# One-dimensional first-order linear hyperbolic systems and their controllability properties

Guillaume Olive

<figure><img src="image_1.png" /></figure>

## To cite this version:

Guillaume Olive. One-dimensional first-order linear hyperbolic systems and their controllability properties. Master. China. 2025. hal-05088994

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Submitted on 28 May 2025

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One-dimensional first-order linear hyperbolic systems
and their controllability properties

Guillaume Olive<fnref n="1" />

May 28, 2025

<footnote n="1">¹Faculty of Mathematics and Computer Science, Jagiellonian University, ul. Łojasiewicza 6, 30-348 Kraków,
Poland. E-mail: math.golive@gmail.com</footnote>
<figure><img src="image_3.png" /></figure>
# Preface

The goal of these lecture notes is to provide accessible material on the analysis of one-dimensional (1D) first-order linear hyperbolic systems of the general form (see the chapters below for the notations)

$$ 
\left\{ 
\begin{array}{ll}
\mathbf{Equ} : & \displaystyle\frac{\partial y}{\partial t}(t, x) + \Lambda(t, x) \frac{\partial y}{\partial x}(t, x) = M(t, x)y(t, x) + f(t, x), \\
\mathbf{IC} : & y(0, x) = y^0(x), \qquad (t, x) \in \mathbb{R}_T, \\
\mathbf{BC} : & \left( \begin{pmatrix} y_{-}(t, 1) \\ y_{+}(t, 0) \end{pmatrix} \right) = Q(t) \left( \begin{pmatrix} y_{-}(t, 0) \\ y_{+}(t, 1) \end{pmatrix} \right) + g(t),
\end{array}
\right. 
$$

and on some control results in the special case where $f = 0$ and the boundary condition takes the form

$$ \begin{pmatrix} y_{-}(t, 1) \\ y_{+}(t, 0) \end{pmatrix} = \begin{pmatrix} u(t) \\ Qy_{-}(t, 0) \end{pmatrix}, $$

in which $u$ is the so-called control.

These notes are also supplemented by several exercises of various difficulty to tickle the readers' curiosity.

They are intended to be expanded in the future.

Any questions or comments can be addressed to: math.golive@gmail.com

**Acknowledgements.** The author thanks Long Hu for proofreading these notes.

<figure><img src="image_4.png" /></figure>
<figure><img src="image_5.png" /></figure>
# Contents

<table><tbody><tr><td>1</td><td>Warm-up: the transport equation</td><td>9</td></tr><tr><td>1.1</td><td>The initial-boundary value problem</td><td>9</td></tr><tr><td>1.2</td><td>The control problem</td><td>11</td></tr><tr><td>1.3</td><td>Some exercises</td><td>14</td></tr><tr><td>2</td><td>Control of linear homogeneous systems</td><td>15</td></tr><tr><td>2.1</td><td>General notations and assumptions</td><td>15</td></tr><tr><td>2.2</td><td>The control problem and first properties</td><td>16</td></tr><tr><td>2.3</td><td>Control of some toy models</td><td>18</td></tr><tr><td>2.3.1</td><td>Control of 1 + 1 systems</td><td>18</td></tr><tr><td>2.3.2</td><td>Control of 2 + 2 systems</td><td>19</td></tr><tr><td>2.4</td><td>Rank revealing minimal canonical form</td><td>21</td></tr><tr><td>2.5</td><td>The minimal control time</td><td>24</td></tr><tr><td>2.6</td><td>Canonical decomposition of boundary coupling matrices</td><td>26</td></tr><tr><td>2.7</td><td>Equivalent systems</td><td>28</td></tr><tr><td>2.8</td><td>Some exercises</td><td>31</td></tr><tr><td>2.9</td><td>Bibliographical notes</td><td>35</td></tr><tr><td>3</td><td>Well-posedness of linear systems</td><td>37</td></tr><tr><td>3.1</td><td>The characteristics</td><td>37</td></tr><tr><td>3.2</td><td>Solution along the characteristics</td><td>38</td></tr><tr><td>3.3</td><td>Continuous solutions</td><td>40</td></tr><tr><td>3.3.1</td><td>A priori estimate</td><td>41</td></tr><tr><td>3.3.2</td><td>Existence</td><td>42</td></tr><tr><td>3.3.3</td><td>Reduction to small boundary coupling matrix</td><td>43</td></tr><tr><td>3.4</td><td>Classical solutions</td><td>44</td></tr><tr><td>3.4.1</td><td>A priori estimates</td><td>45</td></tr><tr><td>3.4.2</td><td>Existence</td><td>47</td></tr><tr><td>3.4.3</td><td>Systems satisfied by the derivatives</td><td>50</td></tr><tr><td>3.5</td><td>Smooth solutions</td><td>52</td></tr><tr><td>3.6</td><td>Some exercises</td><td>55</td></tr><tr><td>3.7</td><td>Bibliographical notes</td><td>55</td></tr><tr><td>4</td><td>A constructive method for the control of linear nonhomogeneous systems</td><td>57</td></tr><tr><td>4.1</td><td>The control problem</td><td>57</td></tr><tr><td>4.2</td><td>The constructive method</td><td>58</td></tr><tr><td>4.3</td><td>A uniqueness result in triangular-like domains</td><td>61</td></tr><tr><td>4.4</td><td>Some exercises</td><td>63</td></tr></tbody></table>

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4.5 Bibliographical notes .................................................... 64

**A Some properties of the characteristics** 65

A.1 Identities .................................................................... 65

A.1.1 Identities for $\chi$ .................................................... 65

A.1.2 Identities for $s^{\text{in}}$ ............................................ 66

A.1.3 Time-independent speed ............................................ 66

A.2 Estimates .................................................................... 67

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# Notation

ODE Ordinary Differential Equation

PDE Partial Differential Equation

**Equ** Equation

**IC** Initial Condition

**BC** Boundary Condition

$$R_T = (0, T) \times (0, 1)$$

$M_k$ For a matrix $M$, denotes its $k$-th row

$$\|y\|_{\mathbb{R}^n} = \max_{1 \le k \le n} |y_k|$$

$\bar{U}$ the closure of the set $U$

$$\|y\|_{C^0([0,1])^n} = \sup_{x \in [0,1]} \|y(x)\|_{\mathbb{R}^n}$$

$$\|y\|_{C^1([0,1])^n} = \|y\|_{C^0([0,1])^n} + \|y'\|_{C^0([0,1])^n}$$

$$\|y\|_{C^1(\bar{U})^n} = \|y\|_{C^0(\bar{U})^n} + \left\| \frac{\partial y}{\partial t} \right\|_{C^0(\bar{U})^n} + \left\| \frac{\partial y}{\partial x} \right\|_{C^0(\bar{U})^n}, U \subset \mathbb{R}^2$$

$y(t, \cdot)$ denotes the function $x \mapsto y(t, x)$.

$C^{0,1}(I)$ The Hölder space with exponent 1, i.e. the Lipschitz continuous functions.

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# Chapter 1

# Warm-up: the transport equation

We start these notes with a presentation of the situation for a single equation, the transport equation:

$$ \frac{\partial y}{\partial t}(t, x) + \lambda(x) \frac{\partial y}{\partial x}(t, x) = 0. $$

## 1.1 The initial-boundary value problem

The function $\lambda: \mathbb{R} \to \mathbb{R}$ is called the speed of the equation, it will be assumed to be Lipschitz (denoted below by $\lambda \in C^{0,1}$) and smooth enough. The main assumption is that $\lambda$ does not change sign. Without loss of generality,

$$ \lambda > 0. $$

We will consider the above equation in the rectangular domain $R_T$ defined by (see Figure 1.1)

$$ R_T = (0, T) \times (0, 1), \quad T > 0. $$

We recall the fundamental property of the transport equation:

$$ y \text{ is constant along the characteristic } s \mapsto (s, \chi(s; t, x)), \qquad (1.1) $$

(as long as it remains in the domain), where $\chi$ is the flow associated with $\lambda$:

$$ \begin{cases} \frac{d}{ds}\chi(s; t, x) = \lambda(\chi(s; t, x)), \\ \chi(t; t, x) = x. \end{cases} $$

Indeed, if $y$ is $C^1$, then

$$ \frac{d}{ds}y(s, \chi(s; t, x)) = \frac{\partial y}{\partial t}(s, \chi(s; t, x)) + \lambda(\chi(s; t, x)) \frac{\partial y}{\partial x}(s, \chi(s; t, x)) = 0. $$

Actually, since $\lambda$ does not depend on the time parameter $t$, the characteristic is in fact explicitly given in function of

$$ \phi(x) = \int_{0}^{x} \frac{1}{\lambda(\xi)} d\xi. $$

More precisely, noting that $\phi(\mathbb{R}) = \mathbb{R}$, we have the explicit formula (see e.g. Appendix A.1.3 for details)

$$ \chi(s; t, x) = \phi^{-1}(s - t + \phi(x)), \qquad (1.2) $$

<figure><img src="image_10.png" /></figure>
<figure><img src="image_11.png" /><figcaption>Figure 1.1: Domain and characteristics.</figcaption></figure>

for all $s, t, x$. We shall make use of this explicit formula to easily verify some properties used below. This is just to simplify the presentation in this introductory chapter. We refer the interested readers to Chapter 3 to see how to avoid the use of this formula (in particular, how to handle a time and space dependent speed $\lambda = \lambda(t, x)$).

**Initial condition.** We consider an initial condition at time $t = 0$:

$$y(0, x) = y^0(x), \quad x \in [0, 1],$$

where $y^0$ is a given function. From the property (1.1), we see that the initial condition $y^0$ uniquely determines the solution $y$ in the closure of the region $V$ defined by (see Figure 1.1)

$$V = \{(t, x) \in \bar{R}_T \mid \chi(0; t, x) \in (0, 1)\}.$$

More precisely, for $(t, x) \in \bar{V}$, the characteristic $(s, \chi(s; t, x))$ stays in $\bar{V}$ for all $s \in [0, t]$ (one can use for instance (1.2)), and we have

$$y(t, x) = y(t, \chi(t; t, x)) = y(0, \chi(0; t, x)) = y^0(\chi(0; t, x)). \quad (1.3)$$

**Boundary condition.** Since $\lambda > 0$, on the other part of the domain, i.e. on

$$\bar{V}^c = \bar{R}_T \setminus \bar{V},$$

the characteristic always touches the boundary $x = 0$ and thus the solution $y$ is uniquely determined by its values there (see Figure 1.1):

$$y(t, 0) = u(t), \quad t > 0,$$

where $u$ is a given function. More precisely, introducing

$$s^{\text{in}}(t, x) = t - \phi(x), \quad (1.4)$$

(it is the solution to $\chi(s^{\text{in}}(t, x); t, x) = 0$) we see that, for $(t, x) \in \bar{V}^c$, the characteristic $(s, \chi(s; t, x))$ stays in $\bar{V}^c$ for all $s \in [s^{\text{in}}(t, x), t]$ (one can use for instance (1.2)), and we have

$$y(t, x) = y(t, \chi(t; t, x)) = y(s^{\text{in}}(t, x), \chi(s^{\text{in}}(t, x); t, x)) = y(s^{\text{in}}(t, x), 0) = u(s^{\text{in}}(t, x)). \quad (1.5)$$

<figure><img src="image_12.png" /></figure>
In summary, because of the sign assumption on the speed, we have to consider a boundary condition at $x = 0$. The initial-boundary value problem is thus

$$ egin{cases} 
\mathbf{Equ}: & \frac{\partial y}{\partial t}(t, x) + \lambda(x) \frac{\partial y}{\partial x}(t, x) = 0, \quad (t, x) \in \overline{R_T}, \\ 
\mathbf{IC}: & y(0, x) = y^0(x), \quad x \in [0, 1], \\ 
\mathbf{BC}: & y(t, 0) = u(t), \quad t \in (0, T]. 
\end{cases} 
\quad (1.6) $$

Moreover, the solution is explicit:

$$ y(t, x) = \begin{cases} y^0(\chi(0; t, x)) & \text{if } (t, x) \in \bar{V}, \\ u(s^{\text{in}}(t, x)) & \text{if } (t, x) \in \bar{V}^c. \end{cases} \quad (1.7) $$

Similarly, if $\lambda < 0$, then the appropriate boundary condition is $y(t, 1) = u(t)$, and $s^{\text{in}}(t, x)$ is the solution to $\chi(s^{\text{in}}(t, x); t, x) = 1$.

**Functional settings.** A classical solution to (1.6) is a function $y \in C^1(\overline{R_T})$ that satisfies the conditions in (1.6) for every $(t, x)$. It is clear that, for a classical solution to exist, $y^0$, $u$ should be $C^1$, $u$ and $u'$ should have a limit when $t \to 0$ (thus, $u \in C^1([0, T])$), and the following $C^1$ compatibility conditions have to hold:

$$ y^0(0) = u(0) \quad (1.8) $$

$$ -\lambda(0)(y^0)'(0) = u'(0). \quad (1.9) $$

If $\lambda, y^0 \in C^1([0, 1])$, $u \in C^1([0, T])$, and if the $C^1$ compatibility conditions (1.8)-(1.9) are satisfied, then there exists a unique classical solution (one can use for instance (1.2)).

There are many other interesting functional settings where (1.7) could also be considered as the solution, in a weaker sense. To name only a few:

* $C^0$ framework: If $\lambda \in C^{0,1}([0, 1])$, $y^0 \in C^0([0, 1])$ and $u \in C^0([0, T])$, and if the $C^0$ compatibility condition (1.8) are satisfied, then $y \in C^0(\overline{R_T})$ (one can use for instance (1.2)).

* $L^2$ framework: If $\lambda \in C^{0,1}([0, 1])$, $y^0 \in L^2(0, 1)$ and $u \in L^2(0, T)$, then $y \in L^2(\overline{R_T})$. In fact, in that case, we even have

$$ y \in C^0([0, T]; L^2(0, 1)), \quad y \in C^0([0, 1]; L^2(0, T)). $$

We recall that the meaning of the first regularity is that, for every $t \in [0, T]$, the function $y(t, \cdot): x \mapsto y(t, x)$ belongs to $L^2(0, 1)$, and then $t \mapsto y(t, \cdot)$ belongs to $C^0([0, T]; L^2(0, 1))$. The meaning of the second one is similar, by changing the roles of $t$ and $x$.

## 1.2 The control problem

The main property discussed in these notes is *controllability*. There are several notions of controllability. We will focus on exact and null controllability. We say that (1.6) is *exactly controllable* in time $T$ if, for every initial condition $y^0$ and every final condition $y^1$, there exists a function $u$ such that the corresponding solution to (1.6) satisfies

$$ y(T, \cdot) = y^1. $$

If this property is satisfied at least for $y^1 = 0$, then we say that (1.6) is *null controllable*. The function $u$ is called *the control*, it is up to us to prove that such a function exists, or not.

<figure><img src="image_13.png" /></figure>
<figure><img src="image_14.png" /><figcaption>Figure 1.2: Control of the transport equation.</figcaption></figure>

The above definitions are valid in any reasonable functional setting, some examples were given at the end of Section 1.1. If the situation is unclear, we shall precise whether we speak of controllability in $L^2, C^0$, etc. Of course, in case we work in classical spaces $(C^0, C^1$, etc.), the control is required to ensure the corresponding compatibility conditions. This also means that it is required to fulfill additional conditions at time $t = T$. For instance, if we work in the $C^0$ framework, then the control for the exact controllability is required to satisfy

$$u(0) = y^0(0), \quad u(T) = y^1(0).$$

Below, we discuss the two functional settings $L^2$ and $C^0$ to highlight some differences.

**Proposition 1.2.1.** Let

$$T_{\inf} = \int_{0}^{1} \frac{1}{\lambda(\xi)} d\xi.$$

1. The transport equation (1.6) is exactly controllable in any time $T > T_{\inf}$.

2. The transport equation (1.6) is not null controllable in any time $T < T_{\inf}$.

3. In time $T = T_{\inf}$, the transport equation (1.6) is exactly controllable in $L^2$, but it is not null controllable in $C^0$.

The quantity $T_{\inf}$ is called the *minimal control time* of the transport equation (1.6).

*Proof of Proposition 1.2.1.* We refer to Figure 1.2 to gain a better understanding of the situation.

1. Assume that $T > T_{\inf}$. Then, the vertical line $\{(T, x) \mid x \in [0, 1]\}$ is entirely contained in the region $\bar{V}^c$ (one can use for instance (1.2)). Using then (1.5), the controllability condition $y(T, x) = y^1(x)$ is equivalent to

$$u(s^{\text{in}}(T, x)) = y^1(x), \quad x \in [0, 1].$$

<figure><img src="image_15.png" /></figure>
We can invert this identity since (using for instance (1.4) and (1.2)):

$$s^{\text{in}}(T, x) = \tau \iff x = \chi(T; \tau, 0).$$

Thus, we obtain the equivalent and explicit condition

$$u(\tau) = y^1(\chi(T; \tau, 0)), \quad \tau \in [s^{\text{in}}(T, 1), T]. \qquad (1.10)$$

2. Assume that $T < T_{\text{inf}}$. Then, there exists $\varepsilon \in (0, 1)$ such that the line $\{(T, x) \mid x \in [1 - \varepsilon, 1]\}$ is entirely contained in the region $\bar{V}$ (one can use for instance (1.2)). Using then (1.3), the controllability condition $y(T, x) = y^1(x)$ is equivalent to

$$y^0(\chi(0; T, x)) = y^1(x), \quad x \in [1 - \varepsilon, 1].$$

This condition already shows that the equation cannot be exactly controllable, but more is true. We can invert this identity (using for instance (1.2)) and obtain the equivalent and explicit condition

$$y^0(\xi) = y^1(\chi(T; 0, \xi)), \quad \xi \in [\chi(0; T, 1 - \varepsilon), \chi(0; T, 1)].$$

This shows that we cannot even reach $y^1 = 0$.

3. Assume that $T = T_{\text{inf}}$.

* In $C^0$, the point $(T, 1)$ is still in the region $\bar{V}$ and we can argue as if in the previous case we had $\varepsilon = 0$.

* In $L^2$, the controllability condition $y(T, \cdot) = y^1$ has to be satisfied in $[0, 1]$, up to a set of measure zero. In particular, we can exclude the point $x = 1$. On the other hand, by assumption, the vertical line $\{(T, x) \mid x \in [0, 1]\}$ is still entirely contained in the region $\bar{V}^c$ and we can argue as in the first case of the proof.

□

**Additional remarks.** We conclude this section with some observations.

* If $\lambda < 0$, then the appropriate boundary condition is $y(t, 1) = u(t)$, and the minimal control time is

$$T_{\text{inf}} = \int_{0}^{1} \frac{1}{|\lambda(\xi)|} d\xi.$$

* The control $u$ can be chosen with at least the same regularity as the final condition $y^1$ (assuming that $\lambda$ is smooth enough). In that regard, controllability in $L^2$ essentially implies controllability in $C^0$.

* $u = 0$ is a natural control to reach $y^1 = 0$ in the $L^2$ framework (but not in the $C^0$ framework).

* At the minimal control time $T = T_{\text{inf}}$, the control $u$ is unique (in the $L^2$ framework).

<figure><img src="image_16.png" /></figure>
1.3 Some exercises

**Exercise 1.3.1.** Show that $L^2$ solutions of the transport equation are limit of classical solutions (assuming that the speed is $C^1$).

**Exercise 1.3.2 (Time-dependent speed).** Consider the transport equation with time-dependent speed

$$
\begin{cases}
\frac{\partial y}{\partial t}(t, x) + e^{-t} \frac{\partial y}{\partial x}(t, x) = 0, & (t, x) \in \overline{R}_T, \\
y(0, x) = y^0(x), & x \in [0, 1], \\
y(t, 0) = u(t), & t \in (0, T].
\end{cases}
$$

Find its unique classical solution (assuming appropriate compatibility conditions). Show that it is never
controllable, whatever how large T is. For t ≥ 0 and x ∈ [0, 1], define

$$
(s^{\text{in}})^{+}(t, x) = \begin{cases} 0 & \text{if } \chi(0; t, x) \in [0, 1], \\ s^{\text{in}}(t, x) & \text{if } \chi(0; t, x) \notin [0, 1], \end{cases}
$$

where $s^{\text{in}}(t, x)$ is the unique solution $s$ to $\chi(s; t, x) = 0$. Show that

$$
\lim_{t \to +\infty} t - (s^{\text{in}})^{+}(t, x) = +\infty, \quad \forall x \in [0, 1].
$$

<figure><img src="image_17.png" /></figure>
# Chapter 2

## Control of linear homogeneous systems

*Prerequisites: Transport equation (Chapter 1), Gaussian elimination.*

In this chapter, we will solve the controllability problem for systems of the following form:

$$ egin{cases} 
\mathbf{Equ}: & \frac{\partial y}{\partial t}(t, x) + \Lambda(x) \frac{\partial y}{\partial x}(t, x) = 0, \\ 
\mathbf{IC}: & y(0, x) = y^0(x), \\ 
\mathbf{BC}: & y_-(t, 1) = u(t), \quad y_+(t, 0) = Qy_-(t, 0), 
\end{cases} 
\quad (t, x) \in \mathcal{R}_T. \quad (2.1) $$

The notations will be precised shortly.

### 2.1 General notations and assumptions

As usual in these notes, we work in the rectangle

$$R_T = (0, T) \times (0, 1).$$

In this chapter and the following ones, we now deal with systems of equations. The state $y: R_T \to \mathbb{R}^n$ is thus a vector-valued function with $n \ge 2$.

The $n \times n$ matrix $\Lambda$ will always be assumed diagonal:

$$\Lambda = \text{diag}(\lambda_1, \dots, \lambda_n),$$

with $m \ge 1$ negative speeds and $p \ge 1$ positive speeds ($m + p = n$):

$$\lambda_1, \dots, \lambda_m < 0 < \lambda_{m+1}, \dots, \lambda_{m+p}.$$

Two positive (or negative) speeds may have common values, or even cross, and the system is therefore not necessarily *strictly hyperbolic*. The speeds are assumed to be Lipschitz continuous:

$$\lambda_1, \dots, \lambda_n \in C^{0,1}([0, 1]).$$

<figure><img src="image_18.png" /></figure>
For each $k$, extending the speed $\lambda_k$ to a function of $C^{0,1}(\mathbb{R})$ if need be, we can define the associated flow $\chi_k$ globally, i.e. for every $(t, x) \in \mathbb{R}^2$, there is a unique function $s \mapsto \chi_k(s; t, x)$ solution to the ODE

$$ \left\{ \begin{array}{l} \frac{d}{ds} \chi_k(s; t, x) = \lambda_k(\chi_k(s; t, x)), \quad \forall s \in \mathbb{R}, \\ \\ \chi_k(t; t, x) = x. \end{array} \right. \qquad (2.2) $$

It is actually given by the explicit formula (1.2).

The structure of $\Lambda$ induces a natural splitting of the state $y$ into components corresponding to negative and positive speeds:

$$ y = \begin{pmatrix} y_- \\ y_+ \end{pmatrix}, \quad y_- \in \mathbb{R}^m, y_+ \in \mathbb{R}^p. $$

Based on what happens for the transport equation (see Chapter 1), for the system to be well-posed we need to consider boundary conditions at $x = 1$ for $y_-$ and at $x = 0$ for $y_+$. The function $u$ is the control. We will also speak of “the controls” when referring to its components $u_1, \dots, u_m$. We have $m$ controls for $n$ equations so that we have less controls than equation. Such systems are called *underactuated*. Since the control acts only on one side of the boundary, we speak of *one-sided controllability*. The system is coupled on the other part of the boundary by the $p \times m$ matrix $Q$. This matrix will be called the *boundary coupling matrix*. We shall refer to systems of the form (2.1) as $m + p$ systems.

**Functional setting.** Several functional settings could be considered for the study of the control problem of system (2.1) (see Chapter 1). All along this chapter we choose to work in $L^2$, that is we assume that

$$ y^0 \in L^2(0, 1)^n, \quad u \in L^2(0, T)^m. $$

The adaption of the results to other functional frameworks (e.g. $C^1$) is left as a (simple) exercise in Section 2.8. In any case, the solution to system (2.1) is explicit and has regularity

$$ y \in C^0([0, T]; L^2(0, 1)^n), \quad y \in C^0([0, 1]; L^2(0, T)^n). $$

Indeed, we can first “solve” the subsystem corresponding to $y_-$ and then “solve” the subsystem corresponding to $y_+$ since $y_-(t, 0)$ becomes a known quantity. We put the word solve in quotes since in this framework it is basically a definition.

