ S}(0,T)} \to0 \quad \mbox{as } \lambda\to0. $$
The operator \(Dy(u^{*})\) denotes the Gâteaux derivative of \(y(u)\) at \(u=u^{*}\), and the function \(Dy(u^{*})w \in{ S}(0,T)\) is called the Gâteaux derivative in the direction \(w\in{\mathcal {U}}\), which plays an important part in the nonlinear optimal control problem.
Theorem 4.2
The map
\(u\to y(u)\)
of
\({\mathcal {U}}\)
into
\({ S}(0,T)\)
is Gâteaux differentiable at
\(u=u^{*}\)
and such the Gâteaux derivative of
\(y(u)\)
at
\(u=u^{*}\)
in the direction
\(u-u^{*}\in{\mathcal {U}}\), say
\(z=Dy(u^{*})(u-u^{*})\), is a unique weak solution of the following problem:
$$ \textstyle\begin{cases} z_{tt} - \Delta z_{tt} + \Delta^{2} z + \int^{t}_{0} k(t-s) \Delta^{2} z(s)\,ds \\ \quad = [z, -G[y(u^{*}),y(u^{*})]] +2 [y(u^{*}), -G[ z,y(u^{*})]] + B(u-u^{*}) \quad \textit{in } Q, \\ z = \frac{\partial z }{\partial\nu} = 0 \quad \textit{on } \Sigma, \\ z (0) = 0,\qquad z_{ t} (0) = 0 \quad \textit{in } \Omega. \end{cases} $$
(4.39)
Proof
Let \(\lambda\in(-1,1)\), \(\lambda\ne0\). We set \(y_{\lambda}:= y(u^{*} +\lambda(u-u^{*}))\) and
$$z_{\lambda}:= \lambda^{-1} \bigl(y_{\lambda} -y \bigl(u^{*} \bigr) \bigr). $$
Then, in the weak sense, \(z_{\lambda}\) satisfies
$$ \textstyle\begin{cases} z_{\lambda, tt} - \Delta z_{\lambda,tt} + \Delta^{2} z_{\lambda} + \int ^{t}_{0} k(t-s) \Delta^{2} z_{\lambda}(s)\,ds = F_{\lambda} + B(u-u^{*}) \quad \mbox{in } Q, \\ z_{\lambda} = \frac{\partial z_{\lambda} }{\partial\nu} = 0 \quad \mbox{on }\Sigma, \\ z_{\lambda} (0) = 0,\qquad z_{\lambda, t} (0) = 0 \quad \mbox{in } \Omega, \end{cases} $$
(4.40)
where
$$F_{\lambda} = \frac{1}{\lambda} \bigl( \bigl[y_{\lambda}, -G[y_{\lambda },y_{\lambda}] \bigr] - \bigl[y \bigl(u^{*} \bigr), -G \bigl[y \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr] \bigr). $$
Here we note that
$$\begin{aligned}& \frac{1}{\lambda} \bigl( \bigl[y_{\lambda}, -G[y_{\lambda},y_{\lambda}] \bigr] - \bigl[y \bigl(u^{*} \bigr), -G \bigl[y \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr] \bigr) \\& \quad = \bigl[z_{\lambda}, -G[y_{\lambda},y_{\lambda}] \bigr] + \bigl[y \bigl(u^{*} \bigr), -G \bigl[ z_{\lambda },y \bigl(u^{*} \bigr)+y_{\lambda} \bigr] \bigr]. \end{aligned}$$
(4.41)
Thus, from (2.5), Theorem 3.1, and (4.41) we deduce
$$\begin{aligned} \Vert F_{\lambda} \Vert _{L^{2}(0, T; L^{2})} \le& C \bigl( \Vert y_{\lambda} \Vert ^{2}_{L^{\infty}(0, T; H^{2}_{0})} + \bigl\Vert y \bigl(u^{*} \bigr) \bigr\Vert _{L^{\infty}(0, T; H^{2}_{0})} \bigl( \Vert y_{\lambda} \Vert _{L^{\infty}(0, T; H^{2}_{0})} \\ & {} + \bigl\Vert y \bigl(u^{*} \bigr) \bigr\Vert _{L^{\infty}(0, T; H^{2}_{0})} \bigr) \bigr) \Vert \Delta z_{\lambda} \Vert _{L^{2}(0, T; L^{2})} \\ \le& C \bigl( \Vert y_{\lambda} \Vert ^{2}_{L^{\infty}(0, T; H^{2}_{0})} + \bigl\Vert y \bigl(u^{*} \bigr) \bigr\Vert ^{2}_{L^{\infty}(0, T; H^{2}_{0})} \bigr) \Vert \Delta z_{\lambda} \Vert _{L^{2}(0, T; L^{2})} \\ \le& C \bigl( \Vert p_{\lambda} \Vert ^{2}_{{\mathcal {P}}} + \bigl\Vert p^{*} \bigr\Vert ^{2}_{{\mathcal {P}}} \bigr) \Vert \Delta z_{\lambda} \Vert _{L^{2}(0, T; L^{2})}, \end{aligned}$$
