Feedback stabilization of a simplified model of fluid

Abstract

In this paper we study the dynamic feedback stability for some simplified model of fluid–structure interaction on a tree. We prove that, under some conditions, the energy of the solutions of the dissipative system decay exponentially to zero when the time tends to infinity. Our technique is based on a frequency domain method and a special analysis for the resolvent.

Keywords

Dynamic feedback asymptotic stabilization resolvent method

1. Introduction

First of all, we introduce some notations needed to formulate the problem under consideration, as introduced in [1] or in [13]. Let $T$ be a tree (i.e. a planar connected graph without closed paths). By the multiplicity of a vertex of $T$ we mean the number of edges that branch out from the vertex. If the multiplicity is equal to one, the vertex is called exterior; otherwise, it is said to be interior. We denote by ${e_{1}, \dots, e_{N}}$ the set of edges of $T$ and ${a_{1}, \dots, a_{N + 1}}$ its set of vertices and we assume that $a_{1}$ is the root of $T$ which will be denoted by $R$ , that $e_{1}$ is the edge containing $R$ and $a_{2}$ is its vertex different from $R$ .

We denote by $M$ the set of the interior vertices of $T$ and by $S$ the set of the exterior vertices, except $R$ and denote $I_{M}$ and $I_{S}$ the sets of indices of interior and exterior vertices, except $R$ , respectively. Then $I = I_{M} \cup I_{S}$ is the set of indices of all vertices, except the root $R$ . We denote by J the set ${1, \dots, N}$ and for $k \in I$ we will denote by $J_{k}$ the set of indices of edges adjacent to $a_{k}$ . If $k \in I_{S}$ , then the index of the unique element of $J_{k}$ will be denoted by $j_{k}$ .

Furthermore, the length of the edge $e_{j}$ will be denoted by $ℓ_{j}$ . Then $e_{j}$ may be parametrized by its arc length by means of the function $π_{j} : [0, ℓ_{j}] ⟶ e_{j}, x_{j} ⟼ π_{j} (x_{j})$ , and sometimes identified with the interval $(0, ℓ_{j})$ .

For a function $y : T ⟶ C$ we set $y^{j} = y \circ π_{j}$ its restriction to the edge $e_{j}$ and we will denote $y^{j} (x) = y^{j} (π_{j} (x))$ for any x in $(0, ℓ_{j})$ .

The incidence matrix $D = {(d_{k j})}_{(N + 1) \times N}$ is defined by $\begin{matrix} d_{k j} = \{\begin{matrix} 1 & if π_{j} (ℓ_{j}) = a_{k}, \\ - 1 & if π_{j} (0) = a_{k}, \\ 0 & otherwise . \end{matrix} \end{matrix}$

We denote by $⟨ \cdot, \cdot ⟩$ and $‖ \cdot ‖$ the inner product and norm in $L^{2}$ -space, respectively.

Fig. 1.

A tree-shaped network.

In this paper, we study the stability of a model of fluid propagating in a 1-d network, under some feedbacks applied at exteriors nodes, with the presence of point mass at inner nodes, see Fig. 1. At rest, the network coincides with the tree $T$ .

More precisely, we consider the following initial and boundary value problem. $\begin{matrix} (1.1) & \{\begin{matrix} y_{t t}^{j} - y_{x x}^{j} = 0 in (0, ℓ_{j}) \times (0, \infty), j \in J, \\ \sum_{j \in J_{k}} d_{k j} y_{x}^{j} (a_{k}, t) = s_{k}^{'} (t), k \in I_{M}, \\ s_{k}^{″} (t) + s_{k} (t) = - y_{t} (a_{k}, t), k \in I_{M}, \\ y^{j} (a_{k}, t) = y^{l} (a_{k}, t), j, l \in J_{k}, k \in I_{M}, \\ y^{1} (a_{1}, t) = 0, \\ d_{k j_{k}} y_{x}^{j_{k}} (a_{k}, t) = - y_{t} (a_{k}, t), k \in I_{S}, \\ s_{k} (0) = s_{k, 0}, s_{k}^{'} (0) = s_{k, 1}, k \in I_{M}, \\ y^{j} (x, 0) = y_{0}^{j} (x), y_{t}^{j} (x, 0) = y_{k}^{1} (x), x \in (0, ℓ_{j}), j \in J, \end{matrix} \end{matrix}$ where $y^{j} : [0, ℓ_{j}] \times (0, \infty) ⟶ R$ , $j \in J$ represents the velocity potential of the fluid on the edge $e_{j}$ and $s_{k} : (0, \infty) ⟶ R$ , $k \in I_{M}$ denotes the movement of the point mass occupant the node $a_{k}$ . These functions allow us to identify the network with its rest graph. See [9] for more details.

This simplified model of fluid–structure interaction draws on the work of Ervedoza and Vanninathan [10], they consider a fluid occupying a domain in two dimensions and a solid immersed in it and prove some results of controllability of such system; see also [18] for more details. Then, we consider a corresponding one dimensional model (where the dimension of space is reduced to 1), with many mass points (which models the structure). We refer to [8] where the authors give some mathematical models of vibrations of fluid–solid structures corresponding to some physical situations, as the tube bundles vibrating inside a moving fluid in a nuclear reactors. The problem of fluid–structure interaction in one dimension has been studied by several authors. In [19] the authors study the asymptotic behavior of a one dimensional model of mass point moving in a fluid. They consider the same system in [20] but with a finite number of mass points floating in the fluid. Recently Tucsnak et al. [12] studied the controllability of a similar system. In [21], the authors modeled a biological problem (an intracranial saccular aneurysm), into a coupled fluid–structure interaction problem, in one dimension, consisting of a wave equation with a dynamical condition at one end.

Note that the point-wise (or boundary) stabilization on the wave equation has been treated during the last few years, see for example [2] for one string, and [3–5] for some networks of strings.

The main result of this paper asserts that, under some conditions, the energy of the solutions of the dissipative system decay exponentially to zero when the time tends to infinity. The method is based on a frequency domain method and a special analysis for the resolvent.

If $(y, s) = ({(y^{j})}_{j \in J}, {(s_{k})}_{k \in M})$ is a solution of (1.1) we define the energy of $(y, s)$ at instant t by $\begin{matrix} E (t) = \frac{1}{2} \sum_{j \in J} \int_{0}^{ℓ_{j}} ({| y_{t}^{j} (x, t) |}^{2} + {| y_{x}^{j} (x, t) |}^{2}) d x + \frac{1}{2} \sum_{k \in I_{M}} ({| s_{k}^{'} (t) |}^{2} + {| s_{k} (t) |}^{2}) . \end{matrix}$ Simple formal calculations show that a sufficiently smooth solution of (1.1) satisfies the energy estimate $\begin{matrix} (1.2) & E (0) - E (t) = \sum_{k \in I_{S}} \int_{0}^{t} {| y_{t}^{j_{k}} (a_{k}, s) |}^{2} d s, \forall t ⩾ 0 . \end{matrix}$ In particular (1.2) implies that $\begin{matrix} E (t) ⩽ E (0), \forall t ⩾ 0 . \end{matrix}$ Define $\begin{matrix} V = {Φ \in \prod_{j \in J} H^{1} (0, ℓ_{j}), Φ^{1} (R) = 0, Φ^{j} (a_{k}) = Φ^{l} (a_{k}), j, l \in J_{k}, k \in I_{M}}, \end{matrix}$ the natural wellposedness space for (1.1) is $\begin{matrix} H = V \times \prod_{j \in J} L^{2} (0, ℓ_{j}) \times {(\prod_{k \in I_{M}} C)}^{2} \end{matrix}$ endowed with the inner product $\begin{matrix} {⟨ Z_{1}, Z_{2} ⟩}_{H} = \sum_{j \in J} (\int_{0}^{ℓ_{j}} \partial_{x} f_{1}^{j} (x) \partial_{x} \overline{f_{2}^{j}} (x) d x + \int_{0}^{ℓ_{j}} g_{1}^{j} (x) \overline{g_{2}^{j}} (x) d x) + \sum_{k \in I_{M}} (c_{1, k} \overline{c_{2, k}} + d_{1, k} \overline{d_{2, k}}), \end{matrix}$ where $Z_{r} = (f_{r}, g_{r}, c_{r}, d_{r})$ for $r = 1, 2$ .

