Coupled linear Schrödinger equations: control and stabilization results

Abstract

This article presents some controllability and stabilization results for a system of two coupled linear Schrödinger equations in the one-dimensional case where the state components are interacting through the Kirchhoff boundary conditions. Considering the system in a bounded domain, the null boundary controllability result is shown. The result is achieved thanks to a new Carleman estimate, which ensures a boundary observation. Additionally, this boundary observation together with some trace estimates, helps us to use the Gramian approach, with a suitable choice of feedback law, to prove that the system under consideration decays exponentially to zero at least as fast as the function $e^{-2\omega t}$ for some $\omega >0$.

Appendices

Appendix A. Auxiliary results

In this first appendix, we briefly discuss the well-posedness of the control system (1.2).

1.1 A.1. Well-posedness results

Consider the following operator associated with the control system (1.2), given by

$$\begin{aligned} \mathcal {A}=\begin{pmatrix} i \gamma _1 \partial _{xx} - i\alpha _1 \mathbb I_d &{} \textbf{0}\\ \\ \textbf{0}&{} i\frac{\gamma _2}{\sigma } \partial _{xx} - i \frac{\alpha _2}{\sigma } \mathbb I_d \end{pmatrix} , \end{aligned}$$

(A.1)

with

$$\begin{aligned} \begin{aligned} \mathcal {D}(\mathcal {A})=\big \{(u_1, u_2)\in [H^2(\Omega )]^2 \mid \,&u_1(0)=u_2(0)=0, \ u_1(1)=u_2(1), \\&\gamma _1 u_{1,x}(1)+\frac{\gamma _2}{\sigma }u_{2,x}(1)+\alpha u_1(1)=0 \big \}. \end{aligned} \end{aligned}$$

(A.2)

With this in hand, the first result shows that the operator in consideration is dissipative.

Proposition A.1

The operator $(\mathcal {A}, \mathcal {D}(\mathcal {A}))$ generates a strongly continuous unitary group in $[L^2(\Omega )]^2$.

Proof

Let us consider $\textbf{U}=(u,v) \in \mathcal D(\mathcal A)$. A simple computation gives that

$$\begin{aligned} \left<{\mathcal {A}\textbf{U}},{\textbf{U}}\right>_{[L^2(\Omega )]^2}= & {} \text {Re}\bigg [i\gamma _1\mathop {\int }\limits _{0}^{1}u_{xx}\overline{u}dx-i\alpha _1\mathop {\int }\limits _{0}^{1}|u|^2dx+i\frac{\gamma _2}{\sigma }\mathop {\int }\limits _{0}^{1}v_{xx}\overline{v}dx-i\frac{\alpha _2}{\sigma }\mathop {\int }\limits _{0}^{1}|u|^2dx\bigg ]\nonumber \\= & {} \text {Re}\bigg [-i\gamma _1\mathop {\int }\limits _{0}^{1}|u_{x}|^2dx-i\frac{\gamma _2}{\sigma }\mathop {\int }\limits _{0}^{1}|v_{x}|^2dx+i\gamma _1u_x(1)\overline{u(1)}+i\frac{\gamma _2}{\sigma }v_x(1)\overline{v(1)}\bigg ]\nonumber \\= & {} \text {Re}\bigg [i\gamma _1u_x(1)\overline{u(1)}+i\frac{\gamma _2}{\sigma }v_x(1)\overline{v(1)}\bigg ]\nonumber \\ {}= & {} \text {Re}\bigg [-i\alpha |u(1)|^2\bigg ]=0. \end{aligned}$$

By using semigroup theory, $\mathcal {A}$ generates a strongly continuous unitary group on $[L^2(\Omega )]^2.$ Moreover it can be easily checked that for all $(u,v)\in \mathcal {D}(\mathcal {A})$,

$$\begin{aligned} \left<{\mathcal {A}(u,v)},{(u,v)}\right>_{[L^2(\Omega )]^2}=-\left<{(u,v)},{\mathcal {A}(u,v)}\right>_{[L^2(\Omega )]^2} \end{aligned}$$

and also $\mathcal {D}(\mathcal {A})=\mathcal {D}(\mathcal {A}^*).$ Therefore $(\mathcal {A}, \mathcal {D}(\mathcal {A}))$ is a skew adjoint operator. $\square$

1.2 A.2. Adjoint system

Let us remember that the adjoint associated with the system (1.2) with (1.3)–(1.4a) or (1.3)–(1.4b), is given by (2.4), with given final data $\zeta :=(\zeta _1, \zeta _2)$ from some suitable Hilbert space. Note that, since $\mathcal {A}=-\mathcal {A}^*$ we have that $\mathcal {D(A^*)} = \mathcal {D(A)}$, so, the following result shows the well-posedness for the system (2.4).

