MATHEMATICAL CONTROL AND doi:10.3934/mcrf.2015.5.721RELATED FIELDSVolume 5, Number 4, December 2015 pp. 721–742

FINITE-TIME STABILIZATION OF A NETWORK OF STRINGS

Fatiha Alabau-Boussouira

Institut Elie Cartan de Lorraine, UMR 7502Universite de Lorraine57045 Metz, France

Vincent Perrollaz

⇤and Lionel Rosier

⇤⇤

⇤Laboratoire de Mathematiques et Physique TheoriqueUniversite de Tours, UFR Sciences et Techniques

Parc de Grandmont37200 Tours, France

⇤⇤Centre Automatique et SystemesMINES ParisTech, PSL Research University

60 Boulevard Saint-Michel75272 Paris Cedex 06, France

(Communicated by Enrique Zuazua)

Abstract. We investigate the finite-time stabilization of a tree-shaped net-work of strings. Transparent boundary conditions are applied at all the externalnodes. At any internal node, in addition to the usual continuity conditions, amodified Kirchho↵ law incorporating a damping term ↵ut with a coe�cient ↵that may depend on the node is considered. We show that for a convenientchoice of the sequence of coe�cients ↵, any solution of the wave equation onthe network becomes constant after a finite time. The condition on the coef-ficients proves to be sharp at least for a star-shaped tree. Similar results arederived when we replace the transparent boundary condition by the Dirichlet(resp. Neumann) boundary condition at one external node. Our results leadto the finite-time stabilization even though the systems may not be dissipative.

1. Introduction. Solutions of certain ODE x = f(x) may reach the equilibriumstate in finite time. This phenomenon, when combined with the stability, wastermed finite-time stability in [4, 11].

A finite-time stabilizer is a feedback control for which the closed-loop system isfinite-time stable around some equilibrium state. In some sense, it satisfies an exactcontrollability objective with a control in feedback form. On the other hand, afinite-time stabilizer may be seen as an exponential stabilizer yielding an arbitrarilylarge decay rate for the solutions to the closed-loop system. Indeed, any solution ofthe closed-loop system can be estimated as

||x(t)|| h(||x0||)1[0,T ](t) h(||x0||)e��(t�T )

2010 Mathematics Subject Classification. Primary: 93D15; Secondary: 34B45, 35L05.Key words and phrases. Finite-time stabilization, network, wave equation, transparent bound-

ary condition, Kirchho↵ law, Riemann invariant.

721

722 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

where h(�) ! 0 as � ! 0, and � > 0 is arbitrarily large. This explains why somee↵orts were made in the last decade to construct finite-time stabilizers for control-lable systems, including the linear ones. See [15] for some recent developments andup-to-date references, and [3] for some connections with Lyapunov theory.

To the best knowledge of the authors, the analysis of the finite-time stabilizationof PDE is not developed yet. However, since [14], it is well-known that solutions ofthe wave equation on certain bounded domains may disappear when using trans-

parent boundary conditions. For instance, the solution of the 1-D wave equation

utt

� c2uxx

= 0, in (0, L) ⇥ (0, T ), (1)

cux

(L, t) = �ut

(L, t), in (0, T ), (2)

cux

(0, t) = ut

(0, t), in (0, T ), (3)

(u(0), ut

(0)) = (u0, u1), in (0, L), (4)

is finite-time stable in the space {(u, v) 2 H1(0, L) ⇥ L2(0, L); c�u(0) + u(L)

�+R

L

0 v(x)dx = 0}, with T = L/c as extinction time (see e.g. [12, Theorem 0.5]for the details.) The condition (2) is “transparent” in the sense that a waveu(x, t) = f(x � ct) traveling to the right satisfies (2) and leaves the domain atx = L without generating any reflected wave. Note that the solution issued fromany state (u0, u1) 2 H1(0, L) ⇥ L2(0, L) is not necessarily vanishing, but constant,for t � L/c. Note also that if we replace (3) by the boundary condition u(0, t) = 0(or u

x

(0, t) = 0), then a finite-time extinction still occurs (despite the fact thatwaves bounce at x = 0) with an extinction time T = 2L/c. We refer to [5] for theanalysis of the finite-time extinction property for a nonhomogeneous string with aviscous damping at one extremity, to [8] for the finite-time stabilization of a stringwith a moving boundary, to [16] (resp. [17]) for the finite-time stabilization of asystem of conservation laws on an interval (resp. on a tree-shaped network).

The finite-time stability of (1)-(4) is easily established when writing (1) as asystem of two transport equations

dt

+ cdx

= 0,

st

� csx

= 0.

where d := ut

�cux

and s := ut

+cux

stand for the Riemann invariants for the waveequation written as a first order hyperbolic system. The boundary conditions (2)and (3) yield d(0, t) = s(L, t) = 0 (and hence d(., t) = s(., t) = 0 for t � L/c), whilethe boundary conditions (2) and u(0, t) = 0 yield s(L, t) = 0 and d(0, t) = �s(0, t)(and hence s(., t) = 0 for t � L/c and d(., t) = 0 for t � 2L/c).

The stabilization of networks of strings has been considered in e.g. [1, 2, 7, 9, 10,19, 22]. In [10], the authors considered a star of vibrating strings, and derived thefinite-time stability (resp. the exponential stability) when transparent boundaryconditions are applied at all external nodes (resp. at all external nodes but one,which is changing as times proceeds). For a more general network, we guess thatthe finite time stability cannot hold without the introduction of additional feedbackcontrols at the internal nodes. Indeed, it is proved here that for a bone-shaped tree,if the feedback controls are applied only at the external nodes, then the finite-timestability fails.

