1 Introduction

The necessary and sufficient conditions of optimality for systems governed by elliptic, parabolic, and hyperbolic operators have been studied by Lions in [7, 8]. The considered systems in these problems are in the scalar case (system of one equation).

The discussion is extended to \(2\times 2\)systems for example in [1, 5, 9, 10, 12] and to \(n\times n\) systems in [11].

Optimal control problem for systems involving Schrödinger operators has been studied for the following elliptic system of distributed type [10]:

$$\begin{aligned} \left\{ {\begin{array}{l} \left( {-\Delta +q} \right) y_1 =a y_1 +b y_2 +f_1 \quad \mathrm{in}\;R^{n}, \\ \left( {-\Delta +q} \right) y_2 =c y_1 +d y_2 +f_2 \quad \mathrm{in}\;R^{n}, \\ y_1 ,y_2 \rightarrow 0\quad \text{ as }\quad \left| \text{ x } \right| \rightarrow \infty . \\ \end{array}} \right. \end{aligned}$$
(1)

and for parabolic system of boundary type in [1].

Systems with different potentials and positive weight function is studied in [12] and with variable coefficients is studied in [9].

The existence of optimal control for systems like (1) has been proved with \(q(x)=0\) in [6], and for semi linear cooperative systems in [5].

Time-optimal control problem for cooperative hyperbolic systems involving the Laplace operator is studied in [2].

Here, we consider the following \(2\times 2\) cooperative hyperbolic systems involving Schrödinger operator:

$$\begin{aligned} \left\{ {\begin{array}{l} \frac{\partial ^{2}y_1 (x)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_1 =a y_1 +b y_2 +f_1 (x,t)\quad \mathrm{in}\; Q, \\ \frac{\partial ^{2}y_2 (x)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_2 =c y_1 +d y_2 +f_2 (x,t)\; \;\mathrm{in}\; Q, \\ y_1 ,y_2 \rightarrow 0\quad \text{ as }\quad \left| \text{ x } \right| \rightarrow \infty , \\ \left. {y_1 } \right| _\Sigma =\left. {y_2 } \right| _\Sigma =0,\;\quad \quad \quad \\ y_1 (x,0)=y_{1,0} (x),\quad y_2 (x,0)=y_{2,0} (x)\quad \quad \mathrm{in}\quad R^{n}, \\ \frac{\partial y_1 (x,0)}{\partial t}=y_{1,1} (x), \quad \frac{\partial y_2 (x,0)}{\partial t}=y_{2,1} (x) \qquad \;\;\mathrm{in}\quad R^{n}. \\ \end{array}} \right. \end{aligned}$$
(2)

with

$$\begin{aligned}&y_1 ,y_2 \in L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) , \\&\quad \quad \quad \quad \quad \frac{\partial y_1 }{\partial t},\frac{\partial y_2 }{\partial t}\in L^{2}\left( {0,T;V_q^{\prime } \left( {R^{n}} \right) } \right) . \\ \end{aligned}$$

where

$$\begin{aligned} a,b,c\, \text{ and }\, d\, \text{ are } \text{ given } \text{ numbers } \text{ such } \text{ that }\, b,c>0, \end{aligned}$$
(3)

(in this case, we say that the system (2) is cooperative)

$$\begin{aligned} q\text{(x) }\, \text{ is } \text{ a } \text{ positive } \text{ function } \text{ and } \text{ tending } \text{ to }\, \infty \, \text{ at } \text{ infinity }, \end{aligned}$$
(4)

and \(Q=R^{n}\times \left] {0,T} \right[\) with boundary \(\Sigma =\Gamma \times \left] {0,T} \right[\).

We first prove the existence and uniqueness of the state for these systems, then we introduce the optimal control of distributed type for these systems.

