1 Introduction

Let us consider an open bounded subset \(\varOmega \) of \(\mathbb {R}^n\), \(n\ge 2\), and let \(\varOmega _1 =\varOmega \setminus \bar{\varOmega _2}\), where \(\varOmega _2\) is an open and bounded set such that \(\varOmega _2\subset \subset \varOmega \). Denote by \(\partial \varOmega \) and \(\varGamma = \partial \varOmega _2\) the external and the interface boundaries respectively (see Fig. 1).

Fig. 1
figure 1

\(\varOmega =\varOmega _1\cup {{\overline{\varOmega }}}_2\) with \(\varOmega _2\subset \subset \varOmega \)

In the above mentioned domain we want to study the boundary exact controllability problem for a hyperbolic system of equations with appropriate boundary and interface conditions on \(\partial \varOmega \) and on \(\varGamma \). This evolution system describes the wave propagation in a composite made up of two components having very different coefficients of propagation. More precisely, we prescribe a homogeneous Dirichlet condition on the exterior boundary and a jump of the displacement proportional to the conormal derivatives on the interface. The discontinuity of the solution is the mathematical interpretation of imperfect interface (see [1, 3, 5,6,7,8,9,10,11,12,13, 15, 16, 21, 22, 25,26,27,28,29,30,31] and references therein).

The boundary exact controllability problem consists in finding a suitable control, acting on the external boundary or even on just a part of it, driving the trajectories of an evolution system to a desired state at a certain time \(T>0\), for all initial data. Eventually, it reduces to prove an estimate for the energy of an uncontrolled system, at time \(t=0\), through partial measurements of its solution done on the boundary control set. This estimate, known as observability inequality, yields an upper bound for the norm of the initial data of the uncontrolled problem.

In general, the observability inequality does not hold for arbitrary T or control regions. Indeed, the part of the boundary where the control is acting has to satisfy certain geometric conditions. Moreover, as usual in the hyperbolic case, due to the finite speed of propagation of waves, one needs to require that T is sufficiently large. For instance, in the class of regular domains, when using the microlocal approach, one can achieve the observability inequality if and only if every ray of geometric optics, propagating into the domain and reflecting on its boundary, enters the control region in time less than the control time T (see [2]).

In this note, we prove the above mentioned observability inequality by means of Lagrange multipliers method. This is done making use of some results of Monsurrò et al. [28], where the exact internal controllability of the same system of equations is studied.

Usually, when dealing with controllability problems, one fixes a point \(x^0\in \mathbb {R}^n\) which can be viewed as an observer, and which determines the control action region. In the framework of boundary controllability, this control region could be either the entire external boundary of \(\varOmega \) or just a part of it, according to the shape of the domain.

In our context, we need to choose \(x^0\) inside \(\varOmega _2\) and to assume that \(\varOmega _2\) is star-shaped with respect to it. This particular choice of \(x^0\) will influence the control time too (see Lemma 5). Moreover, due to the imperfect transmission condition, the control time will depend also on the coefficients matrix and on the proportionality between the jump of the solution and the conormal derivatives on the interface. See Sect. 3 for more details.

The paper is organized as follows. Section 2 is devoted to the introduction of the imperfect interface problem and of the appropriate functional spaces required for its solution (see also [12, 25]). Moreover, as usual when studying controllability problems, since the initial data are in a weak space, we need to apply the so called transposition method (see [20], Chapter 3, Section 9). Finally, we give the definition of exact controllability. In Sect. 3, by using a crucial identity proved in Lemma 3.2 of [28] and by adapting to our case some arguments as in [18, 19], we obtain the observability inequality and find the lower bound for the control time T (see Lemma 5). To this aim, we apply the above mentioned identity in order to establish two fundamental inequalities, given in Lemmas 2 and 3. Lemma 4 justifies the particular choice of the point \(x^0\). In Sect. 4, we find the exact control by using the Hilbert Uniqueness Method (HUM for short) which is a constructive method introduced by Lions [17, 18]. The key point is to define a suitable functional which, as a consequence of the observability estimate, turns out to be an isomorphism. Let us recall that the control obtained by HUM is also the energy minimizing control.

In [18], Chapter 6, J. L. Lions studies for the first time the exact controllability, via HUM, for the wave equation with transmission conditions. More precisely, he considers a Dirichlet problem with a matrix constant on each component of the domain and a control set on part of the external boundary. For the case of a Neumann boundary exact controllability problem in the same framework we quote here [24]. For what concerns the internal exact controllability of hyperbolic problems in composites with imperfect interface we refer to Faella et al. [14], Monsurrò and Perugia [29] and Monsurrò et al. [28].

2 Setting of the problem

Let us consider an open bounded subset \(\varOmega \) of \(\mathbb {R}^n\), \(n\ge 2\), and let \(\varOmega _1 =\varOmega \setminus \bar{\varOmega _2}\), where \(\varOmega _2\) is an open and bounded set such that \(\varOmega _2\subset \subset \varOmega \). Denote by \(\partial \varOmega \) and \(\varGamma = \partial \varOmega _2\) the external and the interface boundaries respectively. Suppose that the interface is Lipschitz continuous. By construction one has

$$\begin{aligned} \partial \varOmega \cap \varGamma =\emptyset . \end{aligned}$$
(2.1)

Let \(T>0\) and define \(Q_1=\varOmega _{1}\times (0,T)\), \(Q_2=\varOmega _{2}\times (0,T)\), \(\varSigma =\partial \varOmega \times (0,T)\) and \(\varSigma _\varGamma =\varGamma \times (0,T)\).

Given a control \(\zeta \), we consider the following boundary exact controllability problem defined in the domain \(\varOmega \)

$$\begin{aligned} \left\{ \begin{array}{@{}ll} u_{1}'' - \text{ div } \left( A(x) \nabla u_{1}\right) = 0 &{}\text{ in } Q_1, \\ u_{2}''- \hbox {div} \left( A(x) \nabla u_{2}\right) = 0 &{}\text{ in } Q_2, \\ A(x) \nabla u_{1}\, n_{1}= -A(x) \nabla u_{2}\, n_{2}&{} \text{ on } \varSigma _\varGamma , \\ A(x) \nabla u_{1}\, n_{1}= -h(x) (u_{1}-u_{2})&{} \text{ on } \varSigma _\varGamma , \\ u_{1}= \zeta &{}\text{ on } \varSigma ,\\ u_{1}(0)= U_{1}^0,\quad u_{1}'(0)= U^1_{1}&{}\text{ in } \varOmega _{1}, \\ u_{2}(0) = U_{2}^0, \quad u_{2}'(0) = U^1_{2}&{}\text{ in } \varOmega _{2}, \end{array} \right. \end{aligned}$$
(2.2)

where \(n_{i}\)’s are the unitary outward normals to \( \varOmega _{i}\), for \(i=1,2\).

