1 Introduction

It is well known that heat equations with memory of the following type

$$\begin{aligned} y_t=\alpha \triangle y+\int _0^tK(t-s)\triangle y(s) ds, \end{aligned}$$
(1)

with \(\alpha >0\), cannot be controlled to rest for large classes of memory kernels and controls, see e.g. [3, 4]. The motivation for that kind of results is due to the smoothing effect of the solutions, because (1) is a parabolic equation when the constant \(\alpha \) before the Laplacian is positive.

On the other hand the class of the partial integro-differential equations changes completely if in the Eq. (1) one takes \(\alpha =0\). The physical model relies on the Cattaneo’s paper [1]. Indeed, in [1] to overcome the fact that the solutions of the heat equation propagate with infinite speed, Cattaneo proposed the following equation

$$\begin{aligned} y_t=\int _0^tK(t-s)\triangle y(s) ds, \end{aligned}$$
(2)

with \(K(t)=e^{-\gamma t}\), \(\gamma \) being a positive constant. The interest for equations of the type (2) is in the property of the solutions to have finite propagation speed, the same property of the solutions of wave equations.

From a mathematical point of view, a natural question is to study integro-differential equations of the type

$$\begin{aligned} u_{t}+\int _0^t\ M(t-s) \triangle ^2 u(s)ds=0\,, \end{aligned}$$

where M(t) is a suitable kernel, locally integrable on \((0,+\infty )\), and \(\triangle ^2\) denotes the biharmonic operator, that is in the N-dimensional case

$$\begin{aligned} \triangle ^2u=\sum _{i=1}^N\sum _{j=1}^N \partial ^2_{ii}\partial ^2_{jj}u\,. \end{aligned}$$

The Hilbert Uniqueness Method has been introduced by Lions, see [7, 8], to study control problems for partial differential systems. That method has been largely used in the literature, see e.g. [5].

Inspired by those problems, the goal of the present paper is to describe the Hilbert Uniqueness Method, for coupled hyperbolic equations of the first order with memory in a general Hilbert space, when the integral kernels involved are general functions \(k_1, k_2\in L^1(0,T)\) and integral terms also occur in the coupling:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{1t}+\displaystyle \int _0^t\ k_1(t-s) {{\mathcal {A}}} u_{1}(s)ds+{{\mathcal {L}}}_1(1*u_2)=0&{} \\ &{}\qquad \text{ in }\ (0,T)\,, \\ u_{2t}+\displaystyle \int _0^t\ k_2(t-s) {{\mathcal {A}}}^2 u_{2}(s)ds+{{\mathcal {L}}}_2(1*u_1)=0 &{} \end{array}\right. } \end{aligned}$$

In another context, in [2] the authors study the exact controllability of the equation

$$\begin{aligned} y_t=\int _0^tK(t-s)\triangle y(s) ds+u\chi _\omega \qquad \text {in} \quad (0,T)\times \varOmega , \end{aligned}$$
(3)

where \(\omega \) is a given nonempty open subset of \(\varOmega \). The hyperbolic nature of (3) allows to show its exact controllability under suitable conditions on the waiting time T and the controller \(\omega \), thanks to observability inequalities for the solutions of the dual system obtained by means of Carleman estimates.

For a different approach leading to solve control problems for hyperbolic systems, we refer to [6, 11].

2 The Hilbert Uniqueness Method

Let H be a real Hilbert space with scalar product \(\langle \cdot \, ,\, \cdot \rangle \) and norm \(\Vert \cdot \Vert \).

We consider a linear operator \(\mathcal {A}:D(\mathcal {A})\subset H\rightarrow H\) with domain \(D(\mathcal {A})\), \(k_1,k_2\in L^1(0,T)\) and \({{\mathcal {L}}}_i\) (\(i=1,2\)) linear operators on H with domain \(D({{\mathcal {L}}}_i)\supset D(\mathcal {A})\). We assume that \({{\mathcal {L}}}_2\) is self-adjoint and \({{\mathcal {L}}}_1\) is self-adjoint on a subset of its domain that will be precised later.

Moreover, let \(H_1\) be another real Hilbert space with scalar product \(\langle \cdot \, ,\, \cdot \rangle _{H_1}\) and norm \(\Vert \cdot \Vert _{H_1}\) and \(\mathcal {B}\in L(H_0;H_1)\), where \(H_0\) is a space such that \(D(\mathcal {A})\subset H_0\subset H\). In the applications \(\mathcal {B} \) could be, for example, a trace operator.