The aim of this paper is to investigate the finite-time stabilization of a tree-shapednetwork of strings. At each internal node n connecting k edges, we assume that theusual continuity condition hold:

FINITE-TIME STABILIZATION IN A NETWORK 723

ui

(n, t) = uj

(n, t), for all indices i, j of edges having n as one end, (5)

while the usual Kirchho↵ law is modified by incorporating a damping term inside:X

i

ci

ui,x

(n, t) � ci0ui0,x(n, t) = �↵(n)u

t

(n, t). (6)

In (6), the sum is over the indices i of the edges having n as initial point, i0 denotesthe index of the edge having n as final point (the tree being oriented), ↵(n) 2 R isa coe�cient depending on the node n, and u(n, t) := u

i

(n, t) (for any i). The case↵ = 0 corresponds to the usual (conservative) Kirchho↵ law.

Note that we can assume without loss of generality that the length of each edgeis one, by scaling the variable x and the coe�cient c

i

along each edge.Even if the finite-time stabilization of 2 ⇥ 2 hyperbolic systems on tree-shaped

networks was already considered in [17] (and applied to the regulation of water flowsin networks of canals, with k � 1 controls at any node connecting k canals), thenovelty (and di�culty) here comes from the fact that only one control is applied ateach internal node. The present work can be seen as a first step in the understandingof the finite-time stabilization of systems of conservation laws with few controls. Inpractice, on account of possible budget limitations, it is important that the numberof interior controls be chosen as small as possible.

A natural guess is that the finite-time stability cannot hold if one can find in thetree a pair of adjacent nodes that are free of any control, because of the (partial butstanding) bounces of waves at these nodes. This conjecture will be demonstratedhere for a star-shaped tree and a bone-shaped tree.

Actually, we shall prove that the finite-time stabilization can be achieved for avery particular choice of the coe�cient ↵ at each internal node. One of the mainresults proved in this paper is the following

Theorem 1.1. Consider any tree-shaped network of strings, with transparent bound-

ary conditions at all the external nodes but one (the root), where a homogeneous

Dirichlet boundary condition is applied, and with the continuity conditions and the

modified Kirchho↵ law at the internal nodes. If at each internal node n connecting

k edges we have

↵(n) = k � 2, (7)

then each solution of the wave equation on the network becomes null after some finite

time. Conversely, the condition (7) is also necessary for the finite-time stability of

the system on a star-shaped tree.

Similar results will be obtained when replacing at the root the homogeneousDirichlet boundary condition by the homogeneous Neumann boundary conditionor by the transparent boundary condition. The necessity of the condition ↵ =k � 2 for the finite-time stability of a star-shaped tree is obtained by doing someexplicit computation of the discrete spectrum of the underlying operator. The sameapproach gives for a bone-shaped tree a necessary and su�cient condition for thefinite-time stability, which di↵ers slightly from those stated in Theorem 1.1.

The condition ↵(n) = k � 2 can be interpreted as follows: the wave emanatingfrom n along any edge is the sum of the waves arriving at n along the k � 1others edges (see below (34)). The node n is transparent in the sense that there istransmission without attenuation nor reflection of all the waves that reach it.

The well-posedness of the system is also completely characterized in terms ofthe coe�cients ↵ (see below Theorem 2.1). We note that the well-posedness of

724 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

the system may hold even if the system is not dissipative, and that our finite-timestability results hold in a class of systems that are not necessarily dissipative.

Finally, we point out that the recent papers [18, 21] contain some results similarto ours. (Note that the finite-time stability is termed super-stability in [18, 21].)The paper [21] is concerned with star-shaped trees, while the paper [18] considersvery general networks. When comparing the results in [18] with ours, we observethe following:

1. The wave equation in [18] takes the form

⇢i

ui,tt

� Ti

ui,xx

= 0, in (0, 1) ⇥ (0, T ) (8)

where Ti

, ⇢i

> 0 are given coe�cients, while the modified Kirchho↵ law at aninternal node of a tree is written

X

i

Ti

ui,x

(n, t) � Ti0ui0,x(n, t) = �↵(n)u

t

(n, t). (9)

Note that (Ti

, ⇢i

) := (ci

, c�1i

) is a possible choice (among others) to yield (12)(see below) on the edge i.

2. The well-posedness of the complete system is derived in [18, Proposition 2.1]under the su�cient condition ↵(n) 0 for all internal nodes n, which ensuresthat the system is dissipative, while a sharp condition is given in our Theorem2.1.

3. When (Ti

, ⇢i

) = (ci

, c�1i

) for all i, the condition (3.7) in [18, Theorem 3.2]which gives the values of ↵(n) at all the internal nodes ensuring the finite-time stability is the same as our condition (92) in Theorem 4.1. However, itis also assumed in [18, Theorem 3.2] that

↵(n) 0 for all internal nodes n. (10)

This imposes some restrictions on the coe�cients Ti

, ⇢i

or on the shape of thetree. For instance, for uniform data (T

i

= ⇢i

= 1 for all i), then the condition(10) (with ↵

n

as in (92)) is satisfied only if kn

= 2 for all internal nodes n,i.e. only if the tree is a line, while our Theorem 4.1 is valid for any tree. Onthe other hand, by choosing other pairs of positive coe�cients (T

i

, ⇢i

) withTi

/⇢i

= c2i

, Theorem 3.2 in [18] gives modified Kirchho↵ laws (9) di↵erentfrom (6) that may lead to the finite-time stability.