## 2.2 The control problem and first properties

Since $y(t, \cdot) \in L^2(0, 1)^n$ for every $t \in [0, T]$, we can consider control problems in the state space $L^2(0, 1)^n$.

**Definition 2.2.1.** Let $T > 0$ be fixed. We say that system (2.1) is:

* exactly controllable in time $T$ (below, (EC) in time $T$) if, for every $y^0, y^1 \in L^2(0, 1)^n$, there exists a control $u \in L^2(0, T)^m$ such that the corresponding solution $y$ to the system (2.1) satisfies $y(T, \cdot) = y^1$.

* null controllable in time $T$ (below, (NC) in time $T$) if the previous property holds at least for $y^1 = 0$.

Clearly, (EC) always implies (NC). In general, the converse implication does not hold. Nevertheless, for our system (2.1), we have equivalence if the boundary coupling matrix is full row rank:

**Proposition 2.2.2.** Assume that rank $Q = p$. Then, for any $T > 0$, if system (2.1) is (NC) in time $T$, then it is (EC) in time $T$.

<figure><img src="image_19.png" /></figure>
**Proof.** Let $y^0, y^1$ be fixed. Since rank $Q = p$, there exists $R \in \mathbb{R}^{m \times p}$ such that $QR = \text{Id}_p$. Let $y^b$ be the solution to the “backward in time” system

$$ 
\left\{ 
\begin{array}{ll}
\mathbf{Equ} : & \displaystyle \frac{\partial y^b}{\partial t}(t, x) + \Lambda(x) \frac{\partial y^b}{\partial x}(t, x) = 0, \\
\mathbf{FC} : & y^b(T, x) = y^1(x), \\
\mathbf{BC} : & y^b(t, 0) = R y^b_+(t, 0), \quad y^b_+(t, 1) = u^b(t),
\end{array}
\right. 
\quad (t, x) \in R_T,
$$

where $u^b$ is arbitrary (this system is well-posed: it can be put in a similar form to (2.1) after using the changes of variables $t \mapsto T-t$ and $x \mapsto 1-x$). Let then $y^f$ be a null controlled solution to the “forward in time” system

$$ 
\left\{ 
\begin{array}{ll}
\mathbf{Equ} : & \displaystyle \frac{\partial y^f}{\partial t}(t, x) + \Lambda(x) \frac{\partial y^f}{\partial x}(t, x) = 0, \\
\mathbf{IC \& FC} : & y^f(0, x) = y^0(x) - y^b(0, x), \quad y^f(T, x) = 0, \\
\mathbf{BC} : & y^f(t, 1) = u^f(t), \quad y^f_+(t, 0) = Q y^f_-(t, 0),
\end{array}
\right. 
\quad (t, x) \in R_T.
$$

Define $u(t) = y^b_-(t, 1) + u^f(t)$ (note that $u \in L^2$ since $y^b \in C^0([0, 1]; L^2(0, T)^n)$) and let $y$ be the corresponding solution to system (2.1). By uniqueness, we have $y = y^b + y^f$. In particular, $y(T, \cdot) = y^1$. $\square$

In case the previous rank condition fails, we cannot hope to have (EC).

**Proposition 2.2.3.** If rank $Q < p$, then system (2.1) is never (EC).

**Proof.** By assumption, there exists a nonzero $\eta \in \mathbb{R}^p$ such that $\eta^T Q = 0$. Thus,

$$ \eta^T y_+(t, 0) = 0, \quad t \in (0, T). \qquad (2.3) $$

Let $k$ be such that $\eta_k \neq 0$. Consider the constant final condition $y_+^1 = e_k$ (the $k$-th canonical vector of $\mathbb{R}^p$) and an arbitrary $y_-^1$. If the system is (EC) in time $T$, then we must have $y_+(t, 0) = e_k$ for $t \in (T - \varepsilon, T)$, for some $\varepsilon > 0$ small enough. But this is of course incompatible with (2.3). $\square$

It follows from the two previous propositions that it is enough to study the (NC) property.

It is not difficult to check that controllability (exact or null) in a time $T_1$ implies controllability in any time $T_2 \ge T_1$. Therefore, it is natural to try to find the smallest possible control time, the so-called *minimal control time*.

**Definition 2.2.4.** Let $T_{\inf} \in [0, +\infty]$ be the minimal time for the (NC) of system (2.1), that is

$$ T_{\inf} = \inf \{ T > 0 \mid (2.1) \text{ is (NC) in time } T \}. \qquad (2.4) $$

We have:

* For every $T > T_{\inf}$, system (2.1) is (NC) in time $T$.

* For every $T < T_{\inf}$, system (2.1) is not (NC) in time $T$.

At the time $T = T_{\inf}$, we cannot conclude from the definition, but we will see that system (2.1) is (NC) in time $T_{\inf}$ (so that the infimum in (2.4) is always minimum for system (2.1)).

It is expected that the minimal control times of the transport equations contained in the system should play an important role in its controllability properties. Therefore, we give it a notation.

<figure><img src="image_20.png" /></figure>
**Definition 2.2.5.** For any $1 \le k \le n$, we denote by $T_k$ the minimal control time of the transport equation with speed $\lambda_k$. Explicitly, since $\Lambda$ only depends on $x$,

$$ T_k = \int_0^1 \frac{1}{|\lambda_k(\xi)|} d\xi, \qquad (2.5) $$

(see Chapter 1).

We also introduce

$$ T_{-} = \max_{k \le m} T_{k}, \quad T_{+} = \max_{l \ge 1} T_{m+l}. $$

We have the following basic remarks:

* The controllability of system (2.1) cannot occur before we have controlled all the transport equations it contains. In other words, $T \ge \max\{T_{-}, T_{+}\}$ is a necessary condition for the (NC) in time $T$ of system (2.1).

* System (2.1) is (NC) in time $T_{-} + T_{+}$. Indeed, taking simply $u = 0$, we have $y_{-}(t, \cdot) = 0$ for $t \ge T_{-}$. After that time, the boundary condition for $y_{+}$ becomes $y_{+}(t, 0) = 0$ and it follows that $y_{+}(t, \cdot) = 0$ for $t \ge T_{-} + T_{+}$.

Therefore, the minimal control time of system (2.1) lies between the max and the sum:

$$ \max\{T_{-}, T_{+}\} \le T_{\inf} \le T_{-} + T_{+}. \qquad (2.6) $$

The question remains: what is the exact value of the minimal control time? The answer depends on $Q$, and we cannot always take $u = 0$ to control in the best time.

## 2.3 Control of some toy models

To understand what happens in general, we propose to have a look at some systems of small size.

We recall that the solution to system (2.1) is explicit. However, similar to what we did for the transport equation (Chapter 1), it is largely preferable to use the geometric meaning of the equations.

### 2.3.1 Control of 1 + 1 systems

The simplest system is obviously a system coupling only two equations. Given the speeds

$$ \lambda_1 < 0 < \lambda_2, $$

we consider the 1 + 1 system

$$ \left\{ \begin{array}{ll} \mathbf{Equ}: & \displaystyle \frac{\partial y_k}{\partial t} + \lambda_k(x) \frac{\partial y_k}{\partial x} = 0, \quad k = 1, 2, \\ \mathbf{IC}: & y_k(0, x) = y_k^0(x), \quad k = 1, 2, \\ \mathbf{BC}: & y_1(t, 1) = u(t), \quad y_2(t, 0) = y_1(t, 0), \end{array} \right. \qquad (t, x) \in \mathcal{R}_T. \qquad (2.7) $$

We already know that this system is (NC) in time $T_1 + T_2 (= T_- + T_+)$ . A simple drawing makes it clear that the system cannot be controllable before that time (see Figure 2.1), i.e. $T_1 + T_2$ is the minimal control time of (2.7). Indeed, for $T \ge T_2$, we see that the null controllability condition $y_2(T, \cdot) = 0$ is

<figure><img src="image_21.png" /></figure>
<figure><img src="image_22.png" /><figcaption>Figure 2.1: Why $T_1 + T_2$ is the minimal control time.</figcaption></figure>

equivalent to the condition $y_2(t, 0) = 0$ for $t \in (T - T_2, T)$ (see (1.10), noting that $s_k^{\text{in}}(T, 1) = T - T_k$). Since $y_2 = y_1$ on this part of the boundary, this is thus also equivalent to

$$y_1(t, 0) = 0, \quad t \in (T - T_2, T).$$

If $\max\{T_1, T_2\} \le T < T_1 + T_2$, then in the nonempty interval $(T - T_2, T_1)$ the previous condition gives

$$y_1^0(\chi_1(0; t, 0)) = 0, \quad t \in (T - T_2, T_1),$$

since $t < T_1$ implies that $(0 <) \chi_1(0; t, 0) < 1$ (see e.g. (1.2)). Finally, this condition is also equivalent to

$$y_1^0(x) = 0, \quad x \in (\chi_1(0; T - T_2, 0), 1).$$

This shows that the initial condition cannot be arbitrary.

### 2.3.2 Control of 2 + 2 systems

The situation in the previous section is a bit too trivial because the boundary coupling “matrix” was just a scalar. To really understand how the information is transmitted by the boundary coupling matrix, we need to consider larger systems. Given the speeds

$$\lambda_1, \lambda_2 < 0 < \lambda_3, \lambda_4,$$

we consider the 2 + 2 system

$$\left\{
\begin{array}{ll}
\text{Equ:} & \displaystyle \frac{\partial y_k}{\partial t} + \lambda_k(x) \frac{\partial y_k}{\partial x} = 0, \quad k = 1, 2, 3, 4, \\
\text{IC:} & y_k(0, x) = y_k^0(x), \quad k = 1, 2, 3, 4, \\
\text{BC:} & \left\{
\begin{array}{ll}
y_1(t, 1) = u_1(t), & \displaystyle \left\{ \begin{array}{l} y_3(t, 0) = ay_1(t, 0) + by_2(t, 0), \\ y_4(t, 0) = cy_1(t, 0) + dy_2(t, 0), \end{array} \right. \\
y_2(t, 1) = u_2(t),
\end{array}
\right.
\quad (t, x) \in R_T,
\end{array}
\right.
\quad (2.8)$$

where $a, b, c, d \in \mathbb{R}$.

We first recall that, taking $u_1 = u_2 = 0$ yields that system (2.8) is (NC), but only in time $T_- + T_+ = \max\{T_1, T_2\} + \max\{T_3, T_4\}$. We would like to see if a smaller control time is possible with a better choice of controls.

We will assume that $T \ge \max_k T_k$, otherwise the system is not controllable (recall (2.6)). Then, the null controllability condition $y_-(T, \cdot) = 0$ is equivalent to

$$u_k = 0 \quad \text{in} \ (T - T_k, T), \quad k = 1, 2,$$

<figure><img src="image_23.png" /></figure>
and the null controllability condition $y_+(T, \cdot) = 0$ is equivalent to

$$y_k(t, 0) = 0 \quad t \in (T - T_k, T), \quad k = 3, 4.$$

Introducing $v_k(t) = y_k(t, 0)$ and using the boundary condition in (2.8), this second pair of conditions is equivalent to

$$\begin{cases} av_1 + bv_2 = 0 & \text{in } (T - T_3, T), \\ cv_1 + dv_2 = 0 & \text{in } (T - T_4, T). \end{cases} \qquad (2.9)$$

The values of $v_k$ are determined by the initial condition in $(0, T_k)$ and free to choose in any sub-interval of $(T_k, T)$ ($k = 1, 2$).

Now we need to know which interval from $(T - T_3, T)$ and $(T - T_4, T)$ is the largest one. Indeed, knowing the values of $v_1$ or $v_2$ on the largest interval will necessarily fix its values on the other interval. We will assume for instance that<fnref n="1" />

$$T_3 \geq T_4,$$

so that $(T - T_3, T)$ is the largest interval. We thus start by solving the first equation in (2.9).

Now to do so, we also need to know if we try to solve the equation with $v_1$ or $v_2$. This is also dictated by the order between the times $T_1$ and $T_2$. We will assume for instance that (see again footnote 1)

$$T_1 \leq T_2.$$

Then, we first observe that, if $T - T_3 < T_1$, then also $T - T_3 < T_2$ and thus $av_1 + bv_2$ is in fact a function depending only on the initial condition. Therefore, it is necessary that $T - T_3 \geq T_1$ (unless $a = b = 0$). Now, for $T - T_3 \geq T_1$, we can solve the equation with $v_1$ (unless $a = 0$).

The rest of the story depends on the values of $a, b, c, d$.

1. Case $a = b = 0$. Then, we only have to take care of the second equation in (2.9).

(a) Case $c = d = 0$. There is nothing to do, and thus no additional condition on $T$.

(b) Case $c \neq 0$. Then, it is necessary that $T - T_4 \geq T_1$ and the second equation becomes

$$v_1 = -\frac{d}{c}v_2 \quad \text{in } (T - T_4, T).$$

(c) Case $c = 0$ and $d \neq 0$. Then, it is necessary that $T - T_4 \geq T_2$ and the second equation becomes

$$v_2 = 0 \quad \text{in } (T - T_4, T).$$

2. Case $a \neq 0$. Then, it is necessary that $T - T_3 \geq T_1$ and the first equation becomes

$$v_1 = -\frac{b}{a}v_2 \quad \text{in } (T - T_3, T). \qquad (2.10)$$

Since $(T - T_3, T)$ is larger than $(T - T_4, T)$, the second equation in (2.9) becomes

$$\left(-\frac{bc}{a} + d\right) v_2 = 0 \quad \text{in } (T - T_4, T).$$

Noting that $-\frac{bc}{a} + d = \frac{ad-bc}{a}$, we then have two cases.

<figure><img src="image_24.png" /></figure>

<footnote n="1">¹ The reason is also that we usually sort the speeds in increasing order if they do not cross: $\lambda_1 \le \dots \le \lambda_m < 0 < \lambda_{m+1} \le \dots \le \lambda_{m+p}$, which implies that $T_1 \le \dots \le T_m$ and $T_{m+1} \ge \dots \ge T_{m+p}$ (see e.g. (2.5)). Note that we can have for instance $T_1 \le T_2$ without having $\lambda_1 \le \lambda_2$.</footnote>
(i) Case $ad - bc = 0$. There is nothing to do, and thus no additional condition on $T$.

(ii) Case $ad - bc \neq 0$. Then, it is necessary that $T - T_4 \geq T_2$ and the second equation becomes

$$v_2 = 0 \quad \text{in } (T - T_4, T).$$

In case $T - T_4 > T_2$, we complete the remaining values of $v_2$ in $(T_2, T - T_4)$ by an arbitrary function. Then, $v_2$ is completely known in $(0, T)$ and it can thus be considered as a right-hand side in (2.10) to obtain $v_1$.

3. Case $a = 0$ and $b \neq 0$. Then, it is necessary that $T - T_3 \geq T_2$ i.e. $T \geq T_- + T_+$. We already know what happens in that case.

In summary, we have shown that, under the assumption

$$T_1 \leq T_2 \quad \text{and} \quad T_3 \geq T_4,$$

system (2.8) is (NC) in time $T$ if, and only if,

$$T \geq \max\{T_2, T_3, T_Q\},$$

where

$$T_Q = \begin{cases} 0 & \text{if } a = b = 0 \text{ and } c = d = 0, \\ T_4 + T_1 & \text{if } a = b = 0 \text{ and } c \neq 0, \\ T_4 + T_2 & \text{if } a = b = 0 \text{ and } c = 0 \text{ and } d \neq 0, \\ T_3 + T_1 & \text{if } a \neq 0 \text{ and } ad - bc = 0, \\ \max\{T_3 + T_1, T_4 + T_2\} & \text{if } a \neq 0 \text{ and } ad - bc \neq 0, \\ T_3 + T_2 & \text{if } a = 0 \text{ and } b \neq 0. \end{cases} \quad (2.11)$$

Moreover, we have seen that we cannot simply take the control to be zero to obtain the minimal control time (Cases 1b and 2).

What we have done in the proof above is nothing but the Gaussian elimination for the system $Qv = 0$ but without swapping rows (this point is crucial). The row we start with is dictated by the order $T_3 \geq T_4$ and the pivot we use is dictated by the order $T_1 \leq T_2$. Thus, we started with the first row and then we proceeded with a downward row substitution (see e.g. Case 2). In a matrix formulation, this corresponds to multiplying $Q$ on the left by a unit lower triangular matrix (we call a lower triangular matrix to be unit if its diagonal entries are all equal to 1). With such a Gaussian elimination process, there is a canonical form that can be obtained. This is discussed next and will be the key tool to find the minimal control time of system (2.1) in the general case.

## 2.4 Rank revealing minimal canonical form

**Definition 2.4.1.** A matrix $Q \in \mathbb{R}^{p \times m}$ is in rank revealing minimal canonical form (in short, (RRMCF)) if $Q = 0$ or it satisfies the following properties for some indices $(r_1, c_1), \dots, (r_\rho, c_\rho)$ with $r_1 < \dots < r_\rho$:

1. $q_{ij} = 0$ for every $i \notin \{r_1, \dots, r_\rho\}$ and every $j$.

2. $q_{r_k j} = 0$ for every $j < c_k$ and $q_{r_k c_k} \neq 0$.

3. $q_{ic_k} = 0$ for every $i > r_k$.

<figure><img src="image_25.png" /></figure>
**Example 2.4.2.** The following matrix is in (RRMCF):

$$ \begin{pmatrix} 0 & \boxed{1} & 2 \\ 0 & 0 & \boxed{1} \\ 0 & 0 & 0 \\ \boxed{4} & 0 & 0 \end{pmatrix} \quad \begin{array}{l} \rightarrow (r_1, c_1) = (1, 2), \\ \rightarrow (r_2, c_2) = (2, 3), \\ \rightarrow \times \textit{ not a } r_k, \\ \rightarrow (r_3, c_3) = (4, 1). \end{array} $$

We will show that any matrix can be put in a unique (RRMCF).

**Theorem 2.4.3.** For every $Q \in \mathbb{R}^{p \times m}$, there exists a unique $Q^0 \in \mathbb{R}^{p \times m}$ which is in (RRMCF) and such that

$$ LQ = Q^0, $$

for some unit lower triangular matrix $L \in \mathbb{R}^{p \times p}$. We call $Q^0$ the (RRMCF) of $Q$.

Let us make some comments before giving the proof.

First of all, with the above result, we can extend the definition of the indices $(r_1, c_1), \dots, (r_\rho, c_\rho)$ to any nonzero matrix:

**Definition 2.4.4.** For any nonzero matrix $Q \in \mathbb{R}^{p \times m}$, we denote by $(r_1, c_1), \dots, (r_\rho, c_\rho)$ the indices of its (RRMCF) $(r_1 < \dots < r_\rho)$.

The existence part in the theorem is just a consequence of the Gaussian elimination, where, as already mentioned at the end of the previous section, the matrix $L$ corresponds to downward row substitutions. Let us detail an example.

**Example 2.4.5.** We illustrate how to find the decomposition of Theorem 2.4.3 in practice. Consider

$$ Q = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 2 & 5 \\ 0 & 1 & 2 \\ 4 & -4 & 4 \end{pmatrix}. $$

We work on the rows with downward substitutions.

1. We look at the first nonzero row. We take the first nonzero entry as pivot and we remove the entries below, by doing the row substitutions $R_2 \leftarrow R_2 - 2R_1$, $R_3 \leftarrow R_3 - R_1$ and $R_4 \leftarrow R_4 + 4R_1$. This gives

$$ L_1 Q = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 4 & 0 & 0 & 1 \end{pmatrix} \quad Q = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 0 & 12 \end{pmatrix}. $$

2. We repeat the previous operation with the next nonzero row. We take the first nonzero entry as pivot and we remove the entries below, by doing the row substitution $R_4 \leftarrow R_4 - 12R_2$. This gives

$$ L_2 L_1 Q = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -12 & 0 & 1 \end{pmatrix} \quad L_1 Q = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 0 & 0 \end{pmatrix}. $$

In conclusion, $L = L_2 L_1$ and we see that the (RRMCF) of $Q$ is the matrix of Example 2.4.2.

<figure><img src="image_26.png" /></figure>
The uniqueness of the decomposition is less straightforward though, because there may be several L that produce the (RRMCF) of Q (note however that, as seen in the previous example, there is in fact a natural candidate L that leads to the (RRMCF) of Q).

*Proof of Theorem 2.4.3.* As explained above, the existence of $Q^0$ and L follows from the Gaussian elimination. Let us now prove the uniqueness of $Q^0$. Assume that we have two decompositions

$$LQ = Q^0, \quad \bar{L}Q = \bar{Q}^0,$$

for some unit lower triangular matrices L, $\bar{L}$ and some matrices $Q, \bar{Q}$ in (RRMCF). Then,

$$L'Q^0 = \bar{Q}^0,$$

with $L' = \bar{L}L^{-1}$ still unit lower triangular. We will prove by induction that $Q^0$ and $\bar{Q}^0$ have the same rows. Below, the k-th row of a matrix M is denoted by $M_k$.

Since $l'_{11} = 1$, we obviously have $Q_1^0 = \bar{Q}_1^0$.

Let us now assume that $Q_k^0 = \bar{Q}_k^0$ for every $k \le N$ for some $N \ge 1$, and let us prove that $Q_{N+1}^0 = \bar{Q}_{N+1}^0$. Since $L'$ is unit lower triangular, we have

$$\sum_{\alpha \le N} l'_{N+1, \alpha} Q_{\alpha}^0 + Q_{N+1}^0 = \bar{Q}_{N+1}^0.$$

We assume that $Q^0 \ne 0$, otherwise the result is trivial. The nonzero rows of $Q^0$ are then $Q_{r_1}^0, \dots, Q_{r_\rho}^0$. The previous identity is thus equivalent to

$$\sum_{k: r_k \le N} l'_{N+1, r_k} Q_{r_k}^0 + Q_{N+1}^0 = \bar{Q}_{N+1}^0.$$

We are going to show that all these $l'_{N+1, r_k}$ are necessarily zero. Let us sort in increasing order the column indices $c_k$ for which $r_k \le N$:

$$c_{k_1} < c_{k_2} < \dots < c_{k_s},$$

for some $s$. The previous identity becomes

$$\sum_{\beta=1}^{s} l'_{N+1, r_{k_{\beta}}} Q_{r_{k_{\beta}}}^{0} + Q_{N+1}^{0} = \bar{Q}_{N+1}^{0}. \qquad (2.12)$$

We first look at the $c_{k_1}$ entry of this identity. We observe that $q_{N+1, c_{k_1}}^0 = \bar{q}_{N+1, c_{k_1}}^0 = 0$ because of property 3 in Definition 2.4.1 (note that the indices $(r_k, c_k)$ with $r_k \le N$ are the same for $Q^0$ and $\bar{Q}^0$ since these matrices have the same first $N$ rows). We also observe that $q_{r_{k_\beta}, c_{k_1}}^0 = 0$ if $\beta \ne 1$ because of property 2 in Definition 2.4.1. As a result, only one term remains: $l'_{N+1, r_{k_1}} q_{r_{k_1}, c_{k_1}}^0 = 0$. Since $q_{r_{k_1}, c_{k_1}}^0 \ne 0$, we deduce that $l'_{N+1, r_{k_1}} = 0$. Identity (2.12) becomes

$$\sum_{\beta=2}^{s} l'_{N+1, r_{k_{\beta}}} Q_{r_{k_{\beta}}}^{0} + Q_{N+1}^{0} = \bar{Q}_{N+1}^{0}.$$

We can then repeating the previous reasoning to obtain that $l'_{N+1, r_{k_2}} = 0$. By induction, we easily see that $l'_{N+1, r_{k_2}} = 0$ for every $\beta$.

□

<figure><img src="image_27.png" /></figure>
## 2.5 The minimal control time

First of all, as we have seen with 2+2 systems in Section 2.3.2, it is important to know the order between the times $T_k$ to find the minimal control time. We will assume that (see again footnote 1)

$$T_1 \le \cdots \le T_m \quad \text{and} \quad T_{m+1} \ge \cdots \ge T_{m+p}. \qquad (2.13)$$

Of course this can always be achieved at the start by simply permuting some rows of system (2.1), if needed.