(4.42)
where \(p_{\lambda} = (y_{0}, y_{1}, B(u^{*} + \lambda(u-u^{*})) ) \) and \(p^{*} = (y_{0}, y_{1}, B u^{*} )\). Hence, by considering the energy equality satisfied by \(z_{\lambda}\) like (3.5) we get from (4.42) and the proof of Theorem 3.1 that the weak solution \(z_{\lambda}\) of Eq. (4.40) satisfies
$$ \Vert z_{\lambda} \Vert _{{ S}(0, T)} \le C \bigl\Vert B \bigl(u-u^{*} \bigr) \bigr\Vert _{L^{2}(0, T; L^{2})}. $$
(4.43)
Therefore, from (4.42) and (4.43) we see that there exists \(z \in{ W}(0,T) \cap L^{\infty} (0, T; H^{2}_{0})\) with \(z' \in L^{\infty}(0, T; H^{1}_{0})\), \(F \in L^{2} (0, T; L^{2})\) and a sequence \(\{\lambda_{k}\} \subset(-1,1)\) tending to 0 such that, as \(k \to \infty\),
$$\begin{aligned}& z_{\lambda_{k}} \to z\quad \mbox{weakly in } { W}(0, T), \end{aligned}$$
(4.44)
$$\begin{aligned}& z_{\lambda_{k}} \to z \quad\mbox{weakly * in } L^{\infty} \bigl(0, T; H^{2}_{0} \bigr), \end{aligned}$$
(4.45)
$$\begin{aligned}& z'_{\lambda_{k}} \to z' \quad \mbox{weakly * in } L^{\infty} \bigl(0, T; H^{1}_{0} \bigr), \end{aligned}$$
(4.46)
$$\begin{aligned}& F_{\lambda_{k}} \to F \quad\mbox{weakly in } L^{2} \bigl(0, T; L^{2} \bigr). \end{aligned}$$
(4.47)
We replace \(z_{\lambda}\) by \(z_{\lambda_{k}}\) and take \(k\to\infty\) in Eq. (4.40). Then by the standard argument in Dautray and Lions ([7], pp.561-565) we conclude that the limit z is a weak solution of
$$ \textstyle\begin{cases} z_{tt} - \Delta z_{tt} + \Delta^{2} z + \int^{t}_{0} k(t-s) \Delta^{2} z(s)\,ds = F + B(u-u^{*}) \quad\mbox{in } Q, \\ z = \frac{\partial z }{\partial\nu} = 0 \quad \mbox{on } \Sigma, \\ z (0) = 0,\qquad z_{ t} (0) = 0 \quad \mbox{in } \Omega. \end{cases} $$
(4.48)
Using (4.44)-(4.47), the respective energy equalities of Eq. (4.40) with \(z_{\lambda}\) replaced by \(z_{\lambda_{k}}\), and Eq. (4.48), we can proceed as in the proof of Theorem 4.1 to obtain
$$ z_{\lambda_{k}} \to z\quad\mbox{strongly in } S(0, T) \mbox{ as } k \to \infty. $$
(4.49)
By Theorem 3.1 and Lemma 2.2 we can verify the following:
$$\begin{aligned}& \bigl\Vert G \bigl[z_{\lambda_{k}}, y \bigl(u^{*} \bigr)+ y_{\lambda_{k}} \bigr] - 2 G \bigl[z, y \bigl(u^{*} \bigr) \bigr] \bigr\Vert _{C([0, T]; W^{2, \infty} )} \\& \quad = \bigl\Vert G \bigl[z_{\lambda_{k}} -z, y \bigl(u^{*} \bigr)+ y_{\lambda_{k}} \bigr] + G \bigl[z, y_{\lambda_{k}} - y \bigl(u^{*} \bigr) \bigr] \bigr\Vert _{C ([0, T]; W^{2, \infty} )} \\& \quad \le C T \bigl( \bigl( \bigl\Vert y \bigl(u^{*} \bigr) \bigr\Vert _{C([0, T]; H^{2}_{0})} + \Vert y_{\lambda_{k}} \Vert _{C([0, T]; H^{2}_{0})} \bigr) \Vert z_{\lambda_{k}} - z \Vert _{C([0, T]; H^{2}_{0})} \\& \qquad {}+ \Vert z\Vert _{C([0, T]; H^{2}_{0})} \bigl\Vert y_{\lambda_{k}} - y \bigl(u^{*} \bigr) \bigr\Vert _{C([0, T]; H^{2}_{0})} \bigr) \\& \quad\le C T \bigl( \bigl( \bigl\Vert p^{*} \bigr\Vert _{{\mathcal {P}}} + \Vert p_{\lambda_{k}} \Vert _{{\mathcal {P}}} \bigr) \Vert z_{\lambda_{k}} - z \Vert _{C([0, T]; H^{2}_{0})} \\& \qquad{} + \bigl\Vert B \bigl(u-u^{*} \bigr) \bigr\Vert _{L^{2}(0, T; L^{2})} \bigl\Vert y_{\lambda_{k}} - y \bigl(u^{*} \bigr) \bigr\Vert _{C([0, T]; H^{2}_{0})} \bigr), \end{aligned}$$
(4.50)
where \(p_{\lambda_{k}} = (y_{0}, y_{1}, B(u^{*} + \lambda_{k} (u-u^{*})) ) \) and \(p^{*} = (y_{0}, y_{1}, B u^{*} )\). Thus, from Lemma 4.1, (4.49), and (4.50), we have
$$ G \bigl[z_{\lambda_{k}}, y \bigl(u^{*} \bigr)+ y_{\lambda_{k}} \bigr] \to 2 G \bigl[z, y \bigl(u^{*} \bigr) \bigr]\quad\mbox{strongly in } C \bigl([0, T]; W^{2, \infty} \bigr) $$
(4.51)
as \(k \to\infty\).
Similarly, we can also show that
$$ G[y_{\lambda_{k}}, y_{\lambda_{k}}] \to G \bigl[y \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr]\quad\mbox{strongly in } C \bigl([0, T]; W^{2, \infty} \bigr) $$
(4.52)
as \(k \to\infty\). Therefore, by (4.51) and (4.52) we can show that
$$ F_{\lambda_{k}} \to \bigl[z, -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr] +2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ z,y \bigl(u^{*} \bigr) \bigr] \bigr]\quad\mbox{strongly in } L^{2} \bigl(0, T; L^{2} \bigr) $$
(4.53)
as \(k \to\infty\).
Consequently, we can infer from (4.47) and (4.52) that
$$ F = \bigl[z, -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr] +2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ z,y \bigl(u^{*} \bigr) \bigr] \bigr]. $$
(4.54)
Hence, it readily follows from (4.49) and (4.54) that \(z_{\lambda_{k}}\to z=Dy(u^{*})(u-u^{*})\) strongly in \({S}(0,T)\) as \(k \rightarrow\infty\), in which z is a weak solution of (4.39).
This completes the proof. □
Theorem 4.2 means that the cost \(J(u)\) is Gâteaux differentiable at \(u^{*}\) in the direction \(u-u^{*}\) and the optimality condition (4.37) is rewritten by
$$\begin{aligned}& \bigl(Cy \bigl(u^{*} \bigr)-Y_{d},C \bigl(Dy \bigl(u^{*} \bigr) \bigl(u-u^{*} \bigr) \bigr) \bigr)_{M}+ \bigl(Ru^{*},u-u^{*} \bigr)_{\mathcal {U}} \\& \quad = \bigl\langle C^{*}\Lambda_{M} \bigl(Cy \bigl(u^{*} \bigr)-Y_{d} \bigr),Dy \bigl(u^{*} \bigr) \bigl(u-u^{*} \bigr) \bigr\rangle _{{ W}(0,T)',{W}(0,T)} \\& \qquad{} + \bigl(Ru^{*},u-u^{*} \bigr)_{\mathcal {U}} \geq0,\quad \forall v\in{ \mathcal {U}}_{\mathrm{ad}}, \end{aligned}$$
(4.55)
where \(\Lambda_{M}\) is the canonical isomorphism M onto \(M'\).