Let us define a class of function spaces which takes into account the geometry of the domain as well as the boundary and interfacial conditions, suitable for the solutions of this kind of interface problems. They were introduced for the first time in [25] in the analogous stationary framework (see also [12] for more details). First of all, let

$$\begin{aligned} V=\ \{ v_1 \in H^1\left( \varOmega _1 \right) |\;v_1=0\, \text {on}\, \partial \varOmega \}, \end{aligned}$$

which, as proved in [4], is a Banach space endowed with the norm

$$\begin{aligned} \Vert v_1 \Vert _{V}=\ \Vert \nabla v_1 \Vert _{L^{2}(\varOmega _{1})}. \end{aligned}$$

By (2.1), \(V\) can be defined as the closure, with respect to the \(H^{1}( \varOmega _{1})\)-norm, of the set of the functions in \(C^{\infty }(\varOmega _{1})\) with a compact support contained in \(\varOmega \). Indeed the condition on \(\partial \varOmega \) in the definition of \(V\) must be interpreted in a density sense, since we do not assume any regularity on \(\partial \varOmega \). We set

$$\begin{aligned} H_{\varGamma }=\ \left\{ v= (v_1,v_2) \mid v_1 \in V \; \text{ and } \; v_2 \in H^{1}(\varOmega _{2})\right\} . \end{aligned}$$
(2.3)

The space \(H_{\varGamma }\) is a separable and reflexive Hilbert space when equipped with the norm

$$\begin{aligned} \Vert v \Vert ^2_{H_{\varGamma }} =\ \Vert \nabla v_1 \Vert ^2 _{L^{2}(\varOmega _{1})} + \Vert \nabla v_2 \Vert ^2_{L^{2}(\varOmega _{2})} + \Vert v_1 - v_2 \Vert ^2 _{L^2(\varGamma )} \end{aligned}$$

and, as evidenced in [7], it can be identified with \(V \times H^{1}( \varOmega _{2})\) endowed with its usual norm. The dual of \(H_{\varGamma }\) is denoted by \((H_{\varGamma })'\) and observe that the norms of \((H_{\varGamma })'\) and \(V'\times (H^1(\varOmega _2))'\) are equivalent (see [9]). Moreover, if \(v=(v_1,v_2)\in (H_{\varGamma })'\) and \(u=(u_1,u_2)\in H_{\varGamma }\), then

$$\begin{aligned}&\left\langle v,u\right\rangle _{(H_{\varGamma })',H_{\varGamma }} =\left\langle v_1,u_1\right\rangle _{V',V}+\left\langle v_2,u_2\right\rangle _{(H^1(\varOmega _{2}))',H^1(\varOmega _{2})}. \end{aligned}$$

As proved in [7, 8], \((H_\varGamma , L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) , \left( H_{\varGamma }\right) ')\) is an evolution triple. Let us set

$$\begin{aligned} W= & {} \left\{ v=\left( v_{1},v_{2}\right) \in L^{2}\left( 0,T;V\times H^{1}\left( \varOmega _{2}\right) \right) \hbox {s.t. }\right. \nonumber \\ v'= & {} \left. \left( v'_1,v'_2\right) \in L^{2}\left( 0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \right\} , \end{aligned}$$
(2.4)

which is a Hilbert space if equipped with the graph norm given by

$$\begin{aligned} \left\| v\right\| _{W}=\left\| v_{1}\right\| _{L^{2}\left( 0,T;V\right) }+\left\| v_{2}\right\| _{L^{2}\left( 0,T;H^{1}\left( \varOmega _{2}\right) \right) }+\left\| v^{\prime }_{1}\right\| _{L^{2}\left( 0,T;L^{2}\left( \varOmega _{1}\right) \right) }+\left\| v^{\prime }_{2}\right\| _{L^{2}\left( 0,T;L^{2}\left( \varOmega _{2}\right) \right) }. \end{aligned}$$

For what concerns the initial data and the control problem (2.2), we suppose that

$$\begin{aligned} \left\{ \begin{array}{@{}l} \text{(i) } U^{0} =\left( U^0_{ 1}, U^0_{ 2 }\right) \in L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) ,\\ \text{(ii) } U^{1} =\left( U^1_{ 1}, U^1_{ 2 }\right) \in \left( H_{\varGamma }\right) ', \end{array}\right. \end{aligned}$$
(2.5)

and

$$\begin{aligned} \zeta \in L^2(\varSigma ). \end{aligned}$$
(2.6)

We also assume that A is a symmetric matrix field such that

$$\begin{aligned} \left\{ \begin{array}{@{}l} (i)\,\,a_{ij}\in \left( W^{1, \infty }\left( \varOmega \right) \right) ^{n^{2}},\,\,\,1\le i,j,k\le n,\\ (ii)\,\,(A(x)\lambda ,\lambda )\ge \alpha {|\lambda |^{2}},\, |A(x) \lambda |\le \beta |\lambda |, \end{array} \right. \end{aligned}$$
(2.7)

for every \(\lambda \in \mathbb {R}^n \) and a.e. in \(\varOmega \), where \(\alpha ,\beta \in \mathbb {R}\), with \(0<\alpha <\beta \), and put

$$\begin{aligned} M=\max _{1\le i,j,k\le n}\,\max _{x\in \varOmega } \left| \dfrac{\partial a_{ ij }}{\partial x_k}\right| . \end{aligned}$$
(2.8)

Furthermore, the function h satisfies

$$\begin{aligned} h \in L^{\infty }(\varGamma ) \;\hbox {and} \; \exists \; h_0 \in \mathbb {R}\; \text{ such } \text{ that } 0<h_0<h(x) \hbox { a.e. in } \>\varGamma . \end{aligned}$$
(2.9)