We now have all the tools to find the minimal control time of system (2.1).

**Theorem 2.5.1.** Assume (2.13). Then, system (2.1) is (NC) in time T if, and only if,

$$T \ge \max\{T_m, T_{m+1}, T_Q\}, \quad T_Q = \max_{k \le p} T_{m+r_k} + T_{c_k}, \qquad (2.14)$$

where the indices $(r_k, c_k)$ refer to $Q$ (see Section 2.4). In case $Q = 0$, we simply omit $T_Q$ in the formula.

For the exact controllability property, since $Q$ has to be full row rank (Proposition 2.2.3), and that $r_k = k$ for all $k$ in that case, we immediately deduce the following result.

**Corollary 2.5.2.** Assume (2.13). Then, system (2.1) is (EC) in time T if, and only if, rank $Q = p$ and

$$T \ge \max\{T_m, T_Q\}, \quad T_Q = \max_{k \le p} T_{m+k} + T_{c_k}, \qquad (2.15)$$

where the indices $c_k$ refer to $Q$ (see Section 2.4).

Let us mention some interesting points before moving to the proof of the theorem.

* If $c_{k_0} = m$ for some $k_0$, then we see that the minimal control time can only be

$$\max \left\{ T_{m+1}, \max_{k \le k_0} T_{m+r_k} + T_{c_k} \right\},$$

since the terms for $k \ge k_0 + 1$ give only smaller times, because of the ordering (2.13). This is for instance what happened in Example 2.4.5 where we found that $c_2 = 3 (= m)$ after a single computation. For the (NC) property we can thus stop the computations. However, pay attention that for the (EC) we still need to finish the computations to check the condition rank $Q = p$.

* There are some sets of boundary coupling matrices of particular importance. These are the sets

$$\mathcal{B}_l = \{ Q \in \mathbb{R}^{p \times m} \mid \det Q_{1:i} \neq 0, \forall i \le l \}, \qquad (2.16)$$

(1 ≤ $l$ ≤ min {$p, m$}), where $Q_{1:i}$ denotes the submatrix of $Q$ obtained by keeping its first $i$ rows and $i$ columns. Then, for any $Q \in \mathcal{B}_l$, we obviously have $r_k = c_k = k$ for all $k \le l$ and it follows from Theorem 2.5.1 and the ordering (2.13) that system (2.1) is (NC) in any time $T$ greater or equal than

$$\max \left\{ \max_{k \le l} T_{m+k} + T_k, \quad T_{m+l+1} + T_m \right\}, \qquad (2.17)$$

$(T_{m+l+1}$ is omitted when $l = p)$. The good point about the set $\mathcal{B}_l$ is that it is stable by small perturbations, and the corresponding control time (2.17) is the same for all boundary coupling matrices within this set. Note however that, in general, (2.17) is a larger control time than the minimal control time, unless $l = \min\{p, m\}$.

<figure><img src="image_28.png" /></figure>
Let us now prove the theorem.

*Proof of Theorem 2.5.1.* We assume $Q \neq 0$, otherwise the proof is trivial. We will assume that $T \ge \max_k T_k$, otherwise the system is not controllable (recall (2.6)). Then, the null controllability condition $y_-(T, \cdot) = 0$ is equivalent to

$$u_k = 0 \quad \text{in } (T - T_k, T), \quad \forall k \le m, \qquad (2.18)$$

and the null controllability condition $y_+(T, \cdot) = 0$ is equivalent to

$$y_{m+i}(t, 0) = 0, \quad t \in (T - T_{m+i}, T), \quad \forall i. \qquad (2.19)$$

Let $Q^0$ be the (RRMCF) of $Q$ and let $L \in \mathbb{R}^{p \times p}$ be a unit lower triangular matrix such that

$$LQ = Q^0.$$

Then, then boundary condition is equivalent to

$$Ly_+(t, 0) = Q^0 y_-(t, 0). \qquad (2.20)$$

By assumption (2.13), the time intervals where we have to solve (2.19) are decreasing:

$$(T - T_{m+1}, T) \supset (T - T_{m+2}, T) \supset \cdots \supset (T - T_{m+p}, T).$$

Since $L$ is an invertible lower triangular matrix, condition (2.19) is then equivalent to

$$L_i y_+(t, 0) = 0, \quad t \in (T - T_{m+i}, T), \quad \forall i,$$

where here, and in what follows, the $i$-th row of a matrix $M$ is denoted by $M_i$. Using the boundary condition (2.20), this is equivalent to

$$Q_i^0 y_-(t, 0) = 0, \quad t \in (T - T_{m+i}, T), \quad \forall i.$$

Using that $Q_i^0 = 0$ for $i \notin \{r_1, \dots, r_\rho\}$, this is also equivalent to

$$Q_{r_k}^0 y_-(t, 0) = 0, \quad t \in (T - T_{m+r_k}, T), \quad \forall k.$$

Let us now write this identity component-wise. Since $Q^0$ is in (RRMCF), using property 2 in Definition 2.4.1, the previous identity is equivalent to

$$y_{c_k}(t, 0) = -\frac{1}{q_{r_k c_k}^0} \sum_{j > c_k} q_{r_k j}^0 y_j(t, 0), \quad t \in (T - T_{m+r_k}, T), \quad \forall k. \qquad (2.21)$$

From this identity the first conclusion we draw is that it is necessary that

$$T - T_{m+r_k} \ge T_{c_k}. \qquad (2.22)$$

Indeed, assume that this is not the case. Then, in the nonempty interval $(T - T_{m+r_k}, T_{c_k})$, we must have

$$y_{c_k}^0(\chi_{c_k}(0; t, 0)) = -\frac{1}{q_{r_k c_k}^0} \sum_{j > c_k} q_{r_k j}^0 y_j^0(\chi_j(0; t, 0)),$$

since $T_{c_k} \le T_j$ for the indices $j$ in the sum by assumption (2.13). This shows that the initial condition $y^0$ cannot be arbitrary.

<figure><img src="image_29.png" /></figure>
Let us now assume that (2.22) holds for all $k$. Then, (2.21) is equivalent to

$$u_{c_k}(t) = -\frac{1}{q_{r_k c_k}^0} \sum_{j > c_k} q_{r_k j}^0 y_j(t + T_{c_k}, 0), \quad t \in (T - T_{m+r_k} - T_{c_k}, T - T_{c_k}), \quad \forall k.$$

This is easily solved as follows. We first fix arbitrary values in these intervals for the controls $u_i$ with $i \notin \{c_1, \dots, c_\rho\}$. Then, we successively fix the values of the remaining controls in decreasing order of indices. Finally, note that all of this is compatible with (2.18) since the time intervals where the values of the controls are prescribed are disjoint intervals.

<figure><img src="image_30.png" /></figure>

The control problem has been completely solved for system (2.1). In the remaining sections we wish to introduce additional concepts that bring complementary viewpoint on the control problem of such systems.

## 2.6 Canonical decomposition of boundary coupling matrices

We start with a key definition.

**Definition 2.6.1.** A boundary coupling matrix $Q \in \mathbb{R}^{p \times m}$ is in canonical form if it has at most one nonzero entry on each row and each column, and this entry is equal to 1. We denote by $(r_1, c_1), \dots, (r_\rho, c_\rho)$ the positions of the corresponding nonzero entries, with $r_1 < \dots < r_\rho$.

The choice of the same notations $(r_k, c_k)$ as in Definition 2.4.1 will be justified shortly.

**Example 2.6.2.** The following matrix is in canonical form:

$$\begin{pmatrix} 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ 0 & 0 & 0 \\ \boxed{1} & 0 & 0 \end{pmatrix} \quad \begin{aligned} & \rightarrow (r_1, c_1) = (1, 2), \\ & \rightarrow (r_2, c_2) = (2, 3), \\ & \rightarrow \times \text{ not a } r_k, \\ & \rightarrow (r_3, c_3) = (4, 1). \end{aligned}$$

Boundary coupling matrices in canonical form yield the simplest possible systems, and the minimal control time of such systems is trivially determined. Indeed, because of this very simple coupling structure, the system consists in uncoupled transport equations and $1+1$ systems. It is then clear that:

**Proposition 2.6.3.** Assume that $Q$ is in canonical form. Then, system (2.1) is

* (NC) in time $T$ if, and only if, $T$ satisfies (2.14). Moreover, $u = 0$ can be used as control.

* (EC) in time $T$ if, and only if, rank $Q = p$ (i.e. $Q$ has no zero rows) and $T$ satisfies (2.15).

We will show that any matrix can be put in a unique canonical form.

**Theorem 2.6.4.** For every $Q \in \mathbb{R}^{p \times m}$, there exists a unique $Q^c \in \mathbb{R}^{p \times m}$ which is in canonical form and such that

$$LQU = Q^c,$$

for some unit lower triangular matrix $L \in \mathbb{R}^{p \times p}$ and some invertible upper triangular matrix $U \in \mathbb{R}^{m \times m}$. We call $Q^c$ the canonical form of $Q$.

Let us make some comments before giving the proof.

First of all, with the above result, we can extend the definition of the indices $(r_1, c_1), \dots, (r_\rho, c_\rho)$ to any nonzero matrix:

<figure><img src="image_31.png" /></figure>
**Definition 2.6.5.** For any nonzero matrix $Q \in \mathbb{R}^{p \times m}$, we denote by $(r_1, c_1), \dots, (r_\rho, c_\rho)$ the positions of the nonzero entries of its canonical form $(r_1 < \dots < r_\rho)$.

The existence part in the theorem is just a consequence of the Gaussian elimination. In fact, we first find $L$ and the (RRMCF) of $Q$ (see Section 2.4). Then, the matrix $U$ corresponds to rightward column substitutions and then normalization to 1 of the remaining nonzero entries. It is clear that the indices $(r_k, c_k)$ in Definition 2.6.5 are thus the same indices as in Definition 2.4.4. Let us detail an example.

**Example 2.6.6.** We illustrate how to find the decomposition of Theorem 2.6.4 in practice. Consider the same matrix as in Example 2.4.5, namely

$$Q = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 2 & 5 \\ 0 & 1 & 2 \\ 4 & -4 & 4 \end{pmatrix}.$$

1. The first step is to find $L$ and the (RRMCF) of $Q$. This was done in Example 2.4.5:

$$Q^0 = LQ = \begin{pmatrix} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 0 & 0 \end{pmatrix}.$$

2. We now construct the matrix $U$. To this end, we work on the columns with rightward substitutions and then we normalize the remaining nonzero entries.

(a) We look at the first nonzero row. We take the first nonzero entry as pivot and we remove the entries to its right, by doing the column substitution $C_3 \leftarrow C_3 - 2C_2$. This gives

$$Q^0 U_1 = Q^0 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 4 & 0 & 0 \end{pmatrix}.$$

(b) Finally, we normalize the remaining nonzero entries to 1 by multiplying on the right by an appropriate invertible diagonal matrix:

$$Q^0 U_1 D = Q^0 U_1 \begin{pmatrix} \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}.$$

As a result, $U = U_1 D$.

In conclusion, we see that the canonical form of $Q$ is the matrix of Example 2.6.2.

The uniqueness of the decomposition is less straightforward though, because there may be several pairs $(L, U)$ that produce the canonical form of $Q$ (note however that, as seen in the previous example, there is in fact a natural pair $(L, U)$ that leads to the canonical form of $Q$).

**Proof of Theorem 2.6.4.** As explained above, the existence of $Q^c$ and $L, U$ follows from the Gaussian elimination. Let us now prove the uniqueness of $Q^c$. Assume that we have two decompositions

$$LQU = Q^c, \quad \bar{L}Q\bar{U} = \bar{Q}^c,$$

<figure><img src="image_32.png" /></figure>
for some unit lower triangular matrices $L, \bar{L}$, some invertible upper triangular matrices $U, \bar{U}$ and some matrices $Q^c, \bar{Q}^c$ in canonical form. Then,

$$L'Q^c = \bar{Q}^c U', \qquad (2.23)$$

with $L' = \bar{L}L^{-1}$ and $U' = \bar{U}^{-1}U$ still unit lower triangular and invertible upper triangular. We will prove by induction that $Q^c$ and $\bar{Q}^c$ have the same rows. Below, the k-th row of a matrix $M$ is denoted by $M_k$.

For the first row, we have

$$L'_{1}Q^{c} = \bar{Q}_{1}^{c}U'_{1}.$$

Since $Q^c, \bar{Q}^c$ are in canonical form, we have $Q_1^c = ae_\alpha^\top$ and $\bar{Q}_1^c = be_\beta^\top$ for some $a, b \in \{0, 1\}$ and some $\alpha, \beta$. Since $L$ is unit lower triangular and $U'$ is upper triangular, the previous identity becomes

$$ae_{\alpha}^{\top} = bu'_{\beta\beta}e_{\beta}^{\top} + b \sum_{j>\beta} u'_{\beta j}e_{j}^{\top}.$$

We discuss three cases.

* Case $\alpha < \beta$. Then, in particular, $a = bu'_{\beta\beta} = 0$. Since $u'_{\beta\beta} \neq 0$, we also have $b = 0$.

* Case $\alpha = \beta$. Then, in particular, $a = bu'_{\beta\beta}$. Since $a, b \in \{0, 1\}$ and $u'_{\beta\beta} \neq 0$, this implies that $a = b$.

* Case $\alpha > \beta$. Then, in particular, $a = bu'_{\beta\alpha}$ and $bu'_{\beta\beta} = 0$. Since $u'_{\beta\beta} \neq 0$, we have $b = 0$. It follows that $a = 0$ as well.

Let us now assume that $Q^c$ and $\bar{Q}^c$ have the same first $N$ rows, for some $N \ge 1$, and let us prove that they must also have the same $(N+1)$-th rows. From (2.23), we have

$$L'_{N+1}Q^c = \bar{Q}_{N+1}^c U'_{1}.$$

Since $Q^c, \bar{Q}^c$ are in canonical form and have the same first $N$ rows, we have $Q_i^c = 0$ for $i \notin \{r_1, \dots, r_\rho\}$, $Q_{r_k}^c = e_{c_k}^\top$, and $Q_{N+1}^c = ae_\alpha^\top$ and $\bar{Q}_{N+1}^c = be_\beta^\top$ for some $a, b \in \{0, 1\}$ and some $\alpha, \beta \notin \{c_k \mid r_k \le N\}$.

$$\sum_{k: r_k \le N} l'_{N+1, r_k} e_{c_k}^{\top} + ae_{\alpha}^{\top} = bu'_{\beta\beta}e_{\beta}^{\top} + b \sum_{j>\beta} u'_{\beta j}e_{j}^{\top}.$$

The conclusion is exactly as before, by discussing the position of $\alpha$ with respect to $\beta$.

□

## 2.7 Equivalent systems

The goal of this final section is to give another proof of Theorem 2.5.1 using the canonical form of the boundary coupling matrix introduced in the previous section. The main idea is to transform our system (2.1) into a system that is equivalent from a controllability point of view and that has a simpler coupling structure (i.e. a simpler $Q$). Let us make this notion precise here. In what follows, it will be convenient to refer to a system of the form (2.1) as

$$(\Lambda, Q).$$

We say that $y$ is a solution to $(\Lambda, Q)$ in $(0, T)$ if it is a solution to system (2.1) for some initial data $y^0$ and some control $u$.

<figure><img src="image_33.png" /></figure>
**Definition 2.7.1.** We say that two systems $(\Lambda, Q_1)$ and $(\Lambda, Q_2)$ are equivalent, and we write

$$(\Lambda, Q_1) \sim (\Lambda, Q_2),$$

if there exists an isomorphism<fnref n="2" />

$$L : L^2(0, 1)^n \to L^2(0, 1)^n,$$

such that, for every $T > 0$, the induced isomorphism $\tilde{L} : C^0([0, T]; L^2(0, 1)^n) \to C^0([0, T]; L^2(0, 1)^n)$ defined by $(\tilde{L}y)(t, \cdot) = L(y(t, \cdot))$ for every $t \in [0, T]$ satisfies

$$\tilde{L}(S_1) = S_2,$$

where $S_k$ ($k = 1, 2$) denotes the space of all the solutions to the system $(\Lambda, Q_k)$ in $(0, T)$.

It is easy to check that $\sim$ is an equivalence relation.

It is also not difficult to check that two equivalent systems share the same controllability properties.

**Proposition 2.7.2.** Assume that $(\Lambda, Q_1) \sim (\Lambda, Q_2)$. Then, for any $T > 0$, $(\Lambda, Q_1)$ is (NC) in time $T$ if, and only if, so is $(\Lambda, Q_2)$.

*Proof.* Assume for instance that $(\Lambda, Q_2)$ is (NC) in time $T$ and let us show that so is $(\Lambda, Q_1)$. Let then $y^0$ be an initial condition for $(\Lambda, Q_1)$. Define $\tilde{y}^0 = Ly^0$. By assumption $(\Lambda, Q_2)$ is (NC) in time $T$ and thus there exists a control $\tilde{u}$ such that the solution $\tilde{y}$ to $(\Lambda, Q_2)$ with initial condition $\tilde{y}^0$ satisfies $\tilde{y}(T, \cdot) = 0$. Since $S_2 \subset \tilde{L}(S_1)$, we have $\tilde{y} = \tilde{L}y$, where $y$ is a solution to $(\Lambda, Q_1)$ for some initial condition and some control. But the initial condition is necessarily $y^0$ since $Ly^0 = \tilde{y}^0 = \tilde{y}(0, \cdot) = (\tilde{L}y)(0, \cdot) = L(y(0, \cdot))$ and $L$ is injective. Similarly, we see that we necessarily have $y(T, \cdot) = 0$. □

The main result of this section is the following.

**Theorem 2.7.3.** Assume (2.13). Then, $(\Lambda, Q) \sim (\Lambda, LQU)$ for any invertible lower triangular matrix $L \in \mathbb{R}^{p \times p}$ and any invertible upper triangular matrix $U \in \mathbb{R}^{m \times m}$.

Applying this result, we have in particular that

$$(\Lambda, Q) \sim (\Lambda, Q^c),$$

where $Q^c$ is the canonical form of $Q$ (see Section 2.6).

As already mentioned in Section 2.6, because of the very simple structure of $Q^c$ it is easy to find the minimal control time of the corresponding system (see Proposition 2.6.3). Therefore, Theorem 2.5.1 and Corollary 2.5.2 are an immediate consequence of Theorem 2.7.3.

For 2+2 systems (see Section 2.3.2), we found out that we had six possibilities of the minimal control time (see (2.11)). This corresponds to six out of the seven possible canonical forms of $Q$:

$$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$

Each matrix on the right or below leads to a larger minimal control time (the last two matrices have the same time).

<figure><img src="image_34.png" /></figure>

<footnote n="2">²Here, isomorphism means an invertible bounded linear operator (with bounded inverse, but we recall that this is automatically guaranteed in Banach spaces by the bounded inverse theorem).</footnote>
*Proof of Theorem 2.7.3.* For any $1 \le k, l \le n$, we denote by $\zeta_{kl}$ the solution to the ODE

$$ \begin{cases} \zeta'_{kl}(x) = \frac{\lambda_l(\zeta_{kl}(x))}{\lambda_k(x)}, & \forall x \in \mathbb{R}, \\ \zeta_{kl}(0) = 0. \end{cases} $$

Explicitly,

$$ \zeta_{kl}(x) = \phi_l^{-1}(\phi_k(x)), \quad \text{where } \phi_r(x) = \int_0^x \frac{1}{\lambda_r(\xi)} d\xi. \qquad (2.24) $$

* We first prove that

$$ (\Lambda, Q) \sim (\Lambda, LQ). $$

To this end, we are going to use the spatial transformation

$$ \tilde{y}_{-} = y_{-}, \quad \tilde{y}_{m+\alpha}(t, x) = \sum_{\beta \le \alpha} l_{\alpha\beta} y_{m+\beta}(t, \zeta_{m+\alpha, m+\beta}(x)) \quad (1 \le \alpha \le p). $$

Using the formula (2.24) and the ordering (2.13) we see that $\zeta_{m+\alpha,m+\beta}$ is a $C^1$-diffeomorphism from $(0, 1)$ to a subset of $(0, 1)$ for $\beta \le \alpha$, and thus the transformation is well defined on $L^2(0, 1)^n$. Besides, using the property $\zeta_{kl}(\zeta_{rk}(x)) = \zeta_{rl}(x)$, we easily check that it is invertible, with inverse given by

$$ y_{-} = \tilde{y}_{-}, \quad y_{m+\beta}(t, x) = \sum_{\gamma \le \beta} l^{\beta\gamma} \tilde{y}_{m+\gamma}(t, \zeta_{m+\beta, m+\gamma}(x)) \quad (1 \le \beta \le p), \qquad (2.25) $$

where $(l^{\beta\gamma})_{1\le\beta,\gamma\le p} = L^{-1}$.

* Assume now that $y$ is a solution to the system $(\Lambda, Q)$ for some $(y^0, u)$ and let us show that $\tilde{y}$ is then a solution to the system $(\Lambda, LQ)$ for some $(\tilde{y}^0, \tilde{u})$. We do it formally but this can be rigorously justified by coming back to the definition of what is a solution.

- **Initial condition.** Obviously, it is given by

$$ \tilde{y}_{-}^{0} = y_{-}^{0}, \quad \tilde{y}_{m+\alpha}^{0}(x) = \sum_{\beta \le \alpha} l_{\alpha\beta} y_{m+\beta}^{0}(\zeta_{m+\alpha, m+\beta}(x)) \quad (1 \le \alpha \le p). $$

- **Boundary conditions.** At $x = 0$ it is clearly satisfied since $\tilde{y}_{-} = y_{-}$ and $\tilde{y}_{+}(t, 0) = Ly_{+}(t, 0)$. At $x = 1$ this just defines the control $\tilde{u} = u$.

- **Equations.** It is clear that $\tilde{y}_k = y_k$ satisfies the desired equation for $k \le m$. Let us now fix $1 \le \alpha \le p$. Using the equations satisfied by $y_+$, we have

$$ \frac{\partial \tilde{y}_{m+\alpha}}{\partial t}(t, x) + \lambda_{m+\alpha}(x) \frac{\partial \tilde{y}_{m+\alpha}}{\partial x}(t, x) = \sum_{\beta \le \alpha} l_{\alpha\beta} \left( -\lambda_{m+\beta}(\zeta_{m+\alpha, m+\beta}(x)) + \lambda_{m+\alpha}(x) \zeta'_{m+\alpha, m+\beta}(x) \right) \frac{\partial y_{m+\beta}}{\partial x}(t, \zeta_{m+\alpha, m+\beta}(x)). $$

The right-hand side is zero by definition of $\zeta_{m+\alpha,m+\beta}$. Consequently, $\tilde{y}_{m+\alpha}$ also satisfies the desired equation.

Similar computations show that, conversely, if $\tilde{y}$ is a solution to the system $(\Lambda, LQ)$ for some $(\tilde{y}^0, \tilde{u})$, then $y$ defined by the inverse transformation (2.25) is a solution to the system $(\Lambda, Q)$ for some $(y^0, u)$.

<figure><img src="image_35.png" /></figure>
• Similarly, we can show that

$$(\Lambda, Q) \sim (\Lambda, QU),$$

by using the spatial transformation

$$\tilde{y}_{\alpha}(t, x) = \sum_{\beta \ge \alpha} u^{\alpha\beta} y_{\beta}(t, \zeta_{\alpha\beta}(x)) \quad (1 \le \alpha \le m), \quad \tilde{y}_{+} = y_{+},$$

where $(u^{\alpha\beta})_{1\le\alpha,\beta\le m} = U^{-1}$. Let us just mention that the control is then

$$\tilde{u}_{\alpha}(t) = \sum_{\beta \ge \alpha} u^{\alpha\beta} y_{\beta}(t, \zeta_{\alpha\beta}(1)) \quad (1 \le \alpha \le m), \qquad (2.26)$$

which is well defined since $y \in C^0([0, 1]; L^2(0, T)^n)$.