In this paper, we consider the following physically important observation. We take \(M=L^{2}(0, T; L^{2})\) and \(C \in{\mathcal {L}}({ W}(0,T), M)\) and observe that \(Cy(u)= y(u; \cdot) \in L^{2}(0, T; L^{2})\).
4.2 Necessary condition of an optimal control for distributive observation
In this subsection, we consider the cost functional expressed by
$$ J(u) = \int^{T}_{0} \bigl\Vert y(u)-Y_{d} \bigr\Vert ^{2} \,dt +(Ru,u)_{\mathcal {U}}\quad \forall u \in{ \mathcal {U}}_{\mathrm{ad}} \subset{ \mathcal {U}}, $$
(4.56)
where \(Y_{d} \in L^{2}(0, T;L^{2})\) is the desired value. Let \(u^{*}\) be the optimal control subject to (4.2) and (4.56). Then the optimality condition (4.55) is represented by
$$ \int^{T}_{0} \bigl(y \bigl(u^{*} \bigr)-Y_{d}, z \bigr)_{2} \,dt + \bigl(Ru^{*},u-u^{*} \bigr)_{\mathcal {U}} \geq0\quad\forall u \in{\mathcal {U}}_{\mathrm{ad}}, $$
(4.57)
where z is the weak solution of Eq. (4.39). Now we formulate the adjoint system to describe the optimality condition:
$$ \textstyle\begin{cases} p_{tt} (u^{*}) -\Delta p_{tt}(u^{*}) + \Delta^{2} p(u^{*}) + \int^{T}_{t} k(\sigma-t) \Delta^{2} p(u^{*}; \sigma)\,d \sigma\\ \quad = [p(u^{*}), -G[y(u^{*}),y(u^{*})]] +2 [y(u^{*}), -G[ p(u^{*}), y(u^{*})]] \\ \qquad{} + y(u^{*})-Y_{d} \quad\mbox{in } Q,\\ p(u^{*})= \frac{\partial p(u^{*})}{\partial\nu} =0 \quad\mbox{on } \Sigma,\\ p(u^{*};T)= p_{t}(u^{*};T) =0 \quad\mbox{in } \Omega. \end{cases} $$
(4.58)
Proposition 4.1
Equation (4.58) admits a unique solution
\(p(u^{*}) \in S(0, T)\).
Proof
Since
$$\int^{T}_{T-t} k(\sigma-T+t) \Delta^{2} p \bigl(u^{*}; \sigma \bigr) \,d\sigma= \int ^{t}_{0} k(t-s) \Delta^{2} p \bigl(u^{*}; T- s \bigr) \,ds, $$
the time reversed equation of Eq. (4.58) (\(t \to T-t\) in Eq. (4.58)) is given by
$$ \textstyle\begin{cases} \psi_{tt} -\Delta \psi_{tt}+ \Delta^{2} \psi+ \int^{t}_{0} k(t-s) \Delta ^{2} \psi(s)\,d s \\ \quad = [ \psi, -G[y(u^{*}),y(u^{*})]] +2 [y(u^{*}), -G[ \psi, y(u^{*})]] + y(u^{*})-Y_{d} \quad\mbox{in } Q,\\ \psi= \frac{\partial \psi}{\partial\nu} =0 \quad\mbox{on } \Sigma,\\ \psi(0)= \psi_{t}(0) =0 \quad\mbox{in } \Omega, \end{cases} $$
(4.59)
where \(\psi(t)= p(u^{*};T-t)\).