4. Our approach is based upon the analysis of the Riemann invariants and thepropagation of waves, while [18] is based upon spectral theory.

The paper is outlined as follows. In Section 2, we provide a sharp condition onthe coe�cients ↵(n) for the system to be well-posed. It is obtained by expressingthe conditions (5)-(6) at the internal nodes in terms of the Riemann invariants.In Section 3, we consider the particular case of a tree with two internal nodes toobtain sharp results for the finite-time stability. In Section 4, we derive the finite-time stability results when the coe�cients ↵ are chosen as in (7). Finally, we provein Section 5 the necessity of that condition for tree-shaped networks.

2. Well-posedness. We introduce some notations inspired by [6]. Let T be a tree,whose vertices (or nodes) are numbered by the index n 2 N = {0, ..., N}, and whoseedges are numbered by the index i 2 I = {1, ..., N}. We choose a simple vertex (i.e.an external node), called the root of T and denoted by R, and which corresponds tothe index n = 0. The edge containing R has i = 1 as index, and its other endpointhas for index n = 1. We choose an orientation of the edges in the tree such that R is

FINITE-TIME STABILIZATION IN A NETWORK 725

the “first” encountered vertex. The depth d of the tree is the number of generations(d = 1 for a tree reduced to a single edge, d = 2 for a star-shaped tree, etc.) Oncethe orientation of the tree is chosen, each point of the i-th edge (of length 1) isidentified with a real number x 2 [0, 1]. The points x = 0 and x = 1 are termedthe initial point and the final point of the i-th edge, respectively. Renumbering theedges if needed, we can assume that the edge of index i has as final point the vertexwith the (same) index n = i for all i 2 I. (See Figure 1.) The set of indices of

11

R�

R

1

2 3

4 5 6

7 8 9 10 11

12 13

8 9 10

6

3

1

2

0

4 5

7

12 13

Figure 1. A tree with 14 nodes, a depth equal to 5, with simplenodes N

S

= {0, 4, 8, 9, 10, 11, 12, 13} and multiple nodes NM

={1, 2, 3, 5, 6, 7}.

simple and multiple nodes are denoted by NS

and NM

, respectively.For n 2 N

M

we denote by In

the set of indices of those edges having the vertexof index n as initial point. As we consider a network of strings whose constants c

i

may vary from one edge to another one, the case #(In

) = 1 (one child) is possible.The number of edges having the vertex of index n as one as their extremities, alsotermed the degree of the vertex n, is

kn

:= #(In

) + 1 � 2. (11)

We consider the following system

ui,tt

� c2i

ui,xx

= 0, t > 0, 0 < x < 1, i 2 I (12)

(ui

(., 0), ui,t

(., 0)) = (u0i

, u1i

), i 2 I (13)

with the following boundary conditions

cn

un,x

(1, t) = �un,t

(1, t), t > 0, n 2 NS

\ {0}, (14)X

i2In

ci

ui,x

(0, t) � cn

un,x

(1, t) = �↵n

un,t

(1, t), t > 0, n 2 NM

, (15)

ui

(0, t) = un

(1, t), t > 0, n 2 NM

, i 2 In

, (16)

where the sequence (↵n

)n2NM is still to be defined. For the boundary condition at

the root R, we shall consider one of the following conditions

u1(0, t) = 0, t > 0 (Dirichlet boundary condition); (17)

u1,x(0, t) = 0, t > 0 (Neumann boundary condition); (18)

c1u1,x(0, t) = u1,t(0, t), t > 0 (Transparent boundary condition). (19)

726 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

Let

H =�(u

i

, vi

)i2I 2

Y

i2I[H1(0, 1) ⇥ L2(0, 1)]; u

i

(0) = un

(1) 8n 2 NM

, 8i 2 In

and H0 =�(u

i

, vi

)i2I 2 H; u1(0) = 0}. The spaces H and H0 are Hilbert spaces

when endowed with the scalar product

�(u

i

, vi

)i2I , (ui

, vi

)i2I�H :=

X

i2I

Z 1

0

�ui

(x)ui

(x) + ui,x

(x)ui,x

(x) + vi

(x)vi

(x)�dx.

Replacing ui,t

by vi

and dropping the variable t, conditions (14) - (19) may berewritten respectively as

cn

un,x

(1) = �vn

(1), n 2 NS

\ {0}, (20)X

i2In

ci

ui,x

(0) � cn

un,x

(1) = �↵n

vn

(1), n 2 NM

, (21)

ui

(0) = un

(1), n 2 NM

, i 2 In

, (22)

u1(0) = 0, (23)

u1,x(0) = 0, (24)

c1u1,x(0) = v1(0). (25)

If t 2 R+ ! (ui

, vi

)i2I 2 D(A

T

) is continuous (for the definition of D(AT

), seebelow), using v

i

= ui,t

, (22) and (23) we obtain

vi

(0) = vn

(1), n 2 NM

, i 2 In

, (26)

v1(0) = 0. (27)

Introduce the operator AD

, AN

and AT

defined as

AD

((ui

, vi

)i2I) = (v

i

, c2i

ui,xx

)i2I ,

AN

((ui

, vi

)i2I) = (v

i

, c2i

ui,xx

)i2I ,

AT

((ui

, vi

)i2I) = (v

i

, c2i

ui,xx

)i2I ,

with respective domains

D(AD

) = {(ui

, vi

)i2I 2

Y

i2I[H2(0, 1) ⇥ H1(0, 1)]; (20) � (22), (23)

and (26) � (27) hold} ⇢ H0,

D(AN

) = {(ui

, vi

)i2I 2

Y

i2I[H2(0, 1) ⇥ H1(0, 1)]; (20) � (22), (24)

and (26) hold} ⇢ H,

D(AT

) = {(ui

, vi

)i2I 2

Y

i2I[H2(0, 1) ⇥ H1(0, 1)]; (20) � (22), (25)

and (26) hold} ⇢ H.