This concludes the proof of Theorem 2.7.3. □

**Explicit formula.** As mentioned before in Proposition 2.6.3, $\tilde{u} = 0$ is a control for the (NC) of $(\Lambda, Q^0)$. Plugging this into formula (2.26) (recall that $(\Lambda, Q)$ and $(\Lambda, LQ)$ have the same controls) and extracting the term for $\beta = \alpha$ from the sum to obtain the control $u_{\alpha}$ (since $\zeta_{\alpha\alpha}(x) = x$), we deduce that

$$u_{\alpha}(t) = -\frac{1}{u^{\alpha\alpha}} \sum_{\beta > \alpha} u^{\alpha\beta} y_{\beta}(t, \zeta_{\alpha\beta}(1)) \quad (1 \le \alpha \le m). \qquad (2.27)$$

This gives a formula of the control only in function of the state $y$ at time $t$: $u(t) = F(y(t, \cdot))$. The linear operator $F$ is called the *feedback law*. Note that it is an unbounded operator in our case (it is not defined on the state space $L^2(0, 1)^n$). Note as well that system (2.1) with $u(t)$ replaced by $F(y(t, \cdot))$ is obviously well-posed because, for this particular $F$, we know that the system is equivalent to $(\Lambda, Q^0)$ with zero control, which is well-posed.

The formula (2.27) can also be used to write the control $u$ explicitly in function of the initial condition $y^0$ (in fact, $y^0_-$). Indeed, for $\alpha = m$, we get $u_m = 0$. Thus, $y_m$ becomes a known quantity, that can be expressed explicitly in function of $y_m^0$. Considering then the formula for $\alpha = m - 1$, we deduce that $u_{m-1}$ can be expressed explicitly in function of $y_{m-1}^0$. Thus, $y_{m-1}$ becomes a known quantity, that can be expressed explicitly in function of $y_m^0$ and $y_{m-1}^0$. We can repeat this reasoning for all rows.

## 2.8 Some exercises

All the systems appearing below are considered with an initial condition at time $t = 0$ and they evolve on the domain $R_T = (0, T) \times (0, 1)$. We also sometimes drop the argument $(t, x)$ for $y$ for brevity.

**Exercise 2.8.1.** Find the minimal time for the (NC) of the system

$$\left\{
\begin{aligned}
\frac{\partial y_1}{\partial t} - 4 \frac{\partial y_1}{\partial x} &= 0, & y_1(t, 1) &= u_1(t), \\
\frac{\partial y_2}{\partial t} - 3 \frac{\partial y_2}{\partial x} &= 0, & y_2(t, 1) &= u_2(t), \\
\frac{\partial y_3}{\partial t} - 2 \frac{\partial y_3}{\partial x} &= 0, & y_3(t, 1) &= u_3(t), \\
\frac{\partial y_4}{\partial t} - \frac{\partial y_4}{\partial x} &= 0, & y_4(t, 1) &= u_4(t), \\
\frac{\partial y_5}{\partial t} + \frac{\partial y_5}{\partial x} &= 0, & y_5(t, 0) &= y_1(t, 0) + y_2(t, 0) - y_3(t, 0) + 2y_4(t, 0), \\
\frac{\partial y_6}{\partial t} + 2 \frac{\partial y_6}{\partial x} &= 0, & y_6(t, 0) &= 3y_1(t, 0) + 5y_2(t, 0) - y_3(t, 0) + 8y_4(t, 0), \\
\frac{\partial y_7}{\partial t} + 3 \frac{\partial y_7}{\partial x} &= 0, & y_7(t, 0) &= y_2(t, 0) + y_3(t, 0) + y_4(t, 0), \\
\frac{\partial y_8}{\partial t} + 4 \frac{\partial y_8}{\partial x} &= 0, & y_8(t, 0) &= -y_1(t, 0) + 3y_2(t, 0) + 6y_3(t, 0) + 4y_4(t, 0),
\end{aligned}
\right.$$

<figure><img src="image_36.png" /></figure>
where $u_1, u_2, u_3, u_4$ are controls. Is this system also (EC)? Same questions for the system:

$$
\left\{
\begin{aligned}
\frac{\partial y_1}{\partial t} - \left(\frac{1}{2} + x\right)^2 \frac{\partial y_1}{\partial x} &= 0, & y_1(t, 1) &= u_1(t), \\
\frac{\partial y_2}{\partial t} - (2-x)^2 \frac{\partial y_2}{\partial x} &= 0, & y_2(t, 1) &= u_2(t), \\
\frac{\partial y_3}{\partial t} + \frac{6}{5} \frac{\partial y_3}{\partial x} &= 0, & y_3(t, 0) &= y_1(t, 0) + 2y_2(t, 0), \\
\frac{\partial y_4}{\partial t} + 2 \frac{\partial y_4}{\partial x} &= 0, & y_4(t, 0) &= 2y_1(t, 0) + 4y_2(t, 0), \\
\frac{\partial y_5}{\partial t} + \frac{3}{4} \frac{\partial y_5}{\partial x} &= 0, & y_5(t, 0) &= u_3(t).
\end{aligned}
\right.
$$

**Exercise 2.8.2** (Non uniqueness of *L*, *U*). Show that the matrix *L* in the (RRMCF) (Theorem 2.4.3) is not necessarily unique. Same question with the pair of matrices (*L*, *U*) in the canonical decomposition (Theorem 2.6.4).

**Exercise 2.8.3** (Two-sided controllability). Find the minimal time for the (NC) of the system

$$
\left\{
\begin{array}{l}
\displaystyle \frac{\partial y}{\partial t} + \Lambda(x) \frac{\partial y}{\partial x} = 0, \\
\displaystyle y_{-}(t, 1) = u(t), \quad y_{+}(t, 0) = v(t),
\end{array}
\right.
$$

where *u* and *v* are controls.

**Exercise 2.8.4** (Two-sided controllability, with less controls). Let $Q \in \mathbb{R}^{(p-r) \times m}$ with $1 \le r < p$ being fixed (assuming $p \ge 2$). Find the minimal time for the (NC) of the system

$$
\left\{
\begin{array}{l}
\displaystyle \frac{\partial y}{\partial t} + \Lambda(x) \frac{\partial y}{\partial x} = 0, \\
\\
\displaystyle y_{-}(t, 1) = u(t), \quad y_{+}(t, 0) = \begin{pmatrix} v(t) \\ Q y_{-}(t, 0) \end{pmatrix},
\end{array}
\right.
$$

where *u* and *v* are controls. Under what condition(s) this system is also (EC)?

**Exercise 2.8.5** (Worst case scenario). Assume that all the inequalities are strict in (2.13). Show that $T_{\inf} = T_m + T_{m+1}$ if, and only if, the first row of $Q$ is a multiple of $(0 \ 0 \ \cdots \ 0 \ 1)$.

**Exercise 2.8.6** (Best value of the minimal control time). Find the value of

$$
\min_{Q:\text{rank} Q=p} T_{\inf}.
$$

Determine the largest set of boundary coupling matrices *Q* for which this minimum is achieved.

**Exercise 2.8.7** (Uniqueness of the control). Let $\{1, \dots, m\} \setminus \{c_1, \dots, c_\rho\} = \{i_1, \dots, i_{m-\rho}\}$. Show that the control is unique at the time *T* if, and only if,

$$
T = T_{m+r_1} + T_{c_1} = \dots = T_{m+r_{\rho}} + T_{c_{\rho}} = T_{i_1} = \dots = T_{i_{m-\rho}}.
$$

**Exercise 2.8.8** (Control on the other side). Find the minimal control time of system (2.1) when the control is acting on the other part of the boundary:

$$
y_{-}(t, 1) = Q y_{+}(t, 1), \quad y_{+}(t, 0) = u(t),
$$

where now $Q \in \mathbb{R}^{m \times p}$.

<figure><img src="image_37.png" /></figure>
**Exercise 2.8.9 (Time-dependent speeds).** Consider system (2.1) with time-dependent speeds $\Lambda = \Lambda(t, x)$, continuous on $\mathbb{R}^2$ and Lipschitz continuous in $x$, uniformly with respect to $t$. For every $k$, let $\chi_k$ be the flow associated with $\lambda_k$ (see (2.2) with right-hand side replaced by $\lambda_k(s, \chi_k(s; t, x))$). Assume that:

1. For every $k$, there exists $T_k > 0$ such that

$$ \chi_k(T_k; 0, 1) = 0 \quad \text{if } \lambda_k < 0, \quad \chi_k(T_k; 0, 0) = 1 \quad \text{if } \lambda_k > 0. $$

2. $T_1, \dots, T_n$ satisfy the order (2.13).

3. For every $l \in \{1, \dots, \rho\}$, there exists $T_{m+r_l, c_l} > 0$ such that

$$ \chi_{m+r_l}(T_{m+r_l, c_l}; T_{c_l}, 0) = 1. $$

Prove that the minimal control time is then

$$ T_{\inf} = \max \{T_m, T_{m+1}, T_Q\}, \quad T_Q = \max_{l \le \rho} T_{m+r_l, c_l}. $$

**Exercise 2.8.10 (C¹ framework).** Adapt the results of this chapter to the C¹ framework.

**Exercise 2.8.11 (Minimal time and bounded feedback).** Consider the system

$$ \left\{ \begin{aligned} \frac{\partial y_1}{\partial t} - \frac{\partial y_1}{\partial x} &= 0, & y_1(t, 1) &= u_1(t), \\ \frac{\partial y_2}{\partial t} - \frac{1}{2} \frac{\partial y_2}{\partial x} &= 0, & y_2(t, 1) &= u_2(t), \\ \frac{\partial y_3}{\partial t} + \frac{\partial y_3}{\partial x} &= 0, & y_3(t, 0) &= y_1(t, 0) + 2y_2(t, 0), \end{aligned} \right. $$

where $u_1, u_2$ are controls. What is the minimal control time $T_{\inf}$ of this system? If the boundary conditions for $y_1$ and $y_2$ are replaced by

$$ \left\{ \begin{aligned} y_1(t, 1) &= \int_0^1 h(\xi)y_2(t, \xi) d\xi, & h \in L^2(0, 1), \\ y_2(t, 1) &= 0, \end{aligned} \right. $$

does the corresponding solution satisfies $y(T, \cdot) = 0$ for $T = T_{\inf}$?

**Exercise 2.8.12 (The formula of N. Weck).** The minimal control time for (NC) was first obtained in [Wec82], however with a different formula<fnref n="3" />. The goal of this exercise is to compare the formula in this article with the formula of this chapter (see Theorem 2.5.1). We first introduce the notations of this article. We consider the system

$$ \left\{ \begin{aligned} \frac{\partial w}{\partial t}(t, x) &= A(x) \frac{\partial w}{\partial x}(t, x), \\ w^+(t, 1) &= u(t), \quad w^-(t, 0) = D_0 w^+(t, 0), & (t, x) \in R_T, \\ w(0, x) &= w_0(x), \end{aligned} \right. \quad (2.28) $$

where the state is $w = \begin{pmatrix} w^- \\ w^+ \end{pmatrix}$, the control is $u$,

$$ A = \operatorname{diag}(\lambda_1^-, \dots, \lambda_p^-, \lambda_1^+, \dots, \lambda_m^+), $$

<figure><img src="image_38.png" /></figure>

<footnote n="3">³Beware that, in [Wec82], the author called “exact controllability” what is in fact null controllability, see Definition 1 therein.</footnote>
with

$$ \lambda_p^-(x) \le \dots \le \lambda_1^-(x) < 0 < \lambda_1^+(x) \le \dots \le \lambda_m^+(x), $$

for every $x \in [0, 1]$ and $D_0 \in \mathbb{R}^{p \times m}$. Then, the author of [Wec82] proved that system (2.28) is (NC) in time T if, and only if,

$$ T \ge \max \{T_1^+, \tilde{T}_c\}, \quad \tilde{T}_c = \max_{1 \le k \le p} T_k^- + T_{l(k)}^+, \qquad (2.29) $$

where:

$$ \bullet \ T_k^\pm = \int_0^1 \frac{1}{|\lambda_k^\pm(\xi)|} d\xi, $$

• $l(k)$ is the first integer $l \in \{1, \dots, m\}$ such that

$$ \ker Q_{l-1}CP_k \subset \ker Q_lCP_k, $$

provided that $CP_k \ne 0$, and $l(k) = m + 1$ otherwise, with then the addition of the artificial time $T_{m+1}^+ = 0$ in the formula (2.29),

$$ \bullet \ C = -(A^+(0))^{-1} D_0^T A^-(0), \text{ with } A^+ = \text{diag}(\lambda_1^+, \dots, \lambda_m^+) \text{ and } A^- = \text{diag}(\lambda_1^-, \dots, \lambda_p^-), $$

• the projection matrices $Q_l \in \mathbb{R}^{m \times m}$ and $P_k \in \mathbb{R}^{p \times p}$ are

$$ Q_l = \begin{pmatrix} 0 & 0 \\ 0 & \text{Id}_{m-l} \end{pmatrix}, \quad P_k = \begin{pmatrix} \text{Id}_k & 0 \\ 0 & 0 \end{pmatrix}, $$

with $Q_0 = \text{Id}_m$, $Q_m = 0$ and $P_p = \text{Id}_p$.

Questions:

1. Write system (2.28) in the form (2.1)-(2.13). Write the boundary coupling matrix $Q$ as a function of $C$.

2. Consider the example presented in [Wec82, Section 4], namely

$$ C = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}. $$

(a) Use the formula (2.29) to find the minimal control time.

(b) Use the formula in Theorem 2.5.1 to find the minimal control time.

Which one of 2a and 2b requires less computations ?

3. Prove the equality $\tilde{T}_c = T_Q$ in the general case.

**Exercise 2.8.13 (Simultaneous control).** Consider the system

$$ \begin{cases} \frac{\partial y_1}{\partial t} - \frac{\partial y_1}{\partial x} = 0, & y_1(t, 1) = ay_2(t, 1) + \alpha u(t), \\ \frac{\partial y_2}{\partial t} + \frac{\partial y_2}{\partial x} = 0, & y_2(t, 0) = by_1(t, 0) + \beta u(t), \end{cases} $$

where $u$ is the control and $a, b, \alpha, \beta \in \mathbb{R}$. Admitting that this system is well-posed, find a necessary and sufficient condition for this system to be (EC) in time $T = 2$.

<figure><img src="image_39.png" /></figure>
## 2.9 Bibliographical notes

**General comments.** Finding the minimal control time for system (2.1) was a problem raised in the celebrated survey paper [Rus78]. It was already observed in that paper that the control time $T_{-} + T_{+}$ is too large for most boundary coupling matrices and that the minimal time should depend on some properties of these matrices. The author could not solve this problem though and it was left as an open problem (see therein the Remark on p.656). This was eventually completely solved few years later in [Wec82] (see Exercise 2.8.12). The problem was also independently solved in [CN19] but assuming in addition that $Q \in \mathcal{B}_l$ with $l = \min\{p, m\}$, and in [HO21] in the general case. The problem was then revisited again in [HO22].

**Section 2.4.** The notion of (RRMCF) is taken from [DJM06]. Theorem 2.4.3 is proven there (see Theorem 1 therein), but our proof is inspired from the one in [HO21, Appendix A].

**Section 2.5.** The (RRMCF) was first used in [HO21] to find the minimal control time for the (EC) of system (2.1) (see [HO21, Section 3]). Our proof is however simpler from the one therein (and it does not use the duality). As we have also mentioned in Exercise 2.8.12, the minimal control time for (NC) was first obtained in [Wec82], however with a different proof, and a more complicated formula. The set $\mathcal{B}_l$ defined in (2.16) was introduced in [CN19] for $l = \min\{p, m-1\}$ ($\mathcal{B}_l = \mathbb{R}^p$ if $m=1$) and $l = \min\{p, m\}$.

**Section 2.6.** The notion of canonical form of boundary coupling matrices was introduced in [HO22] (see Section 1.2 therein).

**Section 2.7.** The notion of equivalent system was introduced in [HO22] (see Section 1.5 therein; see also [HO21, Appendix E]). Theorem 2.7.3 is a special case of [HO22, Proposition 4.1]. ($\Lambda, Q^c$) could be called the canonical form of ($\Lambda, Q$). This is probably one of the simplest meaningful generalization to infinite-dimensional systems of the so-called canonical form of Brunovsky ([Bru70]).

The feedback law (2.27) was obtained in [HO22, Section 5.3] (see also [CN19, p. 1155] when $Q \in \mathcal{B}_l$ for $l = \min\{p, m-1\}$).

Let us conclude by briefly mentioning some related topics.

**General boundary conditions.** Control problems for homogeneous systems with more general boundary conditions of the form

$$ \begin{pmatrix} y_{-}(t, 1) \\ y_{+}(t, 0) \end{pmatrix} = Q \begin{pmatrix} y_{-}(t, 0) \\ y_{+}(t, 1) \end{pmatrix} + Ju(t), \qquad (2.30) $$

with $Q \in \mathbb{R}^{n \times n}$ and $J \in \mathbb{R}^{n \times r}$, are also studied in the literature, see [CFMS24].

**Quasilinear systems.** Controllability of quasilinear versions of system (2.1) are also investigated in the literature, see [CN20, CN22] (see also the bibliographical notes of Chapter 4 below).

**Stability.** Exponential stability of homogeneous systems (linear or quasilinear), with general boundary conditions (2.30) (without control, i.e. $u=0$), was also studied. We refer for instance to [Li94, Chapter 5], [CN15], [BC16, Chapters 2,3 & 4] and the references therein. Finite-time stability charaterizations were also obtained in the paper [KL21].

<figure><img src="image_40.png" /></figure>
<figure><img src="image_41.png" /></figure>
# Chapter 3

## Well-posedness of linear systems

**Prerequisites:** Transport equation (Chapter 1), Banach fixed-point theorem.

In this chapter, we prove the well-posedness of systems of the general form

$$ 
\left\{ 
\begin{array}{ll}
\mathbf{Equ}: & \displaystyle \frac{\partial y}{\partial t}(t,x) + \Lambda(t,x) \frac{\partial y}{\partial x}(t,x) = M(t,x)y(t,x) + f(t,x), \\
\mathbf{IC}: & y(0,x) = y^0(x), \\
\mathbf{BC}: & \left( \begin{pmatrix} y_{-}(t,1) \\ y_{+}(t,0) \end{pmatrix} \right) = Q(t) \left( \begin{pmatrix} y_{-}(t,0) \\ y_{+}(t,1) \end{pmatrix} \right) + g(t),
\end{array}
\right. 
\quad (t,x) \in \mathbb{R}_T. \quad (3.1) 
$$

The notations are the same as in Chapter 2, with the difference that $\Lambda$ now depends on $t$ as well (it is still assumed to be diagonal), that we have a right-hand side in the equations, and that we consider more general boundary conditions. The $n \times n$ matrix $M$, coupling the equations of the system inside the domain, will be called the *internal coupling matrix*. In the literature, systems with $M = 0$ are usually called *homogeneous systems* or *systems of conservation laws*, whereas systems with $M \neq 0$ are usually referred to as *systems of balance laws*. The $n \times n$ matrix $Q$, coupling the equations of the system on the boundary, will be called the *boundary coupling matrix* (in Chapter 2 we considered a particular structure). The functions $f$ and $g$ are given.

### 3.1 The characteristics

We will start by precising the assumptions on the speeds. In what follows, $k \in \{1, \dots, n\}$ is fixed.

**Standing assumptions.** We will always assume that the speed $\lambda_k$ is at least continuous:

$$ \lambda_k \in C^0(\overline{R_T}), $$

and Lipschitz continuous in $x$, uniformly with respect to $t$:

$$ |\lambda_k(t, x) - \lambda_k(t, x')| \le L |x - x'|, $$

for some $L \ge 0$, and for every $t \in [0, T]$ and $x, x' \in [0, 1]$.

<figure><img src="image_42.png" /></figure>
**The k-th characteristic.** It is more convenient to deal with a function $\lambda_k$ defined on $\mathbb{R}^2$. Therefore, we first extend $\lambda_k$ to a function of $\mathbb{R}^2$, still denoted by the same, such that all the previous assumptions hold on $\mathbb{R}^2$. Under these assumptions, by the global Cauchy-Lipschitz theorem (see e.g. [Har02, Chapter II, Theorem 1.1]), we can define the flow $\chi_k$ associated with $\lambda_k$ globally. More precisely, for every $(t, x) \in \mathbb{R}^2$, there is a unique $C^1$ function $s \mapsto \chi_k(s; t, x)$ solution to the ODE

$$ \left\{ \begin{array}{l} \frac{d}{ds} \chi_k(s; t, x) = \lambda_k(s, \chi_k(s; t, x)), \quad \forall s \in \mathbb{R}, \\ \chi_k(t; t, x) = x. \end{array} \right. \qquad (3.2) $$

The curve $s \mapsto (s, \chi_k(s; t, x))$ will be called the k-th characteristic passing through $(t, x)$.

**The set $V_k$, the point $\delta_k$ and the function $s_k^{\text{in}}$.** Let us introduce the sets

$$ V_k = \{(t, x) \in \mathbb{R}_T \mid \chi_k(0; t, x) \in (0, 1)\}, \quad \bar{V}_k^c = \bar{R}_T \setminus \bar{V}_k. $$

Note that the set $V_k$ is open, but the set $\bar{V}_k^c$ is not. The closure of $\bar{V}_k^c$ is $\bar{R}_T \setminus V_k$. It is never empty, since it always contains the boundary line $\{(t, \delta_k) \mid t \in [0, T]\}$, where

$$ \delta_k = \begin{cases} 1 & \text{if } \lambda_k < 0, \\ 0 & \text{if } \lambda_k > 0. \end{cases} $$

Then, for every $(t, x) \in \bar{R}_T \setminus V_k$, by the intermediate value theorem, there exists a solution $s_k^{\text{in}}(t, x) \in [0, t]$ to

$$ \chi_k(s_k^{\text{in}}(t, x); t, x) = \delta_k. \qquad (3.3) $$

This solution is unique by monotonicity of $\chi_k(\cdot; t, x)$. Finally, note that

$$ \bar{V}_k \cap (\bar{R}_T \setminus V_k) = \{(t, x) \in \bar{R}_T \setminus V_k \mid s_k^{\text{in}}(t, x) = 0\}. $$

**The function $(s_k^{\text{in}})^+$.** We define

$$ (s_k^{\text{in}})^+(t, x) = \begin{cases} 0 & \text{if } (t, x) \in \bar{V}_k, \\ s_k^{\text{in}}(t, x) & \text{if } (t, x) \in \bar{V}_k^c. \end{cases} \qquad (3.4) $$

Note that $(s_k^{\text{in}})^+(t, x) \in [0, t]$. Then, observe that, for any $(t, x) \in \bar{R}_T$, the k-th characteristic passing through this point stays in the domain for all $s \in [(s_k^{\text{in}})^+(t, x), t]$. More precisely,

$$ (s, \chi_k(s; t, x)) \in [0, t] \times [0, 1], \quad \forall s \in [(s_k^{\text{in}})^+(t, x), t]. $$

Finally, we have gathered in Appendix A all the additional properties that are related to the characteristics and that will be used in these notes.

## 3.2 Solution along the characteristics

In this section, we introduce the notion of *solution along the characteristics* for systems of the form (3.1). We first recall what it means to be a classical solution.

**Definition 3.2.1.** We say that a function $y$ is a *classical solution* to system (3.1) if $y \in C^1(\bar{R}_T)^n$ and it satisfies the conditions in (3.1) for every $(t, x) \in \bar{R}_T$.