Then, we can rewrite the system (1.1) as a first order differential equation, by putting $z (t) = (y (t), y^{'} (t), s (t), s^{'} (t))$ : $\begin{matrix} z^{'} (t) = A z (t), z (0) = z_{0} = (y_{0}, y_{1}, s^{0}, s^{1}), \end{matrix}$ where $\begin{matrix} A (u, v, p, q) = (v, \frac{d^{2} y}{d x^{2}}, q, - p - v_{M}), \forall (u, v, p, q) \in D (A) \end{matrix}$ with $v_{M} = (v (a_{k}), k \in I_{M})$ , and $\begin{matrix} D (A) = {(y, v, p, q) \in [\prod_{j \in J} H^{2} (0, ℓ_{j}) \cap V] \times V \times {(\prod_{k \in I_{M}} C)}^{2} satisfying (1.3)}, \end{matrix}$ where $\begin{matrix} (1.3) & \{\begin{matrix} \sum_{j \in J_{k}} d_{k j} \frac{d y^{j}}{d x} (a_{k}) = q_{k}, \forall k \in I_{M}, \\ d_{k j} \frac{d y^{j_{k}}}{d x} (a_{k}) = - v^{j_{k}} (a_{k}), \forall k \in I_{S} . \end{matrix} \end{matrix}$ To simplify the notations, sometimes, we will write $y (a_{k})$ instead of $y^{j} (a_{k})$ for y in V.

The outline of this work is the following. In Section 2 we prove the existence and uniqueness of solutions for system (1.1). Section 3 is devoted to prove the exponential stability of the associated semi group. Finally, in Section 4, we prove the lack of exponential stability of a graph containing a circuit and a tree with two uncontrolled exterior nodes (Fig. 2, Fig. 3). The Section 5 is devoted to the study of the case of a chain with non equal mass points (Fig. 4).

2. Wellposedness

Lemma 2.1.
The operator $A$ generates a $C_{0}$ -semigroup of contraction ${(S (t))}_{t \in R^{+}}$ .
Proof.
It is clear that the operator $A$ is dissipative, moreover, for every $z = (y, v, p, q) \in D (A)$ , $\begin{matrix} (2.1) & Re ({⟨ A z, z ⟩}_{H}) = - \sum_{k \in I_{S}} {| v^{j_{k}} (a_{k}) |}^{2} ⩽ 0 . \end{matrix}$

Now we prove that every positive real number λ belongs to $ρ (A)$ , the resolvent set of $A$ . For this, let $Z = (f, g, c, d) \in H$ and we solve the equation $\begin{matrix} (2.2) & (λ - A) z = Z \end{matrix}$ with z in $D (A)$ .

We rewrite (2.2) explicitly $\begin{matrix} (2.3) & \{\begin{matrix} λ y^{j} - v^{j} = f^{j}, j \in J, \\ λ v^{j} - \frac{d^{2} y^{j}}{d x^{2}} = g^{j}, j \in J, \\ λ p_{k} - q_{k} = c_{k}, k \in I_{M}, \\ λ q_{k} + p_{k} + v (a_{k}) = d_{k}, k \in I_{M} . \end{matrix} \end{matrix}$ We eliminate $(v, q)$ in (2.3) to get $\begin{array}{l} (2.4) & λ^{2} y^{j} - \frac{d^{2} y^{j}}{d x^{2}} = g^{j} + λ f^{j}, j \in J, \\ (2.5) & (λ^{2} + 1) p_{k} + v (a_{k}) = d_{k} + λ c_{k}, k \in I_{M} . \end{array}$ Let w in V. Multiplying (2.4) by $w^{j}$ in $L^{2} (0, ℓ_{j})$ and summing over $j \in J$ , $\begin{array}{l} \sum_{j \in J} (λ^{2} \int_{0}^{ℓ_{j}} y^{j} \overline{w^{j}} d x + \int_{0}^{ℓ_{j}} \frac{d y^{j}}{d x} \frac{d \overline{w^{j}}}{d x} d x) - \sum_{k \in I_{M}} \overline{w (a_{k})} q_{k} + \sum_{k \in I_{S}} \overline{w (a_{k})} v (a_{k}) \\ (2.6) & = \sum_{j \in J} \int_{0}^{ℓ_{j}} (g^{j} + λ f^{j}) \overline{w^{j}} d x . \end{array}$ Multiplying (2.5) by $r_{k} \in C$ and summing over $k \in I_{M}$ , we get $\begin{matrix} (2.7) & (λ^{2} + 1) \sum_{k \in I_{M}} p_{k} \overline{r_{k}} + \sum_{k \in I_{M}} v (a_{k}) \overline{r_{k}} = \sum_{k \in I_{M}} (d_{k} + λ c_{k}) \overline{r_{k}} . \end{matrix}$ Summing (2.6) and (2.7) we get $\begin{array}{l} \sum_{j \in J} (λ^{2} ⟨ y^{j}, w^{j} ⟩ + ⟨ \frac{d y^{j}}{d x}, \frac{d w^{j}}{d x} ⟩) + (λ^{2} + 1) \sum_{k \in I_{M}} p_{k} \overline{r_{k}} \\ + \sum_{k \in I_{M}} (v (a_{k}) \overline{r_{k}} - \overline{w (a_{k})} q_{k}) + \sum_{k \in I_{S}} \overline{w (a_{k})} v (a_{k}) \\ = \sum_{j \in J} ⟨ g^{j} + λ f^{j}, w^{j} ⟩ + \sum_{k \in I_{M}} (d_{k} + λ c_{k}) \overline{r_{k}} \end{array}$ to obtain $\begin{matrix} a ((y, p), (w, r)) = f (w, r), \end{matrix}$ where $\begin{array}{rcl} a ((y, p), (w, r)) & = & \sum_{j \in J} (λ^{2} ⟨ y^{j}, w^{j} ⟩ + ⟨ \frac{d y^{j}}{d x}, \frac{d w^{j}}{d x} ⟩) + (λ^{2} + 1) \sum_{k \in I_{M}} p_{k} \overline{r_{k}} \\ + λ \sum_{k \in I_{M}} (y (a_{k}) \overline{r_{k}} - \overline{w (a_{k})} p_{k}) + λ \sum_{k \in I_{S}} \overline{w (a_{k})} y (a_{k}) \end{array}$ and $\begin{array}{rcl} f (w, r) & = & \sum_{j \in J} ⟨ g^{j} + λ f^{j}, w^{j} ⟩ + \sum_{k \in I_{M}} (d_{k} + λ c_{k}) \overline{r_{k}} - \sum_{k \in I_{M}} \overline{w (a_{k})} c_{k} \\ + \sum_{k \in I_{M}} f (a_{k}) \overline{r_{k}} + \sum_{k \in I_{S}} \overline{w (a_{k})} f (a_{k}) . \end{array}$

a is a continuous sesquilinear form on $V \times \prod_{k \in I_{M}} C$ and f is a continuous anti-linear form on $V \times \prod_{k \in I_{M}} C$ . Moreover $\begin{array}{rcl} a ((w, r), (w, r)) & = & \sum_{j \in J} (λ^{2} {‖ w^{j} ‖}^{2} + {‖ \frac{d w^{j}}{d x} ‖}^{2}) + (λ^{2} + 1) \sum_{k \in I_{M}} | r_{k} |^{2} \\ - 2 i λ Im (\sum_{k \in I_{M}} \overline{w (a_{k})} r_{k}) + λ \sum_{k \in I_{S}} {| w (a_{k}) |}^{2} . \end{array}$ We have $\begin{matrix} | a ((w, r), (w, r)) | ⩾ [\sum_{j \in J} (λ^{2} {‖ w^{j} ‖}^{2} + {‖ \frac{d w^{j}}{d x} ‖}^{2}) + λ^{2} \sum_{k \in I_{M}} | r_{k} |^{2}] \end{matrix}$ that is, a is coercive. The conclusion result immediately from the Lax Milgram lemma. □

According to the Lumer Phillips theorem [14] we have the following result: Proposition 2.2.
Suppose that $(y_{0}, y_{1}, s^{0}, s^{1}) \in H$ . Then the problem ( 1.1 ) admits a unique solution $\begin{matrix} (y, y^{'}, s, s^{'}) \in C ([0, + \infty); H) . \end{matrix}$ If $(y_{0}, y_{1}, s^{0}, s^{1}) \in D (A)$ then $\begin{matrix} (y, y^{'}, s, s^{'}) \in C ([0, + \infty), D (A)) \cap C^{1} ([0, + \infty); H) . \end{matrix}$ Moreover $(y, s)$ satisfies the energy estimate ( 1.2 ).

3. Exponential stability

Its clear that if $T$ contains an edge $e_{j}$ , not attached to a leaf, with length $ℓ_{j} \in π N$ then i is eigenvalue of $A$ with eigenvector $z = (y, v, p, q)$ such that $y^{j} = i sin x$ , $v^{j} = - sin x$ , $p_{k} = 1$ and $q_{k} = i$ when $a_{k}$ is the nearest end of $e_{j}$ to the root $R$ , and all the other components of z are null.

In the following, the tree $T$ is said to be a Pi-tree if the length of every edge not attached to a leaf is different from $m π$ for every m in $N^{*}$ . Then we have the following result:

Lemma 3.1.
The spectrum of $A$ contains no point on the imaginary axis if and only if $T$ is a Pi-tree.
Proof.
By the Sobolev embedding theorem, we can deduce that ${(I - A)}^{- 1}$ is a compact operator. Then the spectrum of $A$ only consists of eigenvalues. We will show that the equation $\begin{matrix} (3.1) & A z = i β z \end{matrix}$ with $\begin{matrix} z = (\begin{matrix} y \\ v \\ p \\ q \end{matrix}) \in D (A) and β \in R, \end{matrix}$ has only trivial solution.