Here we note that, like (4.42),
$$\begin{aligned}& \bigl\Vert \bigl[ \psi, -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr] +2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ \psi, y \bigl(u^{*} \bigr) \bigr] \bigr] \bigr\Vert _{L^{2}(0, T; L^{2})} \\& \quad \le C \bigl\Vert p^{*} \bigr\Vert ^{2}_{{\mathcal {P}}} \Vert \Delta\psi \Vert _{L^{2}(0, T; L^{2})}, \end{aligned}$$
(4.60)
where \(p^{*} = (y_{0}, y_{1}, Bu^{*})\). Thus, by Theorem 3.1 and [5], the conditions \(Y_{d} \in L^{2}(0, T;L^{2})\) and (4.60) enable us to deduce that there exists a unique \(\psi\in S(0, T)\).
This completes the proof. □
Now we proceed the calculations. We multiply both sides of the weak form of Eq. (4.58) by z and integrate it over \([0,T]\). Then we have
$$\begin{aligned}& \int^{T}_{0} \bigl\langle p'' \bigl(u^{*} \bigr) -\Delta p'' \bigl(u^{*} \bigr), z \bigr\rangle _{-2, 2} \,dt \\& \qquad{} + \int^{T}_{0} \biggl( \Delta p \bigl(u^{*} \bigr) + \int^{T}_{t} k(\sigma-t) \Delta p \bigl(u^{*}; \sigma \bigr)\,d \sigma, \Delta z \biggr)_{2} \,dt \\& \qquad{} - \int^{T}_{0} \bigl( \bigl[p \bigl(u^{*} \bigr), -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr] +2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ p \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr], z \bigr)_{2} \,dt \\& \quad = \int^{T}_{0} \bigl(y \bigl(u^{*} \bigr)-Y_{d}, z \bigr)_{2} \,dt. \end{aligned}$$
(4.61)
By Fubini’s theorem we have
$$\begin{aligned}& \int^{T}_{0} \biggl( \int^{T}_{t} k(\sigma-t) \Delta p \bigl(u^{*}; \sigma \bigr)\,d \sigma, \Delta z \biggr)_{2} \,dt \\& \quad = \int^{T}_{0} \biggl( \int^{t}_{0} k(t-s) \Delta z (s)\,ds, \Delta p \bigl(u^{*} \bigr) \biggr)_{2} \,dt \\& \quad = \int^{T}_{0} \biggl\langle \int^{t}_{0} k(t-s) \Delta^{2} z (s)\,ds, p \bigl(u^{*} \bigr) \biggr\rangle _{-2,2} \,dt. \end{aligned}$$
(4.62)
By Lemma 2.1 we deduce
$$\begin{aligned}& \int^{T}_{0} \bigl( \bigl[p \bigl(u^{*} \bigr), -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr], z \bigr)_{2} \,dt \\& \quad = \int^{T}_{0} \bigl( \bigl[z, -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr], p \bigl(u^{*} \bigr) \bigr)_{2} \,dt. \end{aligned}$$
(4.63)
We observe that by considering \(\phi, \psi\in H^{2}_{0}\) we have \([\phi, \psi] \in L^{1}\). However, since \(n=2\), we have
$$\begin{aligned} H^{2}_{0} \hookrightarrow L^{\infty}, \end{aligned}$$
(4.64)
and, therefore,
$$\begin{aligned} L^{1} \hookrightarrow H^{-2}. \end{aligned}$$
(4.65)
Thus, since G is a self-adjoint operator, by Lemma 2.1 and (4.65) we have
$$\begin{aligned}& \int^{T}_{0} \bigl( 2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ p \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr], z \bigr)_{2} \,dt \\& \quad = \int^{T}_{0} \bigl\langle 2 \bigl[z, y \bigl(u^{*} \bigr) \bigr], -G \bigl[ p \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr\rangle _{-2,2} \,dt \\& \quad = \int^{T}_{0} \bigl\langle -2 G \bigl[z, y \bigl(u^{*} \bigr) \bigr], \bigl[ p \bigl(u^{*} \bigr), y \bigl(u^{*} \bigr) \bigr] \bigr\rangle _{2,-2} \,dt \\& \quad = \int^{T}_{0} \bigl( 2 \bigl[y \bigl(u^{*} \bigr), - G \bigl[z, y \bigl(u^{*} \bigr) \bigr] \bigr], p \bigl(u^{*} \bigr) \bigr)_{2} \,dt. \end{aligned}$$