<figure><img src="image_43.png" /></figure>
Let now $y \in C^1(\overline{R_T})^n$ be a classical solution to system (3.1). Fix for the moment $1 \le k \le n$ and $(t, x) \in \overline{R_T}$. Since the $k$-th characteristic passing through $(t, x)$ stays in the domain for $s \in [(s_k^{\text{in}})^+(t, x), t]$ (see Section 3.1), we can then consider the solution along this characteristic:

$$Y(s) = y(s, \chi_k(s; t, x)).$$

Below, it will also be convenient to denote the $k$-th row of $M$ and $Q$ by $M_k$ and $Q_k$. Assume temporarily that the interval $[(s_k^{\text{in}})^+(t, x), t]$ is not reduced to a point. Then, taking therein the derivative with respect to $s$ of the identity

$$Y_k(s) = y_k(s, \chi_k(s; t, x)),$$

using the chain rule and then using the conditions in system (3.1), we see that it satisfies the ODE

$$\left\{ \begin{array}{l} \frac{d}{ds} Y_k(s) = M_k(s, \chi_k(s; t, x)) Y(s) + f_k(s, \chi_k(s; t, x)), \quad s \in [(s_k^{\text{in}})^+(t, x), t], \\ Y_k((s_k^{\text{in}})^+(t, x)) = Y_k^0, \end{array} \right. \qquad (3.5)$$

where the initial condition $Y_k^0$ is given by the initial or boundary conditions in (3.1), namely:

$$Y_k^0 = \begin{cases} y_k^0(\chi_k(0; t, x)) & \text{if } (t, x) \in \overline{V}_k, \\ Q_k(s_k^{\text{in}}(t, x)) \begin{pmatrix} y_-(s_k^{\text{in}}(t, x), 0) \\ y_+(s_k^{\text{in}}(t, x), 1) \end{pmatrix} + g_k(s_k^{\text{in}}(t, x)) & \text{if } (t, x) \in \overline{V}_k^c. \end{cases} \qquad (3.6)$$

Note that $Y_k^0$ depends on $t, x$ and also on $y$; however we omit this dependence for brevity reasons. Integrating the ODE (3.5) over $[(s_k^{\text{in}})^+(t, x), t]$, we obtain the system of integral equations

$$y_k(t, x) = Y_k^0 + \int_{(s_k^{\text{in}})^+(t, x)}^{t} (M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) + f_k(s, \chi_k(s; t, x))) ds, \quad 1 \le k \le n. \qquad (3.7)$$

Finally, observe that this identity remains valid even when $(s_k^{\text{in}})^+(t, x) = t$. In conclusion, a classical solution has to satisfy this system of integral equations for every $(t, x) \in \overline{R_T}$.

Conversely, it is not difficult to see that, if $y \in C^1(\overline{R_T})^n$ satisfies the system of integral equations (3.7) in $\overline{R_T}$, then it is a classical solution to system (3.1). Indeed, writing (3.7) for $(\theta, \chi_k(\theta; t, x))$ with $\theta \in [(s_k^{\text{in}})^+(t, x), t]$ instead of $(t, x)$, and using the properties (see Appendix A.1)

$$\chi_k(s; \theta, \chi_k(\theta; t, x)) = \chi_k(s; t, x), \quad (s_k^{\text{in}})^+(\theta, \chi_k(\theta; t, x)) = (s_k^{\text{in}})^+(t, x),$$

we obtain the ODE (3.5). More precisely, we obtain that the initial condition in (3.5) is satisfied for any $(t, x) \in \overline{R_T}$ and that the equation is satisfied for any $(t, x) \in \overline{R_T}$ for which $[(s_k^{\text{in}})^+(t, x), t]$ is not reduced to a point. Noting that this latter restriction happens only for points $(t, x)$ on the boundary of $R_T$, and since $y$ is $C^1$ up to the boundary, the equation of the ODE is thus equivalent to the one in system (3.1).

Now observe that the terms in (3.7) make sense for functions $y$ (and $M, Q, y^0, f, g$) with much weaker regularity than $C^1$. For instance, for the integral to be well-defined, we only need that the function $s \mapsto M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x))$ is integrable in the interval $((s_k^{\text{in}})^+(t, x), t)$. A function $y$ that satisfies the system of integral equations (3.7) in a reasonable sense will be called *solution along the characteristics*. In the next sections, we present various functional settings for this notion of solution.

<figure><img src="image_44.png" /></figure>
3.3 Continuous solutions

In this part, we discuss solutions to the system of integral equations (3.7)-(3.6) which are continuous.
All along this section, we assume that all the data are continuous:

$$
M \in C^0(\overline{R_T})^{n \times n}, \quad Q \in C^0([0, T])^{n \times n},
$$

and

$$
y^0 \in C^0([0, 1])^n, \quad f \in C^0(\overline{R_T})^n, \quad g \in C^0([0, T])^n.
$$

It is clear that all the terms in (3.7)-(3.6) make sense for functions y which are continuous on $\overline{R_T}$.
Therefore, the following notion makes sense.

**Definition 3.3.1.** We say that a function $y$ is a continuous solution to system (3.1) if $y \in C^0(\overline{R_T})^n$ and it satisfies the system of integral equations (3.7)-(3.6) for every $(t, x) \in \overline{R_T}$.

**Compatibility conditions.** It is clear that, for a continuous solution to exist, the following $C^0$ compatibility conditions have to hold:

$$
\left( \begin{array}{c} y_{-}^{0}(1) \\ y_{+}^{0}(0) \end{array} \right) = Q(0) \left( \begin{array}{c} y_{-}^{0}(0) \\ y_{+}^{0}(1) \end{array} \right) + g(0). \qquad (3.8)
$$

**The constants** $L_M$, $L_Q$ and $\alpha_\Lambda$. All along Section 3.3, we work with the following norm on $\mathbb{R}^n$:

$$
\|y\|_{\mathbb{R}^n} = \max_{1 \le k \le n} |y_k|, \quad y \in \mathbb{R}^n. \qquad (3.9)
$$

Let us now introduce some important constants. We denote by $L_M$ the smallest $L \ge 0$ such that

$$
|M_k(t, x)y| \leq L \|y\|_{\mathbb{R}^n},
$$

for every $1 \le k \le n$, $(t, x) \in \overline{R_T}$ and $y \in \mathbb{R}^n$. Similarly, we denote by $L_Q$ the smallest $L \ge 0$ such that

$$
|Q_k(t)y| \le L \|y\|_{\mathbb{R}^n},
$$

for every $1 \le k \le n$, $t \in [0, T]$ and $y \in \mathbb{R}^n$. Their existence are guaranteed since $M$ and $Q$ are in particular bounded.

Let us now point out that $\chi_k$ is $C^0$ in all its variables, i.e. $\chi_k \in C^0(\mathbb{R}^3)$ (see e.g. [Har02, Chapter V, Theorem 2.1]). It follows that $s_k^{\text{in}}$ is $C^0$ on $\overline{R_T} \setminus V_k$ by the implicit function theorem<fnref n="1" />.

Finally, let us introduce

$$
\alpha_{\Lambda} = \sup \{ t - (s_k^{\text{in}})^+(t, x) \mid 1 \le k \le n, (t, x) \in \bar{R}_T \}.
$$

Observe that all of these constants depend a priori also on $T$ and thus they may in particular grow with respect to $T$. Note however that they become bounded with respect to $T$ if we assume for instance that $M, Q$ are defined and bounded on $[0, +\infty) \times [0, 1]$ and $[0, +\infty)$, respectively, and $\Lambda$ satisfies

$$
\inf_{\substack{t \ge 0 \\ x \in [0,1]}} |\lambda_k(t, x)| > 0, \quad \forall k, \tag{3.10}
$$

which gives a uniform bound for $\alpha_\Lambda$ (see Appendix A.2).

One of the goals of this section is to prove the following well-posedness result for continuous solu-
tions.

<figure><img src="image_45.png" /></figure>

<footnote n="1">¹Even though the standard statement of the implicit function theorem requires the function to be $C^1$ in all its variable (which is not the case for $\chi_k$ since we only assume $\lambda_k \in C^0$ all along Section 3.3), the very first step in the proof of implicit function theorem is generally to show that the map is continuous, and for that we only need the continuity in all variables (and of course, the invertibility of the differential with respect to one variable), see e.g. [KP02, Theorem 3.2.1].</footnote>
**Theorem 3.3.2.** For every $T > 0$, $y^0 \in C^0([0, 1])^n$, $f \in C^0(\overline{R_T})^n$ and $g \in C^0([0, T])^n$ satisfying the $C^0$ compatibility conditions (3.8), there exists a unique continuous solution $y$ to system (3.1). Moreover, it satisfies the estimate

$$ \|y\|_{C^0(\overline{R_T})^n} \le Ce^{\omega T} \left( \|y^0\|_{C^0([0,1])^n} + \|f\|_{C^0(\overline{R_T})^n} + \|g\|_{C^0([0,T])^n} \right), \quad (3.11) $$

for some $\omega > 0$ (resp. $C > 0$) depending only on $L_M, L_Q$ (resp. $L_Q, \alpha_\Lambda$).

**The map F.** It is convenient to write the system of integral equations (3.7)-(3.6) in the concise form

$$ y = F(y), $$

where the map $F$ is defined by the corresponding right-hand side. More precisely, if $(t, x) \in \overline{V}_k$, then

$$ F(y)_k(t, x) = y_k^0(\chi_k(0; t, x)) + \int_0^t (M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) + f_k(s, \chi_k(s; t, x))) ds, $$

and, if $(t, x) \in \overline{V}_k^c$, then

$$ F(y)_k(t, x) = Q_k(s_k^{\text{in}}(t, x)) \begin{pmatrix} y_-(s_k^{\text{in}}(t, x), 0) \\ y_+(s_k^{\text{in}}(t, x), 1) \end{pmatrix} + g_k(s_k^{\text{in}}(t, x)) \\ + \int_{s_k^{\text{in}}(t, x)}^t (M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) + f_k(s, \chi_k(s; t, x))) ds. $$

A solution to the system of integral equations (3.7)-(3.6), if it exists, is certainly a fixed-point of the map $F$.

**The norm $|\cdot|_0$.** For the sake of brevity, it will be convenient to use the notation

$$ |h|_0 = \|h\|_{C^0(\bar{\Omega})^n}, $$

for any continuous function $h: \bar{\Omega} \to \mathbb{R}^n$, where $\Omega$ is an open subset of either $\mathbb{R}$ or $\mathbb{R}^2$ (there will be no risk of confusion).

### 3.3.1 A priori estimate

In this section we establish an priori estimate for continuous solutions.

**Estimate of F.** We first estimate $F(y)$ for an arbitrary $y \in C^0(\overline{R_T})^n$. Let $1 \le k \le n$ be fixed. For $(t, x) \in \overline{V}_k$, we have

$$ |F(y)_k(t, x)| \le |y_k^0(\chi_k(0; t, x))| + \int_0^t (|M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x))| + |f_k(s, \chi_k(s; t, x))|) ds. $$

By definition of $L_M$, we have

$$ \begin{align*} \int_0^t |M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x))| ds &\le L_M \int_0^t \|y(s, \chi_k(s; t, x))\|_{\mathbb{R}^n} ds \\ &\le L_M \int_0^t \sup_{\xi \in [0,1]} \|y(s, \xi)\|_{\mathbb{R}^n} ds \\ &= L_M \int_0^t |y(s, \cdot)|_0 ds. \end{align*} $$

<figure><img src="image_46.png" /></figure>
By definition of $\alpha_\Lambda$, we have

$$ \int_0^t |f_k(s, \chi_k(s; t, x))| \, ds \le \alpha_\Lambda |f|_0. $$

Thus,

$$ |F(y)_k(t, x)| \le |y^0|_0 + L_M \int_0^t |y(s, \cdot)|_0 ds + \alpha_\Lambda |f|_0. $$

Similarly, for $(t, x) \in \bar{V}_k^c$, noting that $s_k^{\text{in}}(t, x) \le t$, we can show that

$$ |F(y)_k(t, x)| \le L_Q \sup_{s \le t} |y(s, \cdot)|_0 + |g|_0 + L_M \int_0^t |y(s, \cdot)|_0 ds + \alpha_\Lambda |f|_0, $$

(here we also used that the norm in $\mathbb{R}^n$ is given by (3.9)). Gathering both estimates, we get

$$ \sup_{x \in [0,1]} \|F(y)(t, x)\|_{\mathbb{R}^n} \le L_Q \sup_{s \le t} |y(s, \cdot)|_0 + |y^0|_0 + |g|_0 + \alpha_\Lambda |f|_0 + L_M \int_0^t |y(s, \cdot)|_0 ds, $$

for every $t \in [0, T]$.

**A priori estimate.** Let now $y$ be any continuous solution to system (3.1). Thus, we have $y = F(y)$, and the previous estimate gives

$$ |y(t, \cdot)|_0 \le L_Q \sup_{s \le t} |y(s, \cdot)|_0 + |y^0|_0 + |g|_0 + \alpha_\Lambda |f|_0 + L_M \int_0^t |y(s, \cdot)|_0 ds, \quad (3.12) $$

for every $t \in [0, T]$. To have a chance to deduce an upper bound for $y$, we see that, even in the simple case $M = 0$, for which we have $L_M = 0$, we should have

$$ L_Q < 1. \quad (3.13) $$

Let us then assume this condition and let us show how to deduce an a priori estimate in that case. It will be explained in Section 3.3.3 how to obtain a similar estimate in the general case. Under assumption (3.13), we can “absorb” the first term in estimate (3.12) and obtain that

$$ |y(t, \cdot)|_0 \le \frac{1}{1 - L_Q} (|y^0|_0 + |g|_0 + \alpha_\Lambda |f|_0) + \frac{L_M}{1 - L_Q} \int_0^t |y(s, \cdot)|_0 ds, \quad (3.14) $$

for every $t \in [0, T]$. Applying Grönwall's lemma, we deduce that

$$ |y(t, \cdot)|_0 \le \frac{1}{1 - L_Q} (|y^0|_0 + |g|_0 + \alpha_\Lambda |f|_0) e^{\frac{L_M}{1 - L_Q} t}, $$

for every $t \in [0, T]$.

This a priori estimate gives in particular the estimate in Theorem 3.3.2 and shows the uniqueness of a solution (assuming (3.13) for the moment).

In the next section, we prove the existence a continuous solution.

### 3.3.2 Existence

To show that $F$ has a fixed-point, we are going to apply the Banach fixed-point theorem.

<figure><img src="image_47.png" /></figure>
**Metric space.** Since we look for a solution that is a continuous function, we would like to work in the space $C^0(\overline{R_T})^n$. However, $F$ does not map this space into itself. Recall that the closure of $\overline{V}_k^c$ is $\overline{R_T} \setminus V_k$, and that $\overline{V}_k \cap (\overline{R_T} \setminus V_k) = \{s_k^{\text{in}}(t, x) = 0\}$. Thus, for $F(y)$ to be continuous near this set, we see that $y$ has to satisfy

$$y_k^0(\delta_k) = Q_k(0) \begin{pmatrix} y_-(0,0) \\ y_+(0,1) \end{pmatrix} + g_k(0), \quad 1 \le k \le n. \qquad (3.15)$$

This leads us to work in

$$E_0 = \{y \in C^0(\overline{R_T})^n \mid y \text{ satisfies (3.15)}\}.$$

Then, $F(E_0) \subset C^0(\overline{R_T})^n$, and we easily check that $F$ maps $E_0$ into itself thanks to the $C^0$ compatibility conditions (3.8). Let us also point out that $E_0$ is not empty since, thanks again to the compatibility conditions, it contains any function $y$ satisfying

$$y \in C^0(\overline{R_T})^n, \quad \begin{pmatrix} y_-(0,0) \\ y_+(0,1) \end{pmatrix} = \begin{pmatrix} y_-^0(0) \\ y_+^0(1) \end{pmatrix},$$

(obviously, such functions exist). Clearly, $E_0$ is a complete metric space when equipped with the metric induced by the $C^0$ norm.

**Contraction.** We will now show that $F$ is a contraction for a norm equivalent to the $C^0$ norm. Let $y^1, y^2 \in E_0$ and denote by $y = y^1 - y^2$. Mimicking the estimates of Section 3.3.1, we easily get

$$|F(y^1)(t, \cdot) - F(y^2)(t, \cdot)|_0 \le L_Q \sup_{s \le t} |y(s, \cdot)|_0 + L_M \int_0^t |y(s, \cdot)|_0 ds, \qquad (3.16)$$

for every $t \in [0, T]$. For $F$ to have a chance to be a contraction, we see once again that we should have $L_Q < 1$. Let then $\omega > 0$ and consider the weighted norm

$$|y|_{0,\omega} = \sup_{t \in [0,T]} e^{-\omega t} |y(t, \cdot)|_0. \qquad (3.17)$$

Clearly, it is equivalent to the $C^0$ norm. From (3.16), we deduce that

$$|F(y^1) - F(y^2)|_{0,\omega} \le \rho_{\omega} |y|_{0,\omega}, \qquad (3.18)$$

with

$$\rho_{\omega} = L_Q + \frac{L_M}{\omega}.$$

If $L_Q < 1$, then we can take $\omega$ large enough to make $F$ to be a contraction. By the Banach fixed-point theorem, $F$ has a (unique) fixed-point in $E_0$. The proof of Theorem 3.4.1 is complete, for $Q$ small enough.

It remains to explain how to deduce the same result without the restriction $L_Q < 1$. This is the purpose of the next section.

### 3.3.3 Reduction to small boundary coupling matrix

In this part we explain why we can always assume that the boundary coupling matrix is small. Let $Q$ be arbitrary. Fix $\nu \in \mathbb{R}$ for the moment, and let us consider the spacial transformation mapping any $y \in C^1(\overline{R_T})^n$ to $\tilde{y} \in C^1(\overline{R_T})^n$ defined by

$$\tilde{y}(t, x) = D^{\nu}(x)y(t, x),$$

<figure><img src="image_48.png" /></figure>
where

$$D^{\nu}(x) = \begin{pmatrix} e^{\nu(1-x)}\mathrm{Id}_m & 0 \\ 0 & e^{\nu x}\mathrm{Id}_p \end{pmatrix}.$$

Then, it is easy to see that $y$ is a solution to system (3.1) if, and only if, $\tilde{y}$ is a solution to

$$\left\{
\begin{array}{ll}
\mathbf{Equ}: & \displaystyle \frac{\partial \tilde{y}}{\partial t}(t, x) + \Lambda(t, x) \frac{\partial \tilde{y}}{\partial x}(t, x) = D^{\nu}(x) M(t, x) \tilde{y}(t, x) + D^{\nu}(x) f(t, x), \\
\\
\mathbf{IC}: & \displaystyle \tilde{y}(0, x) = D^{\nu}(0) y^{0}(x), \qquad (t, x) \in \mathbb{R}_{T}. \\
\\
\mathbf{BC}: & \displaystyle \begin{pmatrix} \tilde{y}_{-}(t, 1) \\ \tilde{y}_{+}(t, 0) \end{pmatrix} = e^{-\nu} Q(t) \begin{pmatrix} \tilde{y}_{-}(t, 0) \\ \tilde{y}_{+}(t, 1) \end{pmatrix} + g(t),
\end{array}
\right.
\qquad (3.19)$$

Taking $\nu \to +\infty$, we have

$$L_{e^{-\nu}Q} = e^{-\nu}L_Q < 1.$$

From the previous section, system (3.19) has a unique solution. Therefore, the same is true for system (3.1). Besides, since $\nu$ depends only on $L_Q$, it is clear that the estimate (3.11) remains of the same form.

## 3.4 Classical solutions

In this part, we discuss classical solutions to system (3.1). We recall that a function $y$ is a classical solution if $y \in C^1(\bar{R}_T)^n$ and it satisfies the conditions in (3.1) for every $(t, x) \in \bar{R}_T$. All along this section, we assume that all the data are continuously differentiable:

$$\Lambda, M \in C^1(\bar{R}_T)^{n \times n}, \quad Q \in C^1([0, T])^{n \times n},$$

and

$$y^0 \in C^1([0, 1])^n, \quad f \in C^1(\bar{R}_T)^n, \quad g \in C^1([0, T])^n.$$

**Compatibility conditions.** To state the $C^1$ compatibility condition, it is convenient to write system (3.1) in the form

$$\left\{
\begin{array}{ll}
\displaystyle \frac{\partial y}{\partial t}(t, \cdot) = A(t)y(t, \cdot) + f(t, \cdot), & \\
\\
\displaystyle y(0, \cdot) = y^0, & t \in [0, T], \\
\\
\displaystyle B(t)y(t, \cdot) = g(t), &
\end{array}
\right.
$$

where, for every $t \in [0, T]$, the linear operator $A(t) : C^1([0, 1])^n \to C^0([0, 1])^n$ is defined by

$$(A(t)y)(x) = -\Lambda(t, x)y'(x) + M(t, x)y(x),$$

and the linear operator $B(t) : C^0([0, 1])^n \to \mathbb{R}^n$ is defined by

$$B(t)y = \begin{pmatrix} y_{-}(1) \\ y_{+}(0) \end{pmatrix} - Q(t) \begin{pmatrix} y_{-}(0) \\ y_{+}(1) \end{pmatrix}.$$

It is clear that, for a classical solution to exist, the following $C^1$ compatibility conditions have to hold:

$$\begin{align}
B(0)y^0 &= g(0), \tag{3.20} \\
B'(0)y^0 + B(0)A(0)y^0 &= g'(0) - B(0)f(0, \cdot), \nonumber
\end{align}$$

where

$$B'(t)y = -Q'(t) \begin{pmatrix} y_{-}(0) \\ y_{+}(1) \end{pmatrix}.$$

Note that the first condition in (3.20) is nothing but the $C^0$ compatibility condition (3.8).

<figure><img src="image_49.png" /></figure>
**The constants** $L'_M, L'_Q$ and $\beta_\Lambda, \gamma_\Lambda$. All along Section 3.4, we will work again with the norm (3.9) on $\mathbb{R}^n$. Let us now introduce some important constants.

We denote by $L'_M$ the smallest $L \ge 0$ such that

$$ \left| \frac{\partial M_k}{\partial x}(t, x)y + M_k(t, x)y' \right| \le L(\|y\|_{\mathbb{R}^n} + \|y'\|_{\mathbb{R}^n}), $$

for every $1 \le k \le n$, $(t, x) \in \overline{R_T}$ and $y, y' \in \mathbb{R}^n$. Similarly, we denote by $L'_Q$ the smallest $L \ge 0$ such that

$$ |Q'_k(t)y + Q_k(t)y'| \le L(\|y\|_{\mathbb{R}^n} + \|y'\|_{\mathbb{R}^n}), $$

for every $1 \le k \le n, t \in [0, T]$ and $y, y' \in \mathbb{R}^n$. Their existence are guaranteed since $M, Q$ and their derivatives are in particular bounded.

Let us now point out that $\chi_k$ is $C^1$ in all its variables, i.e. $\chi_k \in C^1(\mathbb{R}^3)$, since $\lambda_k$ is $C^1$ (see e.g. [Har02, Chapter V, Theorem 3.1]). We can then define

$$ \beta_{\Lambda} = \sup \left\{ \left| \frac{\partial \chi_k}{\partial t}(s; t, x) \right|, \left| \frac{\partial \chi_k}{\partial x}(s; t, x) \right| \mid 1 \le k \le n, \ s \in [(s_k^{\text{in}})^+(t, x), t], \ (t, x) \in \overline{R_T} \right\}. $$

It also follows that $s_k^{\text{in}}$ is $C^1$ on $\overline{R_T} \setminus V_k$ by the implicit function theorem. We can then define

$$ \gamma_{\Lambda} = \sup \left\{ \left| \frac{\partial s_k^{\text{in}}}{\partial t}(t, x) \right|, \left| \frac{\partial s_k^{\text{in}}}{\partial x}(t, x) \right| \mid 1 \le k \le n, \ (t, x) \in \overline{V_k}^c \right\}. $$

Observe that all of these constants depend a priori also on $T$ and thus they may in particular grow with respect to $T$. Note however that they become bounded with respect to $T$ if we assume for instance that $M, Q$ and their derivatives are defined and bounded on $[0, +\infty) \times [0, 1]$ and $[0, +\infty)$, respectively, and $\Lambda$ is bounded $[0, +\infty) \times [0, 1]$ and satisfies (3.10) (see Appendix A.2).

One of the goals of this section is to prove the following well-posedness result for classical solutions.

**Theorem 3.4.1.** For every $T > 0$, $y^0 \in C^1([0, 1])^n$, $f \in C^1(\overline{R_T})^n$ and $g \in C^1([0, T])^n$ satisfying the $C^1$ compatibility conditions (3.20), there exists a unique classical solution $y$ to system (3.1). Moreover, it satisfies the estimate

$$ \|y\|_{C^1(\overline{R_T})^n} \le C e^{\omega T} \left( \|y^0\|_{C^1([0,1])^n} + \|f\|_{C^1(\overline{R_T})^n} + \|g\|_{C^1([0,T])^n} \right), \quad (3.21) $$

for some $\omega > 0$ (resp. $C > 0$) depending only on $L'_M, L'_Q, \beta_\Lambda, \gamma_\Lambda$ (resp. $L'_Q, \alpha_\Lambda, \beta_\Lambda, \gamma_\Lambda$).