We take into consideration the following first order coupled system with memory

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{1t}+\displaystyle \int _0^t\ k_1(t-s) {{\mathcal {A}}} u_{1}(s)ds+{{\mathcal {L}}}_1(1*u_2)=0 &{} \\ &{} \qquad \text{ in }\ (0,T), \\ u_{2t}+\displaystyle \int _0^t\ k_2(t-s) {{\mathcal {A}}}^2 u_{2}(s)ds+{{\mathcal {L}}}_2(1*u_1)=0 &{} \end{array}\right. } \end{aligned}$$
(4)

with null initial conditions

$$\begin{aligned} u_1(0)=u_2(0)=0, \end{aligned}$$
(5)

and satisfying

$$\begin{aligned} \mathcal {B} u_1(t)=g_1(t), \quad \mathcal {B} u_2(t)=0, \quad \mathcal {B}\mathcal {A}u_2(t)=g_2(t), \qquad t\in (0,T). \end{aligned}$$
(6)

For a reachability problem we mean the following.

Definition 1

Given \(T>0\) and \(u_{10}\,,u_{20}\in H\), a reachability problem consists in finding \(g_i\in L^2(0,T;H_1)\), \(i=1,2\) such that the weak solution u of problem (4)–(6) verifies the final conditions

$$\begin{aligned} u_1(T)=u_{10},\quad u_2(T)=u_{20}. \end{aligned}$$
(7)

One can solve such reachability problems by means of the Hilbert Uniqueness Method. To show that, we proceed as follows.

To begin with, we assume the following conditions.

Assumptions (H1)

  1. 1.

    There exists a self-adjoint positive linear operator A on H with dense domain D(A) satisfying

    $$\begin{aligned} D(A)\subset D(\mathcal {A}), \qquad \mathcal {A}x=Ax \quad \forall x\in D(A), \qquad D(\sqrt{A})=\mathrm{Ker}(\mathcal {B}). \end{aligned}$$
  2. 2.

    \(\mathcal {L}_2\) is self-adjoint and \(\mathcal {L}_1\) is self-adjoint on \(D(\mathcal {A})\cap \mathrm{Ker}(\mathcal {B})\), that is

    $$\begin{aligned} \langle \mathcal {L}_1\varphi , \xi \rangle =\langle \varphi , \mathcal {L}_1\xi \rangle , \quad \forall \varphi ,\xi \in D(\mathcal {A})\cap \mathrm{Ker}(\mathcal {B}). \end{aligned}$$
    (8)
  3. 3.

    There exists \(D_\nu \in L(H_0;H_1)\) such that the following identity holds

    $$\begin{aligned} \langle \mathcal {A}\varphi , \xi \rangle =\langle \varphi , A\xi \rangle -\langle \mathcal {B}\varphi ,D_\nu \xi \rangle _{H_1}, \quad \forall \varphi \in D(\mathcal {A}),\xi \in D(A). \end{aligned}$$
    (9)

Now, we consider the adjoint system of (4), that is, the following coupled system

$$\begin{aligned} {\left\{ \begin{array}{ll} z_{1t} -\displaystyle \int _t^T\ k_1(s-t)A z_{1}(s)ds-\int _t^T{ L}_2z_2(s)ds= 0 &{} \\ &{} \text{ in }\ (0,T), \\ z_{2t} -\displaystyle \int _t^T\ k_2(s-t)A^2 z_{2}(s)ds-\int _t^T{ L}_1z_1(s)ds= 0 &{} \end{array}\right. } \end{aligned}$$
(10)

with given final data

$$\begin{aligned} z_1(T)=z_{1T},\quad z_2(T)=z_{2T}. \end{aligned}$$
(11)

We assume that for final data sufficiently regular an existence and regularity result for the solution of (10)–(11) holds. Precisely:

Theorem 1

For any \(z_{1T}\in D(A)\) and \(z_{2T}\in D(A^{2})\) there exists a unique solution \((z_1,z_2)\) of (10)–(11) such that \(z_1\in C^{1}([0,T],H)\cap C([0,T],D(A))\) and \(z_2\in C^{1}([0,T],H)\cap C([0,T],D(A^2))\).

That type of result will be true in the applications, taking into account that backward problems are equivalent to forward problems by means of a change of the variable t into \(t-T\).