2 Some concepts and results

In this paper, we shall consider some results introduced in [3], [10] concerning the eigenvalue problem

$$\begin{aligned} \left\{ {\begin{array}{l} \left( {-\Delta + q} \right) \phi = \lambda (q) \phi \qquad \text{ in } \quad R^{n} \\ \phi \left( x \right) \rightarrow 0\quad \text{ as } \quad \left| x \right| \rightarrow \infty ,\quad \phi \succ 0 \\ \end{array}} \right. \end{aligned}$$
(5)

The associated variational space is \(V_q \left( {R^{n}} \right) \), the completion of \(D{}\left( {R^{n}} \right) \), with respect to the norm:

$$\begin{aligned} \left\| y \right\| _q =\left( {\int \limits _{R^{n}} {\left[ {\left| {\nabla y} \right| ^{2}+q{}\left| y \right| ^{2}} \right] dx} } \right) ^{1/2} \end{aligned}$$
(6)

Since the imbedding of \(V_q \left( {R^{n}} \right) \) in to \(L^{2}\left( {R^{n}} \right) \) is compact, then the operator \(\left( {-\Delta +q} \right) \) considered as an operator in \(L^{2}\left( {R^{n}} \right) \) is positive self-adjoint with compact inverse. Hence its spectrum consists of an infinite sequence of positive eigenvalues, tending to infinity; moreover the smallest one which is called the principal eigenvalue denoted by \(\lambda (q)\) is simple and is associated with an eigenfunction which does not change sign in \(R^{n}\). It is characterized by:

$$\begin{aligned} \lambda \left( q \right) \int \limits _{R^{n}} {\left| y \right| ^{2}dx} \le \int \limits _{R^{n}} {\left[ {\left| {\nabla y} \right| ^{2}+q\left| y \right| ^{2}} \right] dx} \;\;\forall y\in V_q \left( {R^{n}} \right) .\nonumber \\ \end{aligned}$$
(7)

We have the following embedding :

$$\begin{aligned}&V_q \left( {R^{n}} \right) \!\times \! V_q \left( {R^{n}} \right) \!\subseteq \! L^{2}\left( {R^{n}} \right) \!\times \! L^{2}\left( {R^{n}} \right) \\&\quad \subseteq V_q^{\prime } \left( {R^{n}} \right) \!\times \! V_q^{\prime } \left( {R^{n}} \right) \end{aligned}$$

which is continuous and compact .

Let us introduce the space \(L^{2}(0,T;V_q (R^{n}))\)of measurable function \( t \rightarrow f(t)\) which is defined on open interval (\(0\),\(T)\), since the variable \( t\in (0,T)\) and \(T < \infty \) denotes the time .

On (\(0\),\(T)\) with Lebesgue measure dt we have the norm:

$$\begin{aligned} \left\| {f(t)} \right\| _{L^{2}(0,T;V_q (R^{n}))} =\left( {\int \limits _{(0,T)} {\left\| {f(t)} \right\| _{V_q (R^{n})}^2 dt} } \right) ^{1/2}\prec \infty \end{aligned}$$

and the scalar product

$$\begin{aligned} \left( {f(t),g(t)} \right) _{L^{2}(0,T;V_q (R^{n}))} =\int \limits _{(0,T)} {\left( {f(t),g(t)} \right) _{V_q (R^{n})} dt} , \end{aligned}$$

the space\(L^{2}\left( 0,T;V_q (R^{n})\right) \)with the scalar product and the norm above is a Hilbert space .

Analogously, we can define the spaces \(L^{2}(0,T;L^{2}(R^{n}))=L^{2}(Q),\)

with the scalar product

$$\begin{aligned} \left( {f(t),g(t)} \right) _{L^{2}(Q)}&=\int \limits _{(0,T)} {\left( {f(t),g(t)} \right) _{L^{2}(R^{n})} dt} \\&=\int \limits _Q {f(t).g(t){}{}{}dx{}dt} \\ \end{aligned}$$

then we have the following embedding

$$\begin{aligned} \left( {L^{2}(0,T;V_q (R^{n}))} \right) ^{2}\subseteq \left( {L^{2}(Q)} \right) ^{2}\subseteq \left( {L^{2}(0,T;V_q^{\prime } (R^{n}))} \right) ^{2} \end{aligned}$$