Recall that a $C^1$ function $y$ that satisfies the system of integral equations $y = F(y)$ is a classical solution (see the end of Section 3.2).

**The norm** $|\cdot|_1$. For the sake of brevity, it will be convenient to use the notation

$$ |h|_1 = \|h\|_{C^1(\bar{\Omega})^n}, $$

for any continuously differentiable function $h: \bar{\Omega} \to \mathbb{R}^n$, where $\Omega$ is an open subset of either $\mathbb{R}$ or $\mathbb{R}^2$ (there will be no risk of confusion).

### 3.4.1 A priori estimates

In this section we establish a priori estimates for classical solutions.

<figure><img src="image_50.png" /></figure>
**Computation of the derivatives of F(y).** We first compute the derivatives of F(y) for an arbitrary $y \in C^1(\overline{R_T})^n$. Note that $F(y)$ is $C^1$ on $\overline{V_k}$ and on the closure of $\overline{V_k}^c$, i.e. $\overline{R_T} \setminus V_k$. Let $1 \le k \le n$ be fixed. For $(t, x) \in \overline{V_k}$, we have

$$ \frac{\partial F(y)_k}{\partial t}(t, x) = (y_k^0)'(\chi_k(0; t, x)) \frac{\partial \chi_k}{\partial t}(0; t, x) \\ + M_k(t, x)y(t, x) + f_k(t, x) \\ + \int_0^t \left( \frac{\partial M_k}{\partial x}(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) + M_k(s, \chi_k(s; t, x)) \frac{\partial y}{\partial x}(s, \chi_k(s; t, x)) + \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) \right) \frac{\partial \chi_k}{\partial t}(s; t, x) ds. $$

For $(t, x) \in \overline{R_T} \setminus V_k$, we have

$$ \frac{\partial F(y)_k}{\partial t}(t, x) = \left( Q'_k(s_k^{\text{in}}(t, x)) \begin{pmatrix} y_-(s_k^{\text{in}}(t, x), 0) \\ y_+(s_k^{\text{in}}(t, x), 1) \end{pmatrix} + Q_k(s_k^{\text{in}}(t, x)) \begin{pmatrix} \frac{\partial y_-}{\partial t}(s_k^{\text{in}}(t, x), 0) \\ \frac{\partial y_+}{\partial t}(s_k^{\text{in}}(t, x), 1) \end{pmatrix} + g'_k(s_k^{\text{in}}(t, x)) \right. \\ \left. - (M_k(s_k^{\text{in}}(t, x), \delta_k)y(s_k^{\text{in}}(t, x), \delta_k) + f_k(s_k^{\text{in}}(t, x), \delta_k)) \right) \frac{\partial s_k^{\text{in}}}{\partial t}(t, x) \\ + M_k(t, x)y(t, x) + f_k(t, x) \\ + \int_{s_k^{\text{in}}(t, x)}^t \left( M_k(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) + M_k(s, \chi_k(s; t, x)) \frac{\partial y}{\partial x}(s, \chi_k(s; t, x)) + \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) \right) \frac{\partial \chi_k}{\partial t}(s; t, x) ds, $$

where we recall that $\delta_k = 1$ if $k \le m$ and $\delta_k = 0$ if $k \ge m + 1$. The space derivative of $F(y)$ has a similar expression.

**Estimate of the derivatives of F(y).** We now estimate the derivatives of $F(y)$, where $y \in C^1(\overline{R_T})^n$ is still arbitrary. Let $1 \le k \le n$ be fixed. For $(t, x) \in \overline{V_k}$, we have

$$ \left| \frac{\partial F(y)_k}{\partial t}(t, x) \right| \le \left| (y_k^0)'(\chi_k(0; t, x)) \right| \left| \frac{\partial \chi_k}{\partial t}(0; t, x) \right| + |M_k(t, x)y(t, x)| + |f_k(t, x)| \\ + \int_0^t \left( \left| \frac{\partial M_k}{\partial x}(s, \chi_k(s; t, x))y(s, \chi_k(s; t, x)) \right| + \left| \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) \right| \right) \left| \frac{\partial \chi_k}{\partial t}(s; t, x) \right| ds. $$

Using the definitions of $\beta_\Lambda$, $L'_M$ and $\alpha_\Lambda$, we obtain

$$ \left| \frac{\partial F(y)_k}{\partial t}(t, x) \right| \le \beta_\Lambda |y^0|_1 + L'_M |y(t, \cdot)|_0 + |f|_0 + L'_M \beta_\Lambda \int_0^t |y(s, \cdot)|_1 ds + \alpha_\Lambda \beta_\Lambda |f|_1. $$

<figure><img src="image_51.png" /></figure>
Similarly, for $(t, x) \in \overline{R_T} \setminus V_k$, noting that $s_k^{\text{in}}(t, x) \le t$, we can show that

$$ \left| \frac{\partial F(y)_k}{\partial t}(t, x) \right| \le \left( L'_Q \left( \sup_{s \le t} |y(s, \cdot)|_0 + \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 \right) + |g|_1 + L'_M \sup_{s \le t} |y(s, \cdot)|_0 + |f|_0 \right) \gamma_\Lambda \\ + L'_M |y(t, \cdot)|_0 + |f|_0 + L'_M \beta_\Lambda \int_0^t |y(s, \cdot)|_1 ds + \alpha_\Lambda \beta_\Lambda |f|_1. $$

Gathering both estimates, we get

$$ \sup_{x \in [0,1]} \left\| \frac{\partial F(y)}{\partial t}(t,x) \right\|_{\mathbb{R}^n} \le \omega_1 \sup_{s \le t} |y(s, \cdot)|_0 + L'_Q \gamma_\Lambda \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 \\ + C_1 (|y^0|_1 + |f|_1 + |g|_1) + L'_M \beta_\Lambda \int_0^t |y(s, \cdot)|_1 ds, \quad (3.22) $$

for every $t \in [0, T]$, with $C_1 = \max \{\beta_\Lambda, \gamma_\Lambda + 1 + \alpha_\Lambda \beta_\Lambda\}$ and $\omega_1 = (L'_Q + L'_M)\gamma_\Lambda + L'_M$.
The space derivative of $F(y)$ satisfies the same estimate, with same constants.

**A priori estimate.** Below, we shall denote by $\omega_2, \omega_3, \dots$ (resp. $C_2, C_3, \dots$) nonnegative numbers that may change from line to line, but that only depends on $L'_M, L'_Q, \beta_\Lambda, \gamma_\Lambda$ (resp. $L'_Q, \alpha_\Lambda, \beta_\Lambda, \gamma_\Lambda$).

Let now $y$ be any classical solution to system (3.1). Thus, we have $y = F(y)$. Assuming that

$$ L'_Q < 1, $$

we can use the $C^0$ a priori estimate (3.14) (note that $L_Q \le L'_Q$ and $L_M \le L'_M$) to bound the first term in (3.22) and obtain

$$ \left| \frac{\partial y}{\partial t}(t, \cdot) \right|_0 \le L'_Q \gamma_\Lambda \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 + C_2 (|y^0|_1 + |f|_1 + |g|_1) + \omega_2 \int_0^t |y(s, \cdot)|_1 ds, $$

for every $t \in [0, T]$. Assuming in addition that

$$ L'_Q \gamma_\Lambda < 1, $$

we can absorb the first term and obtain

$$ \left| \frac{\partial y}{\partial t}(t, \cdot) \right|_0 \le C_3 (|y^0|_1 + |f|_1 + |g|_1) + \omega_3 \int_0^t |y(s, \cdot)|_1 ds, $$

for every $t \in [0, T]$. Now that a bound is known for $\frac{\partial y}{\partial t}$, we can deduce a similar bound for $\frac{\partial y}{\partial x}$. Recalling again (3.14), we deduce the following final estimate:

$$ |y(t, \cdot)|_1 + \left| \frac{\partial y}{\partial t}(t, \cdot) \right|_0 \le C_4 (|y^0|_1 + |f|_1 + |g|_1) + \omega_4 \int_0^t |y(s, \cdot)|_1 ds, $$

for every $t \in [0, T]$. Applying Grönwall's lemma, we deduce estimate (3.21).

As explained in Section 3.3.3, we can always assume that $Q$ is small enough so that $L'_Q$ meets the previous requirements.

### 3.4.2 Existence

We are going to apply the Banach fixed-point theorem.

<figure><img src="image_52.png" /></figure>
**Metric space.** Since we look for a solution that is a continuously differentiable function, we would like to work in the space $C^1(\overline{R}_T)^n \cap E_0$ (recall from Section 3.3.2 that we need to work in a subspace of $E_0$ for $F(y)$ to be continuous). However, $F$ does not map this space into itself. Recall that the closure of $\overline{V}_k^c$ is $\overline{R}_T \setminus V_k$, and that $\overline{V}_k \cap (\overline{R}_T \setminus V_k) = \{s_k^{\text{in}}(t, x) = 0\}$. Thus, for the derivatives of $F(y)$ to be continuous near this set, we see that $y$ has to satisfy, after making use of the identities (see Appendix A.1.2)

$$ \lambda_k(s_k^{\text{in}}(t, x), \delta_k) \frac{\partial s_k^{\text{in}}}{\partial t}(t, x) + \frac{\partial \chi_k}{\partial t}(s_k^{\text{in}}(t, x); t, x) = 0, $$

$$ \lambda_k(s_k^{\text{in}}(t, x), \delta_k) \frac{\partial s_k^{\text{in}}}{\partial x}(t, x) + \frac{\partial \chi_k}{\partial x}(s_k^{\text{in}}(t, x); t, x) = 0, $$

the following relations:

$$ - \lambda_k(0, \delta_k)(y_k^0)'(\delta_k) = Q_k'(0) \begin{pmatrix} y_-(0, 0) \\ y_+(0, 1) \end{pmatrix} + Q_k(0) \begin{pmatrix} \frac{\partial y_-}{\partial t}(0, 0) \\ \frac{\partial y_+}{\partial t}(0, 1) \end{pmatrix} + g_k'(0) \\ - (M_k(0, \delta_k)y(0, \delta_k) + f_k(0, \delta_k)), \quad 1 \le k \le n. \quad (3.23) $$

This leads us to work in

$$ E_1 = \{ y \in C^1(\overline{R}_T)^n \cap E_0 \mid y \text{ satisfies } (3.23) \}. $$

Then, $F(E_1) \subset C^1(\overline{R}_T)^n$, and we easily check that $F$ maps $E_1$ into itself thanks to the $C^1$ compatibility conditions (3.20) and the identity

$$ \frac{\partial \chi_k}{\partial t}(0; 0, x) = -\lambda_k(0, x), $$

(this follows by differentiating the identity $\chi_k(t; t, x) = x$ with respect to $t$). Let us also point out that $E_1$ is not empty since, thanks again to the compatibility conditions and the previous identity, it contains any function $y$ satisfying

$$ y \in C^1(\overline{R}_T)^n, \quad \begin{pmatrix} y_-(0, 0) \\ y_+(0, 1) \end{pmatrix} = \begin{pmatrix} y_0^-(0) \\ y_0^+(1) \end{pmatrix}, \quad \begin{pmatrix} \frac{\partial y}{\partial t} - (0, 0) \\ \frac{\partial y}{\partial t} + (0, 1) \end{pmatrix} = \begin{pmatrix} (A(0)y_0^0) - (0) \\ (A(0)y_0^0) + (1) \end{pmatrix} + \begin{pmatrix} f_-(0, 0) \\ f_+(0, 1) \end{pmatrix}, $$

(obviously, such functions exist). Clearly, $E_1$ is a complete metric space when equipped with the metric induced by the $C^1$ norm.

**Contraction.** We will now show that $F^2 = F \circ F$ is a contraction for a norm equivalent to the $C^1$ norm. Let $y^1, y^2 \in E_1$ and denote by $y = y^1 - y^2$. Mimicking the estimates of Section 3.4.1, we easily get

$$ \left| \frac{\partial F(y^1)}{\partial t}(t, \cdot) - \frac{\partial F(y^2)}{\partial t}(t, \cdot) \right|_0 \le \omega_1 \sup_{s \le t} |y(s, \cdot)|_0 + L'_Q \gamma_\Lambda \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 \\ + L'_M \beta_\Lambda \int_0^t |y(s, \cdot)|_1 ds, \quad (3.24) $$

and

$$ \left| \frac{\partial F(y^1)}{\partial x}(t, \cdot) - \frac{\partial F(y^2)}{\partial x}(t, \cdot) \right|_0 \le \omega_1 \sup_{s \le t} |y(s, \cdot)|_0 + L'_Q \gamma_\Lambda \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 \\ + L'_M \beta_\Lambda \int_0^t |y(s, \cdot)|_1 ds, \quad (3.25) $$

<figure><img src="image_53.png" /></figure>
for every $t \in [0, T]$, where we recall that $\omega_1 = (L'_Q + L'_M)\gamma_\Lambda + L'_M$. Since there is no reason for $\omega_1$ to be small, to bound the first term in the right-hand sides we iterate these estimates and use the $C^0$ estimate of $F$ obtained in Section 3.3.2. More precisely, from (3.24) we have

$$ \left| \frac{\partial F^2(y^1)}{\partial t}(t, \cdot) - \frac{\partial F^2(y^2)}{\partial t}(t, \cdot) \right|_0 \le \omega_1 \sup_{s \le t} \left| F(y^1)(s, \cdot) - F(y^2)(s, \cdot) \right|_0 \\ + L'_Q \gamma_\Lambda \sup_{s \le t} \left| \frac{\partial F(y^1)}{\partial t}(t, \cdot) - \frac{\partial F(y^2)}{\partial t}(t, \cdot) \right|_0 \\ + L'_M \beta_\Lambda \int_0^t \left| F(y^1)(s, \cdot) - F(y^2)(s, \cdot) \right|_1 ds, $$

for every $t \in [0, T]$. Using again (3.24) and (3.25) to estimate the second and third terms in the right-hand side, we get

$$ \left| \frac{\partial F^2(y^1)}{\partial t}(t, \cdot) - \frac{\partial F^2(y^2)}{\partial t}(t, \cdot) \right|_0 \le \omega_1 \sup_{s \le t} \left| F(y^1)(s, \cdot) - F(y^2)(s, \cdot) \right|_0 \\ + L'_M \beta_\Lambda \int_0^t \left| F(y^1)(s, \cdot) - F(y^2)(s, \cdot) \right|_0 ds, \tag{3.26} \\ + \varepsilon \left( \sup_{s \le t} |y(s, \cdot)|_0 + \sup_{s \le t} \left| \frac{\partial y}{\partial t}(s, \cdot) \right|_0 \right) \\ + \omega' \int_0^t \left( \sup_{\tau \le s} |y(\tau, \cdot)|_1 + \sup_{\tau \le s} \left| \frac{\partial y}{\partial t}(\tau, \cdot) \right|_0 \right) ds, $$

for every $t \in [0, T]$, with $\varepsilon = L'_Q \gamma_\Lambda \max \{\omega_1, L'_Q \gamma_\Lambda\}$ and $\omega' = L'_Q \gamma_\Lambda L'_M \beta_\Lambda + L'_M \beta_\Lambda (\omega_1 + L'_Q \gamma_\Lambda + L'_M \beta_\Lambda)$. The space derivative of $F^2(y^1) - F^2(y^2)$ obviously satisfies the same estimate. Clearly,

$$ \varepsilon \to 0 \quad \text{as} \quad L'_Q \to 0. \tag{3.27} $$

Thus, in the right-hand side of the previous estimate we see that there are three kind of terms: the first two terms can be estimated thanks to the $C^0$ estimate of $F$ obtained in Section 3.3.2 (provided that $L'_Q < 1$), the second term is small and the remaining term is an integral term.

We have thus all the ingredients to conclude as in the proof of Theorem 3.3.2. We consider the weighted norm on $C^1(\overline{R_T})^n$ defined by

$$ |y|_{1,\omega} = |y|_{0,\omega} + \left| \frac{\partial y}{\partial t} \right|_{0,\omega} + \left| \frac{\partial y}{\partial x} \right|_{0,\omega}, $$

where $\omega > 0$ will be chosen below and we recall that $|\cdot|_{0,\omega}$ is defined in (3.17). Clearly, this norm is equivalent to the $C^1$ norm. From estimates (3.26) and (3.18), we deduce that

$$ \left| \frac{\partial F^2(y^1)}{\partial t} - \frac{\partial F^2(y^2)}{\partial t} \right|_{0,\omega} \le \rho'_\omega |y|_{1,\omega}, $$

with

$$ \rho'_\omega = \left( \omega_1 + \frac{L'_M \beta_\Lambda}{\omega} \right) \rho_\omega + \varepsilon + \frac{\omega'}{\omega}. $$

The space derivative of $F^2(y^1) - F^2(y^2)$ satisfies the same estimate. From these estimates and (3.18), we deduce the final estimate

$$ |F^2(y^1) - F^2(y^2)|_{1,\omega} \le \rho''_\omega |y|_{1,\omega}, $$

<figure><img src="image_54.png" /></figure>
with $\rho_{\omega}'' = \rho_{\omega}^2 + 2\rho_{\omega}'$. Clearly,

$$ \rho_{\omega}'' \rightarrow \rho_{\infty}'' = (L_Q)^2 + 2(\omega_1 L_Q + \varepsilon) \quad \text{as } \omega \rightarrow +\infty, $$

and (recalling that $L_Q \le L'_Q$ and (3.27))

$$ \rho_{\infty}'' \rightarrow 0 \quad \text{as} \quad L'_{Q} \rightarrow 0. $$

As a result, for any $Q$ small enough so that $\rho_{\infty}'' < 1$, we can take $\omega$ large enough to make $F^2$ to be a contraction. Therefore, by the Banach fixed-point theorem, $F^2$ has a unique fixed-point $y \in E_1$. The uniqueness implies that $y$ is also the unique fixed-point of $F$ in $E_1$. The proof of Theorem 3.4.1 is complete, for $Q$ small enough. Arguing as in Section 3.3.3, we see that the general case follows.

### 3.4.3 Systems satisfied by the derivatives

In this part, we show that the partial derivatives of the classical solution to system (3.1) are continuous solutions to systems of the same form, obtained after formal differentiation.

**Theorem 3.4.2.** Let $y^0 \in C^1([0, 1])^n$, $f \in C^1(\bar{R}_T)^n$ and $g \in C^1([0, T])^n$ satisfy the $C^1$ compatibility conditions (3.20). Let $y \in C^1(\bar{R}_T)^n$ be the classical solution to system (3.1). Then, $\varphi = \frac{\partial y}{\partial t}$ is the continuous solution to the system

$$ \left\{ \begin{array}{ll} \textbf{Equ}: & \displaystyle \frac{\partial \varphi}{\partial t}(t,x) + \Lambda(t,x) \frac{\partial \varphi}{\partial x}(t,x) = M(t,x)\varphi(t,x) + f^1(t,x), \\ \textbf{IC}: & \varphi(0,x) = y^1(x), \\ \textbf{BC}: & \begin{pmatrix} \varphi_{-}(t,1) \\ \varphi_{+}(t,0) \end{pmatrix} = Q(t) \begin{pmatrix} \varphi_{-}(t,0) \\ \varphi_{+}(t,1) \end{pmatrix} + g^1(t), \end{array} \right. \quad (t,x) \in \bar{R}_T, \quad (3.28) $$

with

$$ \begin{align*} y^1(x) &= -\Lambda(0,x)(y^0)'(x) + M(0,x)y^0(x) + f(0,x), \\ f^1(t,x) &= -\frac{\partial \Lambda}{\partial t}(t,x)\frac{\partial y}{\partial x}(t,x) + \frac{\partial M}{\partial t}(t,x)y(t,x) + \frac{\partial f}{\partial t}(t,x), \\ g^1(t) &= Q'(t) \begin{pmatrix} y_{-}(t,0) \\ y_{+}(t,1) \end{pmatrix} + g'(t). \end{align*} $$

Note that $f^1$ depends on $\frac{\partial y}{\partial x}$.

We can show that $\frac{\partial y}{\partial x}$ satisfies a similar system but it will not be needed in the sequel.

*Proof.* 1. Reasoning as in Section 3.2, $\varphi$ is the solution to the system of integral equations corresponding to system (3.28) if, and only if, for every $1 \le k \le n$ and $(t, x) \in \bar{R}_T$,

$$ \Phi(s) = \varphi(s, \chi_k(s; t, x)), $$

satisfies the ODE

$$ \left\{ \begin{array}{l} \displaystyle \frac{d}{ds} \Phi_k(s) = M_k(s, \chi_k(s; t, x)) \Phi(s) + f_k^1(s, \chi_k(s; t, x)), \quad s \in [(s_k^{\text{in}})^+(t, x), t], \\ \Phi_k((s_k^{\text{in}})^+(t, x)) = \Phi_k^0, \end{array} \right. \quad (3.29) $$

where the initial condition $\Phi_k^0$ is given by the initial or boundary conditions in (3.28) (see (3.6) with $y^0, g$ replaced by $y^1, g^1$) and where the equation in (3.29) has to be satisfied only when $[(s_k^{\text{in}})^+(t, x), t]$ is not reduced to a point.

<figure><img src="image_55.png" /></figure>
2. The initial condition in (3.29) is an easy consequence of the conditions in (3.1). Let us now prove the equation. In fact, it will be simpler to establish the equation satisfied by the derivative with respect to x. From the equation in (3.1), we expect that

$$ \Psi(s) = \frac{\partial y}{\partial x}(s, \chi_k(s; t, x)), $$

satisfies

$$ \frac{d}{ds} \Psi_k(s) = -\frac{\partial M_k}{\partial x}(s, \chi_k(s; t, x))Y(s) + M_k(s, \chi_k(s; t, x))\Psi(s) \\ + \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) - \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x))\Psi_k(s), \quad (3.30) $$

for $s \in [(s_k^{in})^+(t, x), t]$, where $Y(s) = y(s, \chi_k(s; t, x))$. Let us first show that $\Phi$ will indeed satisfy the equation in (3.29) if $\Psi$ satisfies the above equation. From the equation satisfies by $y$, we have

$$ \Phi_k(s) = -\lambda_k(s, \chi_k(s; t, x))\Psi_k(s) + M_k(s, \chi_k(s; t, x))Y(s) + f_k(s, \chi_k(s; t, x)). $$

Thus, $\Phi_k$ is $C^1$ and we have

$$ \begin{aligned} \frac{d}{ds} \Phi_k(s) = & - \left( \frac{\partial \lambda_k}{\partial t}(s, \chi_k(s; t, x)) + \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x)) \lambda_k(s, \chi_k(s; t, x)) \right) \Psi_k(s) \\ & - \lambda_k(s, \chi_k(s; t, x)) \left( \frac{\partial M_k}{\partial x}(s, \chi_k(s; t, x)) Y(s) + M_k(s, \chi_k(s; t, x)) \Psi(s) \right. \\ & \left. + \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) - \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x)) \Psi_k(s) \right) \\ & + \left( \frac{\partial M_k}{\partial t}(s, \chi_k(s; t, x)) + \frac{\partial M_k}{\partial x}(s, \chi_k(s; t, x)) \lambda_k(s, \chi_k(s; t, x)) \right) Y(s) \\ & + M_k(s, \chi_k(s; t, x)) \frac{d}{ds} Y(s) \\ & + \frac{\partial f_k}{\partial t}(s, \chi_k(s; t, x)) + \frac{\partial f_k}{\partial x}(s, \chi_k(s; t, x)) \lambda_k(s, \chi_k(s; t, x)). \end{aligned} $$

After straightforward simplifications, we get

$$ \begin{aligned} \frac{d}{ds} \Phi_k(s) = & - \frac{\partial \lambda_k}{\partial t}(s, \chi_k(s; t, x)) \Psi_k(s) \\ & - \lambda_k(s, \chi_k(s; t, x)) M_k(s, \chi_k(s; t, x)) \Psi(s) \\ & + \frac{\partial M_k}{\partial t}(s, \chi_k(s; t, x)) Y(s) \\ & + M_k(s, \chi_k(s; t, x)) \frac{d}{ds} Y(s) \\ & + \frac{\partial f_k}{\partial t}(s, \chi_k(s; t, x)). \end{aligned} $$

Using the definition of $f^1$, this can be rewritten as

$$ \begin{aligned} \frac{d}{ds} \Phi_k(s) = & f^1(s, \chi_k(s; t, x)) \\ & - \lambda_k(s, \chi_k(s; t, x)) M_k(s, \chi_k(s; t, x)) \Psi(s) \\ & + M_k(s, \chi_k(s; t, x)) \frac{d}{ds} Y(s). \end{aligned} $$

<figure><img src="image_56.png" /></figure>
To conclude, we just observe that

$$-\lambda_k(s, \chi_k(s; t, x))\Psi(s) + \frac{d}{ds}Y(s) = \Phi(s).$$

3. Let us now show that $\Psi$ indeed satisfies (3.30). Taking the derivative of the identity (3.7) with respect to $x$, we obtain, for $(t, x) \in \bar{V}_k$,

$$\frac{\partial y_k}{\partial x}(t, x) = (y_k^0)'(\chi_k(0; t, x)) \frac{\partial \chi_k}{\partial x}(0; t, x) + \int_0^t \left( \frac{\partial M_k}{\partial x}(\theta, \chi_k(\theta; t, x)) y(\theta, \chi_k(\theta; t, x)) + M_k(\theta, \chi_k(\theta; t, x)) \frac{\partial y}{\partial x}(\theta, \chi_k(\theta; t, x)) + \frac{\partial f_k}{\partial x}(\theta, \chi_k(\theta; t, x)) \right) \frac{\partial \chi_k}{\partial x}(\theta; t, x) d\theta.$$

Evaluated along the k-th characteristic, this gives, using the identity $\chi_k(\theta; s, \chi_k(s; t, x)) = \chi_k(\theta; t, x)$ (see Appendix A.1.1),

$$\Psi_k(s) = (y_k^0)'(\chi_k(0; t, x)) \frac{\partial \chi_k}{\partial x}(0; s, \chi_k(s; t, x)) + \int_0^s \left( \frac{\partial M_k}{\partial x}(\theta, \chi_k(\theta; t, x)) y(\theta, \chi_k(\theta; t, x)) + M_k(\theta, \chi_k(\theta; t, x)) \frac{\partial y}{\partial x}(\theta, \chi_k(\theta; t, x)) + \frac{\partial f_k}{\partial x}(\theta, \chi_k(\theta; t, x)) \right) \frac{\partial \chi_k}{\partial x}(\theta; s, \chi_k(s; t, x)) d\theta.$$

Using that $s \mapsto \frac{\partial \chi_k}{\partial x}(\theta; s, \chi_k(s; t, x))$ is $C^1$ and (see Appendix A.1.1)

$$\frac{d}{ds} \frac{\partial \chi_k}{\partial x}(\theta; s, \chi_k(s; t, x)) = - \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x)) \frac{\partial \chi_k}{\partial x}(\theta; s, \chi_k(s; t, x)).$$

we obtain that $\Psi_k$ is $C^1$ and satisfies (3.30).

