A hyperbolic system and the cost of the null controllability for the Stokes system

Abstract

This paper is devoted to study the cost of the null controllability for the Stokes system. Using the control transmutation method, we show that the cost of driving the Stokes system to rest at time $T$ is of order e$^{C/T}$ when $T \longrightarrow 0^+$, i.e., the same order of controllability as for the heat equation. For this to be possible, we prove a new exact controllability result for a hyperbolic system with a resistance term, which will be done under assumptions on the control region.

Appendix: boundary observability for the hyperbolic system

This section is devoted to prove the following result.

Theorem 5.1

If we take $T > 2R_0$ then, for every solution of (4.4) with initial data $(\phi ^0,\phi ^1) \in V \times H$, the following estimate holds:

$$\begin{aligned} |\phi ^1|_H^2 + ||\phi ^0||_V^2 \le \frac{R_0}{2(T-2R_0)} \int \!\!\!\!\int _{\Sigma }\biggl ( \frac{\partial \phi }{\partial \nu }\biggl )^2d\Sigma . \end{aligned}$$

(5.1)

For the proof of Theorem 5.1, we need the following two lemmas.

Lemma 5.2

Let $\overline{q} = \overline{q}(x)$ be in $C^1(\bar{\Omega })^N$, then, for every regular solution $u$ of (4.2), the following identity holds:

$$\begin{aligned} \frac{1}{2}\int \! \! \! \int _{\Sigma }\overline{q}_k(x) \nu _k(x)\biggl (\frac{\partial u}{\partial \nu }\biggl )^2d\Sigma&= (u_t(t), \overline{q}(x)\nabla u(t))\bigl |_0^T +\int \! \! \! \int _{Q}\frac{\partial \overline{q}_k}{\partial x_j}\frac{\partial u^i}{\partial x_k}\frac{\partial u^i}{\partial x_j}\mathrm{d}x\mathrm{d}t \nonumber \\&+ \frac{1}{2} \int \! \! \! \int _{Q}\frac{\partial \overline{q}_k}{\partial x_k} \bigl ( |u_t|^2 - |\nabla u|^2 \bigl )\mathrm{d}x\mathrm{d}t \nonumber \\&+\int \! \! \! \int _{Q}\frac{\partial p}{\partial x_i} \overline{q}_k\frac{\partial u^i}{\partial x_k}\mathrm{d}x\mathrm{d}t +\int \! \! \! \int _{Q} h^i \overline{q}_k\frac{\partial u^i}{\partial x_k}\mathrm{d}x\mathrm{d}t. \end{aligned}$$

(5.2)

The proof of Lemma 5.2 is the same as in the case of a single-wave equation, the difference being that here we see the pressure as a force term in the right-hand side.

Lemma 5.3

Let $(u^0, u^1, h) \in V\times H \times L^2(Q)^N$, then the weak solution of (4.2) satisfies:

$$\begin{aligned} \int \! \! \! \int _{\Sigma } \biggl ( \frac{\partial u}{\partial \nu }\biggl )^2 d\Sigma \le C\bigl (|u^1|^2_H + ||u^0||^2_V + ||h||^2_{L^2(Q)^N}\bigl ). \end{aligned}$$

Proof

The proof is obtained exactly as in the case of the wave equation, first showing the result for regular solutions. Indeed, in this case we must take the vector field $\overline{q}$ in Lemma 5.2 to be the vector field $\overline{q}(x) = x$ and use the fact that

$$\begin{aligned} \int \! \! \! \int _{Q}\frac{\partial p}{\partial x_i} \overline{q}_k\frac{\partial u^i}{\partial x_k}\mathrm{d}x\mathrm{d}t = 0. \end{aligned}$$

$\square$

Proof of Lemma 5.1

Without loss of generality, we assume that $\phi$ is regular and then work with the equivalent problem (4.6). Using Lemma 5.2, with $\overline{q}$ being the vector field $m(x) = x$, we have

$$\begin{aligned}&\frac{1}{2}\int \! \! \! \int _{\Sigma }m\cdot \nu \biggl (\frac{\partial \phi }{\partial \nu }\biggl )^2d\Sigma = (\phi _t(.), m(x) \nabla \phi (.))\bigl |_0^T \\&\quad + \int \! \! \! \int _{Q}|\nabla \phi |^2\mathrm{d}x\mathrm{d}t + \frac{N}{2} \int \! \! \! \int _{Q}\bigl ( |\phi _t|^2 - |\nabla \phi |^2 \bigl )\mathrm{d}x\mathrm{d}t. \end{aligned}$$

Next, multiplying (4.6)$_1$ by $\phi$ and integrating by parts, we easily see that

$$\begin{aligned} (\phi _t(.),\phi (.))\bigl |_0^T = \int \! \! \! \int _{Q} |\phi _t|^2\mathrm{d}x\mathrm{d}t - \int \! \! \! \int _{Q} |\nabla \phi |^2\mathrm{d}x\mathrm{d}t. \end{aligned}$$

Then, using this last identity and the fact that

$$\begin{aligned} |\phi _t(t)|_H^2 + ||\phi (t)||_V^2 = |\phi ^1|_H^2 + ||\phi ^0||_V^2 \ \ \ \forall t \in [0,T], \end{aligned}$$

we obtain

$$\begin{aligned} (\phi _t(.),m\nabla u(.) +\frac{N-1}{2}u(.))\bigl |_0^T + T \bigl ( |\phi ^1|_H^2 + ||\phi ^0||_V^2 \bigl ) = \frac{1}{2} \int \! \! \! \int _{\Sigma }m\cdot \nu \biggl (\frac{\partial \phi }{\partial \nu }\biggl )^2d\Sigma . \end{aligned}$$

We also have

$$\begin{aligned} \bigl | m\nabla u(t) +\frac{N-1}{2}u(t) \bigl |^2 \le R_0 |\nabla \phi (t) |^2 \ \ \ \forall t \in [0,T], \end{aligned}$$

which implies, by Gronwall inequality, that

$$\begin{aligned} \biggl | \bigl ( \phi _t(.),m \nabla \phi (.) +\frac{N-1}{2} \phi (.) ) \bigl |_0^T \biggl | \le 2R_0 \bigl (|\phi ^1|_H^2 + ||\phi ^0||_V^2\bigl ). \end{aligned}$$

Finally, combining all the above estimates, we conclude that

$$\begin{aligned} \bigl (T-2R_0\bigl )\bigl (|\phi ^1|_H^2 + ||\phi ^0||_V^2\bigl ) \le \frac{R_0}{2} \ \!\!\!\!\int _{\Sigma }\biggl (\frac{\partial \phi }{\partial \nu }\biggl )^2d\Sigma , \end{aligned}$$

which is exactly (5.1). $\square$