Since the initial data are in a weak space, we need to apply the transposition method to define in an appropriate way the solution of problem (2.2) (see [20], Chapter 3, Section 9). Thus, for every \(g =(g_{1},g_{2})\in L^2(0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )\), we consider the following backward problem

$$\begin{aligned} \left\{ \begin{array}{@{}ll} \psi _{1}'' - \text{ div } (A(x)\nabla \psi _{1} )=g_{1} &{} \text{ in } Q_1,\\ \psi _{2}'' - \text{ div } (A(x)\nabla \psi _{2} )=g_{2} &{} \text{ in } Q_2,\\ A(x) \nabla \psi _{1} \, n_{1}= -A(x) \nabla \psi _{2} \, n_{2}&{} \text{ on } \varSigma _\varGamma , \\ A(x) \nabla \psi _{1} \, n_{1}= -h(x)(\psi _{1}-\psi _{2}) &{} \text{ on } \varSigma _\varGamma , \\ \psi _{1} = 0 &{} \text{ on } \varSigma ,\\ \psi _{1}(T)=\psi '_{1}(T) = 0 &{} \text{ in } \varOmega _{1},\\ \psi _{2}(T)=\psi '_{2}(T) = 0 &{} \text{ in } \varOmega _{2}. \end{array} \right. \end{aligned}$$
(2.10)

The existence and uniqueness of the weak solution in W of problem (2.10) are proved in [7]. For sake of simplicity, in the sequel we omit the explicit dependence on the space variable x in the matrix A and in the function h when considering integrals. Now, we give the following notion of transposition solution.

Definition 1

For any fixed \(\left( U^{0}, U^{1}\right) \in (L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )\times \left( H_{\varGamma }\right) '\), we say that a function \(u =(u_{1},u_{2})\in L^2(0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )\) is a solution of problem (2.2), in the sense of transposition, if it satisfies

$$\begin{aligned} \displaystyle \int _{Q_1}u_{1}g_1dxdt+ \displaystyle \int _{Q_2}u_{2}g_2 dxdt= & {} -\displaystyle \int _{\varOmega _{1}}U^{0}_{1}\psi '_{1}(0)dx+ \left\langle U^{1}_{1},\psi _{1}(0)\right\rangle _{V',V} \nonumber \\&-\,\displaystyle \int _{\varOmega _{2}}U^{0}_{2}\psi '_{2}(0)dx +\left\langle U^{1}_{2},\psi _{2}(0)\right\rangle _{(H^1(\varOmega _{2}))',H^1(\varOmega _{2})}\nonumber \\&-\,\int _\varSigma A \nabla \psi _1\, n_1 \, \zeta \,\; d \sigma _x dt \end{aligned}$$
(2.11)

for all \(g =(g_{1},g_{2})\in L^2(0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )\) and \(\psi \) is the unique solution of problem (2.10) corresponding to g.

Remark 1

Observe that, since \(\psi _1=0\) on \(\varSigma \), one has that \(\nabla \psi _1= n_1 \dfrac{\partial \psi _1}{\partial n_1} \) on \(\varSigma \). Hence (2.11) reads as follows

$$\begin{aligned} \int _{Q_1}u_{1}g_1dxdt+ \displaystyle \int _{Q_2}u_{2}g_2 dxdt= & {} -\displaystyle \int _{\varOmega _{1}}U^{0}_{1}\psi '_{1}(0)dx+ \left\langle U^{1}_{1},\psi _{1}(0)\right\rangle _{V',V} \nonumber \\&-\,\displaystyle \int _{\varOmega _{2}}U^{0}_{2}\psi '_{2}(0)dx +\left\langle U^{1}_{2},\psi _{2}(0)\right\rangle _{(H^1(\varOmega _{2}))',H^1(\varOmega _{2})} \nonumber \\&- \int _{\varSigma } A \> n_1\, n_1 \dfrac{\partial \psi _1}{\partial n_1} \, \zeta \,\; d \sigma _x \;dt.\nonumber \\ \end{aligned}$$
(2.12)

Problem (2.2) admits a unique solution \(u\in C\left( [0,T]; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \cap C^{1}\left( [0,T];\left( H_{\varGamma }\right) '\right) \) satisfying the estimate

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )}+ \Vert u'\Vert _{L^\infty (0,T; (H_{\varGamma })') }\le & {} C( \Vert U^0\Vert _{ L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) } \nonumber \\&+\, \Vert U^1\Vert _{(H_{\varGamma })' }+ \Vert \zeta \Vert _{L^2(\varSigma )}),\nonumber \\ \end{aligned}$$
(2.13)

with C as a positive constant (see [20], Chapter 3, Section 9, Theorems 9.3 and 9.4).

It is clear that the solution of problem (2.2) depends on the control \(\zeta \), i.e. \(u\left( \zeta \right) =\left( u_{1}\left( \zeta \right) ,u_{2}\left( \zeta \right) \right) \). Nevertheless, for sake of simplicity, in the sequel we omit this explicit dependence.

Definition 2

System (2.2) is exactly controllable at time \(T>0\), if for every \(\left( U^{0},U^{1}\right) , \left( Z^{0},Z^{1}\right) \) in \( (L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) ) \times (H_\varGamma )'\), there exists a control \(\zeta \) belonging to \(L^2(\varSigma )\) such that the corresponding solution u of problem (2.2) satisfies

$$\begin{aligned} u(T)=Z^{0},\,\,\,u'(T)=Z^{1}. \end{aligned}$$

Remark 2

It is well known that for a linear system reversible in time, exact controllability is equivalent to null controllability (see, for instance, [32]). Hence, it is enough to prove the existence of a control \(\zeta \in L^2(\varSigma )\) of (2.2) such that \(u(T)=u'(T)=0.\)

From literature it is renowned that it is not possible to achieve controllability without additional requirements. Hence, in this paper, we prove a boundary controllability result under the further geometrical assumption that \(\varOmega _2\) is starshaped with respect to a point \(x^0 \in \varOmega _2\). More precisely, we prove that system (2.2) is exactly controllable, for a suitable time \(T > 0\) sufficiently large, taking a control acting on a part of the external boundary \(\partial \varOmega \), or even on the entire external boundary, according to the shape of the domain (see Theorem 2). We use a constructive method known as the Hilbert Uniqueness Method, HUM for short, introduced by Lions in [17, 18]. Eventually HUM reduces to the derivation of a delicate estimate from below, known as observability estimate, for an uncontrolled problem, see (3.1). This estimate can be obtained by proving some preliminary fundamental results based on the so called Lagrange multipliers method. These results are proved in the following section. We point out that the control obtained by HUM is also the energy minimizing control, hence it is unique.