By taking the inner product of (3.1) with $z \in H$ and using that $\begin{matrix} Re ({⟨ A z, z ⟩}_{H}) = - \sum_{k \in I_{S}} {| v (a_{k}) |}^{2} \end{matrix}$ we obtain that $\begin{matrix} (3.2) & v (a_{k}) = 0 for k \in I_{S} . \end{matrix}$ Furthermore, by the second condition in (1.3) we also have $\begin{matrix} (3.3) & \frac{d y}{d x} (a_{k}) = 0 for k \in I_{S} . \end{matrix}$ Now the equation (3.1) can be rewritten explicitly as $\begin{array}{l} (3.4) & v^{j} = i β y^{j}, j \in J, \\ (3.5) & \frac{d^{2} y^{j}}{d x^{2}} = i β v^{j}, j \in J, \\ (3.6) & q_{k} = i β p_{k}, k \in I_{M}, \\ (3.7) & - p_{k} - v (a_{k}) = i β q_{κ}, k \in I_{M} . \end{array}$

If $β = 0$ then $v = 0$ , $q = 0$ and $p = 0$ .

Multiplying the second equation in the above system by $y^{j}$ and then summing over j, we obtain $\begin{matrix} \sum_{j \in J} {‖ \frac{d y^{j}}{d x} ‖}^{2} = 0, \end{matrix}$ which implies, using continuity condition of y at inner nodes and its Dirichlet condition at $R$ , that $y = 0$ .

Next, we suppose that $β \neq 0$ . From (3.2) and (3.4) we get $\begin{matrix} y^{j_{k}} (a_{k}) = 0 for k \in I_{S} . \end{matrix}$ Using (3.4)–(3.7), we have $\begin{array}{l} (3.8) & β^{2} y^{j} + \frac{d^{2} y^{j}}{d x^{2}} = 0, j \in J, \\ (3.9) & (β^{2} - 1) p_{k} = v (a_{k}), k \in I_{M} . \end{array}$ The function $y^{j}$ , $j \in J$ is then of the form $\begin{matrix} y^{j} = α_{1} sin (β x) + α_{2} cos (β x) . \end{matrix}$ Using (3.2), (3.3) and (3.8) we obtain that $y^{j_{k}} = 0$ , for every k in $I_{S}$ , then using (3.4), $v^{j_{k}} = 0$ , for every k in $I_{S}$ .

For the sequel of the proof we consider two cases:

First case: $β^{2} = 1$ . From (3.9), (3.4) and the Dirichlet condition at $R$ , we deduce that $y^{j} (0) = y^{j} (ℓ_{j}) = 0$ for $j \in J$ . Since $T$ is a Pi-tree we conclude that $y^{j} = 0$ for $j \in J$ . Return back to the balance conditions and (3.6), one can deduce that $q = p = 0$ and hence $z = 0$ .

Second case: $β^{2} \neq 1$ . Let $a_{k}$ the second end of an edge $e_{j}$ attached to a leaf. Since $y^{j}$ and $v^{j}$ are zero, and using (3.9) and (3.6), we deduce that $p_{k} = 0$ and $q_{k} = 0$ . Next, let $e_{l}$ be an internal edge (i.e. not containing a leaf) attached at a node $a_{k}$ to an edge $e^{j}$ ended by a leaf. Using again the balance condition and the continuity of y at $a_{k}$ , we obtain that $y^{l} (a_{k}) = 0$ and $\frac{d y^{l}}{d x} (a_{k}) = 0$ . Then by (3.8) $y^{l} = 0$ . We iterate such procedure from leaves to root to conclude that $z = 0$ . □

The main result of this paper concerns the precise asymptotic behavior of the solution of (1.1). Our technique is based on a frequency domain method and a special analysis for the resolvent.

Recall that the system (1.1) is said to be exponentially stable if there exists two constants $M, ω > 0$ , such that for all $(y_{0}, y_{1}, s^{0}, s^{1}) \in H$ , $\begin{matrix} E (t) ⩽ M e^{- ω t} {‖ (y_{0}, y_{1}, s^{0}, s^{1}) ‖}_{H}^{2}, \forall t ⩾ 0 . \end{matrix}$ Then, our main result is the following: Theorem 3.2.
The system defined by equations ( 1.1 ) is exponentially stable if and only if $T$ is a Pi-tree.
Proof.
The system defined by equations (1.1) is exponentially stable if and only if the $C_{0}$ -semigroup of contraction ${(S (t))}_{t \in R^{+}}$ , generated by $A$ is exponentially stable.

By classical result (see Huang [11] and Prüss [15]) it suffices to show that the operator $A$ satisfies the following two conditions: $\begin{matrix} (3.10) & ρ (A) \supset {i β ∣ β \in R} \equiv i R \end{matrix}$ and $\begin{matrix} (3.11) & lim sup_{| β | \to \infty} ‖ {(i β - A)}^{- 1} ‖ < \infty . \end{matrix}$

By Lemma 3.1 the condition (3.10) is satisfied if and only if $T$ is a Pi-tree. Then we suppose that $T$ is a Pi-tree and by contradiction, we suppose that the condition (3.11) is false. By the Banach–Steinhaus Theorem (see [7]), there exist a sequence of real numbers $β_{n} \to \infty$ ( $β_{n} > 0$ without loss of generality) and a sequence of vector $z_{n} = (y_{n}, v_{n}, p_{n}, q_{n}) \in D (A)$ with $‖ z_{n} ‖_{H} = 1$ such that $\begin{matrix} (3.12) & {‖ (i β_{n} I - A) z_{n} ‖}_{H} ⟶ 0 as n ⟶ \infty, \end{matrix}$ i.e. $\begin{array}{l} (3.13) & i β_{n} y_{n}^{j} - v_{n}^{j} \equiv f_{n}^{j} ⟶ 0 in H^{1} (0, ℓ_{j}), \\ (3.14) & i β_{n} v_{n}^{j} - \frac{d^{2} y_{n}^{j}}{d x^{2}} \equiv g_{n}^{j} ⟶ 0 in L^{2} (0, ℓ_{j}), \\ (3.15) & i β_{n} p_{k, n} - q_{k, n} \equiv h_{k, n} ⟶ 0 in C, \\ (3.16) & i β_{n} q_{k, n} + p_{k, n} + v_{n} (a_{k}) \equiv r_{k, n} ⟶ 0 in C, \end{array}$ for every j in J and every k in $I_{M}$ . Our goal is to derive from (3.12) that $‖ z_{n} ‖_{H}$ converge to zero, thus, a contradiction.

The proof is divided in three steps:

First step. Recall that for every j in J, $\begin{matrix} \frac{‖ v^{j} ‖_{\infty}}{β_{n}^{1 / 2}} ⩽ \frac{C_{1}}{β_{n}^{1 / 2}} {‖ \frac{d v^{j}}{d x} ‖}^{1 / 2} {‖ v^{j} ‖}^{1 / 2} + C_{2} \frac{‖ v^{j} ‖}{β_{n}^{1 / 2}}, \end{matrix}$ for some positives constants $C_{1}$ and $C_{2}$ . This implies, using (3.13), that $\frac{‖ v_{n}^{j} ‖_{\infty}}{β_{n}^{1 / 2}}$ is bounded. Then, for every k in $I_{M}$ , by dividing (3.16) by $β_{n}$ we deduce that $q_{k, n}$ converge to zero, and then $β_{n} p_{k, n}$ converge to zero in view of (3.15). In particular $p_{k, n}$ converge to zero.

Second step. We notice that from (2.1) we have $\begin{matrix} (3.17) & {‖ (i β_{n} I - A) z_{n} ‖}_{H} ⩾ | Re {⟨ (i β_{n} I - A) z_{n}, z_{n} ⟩}_{H} | = \sum_{k \in I_{S}} {| v_{n} (a_{k}) |}^{2} . \end{matrix}$ Then, by (3.12) $\begin{matrix} | v_{n} (a_{k}) | ⟶ 0, \forall k \in I_{S} . \end{matrix}$ This further leads to $\begin{matrix} (3.18) & | β_{n} y_{n} (a_{k}) | ⟶ 0, \forall k \in I_{S} \end{matrix}$ due to (3.13) and the trace theorem.