3 Existence and uniqueness of solution

We introduce the bilinear form

$$\begin{aligned}&\pi \left( {t ;y,\psi } \right) =\left( {A(t){}y,\psi } \right) _{\left( {L^{2}(R^{n})} \right) ^{2}} ,\\&y=\left\{ {y_1 ,y_2 } \right\} ,\quad \psi =\left\{ {\psi _1 ,\psi _2 } \right\} \in \left( {V_q \left( {R^{n}} \right) } \right) ^{2},\\&A\left( t \right) y\in \left( {V_q^{\prime } \left( {R^{n}} \right) } \right) ^{2} \end{aligned}$$

where

$$\begin{aligned}&A\left( t \right) y(x)=\left\{ \left( {-\Delta +q} \right) y_1 -a y_1 -b y_2,\; \right. \\&\left. \left( {-\Delta +q} \right) y_2 -c y_1 -d y_2 \right\} \end{aligned}$$

then

$$\begin{aligned} \pi \left( {t ;y,\psi } \right)&=\frac{1}{b}\int \limits _{R^{n}} {\left[ {\nabla y_1 \nabla \psi _1 +qy_1 \psi _1 } \right] dx} \nonumber \\&\quad +\frac{1}{c}\int \limits _{R^{n}} {\left[ {\nabla y_2 \nabla \psi _2 +qy_2 \psi _2 } \right] dx} \nonumber \\&\quad -\int \limits _{R^{n}} {y_1 \psi _2 \;dx} -\frac{d}{c}\int \limits _{R^{n}} {y_2 \psi _2 \;dx} \nonumber \\&\quad -\frac{a}{b}\int \limits _{R^{n}} {y_1 \psi _1 \;dx} -\int \limits _{R^{n}} {y_2 \psi _1 \;dx} . \end{aligned}$$
(8)

For all \(y,\psi \in \quad \left( {V_q \left( {R^{n}} \right) } \right) ^{2}\), the function \(t\rightarrow \pi \left( {t ;y,\psi } \right) \) is measurable on (\(0\),\(T).\)

By using the necessary and sufficient conditions for having the maximum principle and existence of positive solutions for cooperative system (1) which have been obtained by Fleckinger [4] and take the form

$$\begin{aligned} \left\{ {\begin{array}{l} a\prec \lambda \left( q \right) \;,\;d\prec \lambda \left( q \right) \;, \\ \left( {\lambda \left( q \right) -a} \right) \left( {\lambda \left( q \right) -d} \right) \succ bc\quad , \\ \end{array}} \right. \quad \; \end{aligned}$$
(9)

the coerciveness condition of the bilinear form (8) in \(\left( {V_q \left( {R^{n}} \right) } \right) ^{2}\)has been proved by Serag [10], that means

$$\begin{aligned} \pi \left( {t ;y,y} \right) \ge C \left( {\left\| {y_1 } \right\| _{q,m}^2 +\left\| {y_2 } \right\| _{q,m}^2 } \right) ,\quad C\succ 0 \end{aligned}$$
(10)

Theorm 1

Under the hypotheses (3) and (10), if \(f_1 ,f_2 \in L^{2}(0,T;V_q^{\prime } (R^{n})), y_{1,0} (x),y_{2,0} (x){\in } V_q \left( {R^{n}} \right) \) and \(y_{1,1} (x),y_{2,1} (x)\in V_q^{\prime } \left( {R^{n}} \right) \), then there exists a unique solution \(y=\left\{ {y_1 ,y_2 } \right\} \in \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}\) for system (2).