3 The observability inequality

In this section, we consider an imperfect transmission problem similar to (2.2) presenting a homogeneous Dirichlet condition on the external boundary and more regular initial data. Namely, for \(T>0\), let us introduce the following problem

$$\begin{aligned} \left\{ \begin{array}{@{}ll} z_1'' - \text{ div } \left( A(x) \nabla z_1 \right) = 0&{}\text{ in } Q_1, \\ z_2''- \hbox {div} \left( A(x) \nabla z_2 \right) =0 &{}\text{ in } Q_2, \\ A(x) \nabla z_1 \, n_{1}= -A(x) \nabla z_2\, n_{2}&{} \text{ on } \varSigma _\varGamma , \\ A(x) \nabla z_1 \, n_{1}= -h(x) (z_1-z_2)&{} \text{ on } \varSigma _\varGamma , \\ z_1 = 0&{}\text{ on } \varSigma ,\\ z_1(0)= z_{1}^0,\quad z_1'(0)= z^1_{1}&{}\text{ in } \varOmega _{1}, \\ z_2(0) = z_{2}^0, \quad z_2'(0) = z^1_{2}&{}\text{ in } \varOmega _{2}, \end{array} \right. \end{aligned}$$
(3.1)

with

$$\begin{aligned} \left\{ \begin{array}{@{}l} z^{0} =\left( z^0_{ 1}, z^0_{ 2 }\right) \in H_\varGamma ,\\ z^{1} =\left( z^1_{ 1}, z^1_{ 2 }\right) \in L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) , \end{array}\right. \end{aligned}$$
(3.2)

where \(n_{i}\)’s are the unitary outward normals to \( \varOmega _{i}\), \(i=1,2\). In view of (2.7) and (2.9), as already observed in [7], problem (3.1) admits a unique weak solution in W and its variational formulation is given by

$$\begin{aligned} \left\{ \begin{array}{@{}l} \hbox {Find} \; z = ( z_{1}, z_{2}) \; \text{ in } \,W \text{ such } \text{ that } \\ \langle z_{1}'', v_1 \rangle _{V', V} + \langle z_{2}'', v_2 \rangle _{(H^{1}( \varOmega _{2}))',H^{1}( \varOmega _{2})} \displaystyle + \int _{\varOmega _{1}}{A \nabla z_{1} \nabla v_{1}}\; dx + \int _{\varOmega _{2}}{A \nabla z_{2} \nabla v_{2}} \; dx \\ +\displaystyle \int _{\varGamma }{h (z_{1} - z_{2})(v_1 - v_2) \; d\sigma _x}= 0, \quad \forall \; (v_1,v_2) \in V\times H^{1}( \varOmega _{2}) \text{ in } {{\mathcal {D}}}'(0,T),\\ z_{1}(0)= z_{1}^0, \quad z_{1}'(0)= z^1_{1} \quad \text{ in } \varOmega _{1}, \\ z_{2}(0)= z_{2}^0,\quad z_{2}'(0)= z^1_{2}\quad \text{ in } \varOmega _{2}.\nonumber \end{array} \right. \!\!\!\!\!\!\\ \end{aligned}$$
(3.3)

More precisely it holds the following result.

Theorem 1

[7] Let \(T>0\) and \(H_{\varGamma }\) and W be defined as in (2.3) and (2.4). Under hypotheses (2.7), (2.9) and (3.2), problem (3.1) admits a unique weak solution \(z\in W\). Moreover, there exists a positive constant C, such that

$$\begin{aligned} \left\| z\right\| _{L^{\infty }(0,T;H_{\varGamma })}+\left\| z'\right\| _{L^{\infty }(0,T; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) )}\le C\left( \left\| z^{0}\right\| _{H_{\varGamma } }+\left\| z^{1}\right\| _{ L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) }\right) .\nonumber \\ \end{aligned}$$
(3.4)

In fact, under the same hypotheses of Theorem 1, the solution of problem (3.1) has further regularity

$$\begin{aligned} z\in C\left( \left[ 0,T\right] ;H_{\varGamma }\right) ,\, z'\in C\left( \left[ 0,T\right] ; L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) , \end{aligned}$$

see [20], Chapter 3, Theorem 8.2 for more details. Hence the initial values z(0) and \(z'(0)\) are meaningful in the appropriate spaces. Let us recall a fundamental identity, proved in [28], crucial when establishing the inverse inequalities involved in the exact controllability problem. For clearness sake, from now on we use the repeated index summation convention.

Lemma 1

[28] Let \(q=(q_1, \ldots q_n)\) be a vector field in \((W^{1,\infty }(\varOmega ))^n\) and let \(z=(z_1,z_2)\) be the solution of problem (3.1)–(3.2). Then, the following identity holds

$$\begin{aligned}&\dfrac{1}{2}\int _\varSigma A n_1\, n_1\left( \dfrac{\partial z_1}{\partial n_1}\right) ^2q_k n_{1k}\;d\sigma _x\;dt+\dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{\varGamma }} A n_i \, n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2q_k n_{ik}\;d\sigma _x\;dt\nonumber \\&\qquad -\,\displaystyle \int _{\varSigma _{\varGamma }} h \left( z_1-z_2\right) q_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\nonumber \\&\qquad + \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{\varGamma }} \left( |z'_i|^2-A \nabla _\sigma z_i\nabla _\sigma z_i\right) q_kn_{ik}\;d\sigma _x\;dt\nonumber \\&\quad =\displaystyle \sum _{i=1}^2\left. \left( z'_i,q_k\dfrac{\partial z_i}{\partial x_k}\right) _{\varOmega _i}\right| _{0}^{T} +\dfrac{1}{2}\sum _{i=1}^{2}\int _{Q_i} \left( |z'_i|^2-A \nabla z_i\nabla z_i\right) \dfrac{\partial q_k}{\partial x_k}\;dx\;dt\nonumber \\&\qquad +\,\displaystyle \sum _{i=1}^{2}\int _{Q_i} A \nabla z_i\nabla q_k\dfrac{\partial z_i}{\partial x_k}\;dx\;dt-\dfrac{1}{2}\sum _{i=1}^{2}\int _{Q_i} q_k \sum _{l,j=1}^{n}\dfrac{\partial a_{lj}}{\partial x_k}\dfrac{\partial z_i}{\partial x_l}\dfrac{\partial z_i}{\partial x_j}\;dx\;dt,\nonumber \\ \end{aligned}$$
(3.5)