If Theorem 1 holds true, then the regularity of the solution \((z_1,z_2)\) of (10)–(11) and assumption (H1)-3 allow to obtain the following properties: the functions \(D_\nu z_i\), \(i=1,2\), belong to \(C(0,T;H_1)\), because \(D(A)\subset D(\mathcal {A})\subset H_0\). So, we can consider the nonhomogeneous problem

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \phi _1'(t) +\int _0^t\ k_1(t-s) \mathcal {A}\phi _1(s)ds +{\mathcal L}_1(1*\phi _2)=0 &{}\\ \displaystyle \phi _2'(t) +\int _0^t\ k_2(t-s) \mathcal {A}^2\phi _2(s)ds+{\mathcal L}_2(1*\phi _1)=0 &{} \\ &{} \quad \text{ in }\ (0,T)\\ \displaystyle \mathcal {B}\phi _1(t)=\int _t^T\ k_1(s-t)D_\nu z_1(s)ds, &{} \\ \mathcal {B}\phi _2(t)=0, \quad \displaystyle \mathcal {B}\mathcal {A}\phi _2(t)=\int _t^T\ k_2(s-t)D_\nu z_2(s)ds &{}\\ \phi _1(0)=\phi _2(0)=0. \end{array}\right.&\end{aligned}$$
(12)

If \((\phi _1,\phi _2)\) denotes the solution of problem (12), then we can introduce the following linear operator on \(H\times H\):

$$\begin{aligned} \varPsi (z_{1T},z_{2T})=(\phi _1(T),\phi _2(T)),\qquad (z_{1T},z_{2T})\in D(A)\times D(A^2). \end{aligned}$$

We will prove the next result.

Theorem 2

If \((\xi _{1},\xi _{2})\) is the solution of the system

$$\begin{aligned} \left\{ \begin{array}{ll}\displaystyle \xi '_{1}(t) -\int _t^T\ k_1(s-t) A\xi _{1}(s)ds-\int _t^T{{\mathcal {L}}}_2\xi _2(s)ds= 0, &{} \\ &{} \quad in\ (0,T)\\ \displaystyle \xi '_{2}(t) -\int _t^T\ k_2(s-t) A^2\xi _{2}(s)ds-\int _t^T{{\mathcal {L}}}_1\xi _1(s)ds= 0,&{} \\ \\ \xi _1(T)=\xi _{1T},\quad \xi _2(T)=\xi _{2T}, &{} \end{array}\right. \end{aligned}$$

where \((\xi _{1T},\xi _{2T})\in D(A)\times D(A^2)\), then the identity

$$\begin{aligned} \begin{aligned}&\langle \varPsi (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle \\ =&\int _0^T\ \langle \mathcal {B}\phi _{1}(t),\int _t^T\ k_1(s-t) D_\nu \xi _1(s)\ ds\rangle _{H_1} \ dt \\&+\int _0^T \langle \mathcal {B}\mathcal {A}\phi _2(t),\int _t^Tk_2(s-t) D_\nu \xi _{2}(s)\ ds \rangle _{H_1}\ dt , \end{aligned} \end{aligned}$$
(13)

holds true.

Proof

We multiply the first equation in (12) by \(\xi _1(t)\) and integrate on [0, T], so we have

$$\begin{aligned} \int _0^T\langle \phi '_{1},\xi _1\rangle \ dt&+\int _0^T\langle \int _0^t\ k_1(t-s){{\mathcal {A}}}\phi _{1}(s)\ ds,\xi _1\rangle \ dt \nonumber \\&+\int _0^T \langle {{\mathcal {L}}}_1(1*\phi _{2}), \xi _1 \rangle \ dt=0. \end{aligned}$$
(14)

In the second term of the above identity we change the order of integration and, since \(\xi _1(t)\in D(A)\), we can use (9) to get

$$\begin{aligned} \int _0^T\ \langle \int _0^t k_1(t-s){{\mathcal {A}}}\phi _{1}(s)\ ds,\xi _1(t)\rangle dt&= \int _0^T\ \int _s^Tk_1(t-s)\langle {{\mathcal {A}}}\phi _{1}(s),\xi _1(t)\rangle \ dt \ ds \\&= \int _0^T\ \langle \phi _{1}(s),\int _s^T\ k_1(t-s) A\xi _1(t)\ dt\rangle \ ds \\&\qquad -\int _0^T \langle \mathcal {B}\phi _{1}(s),\int _s^T k_1(t-s) D_\nu \xi _1(t)\ dt\rangle _{H_1} \ ds . \end{aligned}$$