Proof

Let \(\psi \rightarrow L(\psi )\) be a continuous linear form defined on \(\left( {L^{2}\left( Q \right) } \right) ^{2}\)by

$$\begin{aligned} L(\psi )&= \;\frac{1}{b}\int \limits _Q {f_1 (x,t)\;\psi _1 (x) dx{}dt}\nonumber \\&+\frac{1}{c}\int \limits _Q {f_2 (x,t)\;\psi _2 (x) dx{}dt}\nonumber \\&+\frac{1}{b}\int \limits _{R^{n}} {y_{1,1} (x) {}\psi _1 \left( {x,0} \right) dx}\nonumber \\&+\frac{1}{c}\int \limits _{R^{n}} {y_{2,1} (x) {}\psi _2 \left( {x,0} \right) dx} \quad \nonumber \\ \forall \;\psi&= \left\{ {\psi _1 ,\psi _2 } \right\} \in \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}\!, \end{aligned}$$
(11)

then by Lax–Milgram lemma, there exists a unique element \(y=\left\{ {y_1 ,y_2 } \right\} \in \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}\)such that

$$\begin{aligned}&\frac{1}{b}\left( {\frac{\partial ^{2}y_1 }{\partial t^{2}},\psi _1 } \right) +\frac{1}{c}\left( {\frac{\partial ^{2}y_2 }{\partial t^{2}},\psi _2 } \right) +\pi \left( {t ;y,\psi } \right) =L(\psi )\nonumber \\&\forall \psi =\left\{ {{}\psi _1 ,\psi _2 } \right\} \in \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}, \end{aligned}$$
(12)

Now, let us multiply both sides of first equation of system (2) by \(\frac{1}{b}\psi _1 \left( x \right) \), and the second equation by \(\frac{1}{c}\psi _2 \left( x \right) \) then integration over Q, we have:

$$\begin{aligned}&\frac{1}{b}\int \limits _Q {[\frac{\partial ^{2}y_1 (x)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_1 -a y_1 -b y_2 ]\;\psi _1 dx{}dt} \\&\quad =\frac{1}{b}\int \limits _Q {f_1 (x,t)\;\psi _1 dx{}dt} \quad ,\\&\frac{1}{c}\int \limits _Q {[\frac{\partial ^{2}y_2 (x)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_2 -c y_1 -d y_2 ]\;\psi _2 dx{}dt} \\&\quad =\frac{1}{c}\int \limits _Q {f_2 (x,t)\;\psi _2 dx dt} \quad . \end{aligned}$$

By applying Green\(^{{\prime }}\)s formula:

$$\begin{aligned}&\frac{1}{b}\int \limits _Q { \frac{\partial ^{2}y_1 (x)}{\partial t^{2}}\psi _1 \left( x \right) dx} +\frac{1}{b}\int \limits _Q {\nabla y_1 {}\nabla \psi _1 dx dt} \\&-\frac{1}{b}\int \limits _\Sigma {\psi _1 {}\frac{\partial y_1 }{\partial \nu _A }d\Sigma } -\frac{1}{b}\int \limits _{R^{n}} {\psi _1 \left( {x,0} \right) {}\frac{\partial y_1 \left( {x,0} \right) }{\partial t}dx} \\&+\int \limits _Q {(\frac{q}{b}{}y_1 -\frac{a}{b} y_1 - y_2 )\;\psi _1 dx dt} \\&\quad =\frac{1}{b}\int \limits _Q {f_1 (x,t)\;\psi _1 dx dt} \;, \\&\qquad \frac{1}{c}\int _{R^{n}} {\frac{\partial ^{2}y_2 \left( x \right) }{\partial t^{2}}\psi _2 \left( x \right) dx } +\frac{1}{c}\int \limits _Q {\nabla y_2 {}\nabla \psi _2 dx dt} \\&\qquad -\frac{1}{c}\int \limits _\Sigma {\psi _2 {}\frac{\partial y_2 }{\partial \nu _A }d\Sigma } -\frac{1}{c}\int \limits _{R^{n}} {\psi _2 \left( {x,0} \right) {}\frac{\partial y_2 \left( {x,0} \right) }{\partial t}dx} \\&\qquad +\int \limits _Q {(\frac{q}{c}{}y_2 -y_1 -\frac{d}{c} y_2 )\;\psi _2 dx dt} \\&\quad =\frac{1}{c}\int \limits _Q {f_2 (x,t)\;\psi _2 dx dt}. \end{aligned}$$