Proposition A.2

For given $\zeta :=(\zeta _1, \zeta _2)\in \mathcal H$, there exists a unique solution $\varphi :=(\varphi _1, \varphi _2)\in C([0,T]; \mathcal {H})$ to the adjoint system (2.4) such that it satisfies

$$\begin{aligned} \Vert \varphi \Vert _{C([0,T];\mathcal H)} \le C \Vert \zeta \Vert _\mathcal {H} , \end{aligned}$$

(A.3)

for some constant $C>0$.

Appendix B. Key lemma

This second part of this appendix is devoted to presenting an essential lemma that is a key point in ensuring the hypothesis (H3) in Urquiza’s approach.

Lemma B.1

First, consider a function $m \in C^2(0,1)$. Then the solution of (2.4) satisfies the following identity:

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\sum _{j=1}^{2}\nu _j\mathop {\int }\limits _{0}^{T}|\varphi _{j,x}(t,0)|^2m(0) dt=\frac{1}{2}\sum _{j=1}^{2}\nu _j\mathop {\int }\limits _{0}^{T}|\varphi _{j,x}(t,1)|^2m(1) dt\\ {}&-\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{1} \overline{\varphi _{j}(T,x)}m(x)\varphi _{j,x}(T,x)dx\right) +\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{1} \overline{\varphi _{j}(0,x)}m(x)\varphi _{j,x}(0,x)dx\right) \\&+\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{T} \varphi _{j,t}(t,1)m(1)\overline{\varphi _{j}(t,1)}dt\right) -\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j,x} m'(x)\overline{\varphi _{j,x}(t,x)}dxdt\right) \\&-\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j,x} m''(x)\overline{\varphi _{j}(t,x)}dxdt\right) +\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T} \varphi _{j,x}(t,1) m'(1)\overline{\varphi _{j}(t,1)}dt\right) ,\\&-\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \theta _j \mathop {\int }\limits _{0}^{T} m(1)|{\varphi _{j}(t,1)}|^2dt\right) , \end{aligned} \end{aligned}$$

(B.1)

where $\nu _1=\gamma _1, \nu _2=\frac{\gamma _2}{\sigma }, \theta _1=\alpha _1, \theta _2=\frac{\alpha _2}{\sigma }$.

Proof

Multiply the equations (2.4) by $(m\overline{\varphi _{1,x}}+\frac{1}{2}\overline{\varphi _1}m')$ and $(m\overline{\varphi _{1,x}}+\frac{1}{2}\overline{\varphi _1}m')$, respectively, and using integration by parts along with boundary conditions, we have

$$\begin{aligned} \begin{aligned} \text {Re}\sum _{j=1}^{2} i \mathop {\int }\limits _{0}^{T}&\mathop {\int }\limits _{0}^{1} \varphi _{j,t}\left( \frac{1}{2}m'(x)\overline{\varphi _{j}(t,x)}+m(x) \overline{\varphi _{j,x}(t,x)}\right) dxdt=\\ {}&-\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{1} \overline{\varphi _{j}(T,x)}m(x)\varphi _{j,x}(T,x)dx\right) \\&+\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{1} \overline{\varphi _{j}(0,x)}m(x)\varphi _{j,x}(0,x)dx\right) \\ {}&+\frac{1}{2}\text {Im}\left( \sum _{j=1}^{2} \mathop {\int }\limits _{0}^{1} \varphi _{j,t}(t,1)m(1)\overline{\varphi _{j}(t,1)}dt\right) . \end{aligned} \end{aligned}$$

Similarly, we have from the second term of both equations:

$$\begin{aligned} \begin{aligned} \text {Re}\sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j,xx}&\left( \frac{1}{2}m'(x)\overline{\varphi _{j}(t,x)}+m(x) \overline{\varphi _{j,x}(t,x)}\right) dxdt=\frac{1}{2}\sum _{j=1}^{2}\nu _j\mathop {\int }\limits _{0}^{T}|\varphi _{j,x}(t,1)|^2m(1) dt\\&-\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j,x} m'(x)\overline{\varphi _{j,x}(t,x)}dxdt\right) \\ {}&-\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j,x} m''(x)\overline{\varphi _{j}(t,x)}dxdt\right) \\&+\frac{1}{2}\text {Re}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T} \varphi _{j,x}(t,1) m'(1)\overline{\varphi _{j}(t,1)}dt\right) -\frac{1}{2}\left( \sum _{j=1}^{2} \nu _j \mathop {\int }\limits _{0}^{T} m(0)|{\varphi _{j,x}(t,0)}|^2dt\right) . \end{aligned} \end{aligned}$$

Also, we have

$$\begin{aligned} \text {Re}\sum _{j=1}^{2} \theta _j \mathop {\int }\limits _{0}^{T}\mathop {\int }\limits _{0}^{1} \varphi _{j}\left( \frac{1}{2}m'(x)\overline{\varphi _{j}(t,x)}+m(x) \overline{\varphi _{j,x}(t,x)}\right) dxdt=\left( \sum _{j=1}^{2} \theta _j \mathop {\int }\limits _{0}^{T} m(1)|{\varphi _{j}(t,1)}|^2dt\right) . \end{aligned}$$

Putting together the previous relation (B.1) holds and the lemma is finished. $\square$