We have also $\begin{matrix} (3.19) & \frac{d y^{j_{k}}}{d x} (a_{k}) ⟶ 0, \forall k \in I_{S} . \end{matrix}$

Third step. Substituting (3.13) into (3.14) to get $\begin{matrix} (3.20) & - β_{n}^{2} y_{n}^{j} - \frac{d^{2} y_{n}^{j}}{d x^{2}} = g_{n}^{j} + i β_{n} f_{n}^{j}, j \in J . \end{matrix}$ Next, we take the inner product of (3.20) with $\frac{d y_{n}^{j}}{d x} b$ in $L^{2} (0, ℓ_{j})$ for $b \in C^{1} ([0, ℓ_{j}])$ , we get $\begin{array}{l} \frac{1}{2} β_{n}^{2} {[{| y_{n}^{j} (x) |}^{2} b (x)]}_{0}^{ℓ_{j}} + \frac{1}{2} {[{| \frac{d y_{n}^{j}}{d x} (x) |}^{2} b (x)]}_{0}^{ℓ_{j}} + {[Re (i β_{n} f_{n}^{j} (x) \overline{y_{n}^{j}} (x) b (x))]}_{0}^{ℓ_{j}} \\ (3.21) & - \frac{1}{2} \int_{0}^{ℓ_{j}} ({| β_{n} y_{n}^{j} |}^{2} + {| \frac{d y_{n}^{j}}{d x} |}^{2}) \frac{d b}{d x} (x) d x ⟶ 0 . \end{array}$ Using (3.18) and (3.19), this leads to $\begin{matrix} \frac{1}{2} \int_{0}^{ℓ_{j_{k}}} ({| β_{n} y_{n}^{j_{k}} |}^{2} + {| \frac{d y_{n}^{j_{k}}}{d x} |}^{2}) d x ⟶ 0 \end{matrix}$ for every k in $I_{S}$ , by taking $b = x$ or $b = ℓ_{j_{k}} - x$ . It follows that $\begin{matrix} (3.22) & \frac{1}{2} β_{n}^{2} {| y_{n}^{j_{k}} (a_{s}) |}^{2} + \frac{1}{2} {| \frac{d y_{n}^{j_{k}}}{d x} (a_{s}) |}^{2} + Re (i β_{n} f_{n}^{j_{k}} (a_{s}) \overline{y_{n}^{j_{k}}} (a_{s})) ⟶ 0, \end{matrix}$ where $a_{s}$ is the end of $e_{j_{k}}$ , different from $a_{k}$ .

We will show that all the terms in the first members of (3.22) converge to zero.

To do that, we use the following inequality $\begin{matrix} Re (i β_{n} f_{n}^{j_{k}} (a_{s}) \overline{y_{n}^{j_{k}}} (a_{s})) ⩾ - \frac{1}{4} β_{n}^{2} {| y_{n}^{j_{k}} (a_{s}) |}^{2} - {| f_{n}^{j_{k}} (a_{s}) |}^{2} \end{matrix}$ to obtain $\begin{array}{l} \frac{1}{2} β_{n}^{2} {| y_{n}^{j_{k}} (a_{s}) |}^{2} + \frac{1}{2} {| \frac{d y_{n}^{j_{k}}}{d x} (a_{s}) |}^{2} + Re (i β_{n} f_{n}^{j_{k}} (a_{s}) \overline{y_{n}^{j_{k}}} (a_{s})) + {| f_{n}^{j_{k}} (a_{s}) |}^{2} \\ (3.23) & ⩾ \frac{1}{4} β_{n}^{2} {| y_{n}^{j_{k}} (a_{s}) |}^{2} + \frac{1}{2} {| \frac{d y_{n}^{j_{k}}}{d x} (a_{s}) |}^{2} ⩾ 0 . \end{array}$

In addition, we have $f_{n}^{j_{k}} (a_{s}) ⟶ 0$ . Then, (3.23) combined with (3.22) implies that $\begin{matrix} β_{n} y_{n}^{j_{k}} (a_{s}) ⟶ 0 and \frac{d y_{n}^{j_{k}}}{d x} (a_{s}) ⟶ 0 . \end{matrix}$ Furthermore, we have also $\begin{matrix} Re (i β_{n} f_{n}^{j_{k}} (a_{s}) \overline{y_{n}^{j_{k}}} (a_{s})) ⟶ 0 . \end{matrix}$ We then conclude by iteration, as in [17], that for every j in J, $\begin{matrix} \int_{0}^{ℓ_{j}} ({| β_{n} y_{n}^{j} |}^{2} + {| \frac{d y_{n}^{j}}{d x} |}^{2}) d x ⟶ 0 . \end{matrix}$

Finally, in view of (3.13), we also get $\begin{matrix} ‖ v_{n}^{j} ‖ ⟶ 0, for j \in J . \end{matrix}$ In conclusion, $p_{k, n}$ and $q_{k, n}$ converge to zero for every k in $I_{M}$ and, $‖ \frac{d y_{n}^{j}}{d x} ‖$ and $‖ v_{n}^{j} ‖$ converge to zero for every j in J, which implies that $‖ z_{n} ‖_{H} ⟶ 0$ : clearly contradicts (3.12). □

4. Two examples of non exponential stability

In this section we consider two particular cases. In the first, (Fig. 2), there is a circuit in the graph. We prove that, even with much more controls, the exponential stability fails. In the second case (Fig. 3) we eliminate a control of a leaf. We prove that the new system is also non exponentially stable.

Fig. 2.

Circuit.

4.1. A circuit (Fig. 2)

In this part we suppose that $T$ contains a circuit (Fig. 2), with $ℓ_{1} = ℓ_{2} = ℓ_{3} = 1$ and with feedbacks at each inner node. Then the second equation in (1.1) will be $\begin{matrix} \sum_{j \in J_{k}} d_{k j} y_{x}^{j} (a_{k}, t) = s_{1}^{'} (t) - y_{t} (a_{k}, t), a_{k} \in I_{M} \end{matrix}$ and we can rewrite the system (1.1) in the Hilbert space H as $\begin{matrix} z^{'} (t) = A z (t) . \end{matrix}$ The associated state space is $\begin{matrix} H = V \times ({(L^{2} (0, 1))}^{3} \times L^{2} (0, ℓ_{4})) \times {(C^{3})}^{2} \end{matrix}$ with $\begin{array}{rcl} V & = & {Φ \in {(H^{1} (0, 1))}^{3} \times H^{1} (0, ℓ_{4}), Φ^{1} (1) = 0, Φ^{1} (0) = Φ^{2} (0) = Φ^{3} (0), \\ Φ^{2} (1) = Φ^{4} (0), Φ^{3} (1) = Φ^{4} (ℓ_{4})} . \end{array}$ The system (1.1) can be rewritten in the Hilbert space H as $\begin{matrix} z^{'} (t) = A z (t), z (0) = (y_{0}, y_{1}, s^{0}, s^{1}), \end{matrix}$ where $A$ is the operator defined on H by $\begin{matrix} A (y, v, p, q) = (v, \frac{d^{2} y}{d x^{2}}, q, - p - v_{M}), \forall (y, v, p, q) \in D (A) \end{matrix}$ with $v_{M} = (v (a_{2}), v (a_{3}), v (a_{4}))$ and $\begin{matrix} D (A) = {(y, v, p, q) \in [({(H^{2} (0, 1))}^{3} \times H^{2} (0, ℓ_{4})) \cap V] \times V \times {(C^{3})}^{2} satisfying (4.1)}, \end{matrix}$ where $\begin{matrix} (4.1) & \{\begin{matrix} - \sum_{j = 1}^{3} \frac{d y^{j}}{d x} (0) = q_{2} - v (a_{2}), \\ \frac{d y^{2}}{d x} (1) - \frac{d y^{4}}{d x} (0) = q_{3} - v (a_{3}), \\ \frac{d y^{3}}{d x} (1) + \frac{d y^{4}}{d x} (ℓ_{4}) = q_{4} - v (a_{4}) . \end{matrix} \end{matrix}$

The operator $A$ generates a $C_{0}$ -semigroup of contraction ${(S (t))}_{t ⩾ 0}$ satisfying the first result of asymptotic behavior:

Theorem 4.1.
${(S (t))}_{t ⩾ 0}$ is asymptotically stable if and only if $ℓ_{4}$ is irrational and not in $π Z$ .
Proof.
First, if $ℓ_{4}$ is in $π Z$ then i is an eigenvalue of $A$ (as in the case of a tree) and if $ℓ_{4} = \frac{a}{b}$ with a and b integer, then $i b π$ is an eigenvalue of $A$ .

Now, we suppose that $ℓ_{4} \notin Q$ and $ℓ_{4} \notin π Z$ . We only need to prove that $i R \subset ρ (A)$ . For this, we will prove that the equation $\begin{matrix} (4.2) & A z = i β z \end{matrix}$ with $z = (y, v, p, q) \in D (A)$ and $β \in R$ has only trivial solution.

The real part of the inner product of (4.2) with $z \in H$ is $\begin{matrix} (4.3) & Re ({⟨ A z, z ⟩}_{H}) = - \sum_{k \in I_{M}} {| v (a_{k}) |}^{2} = - ({| v (a_{2}) |}^{2} + {| v (a_{3}) |}^{2} + {| v (a_{4}) |}^{2}), \end{matrix}$ then $v (a_{k}) = 0$ for every $k \in 2, 3, 4$ . Now the equation (4.2) can be rewritten as follow, $\begin{array}{l} (4.4) & v^{j} = i β y^{j}, j \in J, \\ (4.5) & \frac{d^{2} y^{j}}{d x^{2}} = i β v^{j}, j \in J, \\ (4.6) & q_{k} = i β p_{k}, k \in 2, 3, 4, \\ (4.7) & - p_{k} - v (a_{k}) = i β q_{κ}, k \in 2, 3, 4 . \end{array}$

If $β = 0$ then, as in for the initial example, we show that $y = 0$ .

