Boundary null-controllability of 1-D coupled parabolic systems with Kirchhoff-type conditions

Abstract

The main concern of this article is to investigate the boundary controllability of some $2\times 2$ one-dimensional parabolic systems with both the interior and boundary couplings: The interior coupling is chosen to be linear with constant coefficient while the boundary one is considered by means of some Kirchhoff-type condition at one end of the domain. We consider here the Dirichlet boundary control acting only on one of the two state components at the other end of the domain. In particular, we show that the controllability properties change depending on which component of the system the control is being applied. Regarding this, we point out that the choices of the interior coupling coefficient and the Kirchhoff parameter play a crucial role to deduce the positive or negative controllability results. Further to this, we pursue a numerical study based on the well-known penalized HUM approach. We make some discretization for a general interior-boundary coupled parabolic system, mainly to incorporate the effects of the boundary couplings into the discrete setting. This allows us to illustrate our theoretical results as well as to experiment some more examples which fit under the general framework, for instance a similar system with a Neumann boundary control on either one of the two components.

A An intermediate result

Lemma A.1

Let $\mathcal {D}$ and $\mathcal {N}$ be two real $d\times d$ matrices such that

$$\begin{aligned} (\mathcal {D},\mathcal {N})\text { is full rank,} \end{aligned}$$

(A.1)

and

$$\begin{aligned} \mathcal {D}\mathcal {N}^*\text { is self-adjoint,} \end{aligned}$$

(A.2)

then $\mathcal {N}+t\mathcal {D}$ is invertible for any $t\in \mathbb {R}$ except perhaps for a finite number of values of t.

Proof

We follow the same computations as in [10, Theorem 1.4.4]. More precisely, we first observe that, under the assumptions of the lemma, we have that $\mathcal {D}+i \mathcal {N}$ is invertible. Indeed,

• by (A.1), we know that $(\ker \mathcal {D}^*) \cap (\ker \mathcal {N}^*)=\{0\}$,

• by (A.2), for any $x\in \mathbb {C}^2$ we have

  $$\begin{aligned} \Vert (\mathcal {D}^*-i \mathcal {N}^*) x \Vert ^2 =\Vert \mathcal {D}^*x\Vert ^2 + \Vert \mathcal {N}^*x\Vert ^2, \end{aligned}$$

  so that $\ker (\mathcal {D}^*-i \mathcal {N}^*)\subset (\ker \mathcal {D}^*) \cap (\ker \mathcal {N}^*)=\{0\}$ and the claim is proved.

We can now can define $\mathcal {U}=-(\mathcal {D}+i\mathcal {N})^{-1}(\mathcal {D}-i\mathcal {N})$ (which is actually a unitary matrix but we do not need this fact here). It satisfies

$$\begin{aligned} 2(\mathcal {D}+i\mathcal {N})^{-1} \mathcal {D}= & {} (\mathcal {D}+i\mathcal {N})^{-1} (\mathcal {D}+i\mathcal {N} + \mathcal {D} - i \mathcal {N}) = (I - \mathcal {U}),\\ 2(\mathcal {D}+i\mathcal {N})^{-1} \mathcal {N}= & {} -i(\mathcal {D}+i\mathcal {N})^{-1} \big ((\mathcal {D} + i\mathcal {N}) - (\mathcal {D} -i \mathcal {N})\big ) = - i (I+\mathcal {U}). \end{aligned}$$

If we assume that $t\in \mathbb {R}$ is such that $\mathcal {N}+t\mathcal {D}$ is not invertible, then there exists $x\in \mathbb {R}^d, x\ne 0$ such that $(\mathcal {N}+t\mathcal {D})x = 0$. Left multiplying this equality by $2(\mathcal {D}+i\mathcal {N})^{-1}$ an using the above relations we end up with

$$\begin{aligned} \left( \frac{t-i}{t+i} I - \mathcal {U}\right) x = 0, \end{aligned}$$

which proves that $(t-i)/(t+i)$ is an eigenvalue of $\mathcal {U}$. This can only happen for a finite number of values of t. $\square$