where

$$\begin{aligned} \left( z'_i,q_k\dfrac{\partial z_i}{\partial x_k}\right) _{\varOmega _i}=\int _{\varOmega _i}z'_i(t)q_k\dfrac{\partial z_i(t)}{\partial x_k}\,dx \end{aligned}$$

and \(\nabla _\sigma z_i=(\sigma _j z_i)_{j=1}^n\) denotes the tangential gradient of \(z_i\) on \(\varGamma \) for \(i=1,2\) (see, for instance, [18], p. 137).

In order to prove the direct inequality, stated in the next Lemma 2, we apply the above identity for a particular choice of the vector field q. At first, let us denote by E(t), the energy of the problem (3.1)–(3.2) which is defined as

$$\begin{aligned} E(t)= & {} \displaystyle \dfrac{1}{2}\left[ \int _{\varOmega _{1}}|z'_{1}(t)|^{2}dx+\int _{\varOmega _{2}}|z'_{2}(t)|^{2}dx+\int _{\varOmega _{1}}A\nabla z_{1}(t)\nabla z_{1}(t)dx\right. \nonumber \\&\left. \displaystyle +\,\int _{\varOmega _{2}}A\nabla z_{2}(t)\nabla z_{2}(t)dx+\int _{\varGamma }h\left| z_{1}(t)-z_{2}(t)\right| ^{2}d\sigma _{x}\right] . \end{aligned}$$
(3.6)

It is easy to see by a direct derivation that the energy defined in (3.6) is conserved (see [8], Lemma 4.1), i.e.

$$\begin{aligned} E(t) = E(0), \quad \forall \, t \in [0,T]. \end{aligned}$$
(3.7)

Lemma 2

Let \(z=(z_1,z_2)\) be the solution of problem (3.1)–(3.2). Then, for any \(T>0\), it holds

$$\begin{aligned} \int _{\varSigma }\left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt \le C E(0), \end{aligned}$$
(3.8)

with C as a positive constant.

Proof

Let us choose \(q_k=\tau _k\) in (3.5) for \(k=1,..,n\), where \(\tau =(\tau _k)_{k=1,..,n}\in (C^1(\overline{\varOmega }))^n\) is such that

$$\begin{aligned} \left\{ \begin{array}{lll} (i)&{} \tau =n_1\quad \text {on }\partial \varOmega ,\\ (ii)&{} \tau =0 \quad \text {in } \varOmega _2,\\ (iii)&{} \Vert \tau \Vert _{(L^\infty (\varOmega _1))^n}\le 1. \end{array} \right. \end{aligned}$$
(3.9)

The existence of such a vectorial field is proved in [18]. By (3.9)(i) and (3.9)(ii) we get

$$\begin{aligned}&\dfrac{1}{2}\int _{\varSigma } A n_1 n_1\left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt\nonumber \\&\quad =\displaystyle \left. \left( z'_1,\tau _k\dfrac{\partial z_1}{\partial x_k}\right) _{\varOmega _1}\right| _{0}^{T} +\dfrac{1}{2}\int _{Q_1} \left( |z'_1|^2-A \nabla z_1\nabla z_1\right) \dfrac{\partial \tau _k}{\partial x_k}\;dx\;dt\nonumber \\&\qquad +\,\displaystyle \int _{Q_1} A \nabla z_1\nabla \tau _k\dfrac{\partial z_1}{\partial x_k}\;dx\;dt-\dfrac{1}{2} \int _{Q_1} \tau _k \sum _{l,j=1}^{n}\dfrac{\partial a_{lj}}{\partial x_k}\dfrac{\partial z_1}{\partial x_l}\dfrac{\partial z_1}{\partial x_j}\;dx\;dt. \end{aligned}$$
(3.10)

Passing to the absolute value, by (2.7), (3.9)(iii), Young inequality, (3.6), the conservation law and since \(\tau \in (C^1(\overline{\varOmega }))^n\) , we obtain

$$\begin{aligned}&\dfrac{1}{2}\left| \int _{\varSigma } A n_1 n_1 \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt\right| \\&\quad \le \displaystyle \dfrac{1}{2}\int _{\varOmega _1} \left| z_1'(0)\right| ^2\,dx+\dfrac{1}{2}\int _{\varOmega _1}\left| z_1'(T)\right| ^2\,dx+\dfrac{1}{2}\int _{\varOmega _1}\left| \nabla z_1(0)\right| ^2\,dx\\&\qquad +\,\dfrac{1}{2}\int _{\varOmega _1}\left| \nabla z_1(T)\right| ^2\,dx+C_1 \int _{Q_1} \left( |z'_1|^2+A \nabla z_1\nabla z_1\right) \;dx\;dt\\&\qquad +\,\displaystyle C_2 \int _{Q_1} A \nabla z_1\nabla z_1\;dx\;dt+C_3 \int _{Q_1}\left| \nabla z_1\right| ^2\;dx\;dt\\&\quad \le \displaystyle 2 \max \left( 1, \dfrac{1}{\alpha }\right) E(0)+C_4 \int _{Q_1} \left( |z'_1|^2+\left| \nabla z_1\right| ^2\right) \;dx\;dt \le C E(0), \end{aligned}$$

where the last inequality is a consequence of estimate (3.4). Then, hypothesis (2.7)(ii) gives the result.\(\square \)

At this point, in order to derive the observability estimate, we adapt to our context some arguments introduced in [18, 19] (see also [28]). Here, for \(x^0\in \mathbb {R}^n\), we set

$$\begin{aligned} m(x)=x-x^0=(x_k-x^0_k)_{k=1}^n. \end{aligned}$$
(3.11)

Lemma 3

Let \(z=(z_1,z_2)\) be the solution of problem (3.1)–(3.2). Then, for any \(T>0\), it holds

$$\begin{aligned}&\dfrac{1}{2}\left| \sum _{i=1}^{2}\int _{\varSigma _{\varGamma }} A n_i n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2 m_k n_{ik}\;d\sigma _x\;dt-\int _{\varSigma _\varGamma } h \left( z_1-z_2\right) m_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\right. \nonumber \\&\qquad +\,\left. \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _\varGamma } \left( |z'_i|^2-A \nabla _\sigma z_i\nabla _\sigma z_i\right) m_k n_{ik}\;d\sigma _x\;dt\right| \le C E(0), \end{aligned}$$
(3.12)

with C as a positive constant.