Note that, in virtue of assumption (H1)-1, one has \(D({A})\subset D(\mathcal {A})\cap \mathrm{Ker}(\mathcal {B})\); so, changing again the order of integration and applying (8), we obtain

$$\begin{aligned} \int _0^T \langle {{\mathcal {L}}}_1(1*\phi _{2}), \xi _1 \rangle \ dt =\int _0^T\ \langle \phi _{2}(s),\int _s^T{{\mathcal {L}}}_1\xi _1(t)\ dt\rangle \ ds . \end{aligned}$$

If we integrate by parts the first term in (14) and take into account the previous two identities, then, in view also of \(\phi _1(0)=0\), we get

$$\begin{aligned} \begin{aligned}&\langle \phi _{1}(T),\xi _{1}(T)\rangle - \int _0^T \langle \phi _1(t),\xi '_{1} (t)\rangle \ dt +\int _0^T\ \langle \phi _{1}(t),\int _t^T\ k_1(s-t) A\xi _1(s)\ ds\rangle \ dt \\ -&\int _0^T\ \langle \mathcal {B}\phi _{1}(t),\int _t^T\ k_1(s-t) D_\nu \xi _1(s)\ ds\rangle _{H_1} \ dt \\ +&\int _0^T\ \langle \phi _{2}(t),\int _t^T{{\mathcal {L}}}_1\xi _1(s)\ ds\rangle \ dt=0. \end{aligned} \end{aligned}$$

As a consequence of the former equation and

$$\begin{aligned} \xi '_{1} (t)-\int _t^T\ k_1(s-t) A\xi _{1}(s)ds=\int _t^T{{\mathcal {L}}}_2\xi _2(s)ds, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned}&\langle \phi _{1}(T),\xi _{1T}\rangle -\int _0^T\ \langle \mathcal {B}\phi _{1}(t),\int _t^T\ k_1(s-t) D_\nu \xi _1(s)\ ds\rangle _{H_1} \ dt \\ +&\int _0^T\ \langle \phi _{2}(t),\int _t^T{{\mathcal {L}}}_1\xi _1(s)\ ds\rangle \ dt - \int _0^T \langle \phi _1(t),\int _t^T{{\mathcal {L}}}_2\xi _2(s)ds\rangle \ dt =0. \end{aligned} \end{aligned}$$
(15)

In a similar way, we multiply the second equation in (12) by \(\xi _2(t)\) and integrate on [0, T]: if we integrate by parts the first term, take into account that \(\phi _2(0)=0\) and change the order of integration in the other two terms, then we have

$$\begin{aligned} \begin{aligned}&\langle \phi _{2}(T),\xi _{2T}\rangle - \int _0^T \langle \phi _2(t),\xi '_{2} (t)\rangle \ dt \\ +&\int _0^T\int _s^Tk_2(t-s) \langle \ \mathcal {A}^2\phi _2(s),\xi _{2}(t) \rangle \ dt\ ds \\ +&\int _0^T \int _s^T\langle {{\mathcal {L}}}_2\phi _1(s),\xi _2(t)\rangle \ dt\ ds=0. \end{aligned} \end{aligned}$$
(16)

Now, we observe that from (9) it follows for any \(\varphi \in D(\mathcal {A}^2)\) and \(\xi \in D(A^2)\)

$$\begin{aligned} \langle \mathcal {A}^2\varphi , \xi \rangle =\langle \varphi , A^2\xi \rangle -\langle \mathcal {B}\varphi ,D_\nu A\xi \rangle _{H_1} -\langle \mathcal {B}\mathcal {A}\varphi ,D_\nu \xi \rangle _{H_1} . \end{aligned}$$

Putting the above equation into (16) and taking into account that the operator \({{\mathcal {L}}}_2\) is self-adjoint yield

$$\begin{aligned} \begin{aligned}&\langle \phi _{2}(T),\xi _{2T}\rangle - \int _0^T \langle \phi _2(t),\xi '_{2} (t)\rangle \ dt \\ +&\int _0^T \langle \phi _2(s),\int _s^Tk_2(t-s) A^2\xi _{2}(t)\ dt \rangle \ ds \\ -&\int _0^T \langle \mathcal {B}\mathcal {A}\phi _2(s),\int _s^Tk_2(t-s) D_\nu \xi _{2}(t)\ dt \rangle _{H_1}\ ds \\ +&\int _0^T \langle \phi _1(s),\int _s^T{{\mathcal {L}}}_2\xi _2(t)\ dt\rangle \ ds=0. \end{aligned} \end{aligned}$$