The main result in this section is concerned with the well-posedness of system(12)-(16) and (17) (or (18), or (19)).

Theorem 2.1. Let T be a tree and let (↵n

)n2NM be a given family of real numbers.

Then AT

generates a strongly continuous semigroup of operators on H if, and only

if,

↵n

6= kn

8n 2 NM

, (28)

FINITE-TIME STABILIZATION IN A NETWORK 727

where kn

is the degree of the vertex n defined in (11). The same conclusion holds

for AN

on H (resp. for AD

on H0).

Remark 1. (1) As a consequence of Theorem 2.1, system (12)-(16) and (19) is notwell-posed when ↵

n

= kn

for some n 2 NM

. The initial data has to fulfill somecompatibility conditions for the solution to exist. Consider the simplest case of atree with two edges (N = 2) and c1 = c2 = 1. From (35) (see below), we have

v01(1 � t) � u01,x(1 � t) + v02(t) + u0

2,x(t) = 0 for a.e. t 2 (0, 1). (29)

This is a necessary condition on the initial data for the condition (15)-(16) to besatisfied. In other words, for initial data not satisfying the condition (29), there isno solution to (12)-(16) and (19).(2) Furthermore, even for initial data satisfying (29), the corresponding solutionsare not unique for positive times. Indeed, taking 0 as initial data, and picking anyfunction h 2 C1(R) with h(t) = 0 for t 0, then the trajectory candidate

u1(x, t) = h(t+ x � 1), 0 x 1, t � 0,

u2(x, t) = h(t � x), 0 x 1, t � 0

satisfies (12)-(16) and (19) with ↵1 = k1 = 2. Thus, there is no uniqueness oftrajectories.(3) On the other hand, system (12)-(16) and (19) is well-posed when ↵

n

6= kn

forall n 2 N

M

. Note that the system is not necessarily dissipative, i.e. the energyis not necessarily nonincreasing (even if the system is finite-time stable). Considerfor instance a star-shaped tree with 3 edges (N = 3), and c1 = c2 = c3 = 1 forsimplicity. Pick ↵1 := k1 � 2 = 1. (According to Theorem 4.1, the system is finite-time stable to constant functions.) Then a direct computation gives for any solution(u

i

, vi

)1i3 of (12)-(16) and (19)

d

dt

1

2

3X

i=1

Z 1

0(u2

i,t

+ u2i,x

)dx = u21,t(1, t) � (u2

1,t(0, t) + u22,t(1, t) + u2

3,t(1, t)).

This expression is positive for t > 0 small enough if the initial data (u0i

, v0i

)1i3 2D(A

T

) is chosen so that

|v01(1)|2 > |v01(0)|2 + |v02(1)|2 + |v03(1)|2.In that case, the energy may increase before reaching the value 0 in finite time.

Proof. We sketch the proof only for AT

. We need a preliminary result about theRiemann invariants around an internal node. Consider any internal node connectingedges whose indices range over {1, ..., k} (to simplify the notations). Consider anysolution of (12) satisfying

u1(1, t) = u2(0, t) = · · · = uk

(0, t) (30)

c2u2,x(0, t) + · · · + ck

uk,x

(0, t) � c1u1,x(1, t) = �↵u1,t(1, t) (31)

Introduce the Riemann invariants

di

(x, t) := ui,t

(x, t) � ci

ui,x

(x, t), (32)

si

(x, t) := ui,t

(x, t) + ci

ui,x

(x, t) (33)

for all i 2 I. Then the following result holds.

728 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

Lemma 2.2. 1. If ↵ 6= k, then s1(1, t), d2(0, t), ..., dk(0, t) can be expressed in a

unique way as functions of d1(1, t), s2(0, t), ..., sk(0, t). In particular, if ↵ =k � 2, we obtain

s1(1, t) =kX

i=2

si

(0, t). (34)

2. If ↵ = k, then the existence of a solution to (12) and (30)-(31) implies

d1(1, t) +kX

i=2

si

(0, t) = 0. (35)

As a consequence, the initial data of a smooth solution should satisfy the

compatibility condition

v01(1 � c1t) � c1u01,x(1 � c1t) +

kX

i=2

[v0i

(ci

t) + ci

u0i,x

(ci

t)] = 0

for 0 t min1ik

c�1i

. (36)

Proof of Lemma 2.2. Using Riemann invariants, we see that (12) and (30)-(31) aretransformed into

di,t

+ ci

di,x

= 0, i = 1, ..., k, (37)

si,t

� ci

si,x

= 0, i = 1, ..., k, (38)

s1(1, t) + d1(1, t) = s2(0, t) + d2(0, t) = · · · = sk

(0, t) + dk

(0, t), (39)kX

i=2

[si

(0, t) � di

(0, t)] � (s1(1, t) � d1(1, t)) = �↵(s1(1, t) + d1(1, t)). (40)

To simplify the notations, we write s1 for s1(1, t), s2 for s2(0, t), etc. Then (39)-(40)can be written

s1 + d1 = di

+ si

, i = 2, ..., k, (41)

(1 � ↵)s1 + d2 + · · · + dk

= (1 + ↵)d1 + s2 + · · · + sk

(42)

We readily infer from (41) that

s1 � d2 = �d1 + s2, (43)

d2 � d3 = �s2 + s3, (44)

...

dk�1 � d

k

= �sk�1 + s

k

. (45)

Adding the k � 1 equations in (41) results in

(k � 1)s1 �kX

i=2

di

= (1 � k)d1 +kX

i=2

si

Subtracting this last equation from (42), we obtain

2kX

i=2

di

= (k + ↵)d1 + (k + ↵ � 2)s1 = 2d1 + (k + ↵ � 2)(d1 + s1).