Proof

Let us choose \(q_k=m_k\,w\) in (3.5) for \(k=1,\ldots ,n\), where \(w\in C^1(\mathbb {R}^n)\) is such that

$$\begin{aligned} \left\{ \begin{array}{lll} (i)&{} \text{ supp } w \subset \overline{\varOmega }_2,\\ (ii)&{} 0\le w \le 1\quad \text {in } \varOmega _2,\\ (iii)&{} w=1\quad \text {on }\varGamma , \\ (iv)&{} \Vert \nabla w\Vert _{L^\infty (\mathbb {R}^n)}\le 1. \end{array}\right. \end{aligned}$$
(3.13)

The existence of such a vectorial field is proved in [18]. By (3.13)(i)–(iii) we get

$$\begin{aligned}&\displaystyle \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{\varGamma }} A n_i n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2 m_k n_{ik}\;d\sigma _x\;dt-\displaystyle \int _{\varSigma _\varGamma } h \left( z_1-z_2\right) m_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\nonumber \\&\qquad +\,\dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _\varGamma } \left( |z'_i|^2-A \nabla _\sigma z_i\nabla _\sigma z_i\right) m_k n_{ik}\;d\sigma _x\;dt\nonumber \\&\quad =\displaystyle \left. \left( z'_2,m_kw\dfrac{\partial z_2}{\partial x_k}\right) _{\varOmega _2}\right| _{0}^{T} +\dfrac{n}{2}\int _{Q_2} \left( |z'_2|^2-A \nabla z_2\nabla z_2\right) w\;dx\;dt\nonumber \\&\qquad +\,\,\displaystyle \dfrac{1}{2}\int _{Q_2} \left( |z'_2|^2-A \nabla z_2\nabla z_2\right) m_k\dfrac{\partial w}{\partial x_k}\;dx\;dt+\int _{Q_2} A \nabla z_2\nabla w\dfrac{\partial z_2}{\partial x_k}m_k \;dx\;dt\nonumber \\&\qquad +\,\displaystyle \int _{Q_2} A \nabla z_2\nabla z_2 w\;dx\;dt-\dfrac{1}{2}\int _{Q_2} m_k w \sum _{l,j=1}^{n}\dfrac{\partial a_{lj}}{\partial x_k}\dfrac{\partial z_2}{\partial x_l}\dfrac{\partial z_2}{\partial x_j}\;dx\;dt. \end{aligned}$$
(3.14)

Passing to the absolute value, by (3.13)(iv), Young inequality, (3.6), the conservation law and (3.4), we obtain

$$\begin{aligned}&\left| \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{1}} A n_i n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2 m_k n_{ik}\;d\sigma _x\;dt-\int _{\varSigma _\varGamma } h \left( z_1-z_2\right) m_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\right. \\&\qquad +\,\left. \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _\varGamma } \left( |z'_i|^2-A \nabla _\sigma z_i\nabla _\sigma z_i\right) m_k n_{ik}\;d\sigma _x\;dt \right| \\&\quad \le \dfrac{1}{2}\int _{\varOmega _{2}}\left| z_2'(0)\right| ^2\,dx+\dfrac{1}{2}\int _{\varOmega _2}\left| z_2'(T)\right| ^2\,dx+C_1\int _{\varOmega _2}\left| \nabla z_2(0)\right| ^2\,dx\\&\qquad +\,C_1\int _{\varOmega _{2}}\left| \nabla z_2(T)\right| ^2\,dx+\dfrac{n}{2}\int _{Q_2} \left( |z'_2|^2+A \nabla z_2\nabla z_2\right) \;dx\;dt\\&\qquad +\, C_2 \int _{Q_2} \left( |z'_2|^2+A \nabla z_2\nabla z_2\right) \;dx\;dt\\&\qquad +\,\displaystyle C_3 \int _{Q_2} A \nabla z_2\nabla z_2\;dx\;dt+C_4 \int _{Q_2}\left| \nabla z_2\right| ^2\;dx\;dt \le C E(0). \end{aligned}$$

where \(C_i\)’s are positive constants independent of T.

Usually, in the context of controllability problems, the point \(x^0\) can be interpreted as an observer and its choice is strictly related to the region where the control is acting. Having this in mind, we define

$$\begin{aligned} \partial \varOmega (x^0)=\{x\in \partial \varOmega :\,m(x) n_1(x)=m_k(x)n_{1k}(x)>0\} \end{aligned}$$
(3.15)

and

$$\begin{aligned} \varSigma (x^0)=\partial \varOmega (x^0)\times (0,T). \end{aligned}$$
(3.16)

Further, we put

$$\begin{aligned} R(x^0)=\max _{x\in \overline{\varOmega }} |m(x)|= \max _{x\in \overline{\varOmega }} \left| \sum _{k=1}^{n} (x_k-x^0_k)^2\right| ^{\frac{1}{2}}. \end{aligned}$$
(3.17)

In our case, as observed in [28], due to the geometry of the domain, \(x^0\) must be chosen such that the set \(\varOmega _2\) is star-shaped with respect to \(x^0\). As we will see later on, this choice will also influence the control time (see Lemma 5). By Lemma 3.3 in [28], we can easily obtain the following result.