(4.66)
Considering from (4.62) to (4.66), the terminal value conditions of p in (4.58), and Eq. (4.39), we can verify by integration by parts that the left-hand side of (4.61) yields
$$\begin{aligned}& \int^{T}_{0} \biggl\langle p \bigl(u^{*} \bigr), z'' - \Delta z'' + \Delta^{2} z + \int^{t}_{0} k(t-s) \Delta^{2} z(s) \,ds \biggr\rangle _{2, -2} \,dt \\& \qquad{} - \int^{T}_{0} \bigl(p \bigl(u^{*} \bigr), \bigl[z, -G \bigl[y \bigl(u^{*} \bigr),y \bigl(u^{*} \bigr) \bigr] \bigr] +2 \bigl[y \bigl(u^{*} \bigr), -G \bigl[ z, y \bigl(u^{*} \bigr) \bigr] \bigr] \bigr)_{2} \,dt \\& \quad = \int^{T}_{0} \bigl(p \bigl(u^{*} \bigr), B \bigl(u-u^{*} \bigr) \bigr)_{2} \,dt. \end{aligned}$$
(4.67)
Therefore, combining (4.61) and (4.67), we deduce that the optimality condition (4.57) is equivalent to
$$\int^{T}_{0} \bigl(p \bigl(u^{*} \bigr), B \bigl(u-u^{*} \bigr) \bigr)_{2}\,dt+ \bigl(Ru^{*}, u-u^{*} \bigr)_{\mathcal {U}} \geq0\quad\forall u\in{ \mathcal {U}}_{\mathrm{ad}}. $$
Hence, we give the following theorem.
Theorem 4.3
The optimal control
\(u^{*}\)
for (4.56) is characterized by the following system of equations and inequality:
$$\begin{aligned}& \textstyle\begin{cases} y_{tt}(u^{*}) - \Delta y_{tt}(u^{*}) + \Delta^{2} y(u^{*}) + \int^{t}_{0} k(t-s) \Delta^{2} y(u^{*};s)\,ds \\ \quad = [y(u^{*}), v(u^{*})] + Bu^{*} \quad\textit{in } Q, \\ \Delta^{2} v(u^{*}) = -[y(u^{*}),y(u^{*})] \quad\textit{in } Q, \\ y(u^{*}) = \frac{\partial y(u^{*}) }{\partial\nu} = v(u^{*}) = \frac {\partial v(u^{*}) }{\partial\nu}=0 \quad\textit{on } \Sigma, \\ y (u^{*};0) = y_{0},\qquad y_{ t}(u^{*}; 0) = y_{1} \quad \textit{in } \Omega, \end{cases}\displaystyle \\& \textstyle\begin{cases} p_{tt} (u^{*}) -\Delta p_{tt}(u^{*}) + \Delta^{2} p(u^{*}) + \int^{T}_{t} k(\sigma-t) \Delta^{2} p(u^{*}; \sigma)\,d \sigma\\ \quad = [p(u^{*}), -G[y(u^{*}),y(u^{*})]] +2 [y(u^{*}), -G[ p(u^{*}), y(u^{*})]] + y(u^{*})-Y_{d} \quad\textit{in } Q,\\ p(u^{*})= \frac{\partial p(u^{*})}{\partial\nu} =0 \quad\textit{on } \Sigma,\\ p(u^{*};T)= p_{t}(u^{*};T) =0 \quad \textit{in } \Omega, \end{cases}\displaystyle \\& \int^{T}_{0} \bigl(p \bigl(u^{*} \bigr), B \bigl(u-u^{*} \bigr) \bigr)_{2}\,dt+ \bigl(Ru^{*}, u-u^{*} \bigr)_{ U} \geq0\quad \forall u \in{ U}_{\mathrm{ad}}. \end{aligned}$$
4.3 Local uniqueness of an optimal control
We note that the uniqueness of an optimal control in nonlinear equation is not ensured. However, it is worth noticing partial results. For instance, we can refer to the result in [12] to obtain the local uniqueness of an optimal control for distributive observation case. For that reason, in this subsection, we take \(M = L^{2}((0, t) \times\Omega)\) and observe that \(y \in L^{2}((0, t) \times\Omega)\). Hence, we consider the following quadratic cost functional:
$$ J(u) = \int^{t}_{0} \bigl\Vert y(u)-Y_{d} \bigr\Vert ^{2} \,ds +(Ru,u)_{ U} \quad\forall u\in{ \mathcal {U}}_{\mathrm{ad}} \subset{\mathcal {U}}, $$
(4.68)
where \(Y_{d} \in L^{2}((0, t) \times\Omega)\).