Next, we suppose that $β \neq 0$ . We have $\begin{matrix} (4.8) & y (a_{k}) = 0 for every k \in 2, 3, 4 \end{matrix}$ in view of (4.3)–(4.5), and $\begin{array}{l} (4.9) & β^{2} y^{j} + \frac{d^{2} y^{j}}{d x^{2}} = 0, j \in J, \\ (4.10) & (β^{2} - 1) p_{k} = 0, k \in 2, 3, 4 . \end{array}$ By using (4.4)–(4.7). The condition (4.8) implies that the solution of (4.9) is of the form: $\begin{matrix} y^{j} = α_{j} sin (β x), \end{matrix}$ where $α_{j} \in R$ . As in the case of a tree we consider two cases: $β^{2} = 1$ and $β^{2} \neq 1$ .

First case: $β^{2} = 1$ . We have, using again (4.5), $α_{1} = α_{2} = α_{3} = 0$ and since $ℓ_{4} \notin π Z$ , $α_{4} = 0$ . Return back to the balance condition and (4.6), at inner nodes, one can deduce that $q = p = 0$ and hence $z = 0$ .

Second case: $β^{2} \neq 1$ . We have $p_{k} = q_{k} = 0$ for every k in $I_{M}$ . Now, if $β \notin π Z$ then (4.7) gives that $α_{1} = α_{2} = α_{3} = 0$ and the balance condition at $a_{3}$ implies that $α_{4} = 0$ . If $β \in π Z$ then (4.7) gives $α_{4} = 0$ , since $ℓ_{4} \notin Q$ . Using again the balance condition respectively at $a_{3}$ , $a_{4}$ and $a_{2}$ we obtain that $α_{2} = α_{3} = α_{1} = 0$ . Then $z = 0$ . □
Theorem 4.2.
The semigroup ${(S (t))}_{t ⩾ 0}$ is not exponentially stable, even if $ℓ_{4}$ is irrational and not in $π Z$ .
Proof.
To prove that ${(S (t))}_{t ⩾ 0}$ is not exponentially stable we consider the sequence $f_{n}$ of vectors of H defined by $f_{n} = (0, g_{n}, 0, 0)$ where $g_{n} = (0, - sin β_{n} x, 0, 0)$ and $β_{n}$ is a sequence of real numbers satisfying $β_{n} ⟼ + \infty$ and that will be defined later. We then prove that the sequence $z_{n} = (y_{n}, v_{n}, p_{n}, q_{n})$ of elements of $D (A)$ such that $\begin{matrix} (i β_{n} - A) z_{n} = f_{n} \end{matrix}$ is not bounded.

The sequences $y_{n}$ and $q_{n}$ should satisfy $\begin{matrix} \{\begin{matrix} β_{n}^{2} y_{n}^{j} + \frac{d^{2} y_{n}^{j}}{d x^{2}} = 0, for j = 1, 3, 4, \\ β_{n}^{2} y_{n}^{2} + \frac{d^{2} y_{n}^{2}}{d x^{2}} = sin β_{n} x, \\ q_{k, n} = - \frac{β_{n}^{2}}{β_{n}^{2} - 1} y_{n} (a_{k}) = - β_{n} c_{n} y_{n} (a_{k}), for k = 2, 3, 4 \end{matrix} \end{matrix}$ with $c_{n} = \frac{β_{n}}{β_{n}^{2} - 1}$ . Then for $j = 1, 3, 4$ , there exists two complex numbers $a_{j}$ and $b_{j}$ (depending of n) such that $\begin{matrix} \{\begin{matrix} y_{n}^{j} = a_{j} sin (β_{n} x) + b_{j} cos (β_{n} x), \\ \frac{d y_{n}^{j}}{d x} = - β_{n} b_{j} sin (β_{n} x) + β_{n} a_{j} cos (β_{n} x) \end{matrix} \end{matrix}$ and there exists two complex numbers $a_{2}$ and $b_{2}$ (depending of n) such that $\begin{matrix} \{\begin{matrix} y_{n}^{2} = a_{2} sin (β_{n} x) + (- \frac{x}{2 β_{n}} + b_{2}) cos (β_{n} x), \\ \frac{d y_{n}^{2}}{d x} = (\frac{x}{2} - β_{n} b_{2}) sin (β_{n} x) + (- \frac{1}{2 β_{n}} + β_{n} a_{2}) cos (β_{n} x) . \end{matrix} \end{matrix}$ The boundary and transmission conditions are expressed as follows $\begin{matrix} (4.11) & \{\begin{matrix} a_{1} sin (β_{n}) + b_{1} cos (β_{n}) = 0, \\ b_{1} = b_{2} = b_{3}, \\ - \frac{1}{2 β_{n}} + β_{n} a_{1} + β_{n} a_{2} + β_{n} a_{3} = (i + c_{n}) β_{n} b_{1}, \\ a_{2} sin (β_{n}) + (- \frac{1}{2 β_{n}} + b_{2}) cos (β_{n}) = b_{4}, \\ β_{n} a_{4} - ((\frac{1}{2} - β_{n} b_{2}) sin (β_{n}) + (- \frac{1}{2 β_{n}} + β_{n} a_{2}) cos (β_{n})) = (i + c_{n}) β_{n} b_{4}, \\ a_{3} sin (β_{n}) + b_{3} cos (β_{n}) = a_{4} sin (β_{n} ℓ_{4}) + b_{4} cos (β_{n} ℓ_{4}), \\ - β_{n} b_{3} sin (β_{n}) + β_{n} a_{3} cos (β_{n}) - β_{n} b_{4} sin (β_{n} ℓ_{4}) + β_{n} a_{4} cos (β_{n} ℓ_{4}) \\ = - (i + c_{n}) β_{n} (a_{3} sin (β_{n}) + b_{3} cos (β_{n})) . \end{matrix} \end{matrix}$ Our goal is to prove that $β_{n} a_{3}$ converge to infinity. A straight-forward calculation leads to $\begin{matrix} (4.12) & (F A + G B) β_{n} a_{3} = B H + F C, \end{matrix}$ where $\begin{array}{l} A = sin β_{n} + ((i + c_{n}) sin β_{n} + cos β_{n}) sin β_{n} ℓ_{4} + sin β_{n} cos (β_{n} ℓ_{4}), \\ \begin{aligned} B = & - cos β_{n} + (\frac{{cos}^{2} β_{n}}{sin β_{n}} + 3 (i + c_{n}) cos β_{n} - (2 - c_{n}^{2} - 2 i c_{n}) sin β_{n}) sin (β_{n} ℓ_{4}) \\ + (2 cos β_{n} - (i + c_{n}) sin β_{n}) cos (β_{n} ℓ_{4}), \end{aligned} \\ C = (\frac{1}{2} (1 + \frac{i + c_{n}}{β_{n}}) sin β_{n} - \frac{i + c_{n}}{2} cos β_{n}) sin (β_{n} ℓ_{4}) + \frac{1}{2} (\frac{sin β_{n}}{β_{n}} - cos β_{n}) cos (β_{n} ℓ_{4}) \end{array}$ and $\begin{array}{l} \begin{aligned} F = & - sin β_{n} + (i + c_{n}) cos β_{n} - (2 cos β_{n} + (i + c_{n}) sin β_{n}) sin (β_{n} ℓ_{4}) \\ + (\frac{{cos}^{2} β_{n}}{sin β_{n}} + 3 (i + c_{n}) cos β_{n} - (2 - c_{n}^{2} - 2 i c_{n}) sin β_{n}) cos (β_{n} ℓ_{4}), \end{aligned} \\ \begin{aligned} G = & cos β_{n} + (i + c_{n}) sin β_{n} + sin β_{n} sin (β_{n} ℓ_{4}) \\ - ((i + c_{n}) sin β_{n} + cos β_{n}) cos (β_{n} ℓ_{4}), \end{aligned} \\ H = \frac{1}{2} (\frac{sin β_{n}}{β_{n}} - cos β_{n}) sin (β_{n} ℓ_{4}) - (\frac{1}{2} (1 + \frac{i + c_{n}}{β_{n}}) sin β_{n} - \frac{i + c_{n}}{2} cos β_{n}) cos (β_{n} ℓ_{4}) . \end{array}$ Now by using the Asymptotic Dirichlet’s theorem [16], there exists $(p_{n}, q_{n}) \in N^{2}$ such that $\frac{p_{n}}{q_{n}}$ converge to $ℓ_{4}$ , $p_{n}$ and $q_{n}$ tend to infinity as n goes to infinity and for every n in $N$ $\begin{matrix} | q_{n} ℓ_{4} - p_{n} | < \frac{1}{q_{n}} . \end{matrix}$ Take $β_{n} = 2 π q_{n} + \frac{2 π}{q_{n}^{1 / 4}}$ , then there exists a positive integer $n_{0}$ such that for every integer $n ⩾ n_{0}$ , $\begin{matrix} 0 < λ_{n} : = - \frac{2 π}{q_{n}} + \frac{2 π ℓ_{4}}{q_{n}^{1 / 4}} < β_{n} ℓ_{4} - 2 π p_{n} < μ_{n} : = \frac{2 π}{q_{n}} + \frac{2 π ℓ_{4}}{q_{n}^{1 / 4}} < \frac{π}{2} \end{matrix}$ and $\begin{matrix} sin (λ_{n}) < sin (β_{n} ℓ_{4}) < sin (μ_{n}), cos (μ_{n}) < cos (β_{n} ℓ_{4}) < cos (λ_{n}) . \end{matrix}$ Moreover $sin (β_{n} ℓ_{4})$ and $cos (β_{n} ℓ_{4})$ satisfy the following asymptotic approximations, $\begin{matrix} sin (β_{n} ℓ_{4}) = \frac{2 π ℓ_{4}}{q_{n}^{1 / 4}} + o (\frac{1}{q_{n}^{1 / 4}}), cos (β_{n} ℓ_{4}) = 1 + o (\frac{1}{q_{n}^{1 / 4}}) . \end{matrix}$ We have also $\begin{array}{rcl} sin (β_{n}) & = & sin (\frac{2 π}{q_{n}^{1 / 4}}) = \frac{2 π}{q_{n}^{1 / 4}} + o (\frac{1}{q_{n}^{1 / 4}}), cos (β_{n}) = 1 + o (\frac{1}{q_{n}^{1 / 4}}), \end{array}$ and $\begin{array}{rcl} cot (β_{n}) & = & \frac{q_{n}^{1 / 4}}{2 π} (1 + o (\frac{1}{q_{n}^{1 / 4}})) . \end{array}$ It follows that $\begin{array}{rcl} A & = & \frac{2 π (2 + ℓ_{4})}{q_{n}^{1 / 4}} + o (\frac{1}{q_{n}^{1 / 4}}), B = 1 + ℓ_{4} + o (1), C = - \frac{1}{2} + o (1), \\ F & = & \frac{q_{n}^{1 / 4}}{2 π} (1 + 4 i \frac{2 π}{q_{n}^{1 / 4}} + o (\frac{1}{q_{n}^{1 / 4}})), G = o (\frac{1}{q_{n}^{1 / 4}}), H = - \frac{i}{2} + o (1) . \end{array}$ Return back to (4.12), we could write $\begin{matrix} (2 + ℓ_{4} + o (1)) β_{n} a_{3} = - \frac{i}{2} (1 + ℓ_{4}) + o (1) - \frac{q_{n}^{1 / 4}}{4 π} (1 + 4 i \frac{2 π}{q_{n}^{1 / 4}} + o (\frac{1}{q_{n}^{1 / 4}})) . \end{matrix}$ Hence $\begin{matrix} β_{n} a_{3} \sim - \frac{1}{4 π (2 + ℓ_{4})} q_{n}^{1 / 4} . \end{matrix}$