Similarly, we can check that $\Psi$ satisfies (3.30) for $(t, x) \in \bar{V}_k^c$ too, by making use of the identity (see Appendix A.1.2)

$$\frac{d}{ds} \frac{\partial s_k^{\text{in}}}{\partial x}(s, \chi_k(s; t, x)) = - \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x)) \frac{\partial s_k^{\text{in}}}{\partial x}(s, \chi_k(s; t, x)).$$

□

## 3.5 Smooth solutions

In this part, we discuss higher regularity for the classical solution to system (3.1). All along this section, we assume that all the data are $C^k$ for some $k \ge 1$:

$$\Lambda, M \in C^k(\overline{R_T})^{n \times n}, \quad Q \in C^k([0, T])^{n \times n}, \qquad (3.31)$$

and

$$y^0 \in C^k([0, 1])^n, \quad f \in C^k(\overline{R_T})^n, \quad g \in C^k([0, T])^n. \qquad (3.32)$$

<figure><img src="image_57.png" /></figure>
**Compatibility conditions.** As we did in Section 3.4, to state the $C^k$ compatibility condition for $k \ge 1$, it is convenient to write system (3.1) in the form

$$ egin{cases} 
\frac{\partial y}{\partial t}(t, \cdot) = A(t)y(t, \cdot) + f(t, \cdot), \\
y(0, \cdot) = y^0, \\
B(t)y(t, \cdot) = g(t), 
\end{cases} 
\quad t \in [0, T], $$

where, for every $t \in [0, T]$, the linear operator $A(t) : C^1([0, 1])^n \to C^0([0, 1])^n$ is defined by

$$ (A(t)y)(x) = -\Lambda(t, x)y'(x) + M(t, x)y(x), $$

and the linear operator $B(t) : C^0([0, 1])^n \to \mathbb{R}^n$ is defined by

$$ B(t)y = \begin{pmatrix} y_{-}(1) \\ y_{+}(0) \end{pmatrix} - Q(t) \begin{pmatrix} y_{-}(0) \\ y_{+}(1) \end{pmatrix}. $$

If $y$ is a classical solution to the system which is $C^k$, then, for any $l \in \{1, \dots, k\}$, we can take $l-1$ times the derivative with respect to $t$ of the first identity and the $l$ times the derivative of the last identity to obtain

$$ \frac{\partial^l y}{\partial t^l}(t, \cdot) = \sum_{i=0}^{l-1} \binom{l-1}{i} A^{(l-1-i)}(t) \frac{\partial^i y}{\partial t^i}(t, \cdot) + \frac{\partial^{l-1} f}{\partial t^{l-1}}(t, \cdot), \quad (3.33) $$

$$ \sum_{i=0}^{l} \binom{l}{i} B^{(l-i)}(t) \frac{\partial^i y}{\partial t^i}(t, \cdot) = g^{(l)}(t), \quad (3.34) $$

where the linear operator $A^{(j)}(t) : C^1([0, 1])^n \to C^0([0, 1])^n$ is defined by

$$ (A^{(j)}(t)y)(x) = -\frac{\partial^j \Lambda}{\partial t^j}(t, x)y'(x) + \frac{\partial^j M}{\partial t^j}(t, x)y(x), $$

and the linear operator $B^{(j)}(t) : C^0([0, 1])^n \to \mathbb{R}^n$ is defined by

$$ B^{(0)}(t) = B(t), \quad B^{(j)}(t)y = -Q^{(j)}(t) \begin{pmatrix} y_{-}(0) \\ y_{+}(1) \end{pmatrix} \quad (j \ge 1). $$

At $t = 0$, $y(0, \cdot) = y^0$ is known, and the identities (3.33) and (3.34) give necessary conditions that $y^0$ has to satisfy. More precisely, this leads us to introduce the sequence of continuous functions

$$ y^0, \quad y^l = \sum_{i=0}^{l-1} \binom{l-1}{i} A^{(l-1-i)}(0)y^i + \frac{\partial^{l-1} f}{\partial t^{l-1}}(0, \cdot) \quad (l \ge 1). $$

Then, the $C^k$ compatibility conditions are:

$$ \sum_{i=0}^{l} \binom{l}{i} B^{(l-i)}(0) y^i = g^{(l)}(0), \quad 0 \le l \le k. \quad (3.35) $$

When $k = 0, 1$, this coincides with the compatibility conditions (3.8) and (3.20), respectively. We will now prove the following result.

<figure><img src="image_58.png" /></figure>
**Theorem 3.5.1.** Let $k \ge 2$. Assume that all the data are $C^k$ (i.e. (3.31) and (3.32) hold) and that the $C^k$ compatibility conditions (3.35) are satisfied. Then, the classical solution $y$ to system (3.1) belongs to $C^k(\overline{R}_T)^n$. Moreover, it satisfies the estimate

$$ \|y\|_{C^k(\overline{R}_T)^n} \le C e^{\omega T} \left( \|y^0\|_{C^k([0,1])^n} + \|f\|_{C^k(\overline{R}_T)^n} + \|g\|_{C^k([0,T])^n} \right), $$

for some $\omega > 0$ (resp. $C > 0$) that does not depend on $y^0$, $f$, $g$.

*Proof.* Let us start with $k = 2$. From the equation

$$ \frac{\partial y}{\partial t}(t, x) + \Lambda(t, x) \frac{\partial y}{\partial x}(t, x) = M(t, x)y(t, x) + f(t, x), \quad (3.36) $$

we only have to prove that $\frac{\partial y}{\partial t} \in C^1$. Let $\varphi = \frac{\partial y}{\partial t}$. Then, $\varphi \in C^0$ and, by Theorem 3.4.2, it is the continuous solution to system

$$ \left\{ \begin{array}{ll} \displaystyle \frac{\partial \varphi}{\partial t}(t, \cdot) = A(t)\varphi(t, \cdot) + f^1(t, \cdot), & \\ \varphi(0, \cdot) = y^1, & t \in [0, T], \\ B(t)\varphi(t, \cdot) = g^1(t), & \end{array} \right. $$

with

$$ \begin{aligned} y^1 &= A(0)y^0 + f(0, \cdot), \\ f^1(t, \cdot) &= A^{(1)}(t)y(t, \cdot) + \frac{\partial f}{\partial t}(t, \cdot), \\ g^1(t) &= -B^{(1)}(t)y(t, \cdot) + g'(t). \end{aligned} $$

Clearly, the $C^1$ compatibility conditions for this system are exactly the $C^2$ compatibility conditions for the system satisfied by $y$ without the $C^0$ compatibility condition. By assumption, all the data of this system are $C^1$, except for $f^1$ since it depends on $\frac{\partial y}{\partial x}$, which is only $C^0$. We replace it by using the equation (3.36). This gives the equivalent system

$$ \left\{ \begin{array}{ll} \displaystyle \frac{\partial \varphi}{\partial t}(t, \cdot) = \tilde{A}(t)\varphi(t, \cdot) + \tilde{f}^1(t, \cdot), & \\ \varphi(0, \cdot) = y^1, & t \in [0, T], \\ B(t)\varphi(t, \cdot) = g^1(t), & \end{array} \right. $$

with

$$ (\tilde{A}(t)\varphi)(x) = (A(t)\varphi)(x) + \frac{\partial \Lambda}{\partial t}(t, x)\Lambda(t, x)^{-1}\varphi(x), $$

and

$$ \tilde{f}^1(t, \cdot) = -\frac{\partial \Lambda}{\partial t}(t, x)\Lambda(t, x)^{-1}(M(t, x)y(t, x) + f(t, x)) + \frac{\partial M}{\partial t}(t, x)y(t, x) + \frac{\partial f}{\partial t}(t, x). $$

Now it is clear that all the data of this system are $C^1$. Obviously, the corresponding compatibility conditions remain satisfied. Therefore, we can apply Theorem 3.4.1 to deduce that $\varphi \in C^1$. In addition, this result also gives the estimate

$$ \|\varphi\|_{C^1(\overline{R}_T)^n} \le C e^{\omega T} \left( \|y^1\|_{C^1([0,1])^n} + \|\tilde{f}^1\|_{C^1(\overline{R}_T)^n} + \|g^1\|_{C^1([0,T])^n} \right), $$

for some $\omega > 0$ (resp. $C > 0$) that does not depend on $y^1$, $\tilde{f}^1$, $g^1$. The desired estimate now follows from the explicit formulas of $y^1$, $\tilde{f}^1$, $g^1$ and the $C^1$ estimate of $y$.

The result for an arbitrary $k \ge 2$ follows by reasoning by induction on $k$ and reproducing the previous arguments. $\square$

<figure><img src="image_59.png" /></figure>
## 3.6 Some exercises

**Exercise 3.6.1 (Less regularity).** Show that there exists a unique classical solution to system (3.1) if for $M, f$ we only assume that

$$M, \frac{\partial M}{\partial x} \in C^0(\overline{R_T})^{n \times n}, \quad f, \frac{\partial f}{\partial x} \in C^0(\overline{R_T})^n,$$

or

$$M, \frac{\partial M}{\partial t} \in C^0(\overline{R_T})^{n \times n}, \quad f, \frac{\partial f}{\partial t} \in C^0(\overline{R_T})^n.$$

**Exercise 3.6.2 (From nonlocal to local).** Transform system (3.1) into an equivalent system with local boundary conditions only, i.e. boundary conditions of the form

$$\begin{aligned} \hat{y}_-(t, 1) &= Q_1(t)\hat{y}_+(t, 1) + g_1(t), \\ \hat{y}_+(t, 0) &= Q_0(t)\hat{y}_-(t, 0) + g_0(t). \end{aligned}$$

**Exercise 3.6.3 (Unstable system).** Show that, for all $a > 0$, the system

$$\left\{ \begin{aligned} \frac{\partial y_1}{\partial t} - \frac{\partial y_1}{\partial x} &= ay_2, & y_1(t, 1) &= \left(1 + \frac{1}{a}\right) y_2(t, 1), \\ \frac{\partial y_2}{\partial t} + \frac{\partial y_2}{\partial x} &= ay_1, & y_2(t, 0) &= 0, \end{aligned} \right.$$

possesses a classical solution satisfying

$$\|y(t)\|_{L^2(0,1)^2} \to +\infty \quad \text{as } t \to +\infty.$$

**Exercise 3.6.4 (L² estimate).** Assume that $Q = 0$. Prove that the continuous solution to (3.1) satisfies

$$\|y\|_{L^2(R_T)^n} \le C \left( \|y^0\|_{L^2(0,1)^n} + \|f\|_{L^2(R_T)^n} + \|g\|_{L^2(0,T)^n} \right),$$

for some $C > 0$ that does not depend on $y^0$, $f$, $g$.

**Exercise 3.6.5 (Semilinear systems).** Prove the existence and uniqueness of classical solutions when we replace the equations in system (3.1) by

$$\frac{\partial y}{\partial t}(t, x) + \Lambda(t, x) \frac{\partial y}{\partial x}(t, x) = f(t, x, y(t, x)),$$

assuming that $f \in C^1(\overline{R_T} \times \mathbb{R}^n)^n$ and $f, \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ are globally Lipschitz in $y \in \mathbb{R}^n$, uniformly with respect to $(t, x) \in \overline{R_T}$.

## 3.7 Bibliographical notes

**Section 3.3.** The “trick” used at the end of this section to reduce the proofs to the case where the boundary coupling matrix is small is not new, see e.g. [AKMK24, Section 8.2]. These exponential weights in space sometimes also appear in inner products, see e.g. [BC16, Appendix A].

<figure><img src="image_60.png" /></figure>
**Section 3.4.** For the existence and uniqueness of classical solutions to the initial-boundary value problem (3.1), there are not many detailed proofs in the literature. Ideas about the proof are for instance mentioned in [LY85, Chapter 4, §3], with details apparently given in [LYS81] (but unfortunately for us, written in Chinese). On the other hand, let us mention that, for the initial value problem only (Cauchy problem), there are plenty of detailed proofs, see for instance [RJ83, Chapter 1, Section 7], [LY85, Chapter 1] and [Bre00, Chapter 3]. The proof presented in Section 3.4, which is solely based on the Banach fixed-point theorem and relies on proving that $F^2$ is a contraction is new up to the author's knowledge.

Let us conclude by mentioning very briefly a related topic.

**Stability.** There are many works devoted to the *stability* properties of such or related systems ([NRL86, XS02, Lic08, ADS23, AKMK24, HWZ24, BABM25]) and their nonlinear versions ([Li94, BC16, Hay21]).

<figure><img src="image_61.png" /></figure>
# Chapter 4

## A constructive method for the control of linear nonhomogeneous systems

*Prerequisites: Well-posedness results (Chapter 3).*

In this chapter, we present the first controllability result about nonhomogeneous systems, that is about systems of the form

$$ egin{cases} 
\mathbf{Equ}: & \frac{\partial y}{\partial t}(t, x) + \Lambda(x) \frac{\partial y}{\partial x}(t, x) = M(x)y(t, x), \\ 
\mathbf{IC}: & y(0, x) = y^0(x), \\ 
\mathbf{BC}: & y_-(t, 1) = u(t), \quad y_+(t, 0) = Qy_-(t, 0), 
\end{cases} 
\quad (t, x) \in \mathbb{R}_T, \quad (4.1) $$

with a nonzero internal coupling matrix $M$. The notations and assumptions are the same as in Chapters 2 and 3. However, all along this chapter, we will assume in addition that the speeds *do not cross*. Since they are continuous functions, we can then always sort them as follows:

$$ \lambda_1 \le \dots \le \lambda_m < 0 < \lambda_{m+1} \le \dots \le \lambda_{m+p}. \quad (4.2) $$

### 4.1 The control problem

Several functional settings could be considered for the control problem of system (4.1). All along this chapter we choose to work in $C^1$, assuming that all the coefficients are $C^1$:

$$ \Lambda, M \in C^1([0, 1])^{n \times n}. $$

By Theorem 3.4.1, system (4.1) is well-posed in this setting: for every $T > 0$, $y^0 \in C^1([0, 1])^n$ and $u \in C^1([0, T])^m$ satisfying the $C^1$ compatibility conditions

$$ y_{-}^{0}(1) = u(0), \qquad (Ay_{-}^{0})_{-}(1) = u'(0), \quad (4.3) $$

$$ y_{+}^{0}(0) - Qy_{-}^{0}(0) = 0, \qquad (Ay_{+}^{0})_{+}(0) - Q(Ay_{-}^{0})_{-}(0) = 0, \quad (4.4) $$

<figure><img src="image_62.png" /></figure>
where $(Ay)(x) = -\Lambda(x)y'(x) + M(x)y(x)$, there exists a unique classical solution $y \in C^1(\overline{R_T})^n$.

Since $y(t, \cdot) \in C^1([0, 1])^n$ for every $t \in [0, T]$, we can consider control problems in the state space $C^1([0, 1])^n$. It is clear that for the exact controllability condition $y(T, \cdot) = y^1$ to hold, it is necessary that $y^1$ and $u$ also satisfy the corresponding $C^1$ compatibility conditions, namely

$$y_{-}^{1}(1) = u(T), \qquad (Ay_{-})_{-}(1) = u'(T), \qquad (4.5)$$

$$y_{+}^{1}(0) - Qy_{-}^{1}(0) = 0, \qquad (Ay_{+})_{+}(0) - Q(Ay_{-})_{-}(0) = 0. \qquad (4.6)$$

**Definition 4.1.1.** Let $T > 0$ be fixed. We say that system (4.1) is:

* exactly controllable in time $T$ (below, (EC) in time $T$) if, for every $y^0, y^1 \in C^1([0, 1])^n$ satisfying the $C^1$ compatibility conditions (4.4) and (4.6), there exists a control $u \in C^1([0, T])^m$ satisfying the $C^1$ compatibility conditions (4.3) and (4.5) such that the corresponding solution $y$ to the system (4.1) satisfies $y(T, \cdot) = y^1$.

* null controllable in time $T$ (below, (NC) in time $T$) if the previous property holds at least for $y^1 = 0$.

Clearly, (EC) always implies (NC). In general, the converse implication does not hold. Nevertheless, for our system (4.1), we have equivalence if the boundary coupling matrix is full row rank, that is if

$$\mathrm{rank} \, Q = p.$$

The proof is exactly the same as that of Proposition 2.2.2.

Let us denote by $T_k$ the minimal control time of the transport equation with speed $\lambda_k$ (Definition 2.2.5). Observe that it follows from assumption 4.2 that

$$T_1 \le \cdots \le T_m \quad \text{and} \quad T_{m+1} \ge \cdots \ge T_{m+p}.$$

## 4.2 The constructive method

The goal of this chapter is to present a powerful method to solve control problems for nonhomogeneous systems. We will present this method in the context of the proof of the following result, and we refer to the exercises section below for more applications.

**Theorem 4.2.1.** Assume (4.2) and rank $Q = p$. Then, system (4.1) is (EC) in any time $T$ such that

$$T > T_m + T_{m+1}. \qquad (4.7)$$

*Proof.* The basic idea is to construct a $C^1$ function $y$ satisfying

$$\left\{ \begin{array}{l} \displaystyle \frac{\partial y}{\partial t}(t, x) + \Lambda(x) \frac{\partial y}{\partial x}(t, x) = M(x)y(t, x), \\ y(0, x) = y^0(x), \quad y(T, x) = 0, \\ y_+(t, 0) = Qy_-(t, 0), \end{array} \right. \qquad (4.8)$$

for every $(t, x) \in R_T$. Once such a $y$ has been constructed, the control is simply

$$u(t) = y_{-}(t, 1).$$

Note that, by construction, $u \in C^1$ and it satisfies the $C^1$ compatibility conditions.

The construction of $y$ is done in several steps as follows.

<figure><img src="image_63.png" /></figure>
1. Forward problem. We first consider the “forward in time” system

$$
\left\{
\begin{array}{ll}
\mathbf{Equ} : & \displaystyle \frac{\partial y^f}{\partial t}(t,x) + \Lambda(x) \frac{\partial y^f}{\partial x}(t,x) = M(x)y^f(t,x), \\
\mathbf{IC} : & y^f(0,x) = y^0(x), \\
\mathbf{BC} : & y^f_{-}(t,1) = u^f(t), \quad y^f_{+}(t,0) = Q y^f_{-}(t,0),
\end{array}
\right.
\qquad (t,x) \in \mathbb{R}^f,
$$

where the domain is

$$
R^f = (0, T_m) \times (0, 1),
$$

and $u^f$ is any function in $C^1([0, T_m])^m$ satisfying the corresponding $C^1$ compatibility conditions.
By Theorem 3.4.1, it has a unique classical solution $y^f$. Then, we set

$$
\varphi^f(t) = y^f(t, 0), \quad t \in (0, T_m).
$$

Clearly, $\varphi^f \in C^1$ and by construction it satisfies

$$
\varphi_{+}^{f}(t) = Q\varphi_{-}^{f}(t).
$$

2. Backward problem. By assumption on $Q$, there exists $S \in \mathbb{R}^{m \times p}$ such that $QS = \text{Id}_p$. We then mimic the previous step, but considering this time the “backward in time” system

$$
\left\{
\begin{array}{ll}
\mathbf{Equ} : & \displaystyle \frac{\partial y^b}{\partial t}(t,x) + \Lambda(x) \frac{\partial y^b}{\partial x}(t,x) = M(x)y^b(t,x), \\
\mathbf{FC} : & y^b(T,x) = y^1(x), \\
\mathbf{BC} : & y^b_{-}(t,0) = S y^b_{+}(t,0), \quad y^b_{+}(t,1) = u^b(t),
\end{array}
\right.
\qquad (t,x) \in \mathbb{R}^b,
$$

where the domain is

$$
R^b = (T - T_{m+1}, T) \times (0, 1),
$$

and $u^b$ is any function in $C^1([0, T])^m$ satisfying the corresponding $C^1$ compatibility conditions.
Note that this system can be put in a similar form to (4.1) after using the changes of variables
$t \mapsto T-t$ and $x \mapsto 1-x$. Thus, by Theorem 3.4.1, it has a unique classical solution $y^b$. Then,
we set

$$
\varphi^b(t) = y^b(t, 0), \quad t \in (T - T_{m+1}, T).
$$

Clearly, $\varphi^b \in C^1$ and by construction it satisfies $\varphi^b(t) = S\varphi^b_+(t)$ so that, multiplying on the left by $Q$,

$$
\varphi_{+}^{b}(t) = Q\varphi_{-}^{b}(t).
$$

3. Gluing the boundary conditions. By assumption (4.7), the interval $(T_m, T - T_{m+1})$ is not empty. We can then define

$$
\varphi(t) = \begin{cases} \varphi^f(t) & \text{if } t \in (0, T_m), \\ \bar{\varphi}(t) & \text{if } t \in [T_m, T - T_{m+1}], \\ \varphi^b(t) & \text{if } t \in (T - T_{m+1}, T), \end{cases}
$$

where $\bar{\varphi}$ is any function in $C^1([T_m, T - T_{m+1}])^m$ satisfying the matching conditions

$$
\begin{align*}
\bar{\varphi}(T_m) &= \varphi^f(T_m), & \bar{\varphi}(T - T_{m+1}) &= \varphi^b(T - T_{m+1}), \\
\bar{\varphi}'(T_m) &= (\varphi^f)'(T_m), & \bar{\varphi}'(T - T_{m+1}) &= (\varphi^b)'(T - T_{m+1}),
\end{align*}
$$

<figure><img src="image_64.png" /></figure>
and the boundary condition at $x = 0$, i.e. such that is such that

$$ \bar{\varphi}_{+}(t) = Q\bar{\varphi}_{-}(t). $$

Clearly, $\varphi \in C^1$ and it satisfies the boundary condition at $x = 0$ for all $t$:

$$ \varphi_{+}(t) = Q\varphi_{-}(t), \quad t \in (0, T). \qquad (4.9) $$

4. **Changing $t$ and $x$.** We now change to role of $t$ and $x$ and consider the function $\varphi$ defined in the previous step as an initial condition. We consider the system

$$ \left\{ \begin{array}{ll} \mathbf{Equ}: & \frac{\partial y}{\partial x}(t,x) + \Lambda(x)^{-1} \frac{\partial y}{\partial t}(t,x) = \Lambda(x)^{-1} M(x)y(t,x), \\ \mathbf{IC}: & y(t,0) = \varphi(t), \\ \mathbf{BC}: & y_{-}(T,x) = y_{-}^{1}(x), \quad y_{+}(0,x) = y_{+}^{0}(x), \end{array} \right. \quad (t,x) \in \mathbb{R}_{T}. $$

We can check that the necessary $C^1$ compatibility conditions are satisfied. Therefore, by Theorem 3.4.1, this ("time" dependent) system is well-posed (note that $(x, t) \mapsto \Lambda(x)^{-1}$ satisfies the assumptions of Section 3.1). This function $y$ is our candidate for satisfying (4.8).