In order to show the local uniqueness of an optimal control by making use of the strict convexity of quadratic cost (see [17]), we consider the following proposition.
Proposition 4.2
The map
\(w\to y(w)\)
of
U
into
\({S}(0,T)\)
is second-order Gâteaux differentiable at
\(w=u\)
and such the second-order Gâteaux derivative of
\(y(w)\)
at
\(w=u\)
in the direction
\(w-u\in{\mathcal {U}}\), say
\(g =D^{2} y(u)(w-u, w-u)\), is a unique solution of the following problem:
$$ \textstyle\begin{cases} g_{tt} - \Delta g_{tt} + \Delta^{2} g + \int^{t}_{0} k(t-s) \Delta^{2} g(s)\,ds \\ \quad = [g, -G[y(u), y(u)]]+2[y(u), -G[g,y(u)]] + F(z, y(u)) \quad \textit{in } Q,\\ g = \frac{\partial g}{\partial\nu} =0 \quad \textit{on } \Sigma,\\ g(0)=g_{t}(0) = 0 \quad\textit{in } \Omega, \end{cases} $$
(4.69)
where
$$F \bigl(z,y(u) \bigr)= 4 \bigl[z, -G \bigl[z, y(u) \bigr] \bigr]+2 \bigl[y(u), -G[z,z] \bigr], $$
and
z
is the weak solution of Eq. (4.39), changing
\(B(u-u^{*})\)
by
\(B(w-u)\).
Proof
The proof is similar to that of Theorem 4.2. □
Lemma 4.2
Let
g
be the weak solution of Eq. (4.69). Then we can show that
$$ \Vert g \Vert _{{ S}(0, T)} \leq C \Vert w- u \Vert ^{2}_{\mathcal {U}}, $$
(4.70)
where
\(C > 0 \)
is a constant depending on the time
T
and the data conditions of the equation of
\(y(u)\).
Proof
Let z be the solution of Eq. (4.39), changing with \(B(u-u^{*})\) to \(B(w-u)\). Then, using the same arguments as in Eq. (3.1), we can deduce that
$$\begin{aligned} \Vert z \Vert _{{S}(0, T)} \leq& C \bigl\Vert B(w-u) \bigr\Vert _{L^{2}(0, T; L^{2})} \\ \leq& C \Vert B \Vert _{{\mathcal {L}}({\mathcal {U}}; L^{2}(0, T; L^{2} ))} \Vert w-u \Vert _{\mathcal {U}} \\ \leq& C \Vert w-u \Vert _{\mathcal {U}}. \end{aligned}$$
(4.71)
Also, for the solution g of Eq. (4.69), we can show that
$$\begin{aligned} \Vert g \Vert _{{ S}(0, T)} \leq& C \bigl\Vert F \bigl(z,y(u) \bigr) \bigr\Vert _{L^{2}(0, T; L^{2})} \\ \leq& C \bigl( \bigl\Vert 4 \bigl[z, -G \bigl[z, y(u) \bigr] \bigr] \bigr\Vert _{L^{2}(0, T; L^{2})} + \bigl\Vert 2 \bigl[y(u), -G[z,z] \bigr] \bigr\Vert _{L^{2}(0, T; L^{2})} \bigr) \\ \leq& C \bigl\Vert y(u) \bigr\Vert _{L^{2}(0, T; H^{2}_{0})} \Vert z \Vert ^{2}_{L^{\infty} (0, T; H^{2}_{0})} \\ \leq& C \sqrt{T} \bigl\Vert y(u) \bigr\Vert _{L^{\infty}(0, T; H^{2}_{0})} \Vert z \Vert ^{2}_{L^{\infty} (0, T; H^{2}_{0})} \\ \leq& C \sqrt{T} \Vert p \Vert _{{\mathcal {P}}} \Vert z \Vert ^{2}_{{S}(0, T)}, \end{aligned}$$
(4.72)
where \(p = (y_{0}, y_{1}, Bu)\). Combining (4.71) with (4.72), we have (4.70). □
We prove the local uniqueness of the optimal control.