Which implies that $‖ y_{n}^{3} ‖$ converges to infinity as n goes to infinity and that consequently $z_{n}$ is not bounded. □
Remark 4.3.
A small change in the proof leads to the conclusion that a polynomial stability, cannot be better than $\frac{1}{t^{2}}$ in the case of this special circuit (by using a frequency domain characterization of polynomial stability of a $C_{0}$ -semigroup of contraction due to Borichev and Tomilov [6]). Precisely we prove that the system is not $\frac{1}{t^{α}}$ -polynomially stable for every α in $(2, \infty)$ .

4.2. A star with two fixed endpoints (Fig. 3)

In this example we have taken $ℓ_{1} = ℓ_{2} = 1$ and the two exterior ends of $e_{1}$ and $e_{2}$ are supposed fixed. More precisely, we consider the following system $\begin{matrix} (4.13) & \{\begin{matrix} y_{t t}^{j} - y_{x x}^{j} = 0 in (0, 1) \times (0, \infty), j \in {1, 2}, \\ y_{t t}^{3} - y_{x x}^{3} = 0 in (0, ℓ_{3}) \times (0, \infty), \\ \sum_{j = 1}^{3} y_{x}^{j} (0, t) = s^{'} (t), s^{″} (t) + s (t) = - y_{t} (0, t), \\ y^{j} (0, t) = y^{l} (0, t), j, l \in {1, 2, 3}, \\ y^{2} (1, t) = y^{3} (ℓ_{3}, t) = 0, y_{x}^{1} (1, t) = - y_{t} (1, t), \\ s (0) = s_{0}, s^{'} (0) = s_{1}, \\ y^{j} (x, 0) = y_{0}^{j} (x), y_{t}^{j} (x, 0) = y_{1}^{j} (x), x \in (0, 1), j \in {1, 2}, \\ y^{3} (x, 0) = y_{0}^{3} (x), y_{t}^{3} (x, 0) = y_{1}^{3} (x), x \in (0, ℓ_{3}) . \end{matrix} \end{matrix}$

Fig. 3.

Star.

We can rewrite the system (4.13) in the Hilbert space H as $\begin{matrix} z^{'} (t) = A z (t), \end{matrix}$ where $\begin{matrix} H = V \times ({(L^{2} (0, 1))}^{2} \times L^{2} (0, ℓ_{3})) \times C^{2} \end{matrix}$ with $\begin{matrix} V = {Φ \in {(H^{1} (0, 1))}^{2} \times H^{1} (0, ℓ_{3}), Φ^{2} (1) = Φ^{3} (ℓ_{3}) = 0, Φ^{j} (0) = Φ^{l} (0), j, l \in {1, 2, 3}} \end{matrix}$ and $\begin{matrix} A (y, v, p, q) = (v, \frac{d y}{d x}, q, - p - v (0)), \forall (y, v, p, q) \in D (A) \end{matrix}$ with $\begin{array}{rcl} D (A) & = & {(y, v, p, q) \in [({(H^{2} (0, 1))}^{2} \times H^{2} (0, ℓ_{3})) \cap V] \times V \times C^{2}; \\ \sum_{j = 1}^{3} \frac{d y^{j}}{d x} (0) = q, and \frac{d y^{1}}{d x} (1) = - v^{1} (1)} . \end{array}$

The operator $A$ generates a $C_{0}$ -semigroup of contraction ${(S (t))}_{t ⩾ 0}$ , and we have the following result.

Theorem 4.4.

The semigroup ${(S (t))}_{t ⩾ 0}$ is asymptotically stable if and only if $ℓ_{3}$ irrational and not in $π Z$ . Even if $ℓ_{3}$ irrational and not in $π Z$ , the semigroup ${(S (t))}_{t ⩾ 0}$ is non exponentially stable.

Proof.

As in the previous case, there exists $(p_{n}, q_{n}) \in N^{2}$ such that $\frac{p_{n}}{q_{n}}$ converge to $ℓ_{4}$ as n goes to infinity. Then $q_{n}$ converge to infinity and there is a subsequence of $q_{n}$ denoted $q_{n}$ such that $\begin{matrix} | q_{n} ℓ_{3} - p_{n} | < \frac{1}{q_{n}} . \end{matrix}$ Then we take $β_{n} = 2 π q_{n} + \frac{2 π}{q_{n}^{1 / 4}}$ and $f_{n} = (0, g_{n}, 0, 0)$ with $g_{n} = (0, - sin β_{n} x, 0)$ . The sequence $β_{n}$ tends to infinity as n goes to infinity, $f_{n} \in H$ and $f_{n}$ is bounded. To conclude, we prove as in the previous case that $z_{n}$ defined by $(A - i β_{n}) z_{n} = f_{n}$ is bounded. □

5. A chain with non equal mass points

In this section, we consider a particular network which is a chain of N edges ( $N ⩾ 2$ ) and $p = N + 1$ vertices such that the $(N - 1)$ interior vertices $a_{j}$ are point masses with mass $m_{j}$ . But the masses $m_{j}$ are not necessarily equal (Fig. 4).

Fig. 4.

A chain of strings.

Precisely, we consider the following system $\begin{matrix} (5.1) & \{\begin{matrix} y_{t t}^{j} - y_{x x}^{j} = 0 in (0, ℓ_{j}) \times (0, \infty), j \in {1, \dots, N}, \\ y_{x}^{j} (0, t) - y_{x}^{j - 1} (ℓ_{j - 1}, t) = s_{j}^{'} (t), j \in {2, \dots, N}, \\ m_{j} s_{j}^{″} (t) + s_{j} (t) = - y_{t} (0, t), j \in {2, \dots, N}, \\ y^{j - 1} (ℓ_{j - 1}, t) = y^{j} (0, t), j \in {2, \dots, N}, \\ y_{x}^{1} (0, t) = y_{t} (0, t), \\ y^{N} (ℓ_{N}, t) = 0, \\ s_{j} (0) = s_{j, 0}, s_{j}^{'} (0) = s_{j, 1}, j \in {2, \dots, N}, \\ y^{j} (x, 0) = y_{0}^{j} (x), y_{t}^{j} (x, 0) = y_{k}^{1} (x), x \in (0, ℓ_{j}), j \in {1, \dots, N} . \end{matrix} \end{matrix}$

Note that the feedback is applied at the vertex $a_{1}$ . We give a necessary and sufficient condition for the exponential stability of system (5.1). The general case of a tree with distinct masses at inner nodes is complicated for the moment, because the calculations are based on some recurrence relations, something we could not do for a general tree.