In virtue of

$$\begin{aligned} \xi '_{2}(t) -\int _t^T\ k_2(s-t) A^2\xi _{2}(s)ds=\int _t^T{{\mathcal {L}}}_1\xi _1(s)ds, \end{aligned}$$

we get

$$\begin{aligned} \begin{aligned}&\langle \phi _{2}(T),\xi _{2T}\rangle -\int _0^T \langle \mathcal {B}\mathcal {A}\phi _2(s),\int _s^Tk_2(t-s) D_\nu \xi _{2}(t)\ dt \rangle _{H_1}\ ds \\ +&\int _0^T \langle \phi _1(t),\int _t^T{{\mathcal {L}}}_2\xi _2(s)ds\rangle \ dt - \int _0^T \langle \phi _2(t),\int _t^T{{\mathcal {L}}}_1\xi _1(s)ds\rangle \ dt =0. \end{aligned} \end{aligned}$$
(17)

If we sum Eqs. (15) and (17), then we have

$$\begin{aligned} \begin{aligned}&\langle \varPsi (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle =\langle \phi _{1}(T),\xi _{1T}\rangle + \langle \phi _{2}(T),\xi _{2T}\rangle \\ =&\int _0^T\ \langle \mathcal {B}\phi _{1}(t),\int _t^T\ k_1(s-t) D_\nu \xi _1(s)\ ds\rangle _{H_1} \ dt \\&+\int _0^T \langle \mathcal {B}\mathcal {A}\phi _2(t),\int _t^Tk_2(s-t) D_\nu \xi _{2}(s)\ ds \rangle _{H_1}\ dt , \end{aligned} \end{aligned}$$
(18)

that is, (13) holds true.    \(\square \)

If we take \((\xi _{1T},\xi _{2T})=(z_{1T},z_{2T})\) in (13), then we have

$$\begin{aligned} \begin{aligned}&\langle \varPsi (z_{1T},z_{2T}),(z_{1T},z_{2T})\rangle \\&= \int _0^T \left( \Big |\int _t^T\ k_1(s-t) D_\nu z_1(s)\ ds\Big |_{H_1}^2 +\Big |\int _t^T\ k_2(s-t) D_\nu z_2(s)\ ds\Big |_{H_1}^2 \right) dt. \end{aligned} \end{aligned}$$

Consequently, we can introduce a semi-norm on the space \(D(A)\times D(A^2)\). Precisely, if we consider, for any \((z_{1T},z_{2T})\in D(A)\times D(A^2)\), the solution \((z_1,z_2)\) of the system (10)–(11), then we define

$$\begin{aligned} \begin{aligned}&\Vert (z_{1T},z_{2T})\Vert _{F}^2:= \\&\int _0^T \left( \Big |\int _t^T\ k_1(s-t) D_\nu z_1(s)\ ds\Big |_{H_1}^2 +\Big |\int _t^T\ k_2(s-t) D_\nu z_2(s)\ ds\Big |_{H_1}^2 \right) dt. \end{aligned} \end{aligned}$$
(19)

We observe that \(\Vert \cdot \Vert _{F}\) is a norm if and only if the following uniqueness theorem holds.

Theorem 3

If \((z_1,z_2)\) is the solution of problem (10)–(11) such that

$$\begin{aligned} \int _t^T\ k_1(s-t) D_\nu z_1(s)\ ds= \int _t^T\ k_1(s-t) D_\nu z_2(s)\ ds=0, \quad on \ [0,T], \end{aligned}$$

then

$$\begin{aligned} z_1=z_2= 0 \quad in \ [0,T]. \end{aligned}$$

The validity of Theorem 3 is the starting point for the application of the Hilbert Uniqueness Method. Indeed, if we assume that Theorem 3 holds true, then we can define the Hilbert space F as the completion of \(D(A)\times D(A^2)\) for the norm \(\Vert \cdot \Vert _F\). Thanks to (13) and (19) we have

$$\begin{aligned} \begin{aligned} \langle \varPsi (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle&=\langle (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle _F \\&\forall (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\in D(A)\times D(A^2) , \end{aligned} \end{aligned}$$
(20)

where \(\langle \cdot ,\cdot \rangle _F\) denotes the scalar product associated with the norm \(\Vert \cdot \Vert _F\).