FINITE-TIME STABILIZATION IN A NETWORK 729

Combined to the relation d1 + s1 = dk

+ sk

, this yields

kX

i=2

di

= d1 + (k + ↵

2� 1)(d

k

+ sk

).

Using this relation in (42) together with the relation s1 = dk

+ sk

� d1, we obtain

(k � ↵)dk

= 2d1 + 2k�1X

i=2

si

+ (↵ � k + 2)sk

. (46)

Thus, if ↵ 6= k, we infer from (43)-(46) that s1(1, t), d2(0, t), ..., dk(0, t) can beexpressed in a unique way as functions of d1(1, t), s2(0, t), ..., sk(0, t). In particular,if ↵ = k � 2, then (46) becomes

dk

= d1 +k�1X

i=2

si

. (47)

Adding (43),(44),...,(45) and (47) yields (34). Finally, if ↵ = k, then (46) gives (35).Computing d

i

and si

using (52)-(53) (see below), and replacing s0i

and d0i

by theirexpressions in terms of u0

i

and v0i

, we obtain (36).Let us proceed to the proof of Theorem 2.1. If (28) is not satisfied, picking some

initial data (u0i

, v0i

)i2I 2 D(A

T

) that does not satisfy the condition

v0n

(1 � cn

t) � cn

u0n,x

(1 � cn

t) +X

i2In

[v0i

(ci

t) + ci

u0i,x

(ci

t)] = 0,

0 t mini2{n}[In

c�1i

(48)

around a node n 2 NM

for which ↵n

= kn

(the existence of such an initial data isobvious, since the conditions (20)-(27) involve only the values of u

i

and vi

or theirderivatives at the nodes), we infer from Lemma 2.2 that system (12)-(16) and (19)does not admit any solution (u

i

, vi

)i2I 2 C(R+;D(A

T

)). This shows that AT

is notthe generator of a continuous semigroup on H.

Conversely, assume that (28) is satisfied. We aim to construct by a fixed-pointprocedure a solution to (12)-(16) and (19). Pick any U0 = (u0

i

, v0i

)i2I 2 H and any

T > sup1iN

c�1i

. Set

d0i

:= v0i

� ci

u0i,x

, s0i

:= v0i

+ ci

u0i,x

, i = 1, ..., N.

Let K :=P

n2NMkn

. Pick any number ⇢ 2 (0, 1), and introduce the Hilbert space

E := L2⇢

tdt

(0, T )K endowed with the norm

||(x1, x2, ...., xK

)||E :=

KX

i=1

ZT

0|x

i

(t)|2⇢tdt! 1

2

.

For n 2 NM

, let mn

:= infj2In j, and let p1 := 1 and

pn

:= 1 +X

m2NM ,m<n

km

, for n 2 NM

.

For instance, with the tree drawn in Figure 1, we have

NM

= {1, 2, 3, 5, 6, 7},(k1, k2, k3, k5, k6, k7) = (3, 3, 2, 4, 3, 3),

730 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

(m1,m2,m3,m5,m6,m7) = (2, 4, 6, 7, 10, 12),

and (p1, p2, p3, p5, p6, p7) = (1, 4, 7, 9, 13, 16).

Let

X(t) := (x1(t), ..., xK

(t)) = (..., dn

(1, t), smn(0, t), ..., smn+kn�2(0, t), ...)

where n ranges over NM

. Note that x1(t) = d1(1, t) and xK

(t) = sN

(0, t). For thetree drawn in Figure 1, we have

X(t) = (x1(t), ..., x18(t))

= (d1(1, t), s2(0, t), s3(0, t), d2(1, t), s4(0, t), s5(0, t), d3(1, t), s6(0, t),

d5(1, t), s7(0, t), s8(0, t), s9(0, t), d6(1, t), s10(0, t), s11(0, t), d7(1, t),

s12(0, t), s13(0, t)).

Finally, let S denote the set of the indices of the components of X(t) correspondingto edges whose final points are simple nodes di↵erent from the root; that is

S := {n 2 {2, ...,K}, 9qn

2 NS

, xn

(t) = sqn(0, t)}.

For the tree in Figure 1, we have S = {5, 11, 12, 14, 15, 17, 18} with

(q5, q11, q12, q14, q15, q17, q18) = (4, 8, 9, 10, 11, 12, 13).

Let

E0 := {(x1, ..., xK

) 2 E ; x1(t) = 0 8t � c�11 and x

n

(t) = 0 8n 2 S, 8t � c�1qn

}.We define a map P : X = (x1, ..., xK

) 2 E0 ! X = (x1, ..., xK

) 2 E0 as follows.Pick any X = (x1, ..., xK

) 2 E0 and any n 2 NM

. By Lemma 2.2, there exists amatrix A

n

2 Rkn⇥kn such that the Riemann invariants associated with the solutionof (12)-(16) and (19) satisfy

0

BBB@

sn

(1, t)dmn(0, t)

...dmn+kn�2(0, t)

1

CCCA= A

n

0

BBB@

dn

(1, t)smn(0, t)

...smn+kn�2(0, t)

1

CCCA.