Lemma 4

Let us suppose that \(\varOmega _2\) is starshaped with respect to a point \(x^0\in \varOmega _2\). Let \(z=(z_1,z_2)\) be the solution of problem (3.1)–(3.2). Then, for any \(T>0\), it holds

$$\begin{aligned}&\dfrac{R(x^0)\beta }{2}\int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt+\dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{\varGamma }} A n_i n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2 m_k n_{ik}\;d\sigma _x\;dt\nonumber \\&\qquad -\,\displaystyle \int _{\varSigma _\varGamma } h(x) \left( z_1-z_2\right) m_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\nonumber \\&\qquad +\, \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _\varGamma } \left( |z'_i|^2-A\nabla _\sigma z_i\nabla _\sigma z_i\right) m_k n_{ik}\;d\sigma _x\;dt\nonumber \\&\quad \ge \left[ T\left( 1-\dfrac{n R(x^0)M}{\alpha }\right) -2\max \left( \dfrac{ R(x^0)}{\sqrt{\alpha }}, \dfrac{(n-1)\sqrt{\alpha }}{2 h_0}\right) \right] E(0), \end{aligned}$$
(3.18)

with M given in (2.8).

Combining the results of Lemmas 4 and 3, we obtain the claimed observability inequality.

Lemma 5

Let us suppose that \(\varOmega _2\) is starshaped with respect to a point \(x^0\in \varOmega _2\) such that

$$\begin{aligned} R(x^0)<\frac{\alpha }{nM}, \end{aligned}$$
(3.19)

with \(\alpha \) and M respectively given by (2.7) and (2.8). Let \(z=(z_1,z_2)\) be the solution of the problem (3.1)–(3.2). Then, there exists \(T_0>0\) such that

$$\begin{aligned} E(0)\le C(T) \int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt , \end{aligned}$$
(3.20)

for T large enough so that

$$\begin{aligned} \dfrac{T-T_0}{T}>\dfrac{n R(x^0)M}{ \alpha }. \end{aligned}$$
(3.21)

Proof

By (3.18) and (3.12), we get

$$\begin{aligned}&\left[ T\left( 1-\dfrac{n R(x^0)M}{\alpha }\right) -2\max \left( \dfrac{ R(x^0)}{\sqrt{\alpha }}, \dfrac{(n-1)\sqrt{\alpha }}{2 h_0}\right) \right] E(0) \\&\quad \le \dfrac{R(x^0)\beta }{2}\int _{\varSigma (x_0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt+\dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _{1}} A n_i n_i\left( \dfrac{\partial z_i}{\partial n_i}\right) ^2 m_k n_{ik}\;d\sigma _x\;dt\\&\qquad -\,\displaystyle \int _{\varSigma _\varGamma } h(x)\left( z_1-z_2\right) m_k\left( \nabla _\sigma (z_1-z_2)\right) _k\;d\sigma _x\;dt\\&\qquad +\, \dfrac{1}{2}\sum _{i=1}^{2}\int _{\varSigma _\varGamma } \left( |z'_i|^2-A\nabla _\sigma z_i\nabla _\sigma z_i\right) m_k n_{ik}\;d\sigma _x\;dt\\&\quad \le \displaystyle \dfrac{R(x^0)\beta }{2}\int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt + C E(0), \end{aligned}$$

where C is the positive constant given in Lemma 3. Denoting

$$\begin{aligned} T_0=2\max \left( \dfrac{R(x^0)}{\sqrt{\alpha }}, \dfrac{(n-1)\sqrt{\alpha }}{2 h_0}\right) + C, \end{aligned}$$
(3.22)

we obtain

$$\begin{aligned} \begin{array}{c} \left[ T\left( 1-\dfrac{n R(x^0)M}{\alpha }\right) -T_0\right] E(0)\le \displaystyle \dfrac{R(x^0)\beta }{2}\int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt. \end{array} \end{aligned}$$

Following an argument similar as in [23], if (3.19) is satisfied and if T is large enough so that (3.21) holds, then \(T\left( 1-\dfrac{n R(x^0)M}{\alpha }\right) -T_0\) is positive and we get the required result. \(\square \)

As a consequence of the direct and inverse inequalities, we immediately get the following equivalence.

Corollary 1

Under the hypotheses of Lemma 5, there exists \(T_0>0\) such that

$$\begin{aligned} E(0)\le C_1(T) \int _{\varSigma (x_0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt \le C_2 (T) \, E(0), \end{aligned}$$
(3.23)

for T large enough so that

$$\begin{aligned} \dfrac{T-T_0}{T}>\dfrac{n R(x^0)M}{ \alpha }. \end{aligned}$$

4 The boundary exact controllability result

In this section, let us prove the boundary exact controllability of system (2.2) by means of the Hilbert Uniqueness Method introduced by Lions (see [17, 18]).

Theorem 2

Let us suppose that \(\varOmega _2\) is starshaped with respect to a point \(x^0\in \varOmega _2\) such that

$$\begin{aligned} R(x^0)<\frac{\alpha }{nM}, \end{aligned}$$

with \(\alpha \), M and \(R(x^0)\) respectively given by (2.7), (2.8) and (3.17). Under assumptions (2.7) and (2.9), for any given \(\left( U^{0},U^{1}\right) \) in \( \left( L^{2}\left( \varOmega _1\right) \times L^{2}\left( \varOmega _2\right) \right) \times (H_\varGamma )'\), there exist a control \(\zeta \in L^2(\varSigma (x^0))\) and a time \(T_0>0\) such that the corresponding solution of problem (2.2) satisfies

$$\begin{aligned} u(T)=u'(T)=0,\end{aligned}$$
(4.1)

for T large enough so that

$$\begin{aligned} \dfrac{T-T_0}{T}>\dfrac{n R(x^0)M}{ \alpha }. \end{aligned}$$
(4.2)

