Local null controllability of the three-dimensional Navier–Stokes system with a distributed control having two vanishing components

Abstract

In this paper, we prove a local null controllability result for the three-dimensional Navier–Stokes equations on a (smooth) bounded domain of ${\mathbb R}^3$ with null Dirichlet boundary conditions. The control is distributed in an arbitrarily small nonempty open subset and has two vanishing components. Lions and Zuazua proved that the linearized system is not necessarily null controllable even if the control is distributed on the entire domain, hence the standard linearization method fails. We use the return method together with a new algebraic method inspired by the works of Gromov and previous results by Gueye.

Appendix A: Creation of the matrix \({L_0}\)

In this appendix, we explain how the matrix ${L_0}$ at point $\xi ^0$ (which represents all the differentiated equations of System (3.27) up to the order $19$) was created. The program is written in $C^{++}$, using the library uBLAS which is well-adapted to the manipulation of sparse matrices. It is a parallel openMP algorithm, using 8 cores. We are not going to give all the technical details but just explain rapidly the spirit of the algorithm. To simplify, we will assume that the following “black boxes” (that had to be created) are at our disposal:

• 1. An evaluation function ep which evaluates a polynomial (represented by a vector) at $\xi ^0$. This evaluation function can be created so that it can verify that $\xi ^0$ is not a root of the polynomial $P(0,.,.)$. (one just has to see if the evaluation is equal to 0 whereas the polynomial has nonzero coefficients).

• 2. A derivation function deqex which differentiates an equation of level $m$ with respect to $x_1,x_2 ,x_3$ or $t$.

A partial differential equation which is a derivative of order $m$ of some of the equations of (3.27) will be represented in a matricial form in the following way: We know that there are at most $F(m+3)$ derivatives appearing, and we observe that the coefficients are polynomials in $(x_1,x_2,x_3)$ of an order less than 4 (it is a vector space of dimension $35$). Hence an equation of order $m$ is represented by a matrix with $F(m+3)$ lines and $35$ columns, where on each line one can find the coefficient of the partial derivatives of $z^1$ (or $z^2$ appearing) corresponding to the number of this line, thanks to the natural bijection between ${\mathbb N}^4$ and ${\mathbb N}$. Since we have three equations in (3.27) and two unknowns ($z^1$ and $z^2$), one can write the matrix $M$ in the following way:

$$\begin{aligned} \begin{pmatrix} A_1&{}\quad \! B_1\\ A_2&{}\quad \! B_2\\ A_3&{}\quad \! B_3 \end{pmatrix}. \end{aligned}$$

(4.8)

For $i=1,2,3, A_i$ represents the derivatives of $z^1$ appearing in the derivatives of the $i$-th equation of (3.27) and $B_i$ those of $z^2$. Hence, we can compute these $A_i$ and $B_i$ separately and then gather them to obtain ${L_0^0}$.

The algorithm is the following. We explain it for the first equation of (3.27) and for the unknown $z^1$ (i.e. for $A_1$, but it is the same for the other matrices).

• 1. We create a matrix $e$ that represents the equation. We use ep to fill the line of ${L_0^0}$ corresponding to the equation in a .txt file under the form $i$ $j$ $A_1(i,j)$. We create a matrix $h$ which is empty for the moment. In fact in $e$ we will keep the equations of level $m-1$ and in $h$ we will fill the equations of level $m$.

• 2. We create a “for” loop on $m$ which will represent the level of equations we are creating. The integer $m$ goes from 1 to $19$ since we differentiate $19$ times at most.

• 3. We create a second “for” loop in the interior of the first loop on a number $n$ which represents one of the equations of level $m$. Thanks to the definition of the function $F$ given in Sect. 3.2.2, we have $F(m-1)+1\leqslant n\leqslant F(m)$. If $m=1$, then $n$ goes from $F(0)+1=2$ to $F(1)=5$ ($n$ represents $\partial _1$, $\partial _2$, $\partial _3$ or $\partial _t$). If $m=2$, then $n$ goes from $F(1)+1=6$ to $F(2)=15$ ($n$ represents $\partial ^2_{11},\partial ^2_{12},\partial ^2_{13},\partial ^2_{1t},\partial ^2_{22},\partial ^2_{23},\partial ^2_{2t},\partial ^2_{33},\partial ^2_{3t}$ or $\partial ^2_{tt}$), etc. This loop is parallelized on our 8 cores. In this loop, we want to create the $n$-th equation denoted $E_n$, which is of level $m$. Hence we take a suitable equation of level $m-1$ denoted $E_r$ which is so that if we differentiate $E_r$ with respect to $1,2,3$ or $t$, we obtain $E_n$. For example, if we consider $m=2$ and if we want to obtain the first equation of (3.27) differentiated two times with respect to 1, then we consider the equation $E_r$ to be the first equation of (3.27) differentiated one time with respect to 1 and differentiated with respect to 1 to obtain $E_n$.

• 4. Once the loop on $n$ is ended, we have in our matrix $e$ all the equations of level $m-1$ and in $h$ we have just created all the equations of level $m$. Now we just have to use our evaluation function ep on $h$ to obtain the coefficients of the lines of $A_1$ corresponding to the equations that are of level $m$, i.e. the equations numbered from $F(m-1)+1$ to $F(m)$. We write these coefficients in our .txt file under the form $i$ $j$ $A_1(i,j)$.

• 5. We update now $e$, take $e=h$, we empty $h$ and we can go to the following loop $m+1$.

At the end we have created a file containing the coefficients of a sparse matrix $A_1$ of size $(8855,14950)$. Using the same program with $z^2$ and the two others equation we obtain five other files representing five matrices that we gather as in (4.8) to obtain the matrix ${L_0}(\xi ^0)={L_0^0}$. Our matrix ${L_0^0}$, which represents all the equations, is of size $(30360,29900)$ and has 651128 nonzero coefficients. Only 0.07 % of the coefficients are different from 0, with an average of 21.44 nonzero coefficients on each row, which is logical since we are working with coefficients that are polynomials of small degree, so we do not create many terms on each line when we differentiate the equations. In the above figure, one can observe how the nonzero coefficients of ${L_0^0}$ are distributed (Fig. 3).

Fig. 3

Distribution of the nonzero coefficients of ${L_0^0}$