Then, we set for all t � 00

BBB@

sn

(1, t)dmn(0, t)

...dmn+kn�2(0, t)

1

CCCA:= A

n

0

BBB@

xpn(t)

xpn+1(t)

...xpn+kn�1(t)

1

CCCA. (49)

(Recall that xpn(t) stands for dn(1, t).) Next we set

d1(0, t) = 0, t � 0, (50)

sm

(1, t) = 0, m 2 NS

\ {0}, t � 0. (51)

Note that di

(0, t) and si

(1, t) have been defined for all i 2 {1, ..., N} and all t � 0.Next, solving (37)-(38), we set for all i 2 {1, ..., N} and all (x, t) 2 [0, 1] ⇥ R+

di

(x, t) :=

⇢d0i

(x � ci

t) if 0 x � ci

t 1,di

(0, t � c�1i

x) if t > x/ci

,(52)

and

si

(x, t) :=

⇢s0i

(x+ ci

t) if 0 x+ ci

t 1,si

(1, t � c�1i

(1 � x)) if t > (1 � x)/ci

.(53)

FINITE-TIME STABILIZATION IN A NETWORK 731

Finally, we set for n 2 NM

and t � 00

BBB@

xpn(t)

xpn+1(t)

...xpn+kn�1(t)

1

CCCA:=

0

BBB@

dn

(1, t)smn(0, t)

...smn+kn�2(0, t)

1

CCCA. (54)

and P (X) := X = (x1, ..., xK

).From (50)-(53), we have that

d1(1, t) = 0, for t � c�11 , (55)

sm

(0, t) = 0, for m 2 NS

\ {0}, t � c�1m

. (56)

It follows that P is a map from E0 into itself. Let us check that it is a contractionfor ⇢ small enough. Let X1 = (x1

1, ..., x1K

) and X2 = (x21, ..., x

2K

) be given in E0.In what follows, c denotes a constant that may vary from line to line. Then, using(49)-(54), we have that

||P (X1) � P (X2)||2E c

KX

j=1

NX

i=1

ZT

c

�1i

|x1j

(t � c�1i

) � x2j

(t � c�1i

)|2⇢tdt (57)

c(maxi2I

⇢c�1i )||X1 � X2||2E . (58)

This proves that P is a contraction in E0 for ⇢ > 0 small enough. It follows fromthe contraction principle that P has a (unique) fixed-point in E0. It is then easy tocheck that the Riemann invariants d

i

, si

, 1 i N , defined along (52)-(53), solve(37)-(38) in the distributional sense and satisfy (39)-(40) almost everywhere. Usingagain (52)-(53), one has that for any i 2 I

si

(x, 0) = s0i

(x), di

(x, 0) = d0i

(x), for a.e. x 2 [0, 1].

We can therefore define for all i 2 I and all T > sup1iN

c�1i

a function ui

2H1((0, 1) ⇥ (0, T )) by

ui,t

=1

2(s

i

+ di

) =: vi

, ui,x

=1

2ci

(si

� di

),

the constant of integration being chosen so that

ui

(x, t) = u0i

(x) +

Zt

0vi

(x, s)ds for a.e. (x, t) 2 (0, 1) ⇥ (0, T ).

Then (ui

, vi

) 2 C(R+, H1(0, 1) ⇥ L2(0, 1)), and (22) follows from (39). We inferthat (u

i

, vi

)i2I is a (weak) solution of (12)-(16) and (19) which is continuous in

time with values in H. Set S(t)U0 := (ui

(t), vi

(t))i2I . Then it can be seen that�

S(t)�t�0

is a strongly continuous semigroup in H whose generator is AT

. Theproof of Theorem 2.1 is complete.

3. The example of a tree with two internal nodes. In this section, we con-sider a particular case which can be treated in a direct way, and which shows theimportance of condition (7) in Theorem 1.1.

To prove on a given example that the finite-time stability (to 0 or to constantfunctions) does not occur, a way consists in finding an eigenvalue of the underlyingoperator. Indeed, if we can find an eigenvalue, then the corresponding exponentialsolution does not steer 0 in finite time. If, in addition, the eigenvalue is di↵erentfrom 0, then the corresponding exponential solution does not take a constant value

732 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

after some finite time. The explicit computation of the eigenvalues can be achievedwhen the geometry is su�ciently simple, typically d 3.

In this section, we assume now T is a tree with N +1 nodes, two of which beingmultiple (d = 3, N

M

= {1, 2}, k1 � 2, k2 � 2, k1 + k2 = N + 1), and we considerthe boundary conditions (14)-(16) and (19). (See Figure 2.) We will let ↵1 and

2

Transparent boundary condition

3 5

1 4

04

53

1 2

Figure 2. A bone-shaped tree with k1 = k2 = 3.

↵2 range over R, assuming only that (28) holds. In particular, when ↵1 = ↵2 = 0,there is no damping at the internal nodes n = 1, 2. We shall show that the finite-time stabilization cannot hold in that case, because of the (partial but continuous)bounces of waves at the internal nodes. Note that for this geometry, condition (7)reads

↵1 = k1 � 2, ↵2 = k2 � 2. (59)

Here, we shall show that there is an eigenvalue (and actually a sequence of eigen-values, so that the finite-time stability to constant functions fails) if, and only if,both ↵1 6= k1 � 2 and ↵2 6= k2 � 2. Notice that this condition is stronger than(↵1,↵2) 6= (k1 � 2, k2 � 2). We shall also prove that, when

(↵1,↵2) 2 {k1 � 2} ⇥ (R \ {k2}) [ (R \ {k1}) ⇥ {k2 � 2}, (60)

then the finite-time stability (to constant functions) occurs. We conclude that,when d = 3 and transparent boundary conditions are imposed at all the externalnodes, a necessary and su�cient condition for the finite-time stability (to constantfunctions) is (60). The interpretation is that the nodes satisfying (7) and for whichall the adjacent nodes but one are external, are “transparent” and can be “removed”from the tree.