Consequently,

$$\begin{aligned} \begin{aligned} \big |\langle \varPsi (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle \big |&\le \Vert (z_{1T},z_{2T})\Vert _F\ \Vert (\xi _{1T},\xi _{2T})\Vert _F \\&\forall (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\in D(A)\times D(A^2) . \end{aligned} \end{aligned}$$

Thanks to the above inequality, the operator \(\varPsi \) can be extended uniquely to a linear continuous operator, denoted again by \(\varPsi \), from F into its dual space \(F'\). By (20) it follows that

$$\begin{aligned} \begin{aligned} \langle \varPsi (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle&=\langle (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\rangle _F \\&\qquad \forall (z_{1T},z_{2T}),(\xi _{1T},\xi _{2T})\in F , \end{aligned} \end{aligned}$$

and, as a consequence, we have that the operator \(\varPsi :F\rightarrow F'\) is an isomorphism.

Moreover, the key point to characterize the space F is to establish observability estimates of the following type

$$\begin{aligned} \begin{aligned}&\int _0^T \left( \Big |\int _t^T\ k_1(s-t) D_\nu z_1(s)\ ds\Big |_{H_1}^2 +\Big |\int _t^T\ k_2(s-t) D_\nu z_2(s)\ ds\Big |_{H_1}^2 \right) dt \\&\qquad \qquad \asymp \Vert z_{1T}\Vert ^2_{F_1}+\Vert z_{2T}\Vert ^2_{F_2} \end{aligned} \end{aligned}$$
(21)

for suitable spaces \(F_1\,,F_2\). In that case, the uniqueness result stated by Theorem 3 holds true, so the operator \(\varPsi :F\rightarrow F'\) is an isomorphism, and in virtue of (19) and (21) we get

$$\begin{aligned}F=F_1\times F_2\end{aligned}$$

with the equivalence of the respective norms. Finally, we are able to solve the reachability problem (4)–(7) for \((u_{10},u_{20})\in F'_1\times F'_2\).

3 Applications

Example 1

Let \(H=L^2(0,\pi )\) be endowed with the usual scalar product and norm. In [9] we take \({\mathcal {A}}=\frac{d^2}{dx^2}\) with null Dirichlet boundary conditions, \(k_1(t)=\frac{\beta }{\eta }e^{-\eta t}+1-\frac{\beta }{\eta }\), \(k_2\equiv 1\). We examine the case in which \({\mathcal {L}}_i=a_i \mathrm{I}\), with \(a_i\in {{\mathbb {R}}}\), \(i=1,2\) and \(\mathrm{I}\) the identity operator on H.

By writing the solutions as Fourier series, we are able to prove Theorems 1 and 3, thanks also to some properties of the solutions of integral equations. In particular, by showing suitable Ingham type estimates, we prove observability estimates of the type (21) where \(F=H^1_0(0,\pi )\times H^1_0(0,\pi )\). Therefore, we can deduce reachability results by means of the Hilbert Uniqueness Method.

Example 2

We consider \(H=L^2(0,\pi )\) endowed with the usual scalar product and norm. In [9] we take \({\mathcal {A}}=\frac{d^2}{dx^2}\) with null Dirichlet boundary conditions, \(k_1(t)=\frac{\beta }{\eta }e^{-\eta t}+1-\frac{\beta }{\eta }\), \(k_2\equiv 1\), \({\mathcal {L}}_1=a_1\frac{d^2}{dx^2}\) and \({\mathcal {L}}_2=a_2\mathrm{I}\) with \(a_i\in {{\mathbb {R}}}\), \(i=1,2\).

Example 3

Let \(H=L^2(\varOmega )\) be endowed with the usual scalar product and norm. In [10] we take \({\mathcal {A}}=\triangle \) with null Dirichlet boundary conditions, \(k_1(t)=\frac{\beta }{\eta }e^{-\eta t}+1-\frac{\beta }{\eta }\), \(k_2\equiv 1\), \({\mathcal {L}}_1=a_1\triangle \) and \({\mathcal {L}}_2=a_2\mathrm{I}\) with \(a_i\in {{\mathbb {R}}}\), \(i=1,2\).