To start we quickly redefine the associated state space H and the operator $A$ as follow: $\begin{matrix} H = V \times \prod_{j = 1}^{N} L^{2} (0, ℓ_{j}) \times {(\prod_{j = 1}^{N - 1} C)}^{2} \end{matrix}$ with $\begin{matrix} V = {Φ \in \prod_{j = 1}^{N} H^{1} (0, ℓ_{j}), Φ^{N} (ℓ_{N}) = 0, Φ^{j - 1} (ℓ_{j - 1}) = Φ^{j} (0), j = 2, \dots, N} \end{matrix}$ and $\begin{matrix} A (y, v, p, q) = (y, \frac{d^{2} y}{d x^{2}}, q, - m^{- 1} (p + v_{M})), \forall (y, v, p, q) \in D (A) \end{matrix}$ with $- m^{- 1} (p + v_{M}) = {(- \frac{1}{m j} p_{j} - \frac{1}{m_{j}} v (ℓ_{j}))}_{j \in {2, \dots, N - 1}}$ and $\begin{matrix} D (A) = {(y, v, p, q) \in [\prod_{j \in J} H^{2} (0, ℓ_{j}) \cap V] \times V \times {(\prod_{k \in I_{M}} C)}^{2} satisfying (5.2)}, \end{matrix}$ where $\begin{matrix} (5.2) & \{\begin{matrix} \frac{d y^{j}}{d x} (0, t) - \frac{d y^{j - 1}}{d x} (ℓ_{j - 1}) = q_{j}, j \in {2, \dots, N}, \\ \frac{d y^{1}}{d x} (0) = v^{1} (0) . \end{matrix} \end{matrix}$

Then the operator $A$ generates a $C_{0}$ -semigroup of contraction ${(S (t))}_{t \in R^{+}}$ . Moreover, $σ (A) = σ_{p} (A)$ .

For every mass point m we denote by $i_{1} (m), \dots, i_{k_{m}} (m)$ the indices of the interior nodes with masses equal to m and ordered as follow, $i_{1} (m) < i_{2} (m) < \dots < i_{k_{m}} (m)$ .

For $r = r (m) \in {1, \dots, k_{m}}$ we define the scalars $\begin{matrix} Π_{m, r (m), s} = \sum_{i_{r} = j_{0} < j_{1} < \dots < j_{s} < i_{r + 1}} (\prod_{i = 0}^{s - 1} c_{j_{i + 1}} sin (β ℓ_{j_{i}} + \dots + β ℓ_{j_{i + 1} - 1})) sin (β ℓ_{j_{s}} + \dots + β ℓ_{i_{r + 1} - 1}), \end{matrix}$ where $β = \pm \frac{1}{m}$ and $c_{j_{i}} = \frac{1}{β (m_{j_{i}} - m)}$ for $i = 1, \dots, s$ and $\begin{array}{rcl} Δ_{r (m)} & : = & {(- 1)}^{i_{r + 1} - i_{r}} sin (β ℓ_{i_{r}} + \dots + β ℓ_{i_{r + 1} - 1}) \\ + \sum_{s = 0}^{i_{r + 1} - i_{r} - 1} {(- 1)}^{i_{r + 1} - i_{r} + s} Π_{m, r (m), s} \end{array}$ with $Δ_{r (m)} = sin (β ℓ_{r})$ if $k_{m} = 1$ .

Then we have the following result of asymptotic behavior of the system (5.1):

Lemma 5.1.

The system defined by ( 5.1 ) is exponentially stable if and only if for every mass point m and for every $r (m)$ , $Δ_{r (m)} \neq 0$ .

Proof.

The first question is whether $i R$ belongs to $ρ (A)$ . Thus, we will solve the equation $\begin{matrix} (5.3) & A z = i β z \end{matrix}$ of unknown $z = (y, v, p, q) \in D (A) - {0}$ and $β \in R$ .

If $β = 0$ then $z = 0$ . Thus suppose that $β \neq 0$ . We have $\begin{array}{l} (5.4) & v^{j} = i β y^{j}, β^{2} y^{j} + \frac{d^{2} y^{j}}{d x^{2}} = 0, j \in {1, \dots, N}, \\ (5.5) & q^{j} = i β p^{j}, (m_{j} β^{2} - 1) p_{j} = v (a_{j}), j \in {2, \dots, N} . \end{array}$ The function $y^{j}$ , $j \in {1, \dots, N}$ is then of the form $\begin{matrix} y^{j} = α_{j} cos (β x) + γ_{j} sin (β x) . \end{matrix}$ If $m_{j} β^{2} \neq 1$ for every j in ${2, \dots, N}$ then we prove by iteration, starting with $j = 1$ , that $y^{j} = 0$ for every j in ${2, \dots, N}$ and consequently $z = 0$ .

Now we suppose the existence of a mass point m with $m β^{2} = 1$ . Let $i_{1} (m) < \dots < i_{k_{m}} (m)$ the indices of inner nodes with masses equal to m. Then as in the first case $y^{j} = 0$ for every $j < i_{1}$ . Let $r = r (m) \in {1, \dots, k_{m}}$ we have the following system $\begin{matrix} (5.6) & \{\begin{matrix} y^{r} (0) = 0, \\ for j = i_{r} to j = i_{r + 1} - 2, y^{j} (ℓ_{j}) = y^{j + 1} (0), \\ and - \frac{d y^{j}}{d x} (ℓ_{j}) + \frac{d y^{j + 1}}{d x} (0) = q^{j + 1}, \\ y^{i_{r + 1} - 1} (ℓ_{i_{r + 1} - 1}) = 0 . \end{matrix} \end{matrix}$ It is obvious that the system (5.3) has trivial solution if and only if for every m and every $r (m)$ the system (5.6) has a trivial solution.

By changes of indices we can suppose that $i_{r} = 2$ , $i_{r + 1} = N + 1$ . The system (5.6) can be rewritten as $\begin{matrix} \{\begin{matrix} α_{2} = 0, \\ for j = 2 to j = N - 1, α_{j} cos (β ℓ_{j}) + γ_{j} sin (β ℓ_{j}) = α_{j + 1}, \\ and α_{j} sin (β ℓ_{j}) - γ_{j} cos (β ℓ_{j}) + \frac{1}{β} \frac{1}{m_{j} - m} α_{j + 1} + γ_{j + 1} = 0, \\ α_{N} cos (β ℓ_{N}) + γ sin (β ℓ_{N}) = 0 . \end{matrix} \end{matrix}$ The matrix of such system is $\begin{matrix} S_{N} = (\begin{matrix} 1 & 0 & 0 \\ cos x_{2} & sin x_{2} & - 1 & 0 \\ sin x_{2} & - cos x_{2} & c_{3} & 1 & 0 \\ 0 & 0 & cos x_{3} & sin x_{3} & - 1 & 0 & (0) \\ 0 & 0 & sin x_{3} & - cos x_{3} & c_{4} & 1 & 0 \\ \dots & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots & 0 \\ (0) & \dots & cos x_{N - 1} & sin x_{N - 1} & - 1 & 0 \\ \dots & sin x_{N - 1} & - cos x_{N - 1} & c_{N} & 1 \\ \dots & 0 & 0 & cos x_{N} & sin x_{N} \end{matrix}), \end{matrix}$ where $x_{j} = β ℓ_{j}$ and $c_{j} = \frac{1}{β (m_{j} - m)}$ . We want to calculate the determinant $Δ_{N}$ of $S_{N}$ . For this, let $M_{N}$ the determinant of the matrix obtain from $S_{N}$ by replacing $cos x_{N}$ and $sin x_{N}$ in the last line by $sin x_{N}$ and $- cos x_{N}$ respectively.