By sum the two equations, then comparing the summation with (8), (11) and (12) we get:

$$\begin{aligned}&-\frac{1}{b}\int \limits _\Sigma {\psi _1 {}\frac{\partial y_1 }{\partial \nu _A }d\Sigma } -\frac{1}{c}\int \limits _\Sigma {\psi _2 \frac{\partial y_2 }{\partial \nu _A }d\Sigma } \\&\!-\!\frac{1}{b}\int \limits _{R^{n}} {\psi _1 \left( {x,0} \right) \frac{\partial y_1 \left( {x,0} \right) }{\partial t}dx} \!-\!\frac{1}{c}\int \limits _{R^{n}} {\psi _2 \left( {x,0} \right) \frac{\partial y_2 \left( {x,0} \right) }{\partial t}dx} \\&=\frac{1}{b}\int \limits _{R^{n}} {y_{1,1} (x){}{}\psi _1 \left( {x,0} \right) dx} +\frac{1}{c}\int \limits _{R^{n}} {y_{2,1} (x){}{}\psi _2 \left( {x,0} \right) dx}, \end{aligned}$$

then we deduce that:

$$\begin{aligned}&\left. {y_1 } \right| _\Sigma =\left. {y_2 } \right| _\Sigma =0\\&\frac{\partial y_1 (x,0)}{\partial t}=y_{1,1} (x), \quad \frac{\partial y_2 (x,0)}{\partial t}=y_{2,1} (x)\quad \quad \mathrm{in} \quad R^{n}. \end{aligned}$$

which completes the proof.\(\square \)

4 Formulation of the control problem

The space \(L^{2}(Q)\times L^{2}(Q)\) is the space of controls. For a control \(u=\left\{ {u_1 ,u_2 } \right\} \in (L_2 (Q))^{2}\), the state \(y\left( u \right) =\left\{ {y_1 \left( u \right) ,y_2 (u)} \right\} \in \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}\) of the system (2) is given by the solution of

$$\begin{aligned} \left\{ {\begin{array}{l} \frac{\partial ^{2}y_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_1 (u)=a y_1 (u)+b y_2 (u)+f_1 +u_1 \;\; {}\mathrm{in}\; Q,\\ \frac{\partial ^{2}y_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_2 (u)=c y_1 (u)+d y_2 (u)+f_2 +u_2 \;\;\mathrm{in}\; Q, \\ y_1 {} ,y_2 \rightarrow 0\quad \text{ as }\quad \left| \text{ x } \right| \rightarrow \infty , \\ \left. {y_1 (u)} \right| _\Sigma =\left. {y_2 (u)} \right| _\Sigma =0,\;\quad \quad \quad \\ y_1 (x,0,u)=y_{1,0} (x),\quad y_2 (x,0,u)=y_{2,0} (x)\quad \mathrm{in} \quad R^{n}, \\ \frac{\partial y_1 (x,0,u)}{\partial t}=y_{1,1} (x), \quad \frac{\partial y_2 (x,0,u)}{\partial t}=y_{2,1} (x)\quad \quad \,\, \mathrm{in}\quad R^{n}. \\ \end{array}} \right. \end{aligned}$$
(13)

with

$$\begin{aligned}&y_1 (u),y_2 (u)\in L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) \;, \\&\quad \frac{\partial y_1 (u)}{\partial t},\frac{\partial y_2 (u)}{\partial t}\in L^{2}\left( {0,T;V_q^{\prime } \left( {R^{n}} \right) } \right) \\ \end{aligned}$$

The observation equation is given by \(z(u) = \{z_{1}(u),z_{2}(u)\} = y(u) = \{y_{1}(u),y_{2}(u)\}.\)