Proof

Let \(T_0\) be as in Lemma 5 and T as in (4.2). Let z be the solution of (3.1) and consider the following backward problem

$$\begin{aligned} \left\{ \begin{array}{@{}ll} \theta _1'' - \text{ div } (A(x)\nabla \theta _{1} )=0&{} \text{ in } Q_1,\\ \theta _2'' - \text{ div } (A(x)\nabla \theta _{2} )=0&{} \text{ in } Q_2,\\ A(x) \nabla \theta _{1} n_{1}= -A(x) \nabla \theta _{2} n_{2}&{} \text{ on } \varSigma _\varGamma , \\ A(x) \nabla \theta _{1} n_{1}= -h(x)(\theta _{1}-\theta _{2}) &{} \text{ on } \varSigma _\varGamma , \\ \theta _{1} = \left\{ \begin{array}{@{}ll}\dfrac{\partial z_1}{\partial n_1} &{} \text{ on } \varSigma (x^0),\\ 0 &{} \text{ on } \varSigma {\setminus } \varSigma (x^0) \end{array} \right. \\ \theta _{1}(T)=\theta '_{1}(T) = 0 &{} \text{ in } \varOmega _{1},\\ \theta _{2}(T)=\theta '_{2}(T) = 0 &{} \text{ in } \varOmega _{2}. \end{array} \right. \end{aligned}$$
(4.3)

Similar to the solution of problem (2.2), the solution \(\theta =(\theta _1,\theta _2)\) of problem (4.3) is also defined by the transposition method. Hence, it satisfies

$$\begin{aligned} \int _{\varSigma (x^0)} A \nabla z_1 n_1 \dfrac{\partial z_1}{\partial n_1}\,\; d \sigma _x \;dt= & {} \displaystyle \left\langle \theta '_{1}(0), z^{0}_{1} \right\rangle _{V',V} - \int _{\varOmega _{1}} z^{1}_{1}\theta _{1}(0) dx\nonumber \\&+\,\displaystyle \left\langle \theta '_{2}(0), z^{0}_{2} \right\rangle _{(H^1(\varOmega _2))',H^1(\varOmega _2)} - \int _{\varOmega _{2}} z^{1}_{2}\theta _{2}(0) dx,\nonumber \\ \end{aligned}$$
(4.4)

where z is the solution of (3.1) for \(z^0, z^1\) as in (3.2). Now, since \(z_1=0\) on \(\varSigma \), we see that \(\nabla z_1= n_1 \dfrac{\partial z_1}{\partial n_1} \) on \(\varSigma \). Hence (4.4) can be rewritten as

$$\begin{aligned} \int _{\varSigma (x^0)} A n_1 \dfrac{\partial z_1}{\partial n_1} n_1 \dfrac{\partial z_1}{\partial n_1}\,\; d \sigma _x \;dt= & {} \displaystyle \left\langle \theta '_{1}(0), z^{0}_{1} \right\rangle _{V',V} - \int _{\varOmega _{1}} z^{1}_{1}\theta _{1}(0) dx\nonumber \\&+\,\displaystyle \left\langle \theta '_{2}(0), z^{0}_{2} \right\rangle _{(H^1(\varOmega _2))',H^1(\varOmega _2)} - \int _{\varOmega _{2}} z^{1}_{2}\theta _{2}(0) dx.\nonumber \\ \end{aligned}$$
(4.5)

Inspired by HUM method, we introduce the linear operator

$$\begin{aligned} \varLambda :H_{\varGamma } \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \rightarrow \left( H_{\varGamma }\right) ' \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \end{aligned}$$
(4.6)

by setting for all \(\left( z^{0},z^{1}\right) \in H_\varGamma \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \),

$$\begin{aligned} \varLambda \left( z^{0},z^{1}\right) = \left( \theta '(0),-\theta (0)\right) , \end{aligned}$$
(4.7)

where \(\theta \) is the unique solution of problem (4.3). It is easy to verify that

$$\begin{aligned} \left\langle \varLambda \left( z^{0},z^{1}\right) , \left( z^{0},z^{1}\right) \right\rangle= & {} \left\langle \left( \theta '(0),-\theta (0)\right) , \left( z^{0},z^{1}\right) \right\rangle \nonumber \\= & {} \displaystyle \left\langle \theta '_{1}(0), z^{0}_{1} \right\rangle _{V',V} - \int _{\varOmega _{1}} z^{1}_{1}\theta _{1}(0) dx\nonumber \\&+\,\left\langle \theta '_{2}(0), z^{0}_{2} \right\rangle _{(H^1(\varOmega _2))',H^1(\varOmega _2)} - \int _{\varOmega _{2}} z^{1}_{2}\theta _{2}(0) dx, \end{aligned}$$
(4.8)

for every \(\left( z^{0},z^{1}\right) \in H_\varGamma \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \).

Putting together (4.5) and (4.8), we can obtain the following explicit formula for the operator \(\varLambda \)

$$\begin{aligned} \left\langle \varLambda \left( z^{0},z^{1}\right) , \left( z^{0},z^{1}\right) \right\rangle = \int _{\varSigma (x^0)} A n_1 n_1 \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\,\; d \sigma _x \;dt. \end{aligned}$$
(4.9)

Observe that, by (2.7),

$$\begin{aligned} \alpha \int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\;d\sigma _x\;dt\le \int _{\varSigma (x^0)} A n_1 n_1 \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2\,\; d \sigma _x \;dt \le \beta \int _{\varSigma (x^0)} \left( \dfrac{\partial z_1}{\partial n_1}\right) ^2 d\sigma _x\;dt. \end{aligned}$$

In view of Corollary 1, the right hand side of (4.9) defines a norm on \( H_\varGamma \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \) equivalent to the inner one. Therefore, \(\varLambda \) is an isomorphism between \( H_\varGamma \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \) and \( \left( H_{\varGamma }\right) ' \times \left( L^{2}\left( \varOmega _{1}\right) \times L^{2}\left( \varOmega _{2}\right) \right) \). Thus, if \((U^0, U^1)\) are the initial conditions of problem (2.2), the equation

$$\begin{aligned} \varLambda \left( z^{0},z^{1}\right) =(U^1, - U^0) \end{aligned}$$

has a unique solution. This motivates us to take the control \(\zeta \) in (2.2) as

$$\begin{aligned} \zeta = \left\{ \begin{array}{@{}ll}\dfrac{\partial z_1}{\partial n_1} &{} \text{ on } \varSigma (x^0),\\ 0 &{} \text{ on } \varSigma {\setminus } \varSigma (x^0). \end{array} \right. \end{aligned}$$
(4.10)

By uniqueness, we see that \(u=\theta \) and therefore, we have the null controllability of problem (2.2) and hence its exact controllability. \(\square \)