One can verifies easily that $\begin{array}{l} Δ_{2} = sin x_{2}, Δ_{3} = - sin (x_{2} + x_{3}) + c_{3} sin x_{2} sin x_{3}, \\ \begin{aligned} Δ_{4} = & sin (x_{2} + x_{3} + x_{4}) - c_{3} sin x_{2} sin (x_{3} + x_{4}) - c_{4} sin (x_{2} + x_{3}) sin x_{4} \\ + c_{3} c_{4} sin x_{2} sin x_{3} sin x_{4}, \end{aligned} \\ M_{2} = - cos x_{2}, M_{3} = cos (x_{2} + x_{3}) - c_{3} sin x_{2} cos x_{3}, \\ \begin{aligned} M_{4} = & - cos (x_{2} + x_{3} + x_{4}) + c_{3} sin x_{2} cos (x_{3} + x_{4}) + c_{4} sin (x_{2} + x_{3}) cos x_{4} \\ - c_{3} c_{4} sin x_{2} sin x_{3} cos x_{4} . \end{aligned} \end{array}$

We will prove by induction that for every integer $N ⩾ 2$ , $\begin{array}{rcl} Δ_{N} & = & {(- 1)}^{N} sin (x_{2} + \dots + x_{N}) + {(- 1)}^{N - 1} \sum_{j = 2}^{N - 1} c_{j + 1} sin (x_{2} + \dots + x_{j - 1}) sin (x_{j} + \dots + x_{N}) \\ + \sum_{s = 2}^{N - 2} {(- 1)}^{N - 1 - s} \sum_{2 = j_{0} < j_{1} < \dots < j_{s} ⩽ N} (\prod_{i = 0}^{s - 1} c_{j_{i + 1}} sin (x_{j_{i}} + \dots + x_{j_{i + 1} - 1})) \\ (5.7) & \times sin (β ℓ_{j_{s}} + \dots + β ℓ_{N}) \end{array}$ and $\begin{array}{rcl} M_{N} & = & {(- 1)}^{N + 1} cos (x_{2} + \dots + x_{N}) + {(- 1)}^{N} \sum_{j = 2}^{N - 1} c_{j + 1} sin (x_{2} + \dots + x_{j - 1}) cos (x_{j} + \dots + x_{N}) \\ + \sum_{s = 2}^{N - 2} {(- 1)}^{N - s} \sum_{2 = j_{0} < j_{1} < \dots < j_{s} ⩽ N} (\prod_{i = 0}^{s - 1} c_{j + 1} sin (x_{j_{i}} + \dots + x_{j_{i + 1} - 1})) \\ (5.8) & \times cos (x_{j_{s}} + \dots + x_{N}) . \end{array}$ Such rules are true for $N = 2$ and $N = 3$ . Let $N \in N$ with $N ⩾ 3$ and suppose that (5.7) and (5.8) are true. Some calculations leads to $\begin{array}{rcl} Δ_{N} & = & (- cos x_{N} + c_{N} sin x_{N}) Δ_{N - 1} + (sin x_{N}) M_{N - 1}, \\ M_{N} & = & (- sin x_{N} - c_{N} cos x_{N}) Δ_{N - 1} - (cos x_{N}) M_{N - 1} . \end{array}$ We can now verify the rule (5.7) to order N: $\begin{array}{rcl} Δ_{N} & = & ({(- 1)}^{N} sin (x_{2} + \dots + x_{N}) + {(- 1)}^{N - 1} \sum_{j = 2}^{N - 1} c_{j + 1} sin (x_{2} + \dots + x_{j - 1}) sin (x_{j} + \dots + x_{N}) \\ + \sum_{s = 2}^{N - 3} {(- 1)}^{N - 1 - s} \sum_{2 = j_{0} ⩽ j_{1} < \dots < j_{s} ⩽ N - 1} (\prod_{i = 0}^{s - 1} c_{j_{i + 1}} sin (x_{j_{i}} + \dots + x_{j_{i + 1} - 1})) sin (x_{j_{s}} + \dots + x_{N})) \\ + \sum_{s = 2}^{N - 3} {(- 1)}^{N - 2 - s} \sum_{2 = j_{0} ⩽ j_{1} < \dots < j_{s} ⩽ N - 1} c_{N} (\prod_{i = 0}^{s - 1} c_{j_{i + 1}} sin (x_{j_{i}} + \dots + x_{j_{i + 1} - 1})) \\ \times sin (x_{j_{s}} + \dots + x_{N - 1}) sin x_{N} \\ = & ({(- 1)}^{N} sin (x_{2} + \dots + x_{N}) + {(- 1)}^{N - 1} \sum_{j = 2}^{N - 1} c_{j + 1} sin (x_{2} + \dots + x_{j - 1}) sin (x_{j} + \dots + x_{N}) \\ + \sum_{s = 2}^{N - 2} {(- 1)}^{N - 1 - s} \sum_{2 = j_{0} ⩽ j_{1} < \dots < j_{s} ⩽ N} (\prod_{i = 0}^{s - 1} c_{j_{i + 1}} sin (x_{j_{i}} + \dots + x_{j_{i + 1} - 1})) sin (x_{j_{s}} + \dots + x_{N})) . \end{array}$ A similar calculus, using (5.8) proves that (5.8) is verified in order N.

We can now state the following result:

The associated semigroup $S (t)$ is asymptotically stable if and only if $Δ_{r (m)}$ is different from zero for every mass point m and every $r (m)$ . To conclude that ${(S (t))}_{t ⩾ 0}$ is exponentially stable, it suffices to prove that (3.11) is satisfied by ${(S (t))}_{t ⩾ 0}$ , exactly as in the proof of Theorem 3.2. □

Footnotes

Acknowledgements

This work was carried out under Airbus Chair Grant. Financial support from Airbus Chair Grant is gratefully acknowledged.

We thank the anonymous reviewers for their careful reading of our manuscript and their many insightful comments and suggestions.

References

A.B.

Abdallah and

Shel, Exponential stability of a general network of 1-d thermoelastic materials, Mathematical Control and Related Fields 2 (2012), 1–16. doi:10.3934/mcrf.2012.2.1.

Ammari,

Henrot and

Tucsnak, Asymptotic behaviour of the solutions and optimal location of the actuator for the pointwise stabilization of a string, Asymptotic Analysis 28 (2001), 215–240.

Ammari and

Jellouli, Stabilization of star-shaped networks of strings, Differential and Integral Equations 17 (2004), 1395–1410.

Ammari,

Jellouli and

Khenissi, Stabilization of generic trees of strings, J. Dynamical Control Systems 11 (2005), 177–193. doi:10.1007/s10883-005-4169-7.

Ammari and

Nicaise, Stabilization of Elastic Systems by Collocated Feedback, Lecture Notes in Mathematics, Vol. 2124 Springer, Cham, 2015.

Batty and

Tomilov, Optimal polynomial decay of functions and operator semigroups, Math. Ann. 347 (2010), 455–478. doi:10.1007/s00208-009-0439-0.

Brezis, Analyse Fonctionnelle, Théorie et Applications, Masson, Paris, 1983.

Conca,

Planchard and

Vanninathan, Fluids and Periodic Structures, Research in Applied Mathematics, Vol. 38, John Wiley & Sons, Masson, 1995.

Dàger and

Zuazua, Wave Propagation, Observation and Control in 1-d Flexible Multi-Structures, Mathématiques & Applications, Springer-Verlag, Berlin, 2006.

10.

Ervedoza and

Vanninathan, Controllability of a simplified model of fluid–structure interaction, ESAIM Control Optim. Calc. Var. 20 (2014), 547–575. doi:10.1051/cocv/2013075.

11.

F.L.

Huang, Characteristic condition for exponential stability of linear dynamical systems in Hilbert spaces, Ann. Diff. Eqs. 1 (1985), 43–56.

12.

Liu,

Takahashi and

Tucsnak, Single input of a simplified fluid–structure interaction model, ESAIM Control Optim. Calc. Var. 19 (2013), 20–42. doi:10.1051/cocv/2011196.

13.

Mercier and

Régnier, Control of a network of Euler–Bernoulli beams, J. Math. Anal. Appl. 342 (2008), 874–894. doi:10.1016/j.jmaa.2007.12.062.

14.

Pazy, Semigroup of Linear Operators and Applications to Partial Differential Equations, Springer, 1983.

15.

Prüss, On the spectrum of

C_{0}

-semigroups, Trans. Amer. Math. Soc. 284 (1984), 847–857. doi:10.2307/1999112.

16.

W.M.

Schmidt, Diophantine Approximation, Lecture Notes in Mathematics, Vol. 785, Springer-Verlag, Berlin, 1980.

17.

Shel, Exponential stability of a network of elastic and thermoelastic materials, Math. Meth. in Appl. Sci. 36 (2013), 869–879. doi:10.1002/mma.2644.

18.

Tucsnak and

Vanninathan, Locally distributed control for a model of fluid–structure interaction, System and Control Letters 58 (2009), 547–552. doi:10.1016/j.sysconle.2009.03.002.

19.

J.L.

Vazquez and

Zuazua, Large time behavior for a simplified 1D model of fluid–solid interaction, Comm. Partial Differential Equations 28 (2003), 1705–1738. doi:10.1081/PDE-120024530.

20.

J.L.

Vazquez and

Zuazua, Lack of collision in a simplified 1D model for fluid–solid interaction, Math. Models Methods Appl. Sci. 16 (2006), 637–678. doi:10.1142/S0218202506001303.

21.

S.M.

Venuti, Modeling, analysis and computation of fluid structure interaction models for biological systems, SIAM Undergraduate Research Online 3 (2010), 1–17. doi:10.1137/09S010496.