For a given \(z_{d} = \{z_{d1},z_{d2}\} \in \left( {L^{2}\left( Q \right) } \right) ^{2}\), the cost function is given by

$$\begin{aligned}&J(v)=\left\| {y_1 (v)-z_{d1} } \right\| _{L^{2}(Q)}^2 +\left\| {y_2 (v)-z_{d2} } \right\| _{L^{2}(Q)}^2 \nonumber \\&\qquad \qquad +\left( {Nv , v} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \end{aligned}$$
(14)

where \(N\in \;L\left( {\left( L^{2}(Q)\right) ^{2} ,\left( L^{2}(Q)\right) ^{2}} \right) \) is a Hermitian positive definite operator:

$$\begin{aligned} \left( {Nu , u} \right) _{\left( {L^{2}(Q)}\right) ^{2}} \ge \;\gamma \;\left\| {{}u{}} \right\| _{\left( {L^{2}(Q)} \right) ^{2}}^2,\quad \gamma \succ 0. \end{aligned}$$
(15)

The control problem then is to find \(u =\{u_{1},u_{2}\} \in U_{ad}\) such that \(J(u)\le J(v)\),

where \(U_{ad}\) is a closed convex subset of \(\left( {L^{2}\left( Q \right) } \right) ^{2}\).

Since the cost function (14) can be written as (see [7]):

$$\begin{aligned} J(v)=a(v,v)-2L(v)+\left\| {y(0)-z_d } \right\| _{\left( {L^{2}(Q)} \right) ^{2}}^2 , \end{aligned}$$

where \(a(v,v)\) is a continuous coercive bilinear form and \(L(v)\) is a continuous linear form on \(\left( {L^{2}\left( Q \right) } \right) ^{2}\). Then using the general theory of Lions [7], there exists a unique optimal control \(u\in U_{ad}\) such that J(u) = inf J(v) for all \(v\in U_{ad}\) . Moreover, we have the following theorem which gives the necessary and sufficient conditions of optimality :

Theorm 2

Assume that (10) and (15) hold. If the cost function is given by (14), the optimal control \(u=\left\{ {u_1 ,u_2 } \right\} \in (L_2 (Q))^{2}\)is then characterized by the following equations and inequalities:

$$\begin{aligned} \left\{ {\begin{array}{l} \frac{\partial ^{2}p_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_1 (u)-a p_1 (u)-c p_2 (u)=y_1 \left( u \right) -z_{d1} \;\;\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{in}\; Q, \\ \frac{\partial ^{2}p_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_2 (u)-b p_1 (u)-d p_2 (u)=y_2 \left( u \right) -z_{d2} \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\mathrm{in}\; Q, \\ p_1 {} ,p_2 \rightarrow 0\quad \text{ as }\quad \left| \text{ x } \right| \rightarrow \infty , \\ \left. {p_1 (u)} \right| _\Sigma =\left. {p_2 (u)} \right| _\Sigma =0,\;\quad \quad \quad \\ p_1 (x,T,u)=p_2 (x,T,u)=0\quad \quad \quad \quad \quad \quad \; \mathrm{in}\quad R^{n}, \\ \frac{\partial p_1 (x,T,u)}{\partial t}=\frac{\partial p_2 (x,T,u)}{\partial t}=0\qquad \qquad \qquad \mathrm{in}\quad R^{n}. \\ \end{array}} \right. \end{aligned}$$
(16)

with

$$\begin{aligned}&p_1 (u),p_2 (u)\in L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) \;, \nonumber \\&\quad \quad \frac{\partial p_1 (u)}{\partial t},\frac{\partial p_2 (u)}{\partial t}\in L^{2}\left( {0,T;V_q^{\prime } \left( {R^{n}} \right) } \right) \nonumber \\&\left( {p_1 \left( { u} \right) \; ,v_1 -u_1 } \right) _{L^{2}(Q)} +\left( {p_2 \left( { u} \right) \; ,v_2 -u_2 } \right) _{L^{2}(Q)} \nonumber \\&+\left( {N u\; ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0\quad \forall \;v=\{v_1 ,v_2 \}\in \;U_{ad} \end{aligned}$$
(17)

together with (13), where \(p(u) = \{p_{1}(u),p_{2}(u)\}\) is the adjoint state .