Let (↵1,↵2) 2 R2 be given. The operator AT

reads then

AT

�(u

i

, vi

)i2I�= (v

i

, c2i

u00i

)i2I

with domain

D(AT

) = {(ui

, vi

)i2I 2 H; (u

i

, vi

)i2I 2

Y

i2I[H2(0, 1) ⇥ H1(0, 1)],

c1u01(0) = v1(0), c

i

u0i

(1) = �vi

(1) for i 2 {3, ..., N},X

2ik1

ci

u0i

(0) � c1u01(1) = �↵1v1(1),

X

k1+1iN

ci

u0i

(0) � c2u02(1) = �↵2v2(1),

FINITE-TIME STABILIZATION IN A NETWORK 733

(ui

(0), vi

(0)) = (u1(1), v1(1)) for 2 i k1,

(ui

(0), vi

(0)) = (u2(1), v2(1)) for k1 + 1 i N}.where 0 = d/dx, 00 = d2/dx2, etc.

Setting U = (ui

, vi

)i2I , we see that (12)-(16) and (19) may be written as

Ut

= AT

U, (61)

U(0) = U0 = (u0i

, u1i

)i2I . (62)

If AT

U0 = �U0 with U0 6= 0 and � 6= 0, then the solution U of (61)-(62)reads U(t) = e�tU0 (exponential solution), and then the finite-time stabilization toconstant functions cannot hold.

Proposition 1. Let T denote a tree with N edges and two internal nodes (NM

={1, 2}), and assume that

↵1 6= k1 and ↵2 6= k2. (63)

Then the operator AT

has at least one eigenvalue if, and only if,

↵1 6= k1 � 2 and ↵2 6= k2 � 2. (64)

Furthermore, if (64) holds, then the discrete spectrum of AT

is �d

(AT

) = {�k

; k 2Z} where

�k

=c22log�⇡

2

� (2 + ↵1 � k1)(2 + ↵2 � k2)

(↵1 � k1)(↵2 � k2)

�+ ic2k⇡. (65)

In particular, the finite-time stability to constant functions does not hold for (12)-(16) and (19). Finally, if (60) is satisfied, then the finite-time stability to constant

functions holds.

In (65), log�⇡2denotes the usual determination of the logarithm in C\iR�. Thus

log�⇡2(z) =

⇢log |z| if z 2 (0,+1),log |z| + i⇡ if z 2 (�1, 0).

Proof. First, AT

generates a strongly continuous semigroup of operators in H by(63) and Theorem 2.1. Let � 2 C and U = (u

i

, vi

)i2I 2 D(A

T

). Then the equationA

T

U = �U is equivalent to the following system

(vi

, c2i

u00i

) = �(ui

, vi

) (66)

c1u01(0) = v1(0) (67)

ci

u0i

(1) = �vi

(1), 3 i N (68)X

2ik1

ci

u0i

(0) � c1u01(1) = �↵1v1(1) (69)

X

k1+1iN

ci

u0i

(0) � c2u02(1) = �↵2v2(1) (70)

ui

(0) = u1(1), 2 i k1, (71)

ui

(0) = u2(1), k1 + 1 i N. (72)

Note that the conditions vi

(0) = v1(1) for 2 i k1 and vi

(0) = v2(1) fork1 + 1 i N are satisfied whenever (66) and (71)-(72) hold. (66) is easily solvedas

ui

(x) = ai

e�x/ci + bi

e��x/ci , vi

= �ui

, i 2 I, (73)

734 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

where ai

, bi

2 C are constants to be determined. Substituting the above expressionof u

i

(x) in (67)-(72) yields the system

�b1 = 0, (74)

�ai

= 0, 3 i N, (75)

�X

2ik1

(ai

� bi

) � �(a1e�/c1 � b1e

��/c1) = �↵1�(a1e�/c1 + b1e

��/c1), (76)

�X

k1+1iN

(ai

� bi

) � �(a2e�/c2 � b2e

��/c2) = �↵2�(a2e�/c2 + b2e

��/c2), (77)

ai

+ bi

= a1e�/c1 + b1e

��/c1 , 2 i k1, (78)

ai

+ bi

= a2e�/c2 + b2e

��/c2 , k1 + 1 i N. (79)

If � = 0, we infer from (78)-(79) and (73) that ui

(x) = a1 + b1 for all i 2 I, i.e.U = const, which is excluded. Assume from now on that � 6= 0. Then (74)-(79) isequivalent to the system

b1 = 0, (80)

ai

= 0, 3 i N, (81)

b2 = a1e�/c1 � a2, (82)

bi

= a1e�/c1 , 3 i k1, (83)

bi

= a2e�/c2 + b2e

��/c2 , k1 + 1 i N, (84)

2a2 + (↵1 � k1)e�/c1a1 = 0, (85)

[(�N + k1 � 1 + ↵2)e�/c2 + (N � k1 � 1 � ↵2)e

��/c2 ]a2

+(�N + k1 + 1 + ↵2)e��/c2e�/c1a1 = 0. (86)

The existence of a nontrivial solution ((a1, a2) 6= (0, 0)) holds if, and only if, thedeterminant of the system (85)-(86) in e�/c1a1 and a2 vanishes, i.e.