5. **Conclusion.** The function $y$ constructed in the previous step obviously satisfies the equations in (4.8), the boundary condition at $x = 0$ is satisfied because of (4.9) and parts of the initial and final conditions are also satisfied. It remains to be sure that

$$ y_{-}(0,x) = y_{-}^{0}(x), \quad y_{+}(T,x) = y_{+}^{1}(x). $$

From a uniqueness result in triangular-like domains (proved in Section 4.3 below), we necessarily have

$$ y = y^f \quad \text{in} \quad \mathcal{T}^f, $$

where

$$ \mathcal{T}^f = \{ (t,x) \in \mathbb{R}^f \mid x < \chi_m(t; 0, 1) \}. \qquad (4.10) $$

Clearly, $(0, x) \in \overline{\mathcal{T}^f}$. Therefore, $y(0, x) = y^f(0, x) = y^0(x)$. Similarly, the final condition is obtained from

$$ y = y^b \quad \text{in} \quad \mathcal{T}^b, $$

where

$$ \mathcal{T}^b = \{ (t,x) \in \mathbb{R}^b \mid x < \chi_{m+1}(t; T, 1) \}. \qquad (4.11) $$

□

**Null controllability.** We have established a result for the (EC) property but it is straightforward to adapt the method presented in its proof to obtain a similar result for the (NC) property:

**Theorem 4.2.2.** Assume (4.2). Then, system (4.1) is (NC) in any time $T$ such that

$$ T > T_m + T_{m+1}. $$

*Proof.* We simply reproduce the proof of Theorem 4.2.1 with $y^1 = 0$ and taking this time $y^b = 0$. Note that the assumption rank $Q = p$ was used before only to construct $y^b$. □

<figure><img src="image_65.png" /></figure>
## 4.3 A uniqueness result in triangular-like domains

In this part we prove the following result, used at the end of the proof of Theorem 4.2.1.

**Theorem 4.3.1.** Assume (4.2).

* Let $\mathcal{T}^f$ be the domain defined by (4.10). Assume that $y \in C^1(\overline{\mathcal{T}^f})^n$ satisfies

$$ 
\left\{ 
\begin{array}{l} 
\displaystyle \frac{\partial y}{\partial t}(t, x) + \Lambda(x) \frac{\partial y}{\partial x}(t, x) = M(x)y(t, x), \\ 
y(t, 0) = 0, \\ 
y_+(0, x) = 0, 
\end{array} 
\right. 
\quad (t, x) \in \mathcal{T}^f. 
$$

Then, $y = 0$ in $\mathcal{T}^f$.

* Let $\mathcal{T}^b$ be the domain defined by (4.11). Assume that $y \in C^1(\overline{\mathcal{T}^b})^n$ satisfies

$$ 
\left\{ 
\begin{array}{l} 
\displaystyle \frac{\partial y}{\partial t}(t, x) + \Lambda(x) \frac{\partial y}{\partial x}(t, x) = M(x)y(t, x), \\ 
y(t, 0) = 0, \\ 
y_-(T, x) = 0, 
\end{array} 
\right. 
\quad (t, x) \in \mathcal{T}^b. 
$$

Then, $y = 0$ in $\mathcal{T}^b$.

*Proof.* We only prove the first point, the proof of the second one being similar. We refer to Figure 4.1 to gain a better understanding of the geometric situation.

1. We first show that, if $(t, x)$ is in the domain $\overline{\mathcal{T}^f}$, then so are all the characteristics for $k \le m$:

$$ (s, \chi_k(s; t, x)) \in \overline{\mathcal{T}^f}, \quad \forall s \in [t, s_k^{\text{out}}(t, x)], $$

where $s_k^{\text{out}}(t, x)$ is the unique solution $s$ to $\chi_k(s; t, x) = 0$. This is where we use assumption (4.2). Indeed, thanks to this assumption and a standard comparison principle for ODE (see e.g. [Har02, Chapter 3, Corollary 4.2]), or using the explicit formula (1.2), we have (see also Figure 4.1)

$$ \chi_k(s; t, x) \le \chi_m(s; t, x), \quad \forall s \ge t, $$

as well as

$$ x \le \chi_m(t; 0, 1) \implies \chi_m(s; t, x) \le \chi_m(s; 0, 1), \quad \forall s \ge t. $$

Therefore, for $(t, x) \in \overline{\mathcal{T}^f}$,

$$ \chi_k(s; t, x) \le \chi_m(s; 0, 1), \quad \forall s \ge t. $$

The first conclusion we draw is the existence of $s_k^{\text{out}}$. Indeed, the function $s \mapsto \chi_k(s; t, x)$ is continuous and, for every $(t, x) \in \overline{\mathcal{T}^f}$, we have

$$ \chi_k(t; t, x) = x \ge 0, \quad \chi_k(T_m; t, x) \le \chi_m(T_m; 0, 1) = 0, $$

(the last equality is basically the definition of $T_m$). Thus, by the intermediate value theorem, there exists a (unique, by monotonicity) solution $s_k^{\text{out}}(t, x) \in [t, T_m]$ to $\chi_k(s_k^{\text{out}}(t, x); t, x) = 0$. Finally, it is clear that $(s, \chi_k(s; t, x)) \in \overline{\mathcal{R}^f} = [0, T_m] \times [0, 1]$ since $s \in [t, s_k^{\text{out}}(t, x)] \subset [0, T_m]$ and $\chi_k(s; t, x) \in [0, x] \subset [0, 1]$ for $s \in [t, s_k^{\text{out}}(t, x)]$ by monotonicity.

<figure><img src="image_66.png" /></figure>
2. For $k \le m$ and $(t, x) \in \overline{T^f}$, we can then consider the solution along the $k$-th characteristic:

$$Y_l(s) = y_l(s, \chi_k(s; t, x)), \quad 1 \le l \le n.$$

For $l = k$ it satisfies the ODE

$$\begin{cases} \frac{d}{ds} Y_k(s) = \sum_{l=1}^{n} m_{kl}(\chi_k(s; t, x)) Y_l(s), & s \in [t, s_k^{\text{out}}(t, x)], \\ Y_k(s_k^{\text{out}}(t, x)) = 0, & \end{cases}$$

provided that the interval $[t, s_k^{\text{out}}(t, x)]$ is not reduced to a point. In that case, integrating we deduce that

$$y_k(t, x) = \sum_{l=1}^{n} \int_{t}^{s_k^{\text{out}}(t, x)} m_{kl}(\chi_k(s; t, x)) y_l(s, \chi_k(s; t, x)) ds.$$

Now observe that this identity remains also true when $[t, s_k^{\text{out}}(t, x)]$ is a point, since it happens only if $x = 0$ by very definition of $s_k^{\text{out}}$. Doing the change of variable $\xi = \chi_k(s; t, x)$, denoting by $s = \psi_k(\xi; t, x)$ the inverse change of variables, the previous equality becomes

$$y_k(t, x) = \sum_{l=1}^{n} \int_{0}^{x} m_{kl}(\xi) y_l(\psi_k(\xi; t, x), \xi) \frac{1}{\lambda_k(\xi)} d\xi. \quad (4.12)$$

On the other hand, for $k \ge m + 1$, we easily check that

$$(s, \chi_k(s; t, x)) \in \overline{T^f}, \quad \forall s \in [(s_k^{\text{in}})^+(t, x), t],$$

for every $(t, x) \in \overline{T^f}$ (see (3.4) for the definition of $(s_k^{\text{in}})^+$) and, using that $y_k(0, x) = y_k(t, 0) = 0$, we get

$$y_k(t, x) = \sum_{l=1}^{n} \int_{\chi_k((s_k^{\text{in}})^+(t, x); t, x)}^{x} m_{kl}(\xi) y_l(\psi_k(\xi; t, x), \xi) \frac{1}{\lambda_k(\xi)} d\xi. \quad (4.13)$$

This leads us to introduce the function

$$\bar{y}(x) = \max_{s:(s,x) \in \overline{T^f}} \|y(s,x)\|_{\mathbb{R}^n}, \quad x \in [0, 1].$$

We can check that $\bar{y} \in C^0([0, 1])$. From the equalities (4.12) and (4.13) we easily deduce that

$$\bar{y}(x) \le C \int_{0}^{x} \bar{y}(\xi) d\xi, \quad \forall x \in [0, 1],$$

with $C = \max_{k, \xi} \sum_{l=1}^{n} \frac{|m_{kl}(\xi)|}{|\lambda_k(\xi)|}$. This inequality implies that $\bar{y} = 0$ (by iterating the inequality, or using Grönwall's lemma). It follows that $y = 0$ in $\overline{T^f}$ as well.

□

<figure><img src="image_67.png" /></figure>
<figure><img src="image_68.png" /><figcaption>Figure 4.1: Example of domain $\mathcal{T}^f$.</figcaption></figure>

## 4.4 Some exercises

All the systems appearing below are considered with an initial condition at time $t = 0$ and they evolve on the domain $R_T = (0, T) \times (0, 1)$. We also sometimes drop the argument $(t, x)$ for $y$ for brevity. We recall that we always assume (4.2). The framework is $C^1$.

**Exercise 4.4.1 (Two-sided controllability).** *Show that the system*

$$ egin{cases} \frac{\partial y}{\partial t} + \Lambda(x) \frac{\partial y}{\partial x} = M(x)y, \\ y_{-}(t, 1) = u(t), \quad y_{+}(t, 0) = v(t), \end{cases} $$

where $u$ and $v$ are controls, is (EC) in any time $T$ such that

$$ T > \max\{T_m, T_{m+1}\}. $$

**Exercise 4.4.2 (Two-sided controllability, with less controls).** Let $Q \in \mathbb{R}^{(p-r) \times m}$ with $1 \le r < p$ being fixed (assuming $p \ge 2$). Show that the system

$$ egin{cases} \frac{\partial y}{\partial t} + \Lambda(x) \frac{\partial y}{\partial x} = M(x)y, \\ y_{-}(t, 1) = u(t), \quad y_{+}(t, 0) = \begin{pmatrix} v(t) \\ Qy_{-}(t, 0) \end{pmatrix}, \end{cases} $$

where $u$ and $v$ are controls, is (NC) in any time $T$ such that

$$ T > \max\{T_{m+r+1} + T_m, T_{m+1}\}. $$

Find additional condition(s) to ensure that this system is also (EC).

**Exercise 4.4.3 (Time-dependent systems).** Consider system (4.1) with time-dependent coefficients $\Lambda = \Lambda(t, x)$, $M = M(t, x)$ and $Q = Q(t)$. Assume that there exist $T_m > 0$ and $T_{m+1,m} > 0$ such that

$$ \chi_m(T_m; 0, 1) = 0, \quad \chi_{m+1}(T_{m+1,m}; T_m, 0) = 1. $$

Show that the system is (NC) in any time $T > T_{m+1,m}$. Find additional condition(s) to ensure that this system is also (EC).

<figure><img src="image_69.png" /></figure>
## 4.5 Bibliographical notes

**Section 4.2.** The control result in Theorem 4.2.1 was essentially first obtained in [Rus78, Section 3] (in the $L^2$ framework, and assuming that $Q$ is invertible), see Theorem 3.2 therein. The proof also relies on a constructive method, but different from the one presented in this chapter. The proof we have presented here is the one from [Li10], see Section 3.4 therein. It is based on the constructive method introduced in the papers [LR02, LR03] and developed afterwards, see [Li10].

The condition rank $Q = p$ is in fact also necessary for the (EC) to hold in some time. For $M = 0$ this was proved in Proposition 2.2.3. When $M \neq 0$ the proof is far from being trivial. We refer to [HO21].

It is often said that the control time $T_m + T_{m+1}$ is “sharp” (see e.g. [Li10, Remark 3.4]). The meaning is that this control time works for any coupling matrices $M$ and $Q$ and, at the same time, there exist particular $M$ and $Q$ such that the corresponding system is not controllable before this time, which is easy to see by simply considering

$$M = 0, \quad Q = \begin{pmatrix} 0 & \cdots & 0 & 1 \\ \vdots & & \vdots & \vdots \\ \vdots & & \vdots & \vdots \\ 0 & \cdots & 0 & 0 \end{pmatrix}.$$

In other words, $T_m + T_{m+1}$ is the infimum of control times with respect to both $M$ and $Q$. It is however important to point out that this does not mean that we cannot have a better control time for given matrices $M$ and $Q$, i.e. $T_m + T_{m+1}$ is not always the minimal control time. In fact, it is “rarely” the case, see e.g. Exercise 2.8.5.

**Section 4.3.** The uniqueness result of this section can apparently be found in [LY85], according to [Li10].

Let us now briefly mention other contexts where the constructive has proven effective.

**Quasilinear systems.** Even though we have presented the constructive method for systems that are linear, it has been originally introduced for *quasilinear* systems, see the references [LR02, LR03, Wan06, Li10].

**Internal controllability.** Everywhere in these notes, the controls are acting at the boundary. The situation where they act inside the equations, called *internal controllability*, was also studied in the literature. The constructive method was notably applied in that context in [LLQ24]. On this subject, we also refer to [HO24] for optimal results but with a different method, and to [ABCO17] for results about underactuated systems.

**Entropy solutions.** The constructive method has also proven effective in the context of so-called *entropy solutions*. We refer to the series of works [LY17, LY18, LY19] for more details.

<figure><img src="image_70.png" /></figure>
# Appendix A

## Some properties of the characteristics

In this appendix, we gather some useful properties of the characteristics. They were used all along these notes. Below, we use the notations of Section 3.1, but where we drop the index $k$ for the sake of the presentation. Unless otherwise mentioned, all the properties below are valid for any arguments ($s, t, x$, etc.).

### A.1 Identities

#### A.1.1 Identities for $\chi$

**Group property.** By uniqueness of the solution to the ODE (3.2), we have the identity

$$ \chi(s; \theta, \chi(\theta; t, x)) = \chi(s; t, x). \qquad (A.1) $$

**Derivatives.** Taking formally the derivative of the ODE (3.2) with respect to $x$, we expect that the derivative of $\chi(s; t, x)$ with respect to $x$ is solution to the linear ODE

$$ \left\{ \begin{aligned} \frac{d}{ds} \frac{\partial \chi}{\partial x}(s; t, x) &= \frac{\partial \lambda}{\partial x}(s, \chi(s; t, x)) \frac{\partial \chi}{\partial x}(s; t, x), \\ \frac{\partial \chi}{\partial x}(t; t, x) &= 1, \end{aligned} \right. $$

that is

$$ \frac{\partial \chi}{\partial x}(s; t, x) = e^{\int_{t}^{s} \frac{\partial \lambda}{\partial x}(\tau, \chi(\tau; t, x)) d\tau}. \qquad (A.2) $$

This is indeed true, provided that

$$ \frac{\partial \lambda}{\partial x} \text{ exists and is continuous,} \qquad (A.3) $$

we refer for instance to [Har02, Chapter V, Theorem 3.1] for a rigorous proof. Under this assumption, we also deduce from (A.2) and (A.1) the identity

$$ \frac{d}{d\theta} \frac{\partial \chi}{\partial x}(s; \theta, \chi(\theta; t, x)) = - \frac{\partial \lambda}{\partial x}(s, \chi(s; t, x)) \frac{\partial \chi}{\partial x}(s; \theta, \chi(\theta; t, x)). \qquad (A.4) $$

<figure><img src="image_71.png" /></figure>
Finally, taking the derivative of (A.1) with respect to $\theta$, using the chain rule, and then taking $\theta = t$, we get the relation

$$ \frac{\partial \chi}{\partial t}(s; t, x) + \lambda(t, x) \frac{\partial \chi}{\partial x}(s; t, x) = 0. \qquad (A.5) $$

Therefore, most properties for the derivative with respect to $t$ can be deduced from the derivatives with respect to $x$.

### A.1.2 Identities for $s^{\text{in}}$

Most properties for $s^{\text{in}}$ follow from its definition (3.3) and the properties of $\chi$. Below, $(t, x)$ is an arbitrary point of the closure of $\bar{V}^c$, i.e. $(t, x) \in \bar{R}_T \setminus V$.

**Group property.** From the group property (A.1), we have

$$ (s, \chi(s; t, x)) \in \bar{R}_T \setminus V, \quad \forall s \in [s^{\text{in}}(t, x), t], $$

and $s^{\text{in}}$ is constant along the characteristic:

$$ s^{\text{in}}(s, \chi(s; t, x)) = s^{\text{in}}(t, x), \quad \forall s \in [s^{\text{in}}(t, x), t]. \qquad (A.6) $$

**Derivatives.** Taking the derivative of (3.3) with respect to $x$, we get

$$ \lambda(s^{\text{in}}(t, x), \delta) \frac{\partial s^{\text{in}}}{\partial x}(t, x) + \frac{\partial \chi}{\partial x}(s^{\text{in}}(t, x); t, x) = 0. \qquad (A.7) $$

If we assume (A.3), then from (A.7), (A.6) and (A.4), we also have the identity

$$ \frac{d}{ds} \frac{\partial s_k^{\text{in}}}{\partial x}(s, \chi_k(s; t, x)) = - \frac{\partial \lambda_k}{\partial x}(s, \chi_k(s; t, x)) \frac{\partial s_k^{\text{in}}}{\partial x}(s, \chi_k(s; t, x)), \quad \forall s \in [s^{\text{in}}(t, x), t]. $$

Finally, taking the derivative of (A.6) with respect to $s$, using the chain rule, and then taking $s = t$, we get the relation

$$ \frac{\partial s^{\text{in}}}{\partial t}(t, x) + \lambda(t, x) \frac{\partial s^{\text{in}}}{\partial x}(t, x) = 0. \qquad (A.8) $$

Therefore, most properties for the derivative with respect to $t$ can be deduced from the derivatives with respect to $x$.

### A.1.3 Time-independent speed

Assume here only that the speed depends only on the space variable: $\lambda(t, x) = \lambda(x)$. Then, the ODE is

$$ \begin{cases} \frac{d}{ds} \chi(s; t, x) = \lambda(\chi(s; t, x)), & \forall s \in \mathbb{R}, \\ \chi(t; t, x) = x. \end{cases} $$

Formally, we have

$$ \frac{d\chi}{\lambda(\chi)} = ds. $$

Integrating, this gives

$$ \int_{t}^{\tau} \frac{d\chi}{\lambda(\chi)} = \int_{t}^{\tau} ds. $$

<figure><img src="image_72.png" /></figure>
Using the change of variable $\xi = \chi$, we get

$$ \int_x^\chi \frac{d\xi}{\lambda(\xi)} = \int_t^\tau ds. $$

Introducing

$$ \phi(x) = \int_0^x \frac{d\xi}{\lambda(\xi)}, $$

we thus obtain

$$ \phi(\chi) = \tau - t + \phi(x), $$

which gives

$$ \chi(=\chi(\tau; t, x)) = \phi^{-1}(\tau - t + \phi(x)), $$

provided that $\phi$ is invertible. Clearly, $\phi$ is strictly monotonic. It is invertible if $\phi(\mathbb{R}) = \mathbb{R}$. Since $\lambda$ is assumed Lipschitz on $\mathbb{R}$, we have

$$ |\lambda(\xi) - \lambda(0)| \le L |\xi|, $$

for some $L > 0$, for every $\xi \in \mathbb{R}$. Denote by $a = \lambda(0)$. By assumption $a \ne 0$. Assume for instance that $\lambda > 0$, the other case being similar. We have

$$ 0 < \lambda(\xi) \le L |\xi| + a, $$

so that

$$ \frac{1}{\lambda(\xi)} \ge \frac{1}{L |\xi| + a}. $$

Let $x \ge 0$. Integrating, we get

$$ \phi(x) \ge \int_0^x \frac{1}{L\xi + a} d\xi = \frac{1}{L} \ln \left( \frac{Lx + a}{a} \right). $$

Therefore, $\phi(x) \to +\infty$ as $x \to +\infty$. Similarly, we can show that $\phi(x) \to -\infty$ as $x \to -\infty$. Since $\phi$ is continuous, we get $\phi(\mathbb{R}) = \mathbb{R}$.

Finally, note that $s^{\text{in}}$ becomes also explicit in that case. Indeed, from its definition $\chi(s^{\text{in}}(t, x); t, x) = \delta$, we deduce that

$$ s^{\text{in}}(t, x) = t - \phi(x) + \phi(\delta), $$

(which actually makes sense for all $(t, x)$, not only for $(t, x)$ in the closure of $\overline{V}^c$).

## A.2 Estimates

All along this section, we assume that $\lambda$ is defined on the closure of $R = (0, +\infty) \times (0, 1)$. Note that we can in fact define $s^{\text{in}}(t, x)$ for all $(t, x) \in \overline{R}$ such that $\chi(0; t, x) \notin (0, 1)$. Thus, $(s^{\text{in}})^+)$ is also defined on $\overline{R}$.

In this section, we will assume that

$$ \lambda_{\min} = \inf_{(t,x) \in \overline{R}} |\lambda(t,x)| > 0. \qquad (A.9) $$

Let also $L_{\Lambda} \ge 0$ be such that

$$ |\lambda_k(t, x) - \lambda_k(t, x')| \le L_{\Lambda} |x - x'|, $$

<figure><img src="image_73.png" /></figure>
for every $t \ge 0$ and $x, x' \in [0, 1]$.

From now on, $(t, x) \in \bar{R}$ is arbitrary, unless otherwise mentioned.

Integrating the ODE (3.2) over $[(s^{\text{in}})^{+}(t, x), t]$, we deduce the following bound:

$$t - (s^{\text{in}})^{+}(t, x) \le \frac{1}{\lambda_{\min}}. \qquad (A.10)$$

Note that, without (A.9), the map $t \mapsto t - (s^{\text{in}})^{+}(t, x)$ can be unbounded (see Exercise 1.3.2).

Assume now that $\lambda$ is $C^1$. From the explicit formula (A.2), we have

$$\left| \frac{\partial \chi}{\partial x}(s; t, x) \right| \le e^{L_{\lambda}|s-t|},$$

for all $s$. Using (A.9) and (A.10), we get

$$\left| \frac{\partial \chi}{\partial x}(s; t, x) \right| \le e^{\frac{L_{\lambda}}{\lambda_{\min}}}, \quad \forall s \in [(s^{\text{in}})^{+}(t, x), t].$$

Combined with (A.7), we also deduce that

$$\left| \frac{\partial s^{\text{in}}}{\partial x}(t, x) \right| \le \frac{1}{\lambda_{\min}} e^{\frac{L_{\lambda}}{\lambda_{\min}}},$$

for every $(t, x)$ such $s^{\text{in}}(t, x)$ is defined.

Similar bounds for the derivatives with respect to $t$ can be obtained from the identities (A.5) and (A.8), if we assume in addition that $\lambda$ is bounded on $[0, +\infty) \times [0, 1]$.

<figure><img src="image_74.png" /></figure>
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