Proof

The optimal control \(u=\left\{ {u_1 ,u_2 } \right\} \in (L_2 (Q))^{2}\)is characterized by (see[7])

$$\begin{aligned}&J{\prime }(u)\;(v-u)\ge 0\quad \forall \;v\in U_{ad}, \\&\text{ which } \text{ is } \text{ equivalent } \text{ to: } \\&\left( {y(u)-z_d ,y(v)-y(u)} \right) _{\left( {L^{2}(Q)} \right) ^{2}}\\&+\;\left( {N u ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0 \end{aligned}$$

i.e.

$$\begin{aligned}&\left( {y_1 (u)-z_{d1} ,y_1 (v)-y_1 (u)} \right) _{L^{2}(Q)} \\&\quad +\left( {y_2 (u)-z_{d2} ,y_2 (v)-y_2 (u)} \right) _{L^{2}(Q)}\\&\quad +\;\left( {N u ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0 \\ \end{aligned}$$

this inequality can be written as

$$\begin{aligned}&\int \limits _0^T {\left( {y_1 (u)-z_{d1} ,y_1 (v)-y_1 (u)} \right) _{L^{2}(R^{n})} dt} \nonumber \\&\quad +\int \limits _0^T {\left( {y_2 (u)-z_{d2} ,y_2 (v)-y_2 (u)} \right) _{L^{2}(R^{n})} dt} \nonumber \\&\qquad \qquad +\;\left( {N u ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0 \end{aligned}$$
(18)

Now, since

$$\begin{aligned}&\left( {p,B y} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \\&=\int \limits _0^T \left( p_1 \left( u \right) ,\frac{\partial ^{2}y_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_1 (u)\right. \\&\qquad \left. -a y_1 (u)-b y_2 (u) \right) _{L^{2}(R^{n})} dt \\&+\int \limits _0^T \left( p_2 \left( u \right) ,\frac{\partial ^{2}y_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_2 (u)\right. \\&\qquad \left. -cy_1 (u)-d y_2 (u) \right) _{L^{2}(R^{n})} dt \\ \end{aligned}$$

where

$$\begin{aligned}&By\left( u \right) =B\left\{ {y_1 \left( u \right) ,y_2 (u)} \right\} \\&=\left\{ {\frac{\partial ^{2}y_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_1 (u)-a y_1 (u)-b y_2 (u)\;,\;} \right. \\&\quad \left. {\frac{\partial ^{2}y_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) y_2 (u)-c y_1 (u)-d y_2 (u)} \right\} \\ \end{aligned}$$

by using Green formula and (13), we get

$$\begin{aligned}&\left( {p,B y} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \\&=\int \limits _0^T \bigg ( \frac{\partial ^{2}p_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_1 (u)\\&\quad -a p_1 (u)-c p_2 (u) ,y_1 \left( u \right) \bigg )_{L^{2}(R^{n})} dt \\&\quad +\int \limits _0^T \bigg ( \frac{\partial ^{2}p_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_2 (u)-b p_1 (u)\\&\quad -d p_2 (u) ,y_2 \left( u \right) \bigg )_{L^{2}(R^{n})} dt \\&=\left( {B^{{*}}p,y} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \\ \end{aligned}$$

then

$$\begin{aligned}&B^{{*}}p\left( u \right) =B^{{*}}\left\{ {p_1 \left( u \right) ,p_2 (u)} \right\} \\&=\left\{ {\frac{\partial ^{2}p_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_1 (u)-a p_1 (u)-c p_2 (u)\;,} \right. \\&\;\quad \quad \left. {\frac{\partial ^{2}p_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_2 (u)-b p_1 (u)-d p_2 (u)} \right\} \\ \end{aligned}$$

and

$$\begin{aligned}&A^{{*}}p\left( u \right) =A^{{*}}\left\{ {p_1 \left( u \right) ,p_2 (u)} \right\} \\&=\left\{ {\left( {-\Delta +q} \right) p_1 (u)-a p_1 (u)-c p_2 (u)\;,\;} \right. \\&\quad \quad \quad \left. {\left( {-\Delta +q} \right) p_2 (u)-b p_1 (u)-d p_2 (u)} \right\} \\ \end{aligned}$$