Theorem 4.4
When
t
is small enough, there is a unique optimal control for the cost (4.68).
Proof
We show the local uniqueness by proving the strict convexity of the map \(u \in{\mathcal {U}}_{\mathrm{ad}} \to J(u)\). Therefore, as in [17], we need to show, for all \(u, w \in{\mathcal {U}}_{\mathrm{ad}}\) (\(u \ne w\)),
$$ D^{2}J \bigl(u + \xi(w-u) \bigr) (w-u, w-u) > 0 \quad ( 0 < \xi< 1 ). $$
(4.73)
For simplicity, we denote \(y(u + \xi(w-u))\), \(z(u + \xi(w-u))\), and \(g(u + \xi(w-u))\) by \(y(\xi)\), \(z(\xi)\), and \(g(\xi)\), respectively. We calculate
$$\begin{aligned}& DJ \bigl(u + \xi(w-u) \bigr) (w-u) \\& \quad = \lim_{l \to0} \frac{J(u + (\xi+ l)(w-u)) - J(u + \xi(w-u))}{l} \\& \quad = 2 \int^{t}_{0} \bigl(y(\xi) - Y_{d}, z(\xi) \bigr)_{2} \,ds + 2 \bigl(R \bigl(u+ \xi(w-u) \bigr), w-u \bigr)_{\mathcal {U}}. \end{aligned}$$
(4.74)
From (4.74) we obtain the second Gâteaux derivative of J:
$$\begin{aligned}& D^{2}J \bigl(u + \xi(w-u) \bigr) (w-u, w-u ) \\& \quad = \lim_{k \to0} \frac{DJ(u + (\xi+ k)(w-u))(w-u) - DJ(u + \xi (w-u))(w-u)}{k} \\& \quad = 2 \int^{t}_{0} \bigl(y(\xi) - Y_{d}, g(\xi) \bigr)_{2} \,ds + 2 \int^{t}_{0} \bigl\Vert z(\xi) \bigr\Vert ^{2} \,ds \\& \qquad{} + 2 \bigl(R (w-u), w-u \bigr)_{\mathcal {U}}. \end{aligned}$$
(4.75)
By Lemma 4.2 and (4.75) we deduce that
$$\begin{aligned}& D^{2}J \bigl(u + \xi(w-u) \bigr) (w-u, w-u ) \\& \quad \ge - 2 \bigl\Vert g(\xi) \bigr\Vert _{L^{\infty}(0, t; L^{2})} \int^{t}_{0} \bigl\Vert y(\xi) - Y_{d} \bigr\Vert \,ds \\& \qquad{} + 2 \int^{t}_{0} \bigl\Vert z(\xi) \bigr\Vert ^{2} \,ds + 2 d \Vert w-u \Vert ^{2}_{\mathcal {U}} \\& \quad \ge - 2 C \sqrt{t} \bigl\Vert g(\xi) \bigr\Vert _{{ S}(0, t)} \bigl\Vert y(\xi) - Y_{d} \bigr\Vert _{L^{2}(0, t; L^{2})} \\& \qquad{} + 2 \int^{t}_{0} \bigl\Vert z(\xi) \bigr\Vert ^{2} \,ds + 2 d \Vert w-u \Vert ^{2}_{\mathcal {U}} \\& \quad \ge 2 \bigl( d - C \sqrt{t} \bigl\Vert y(\xi) - Y_{d} \bigr\Vert _{L^{2}(0, t; L^{2})} \bigr) \Vert w-u \Vert ^{2}_{\mathcal {U}} \\& \qquad{} + 2 \int^{t}_{0} \bigl\Vert z(\xi) \bigr\Vert ^{2} \,ds. \end{aligned}$$
(4.76)
Here we can take \(t > 0 \) small enough so that the right-hand side of (4.76) is strictly greater than 0. Therefore, we obtain the strict convexity of the quadratic cost \(J(u)\), \(u \in{\mathcal {U}}_{\mathrm{ad}}\), which proves this theorem. □
Remark 4.1
If we assume that d is large enough, then we can obtain the strict convexity of the quadratic cost (4.68) in the global sense. Therefore, we can obtain the desired result of Theorem 4.4 in the global sense for the cost (4.68).