since the adjoint equation takes the form:

$$\begin{aligned} \frac{\partial ^{2}p(u)}{\partial t^{2}}+A^{{*}}p(u)=y(u)-z_d \end{aligned}$$

and from Theorem1, we get a unique solution \(p{}\left( u \right) {\in } \left( {L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) } \right) ^{2}\)which satisfies

$$\begin{aligned}&p_1 (u),p_2 (u)\in L^{2}\left( {0,T;V_q \left( {R^{n}} \right) } \right) \;, \\&\quad \quad \quad \quad \quad \quad \frac{\partial p_1 (u)}{\partial t},\frac{\partial p_2 (u)}{\partial t}\in L^{2}\left( {0,T;V_q^{\prime } \left( {R^{n}} \right) } \right) . \\ \end{aligned}$$

This proves system (16).

Now, we transform (18) by using (16) as follows:

$$\begin{aligned}&\int \limits _0^T \left( \frac{\partial ^{2}p_1 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_1 (u)-a p_1 (u)\right. \\&\qquad \left. -c p_2 (u) ,y_1 (v)-y_1 (u) \right) _{L^{2}(R^{n})} dt \\&+\int \limits _0^T \left( \frac{\partial ^{2}p_2 (u)}{\partial t^{2}}+\left( {-\Delta +q} \right) p_2 (u)-b p_1 (u)\right. \\&\qquad \left. -d p_2 (u) ,y_2 (v)-y_2 (u) \right) _{L^{2}(R^{n})} dt \\&+\left( {N u ,v -u } \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0. \end{aligned}$$

Using Green formula, we obtain

$$\begin{aligned}&\int \limits _0^T {\left( {p_1 \left( u \right) ,\left( {\frac{\partial ^{2}}{\partial t^{2}}+\left( {-\Delta +q} \right) } \right) y_1 (v)-y_1 (u) } \right) _{L^{2}(R^{n})} dt} \\&+\int \limits _0^T {-a \left( {p_1 \left( u \right) , y_1 (v)-y_1 (u) } \right) _{L^{2}(R^{n})} dt} \\&+\int \limits _0^T {-c \left( {p_2 \left( u \right) , y_1 (v)-y_1 (u) } \right) _{L^{2}(R^{n})} dt} \\&\!+\!\int \limits _0^T {\left( {p_2 \left( u \right) ,\left( {\frac{\partial ^{2}}{\partial t^{2}}\!+\!\left( {\!-\!\Delta \!+\!q} \right) } \right) y_2 (v)-y_2 (u) } \right) _{L^{2}(R^{n})} dt} \\&+\int \limits _0^T {-b \left( {p_1 \left( u \right) , y_2 (v)-y_2 (u) } \right) _{L^{2}(R^{n})} dt} \\&+\int \limits _0^T {-d \left( {p_2 \left( u \right) , y_2 (v)-y_2 (u) } \right) _{L^{2}(R^{n})} dt} \\&+\;\left( {N u ,v -u } \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0. \\ \end{aligned}$$

Using (13), we have

$$\begin{aligned}&\int \limits _0^T {\left( {p_1 \left( { u} \right) \; ,v_1 -u_1 } \right) _{L^{2}(R^{n})} dt} \\&+\int \limits _0^T {\left( {p_2 \left( { u} \right) \; ,v_2 -u_2 } \right) _{L^{2}(R^{n})} dt}\\&+\left( {N u\; ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0\quad \\ \end{aligned}$$

which is equivalent to

$$\begin{aligned}&\left( {p_1 \left( { u} \right) \; ,v_1 -u_1 } \right) _{L^{2}(Q)} +\left( {p_2 \left( { u} \right) \; ,v_2 -u_2 } \right) _{L^{2}(Q)} \\&\quad \quad \quad \quad +\left( {N u\; ,v-u} \right) _{\left( {L^{2}(Q)} \right) ^{2}} \ge 0 \\ \end{aligned}$$

Thus the proof is complete.\(\square \)