(2 + ↵1 � k1)(�N + k1 + 1 + ↵2)e��/c2 � (↵1 � k1)(�N + k1 � 1 + ↵2)e

�/c2 = 0.

Since �N + k1 = 1 � k2, this can be expressed as

(2 + ↵1 � k1)(2 + ↵2 � k2)e��/c2 � (↵1 � k1)(↵2 � k2)e

�/c2 = 0.

Using (63), the last equation is equivalent to

e2�c2 =

(2 + ↵1 � k1)(2 + ↵2 � k2)

(↵1 � k1)(↵2 � k2)· (87)

(87) has a solution � 2 C if and only if (2 + ↵1 � k1)(2 + ↵2 � k2) 6= 0, and in thatcase the solutions of (87) read

�k

=c22log�⇡

2

� (2 + ↵1 � k1)(2 + ↵2 � k2)

(↵1 � k1)(↵2 � k2)

�+ ic2k⇡, k 2 Z.

Assume finally that (60) holds, e.g. ↵1 = k1�2 and ↵2 2 R\{k2}. Since transparentboundary conditions are applied at all the external nodes, we have

si

(1, t) = 0, i = 3, ..., N, t � 0,

d1(0, t) = 0, t � 0.

FINITE-TIME STABILIZATION IN A NETWORK 735

This implies

si

(x, t) = 0, i = 3, ..., N, x 2 [0, 1], t � c�1i

, (88)

d1(x, t) = 0, x 2 [0, 1], t � c�11 . (89)

It follows from (34) and (88) that

s2(0, t) = s1(1, t), t � max3ik1

c�1i

.

Combined with the continuity condition u1(1, t) = u2(0, t), this yields

d2(0, t) = d1(1, t) = 0, t � maxi2{1}[[3,k1]

c�1i

.

We infer then from Lemma 2.2 (see (43)-(45) and (46)) that s2(1, t) = dk1+1(0, t) =

· · · = dN

(0, t) = 0 for t large enough. This in turn implies s1(1, t) = d3(0, t) = · · · =dk1(0, t) = 0 for t large enough. We conclude that for some constants C and T ,

ui

(x, t) = C for all i 2 I, all x 2 [0, 1] and all t � T .

4. Finite-time extinction. Pick any tree of depth d � 1, and define the sequence(t

i

)i2I as follows

ti

= c�1i

if i 2 NS

\ {0}, (90)

ti

= c�1i

+maxj2Ii

tj

if i 2 NM

. (91)

Set T (R) = t1. Then it is easily seen that T (R) is the maximum of the quantities

c�1i1

+ c�1i2

+ · · · + c�1ip

,

where p � 1, i1 = 1, iq+1 2 I

iq for 1 q p � 1, and the final point of the edgeof index i

q

is an external node (di↵erent from R). Define T (T ) as the largest ofthe T (R)’s when the root R ranges over N

S

; that is, we take as root of the treeany external node, change the numbering of the edges and nodes, and define thecorresponding sequences (I

i

)i2I and (t

i

)i2I . Obviously, T (R) T (T ) 2T (R).

Example. Consider again the tree drawn in Figure 1, and assume for simplicitythat c

i

= 1 for all i 2 [1, 13]. Then T (R) = 5 and T (T ) = 7. Indeed, if we takethe node of index n = 12 as (new) root, we obtain T (R

n=12) = 7. Similarly, wesee that T (R

n=13) = 7, T (Rn=8) = T (R

n=9) = 6, T (Rn=4) = 5, and T (R

n=10) =T (R

n=11) = 7.

Theorem 4.1. Let T be a tree of root R, and let T (R) and T (T ) be as above.

Assume that the sequence (↵n

)n2NM satisfies the condition

↵n

= kn

� 2 n 2 NM

. (92)

Pick any initial data U0 = {(u0i

, u1i

}i2I 2 H.

(i) If U0 2 H0, then the solution (ui

)i2I of (12)-(16) and (17) satisfies

ui

(., t) ⌘ 0, 8t � 2T (R), 8i 2 I; (93)

(ii) The solution (ui

)i2I of (12)-(16) and (18) satisfies for some number C 2 R

ui

(., t) ⌘ C, 8t � 2T (R), 8i 2 I. (94)

(iii) The solution (ui

)i2I of (12)-(16) and (19) satisfies for some number C 2 R

ui

(., t) ⌘ C, 8t � T (T ), 8i 2 I. (95)

736 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

Remark 2. It is likely that the extinction time Te

(i.e. the least time after whichsolutions remain constant) is given by 2T (R) in the cases (i) and (ii), and T (T )in case (iii), so that the above results are sharp. Actually, for one string, it is wellknown that T

e

= 2/c1 for the solutions of (12)-(16) and (17) (or for the solutionsof (12)-(16) and (18)), while T

e

= 1/c1 for the solutions of (12)-(16) and (19) (seee.g. [12, 14, 17] and the references therein).

Proof. We use again the Riemann invariants di

, si

defined in (32)-(33) that satisfythe transport equations (37)-(38). We need the following

Lemma 4.2. Let T be a tree, and let the sequence (ti

)i2I be as in (90)-(91).

Assume that the sequence (↵n

)n2NM satisfies (92). Then for any U0 2 H and any

solution (ui

)i2I of (12)-(16), with corresponding Riemann invariants d

i

, si

, we have

for all i 2 Isi

(x, t) = 0 8x 2 [0, 1], 8t � ti

. (96)

Proof of Lemma 4.2. We argue by induction on the depth d of the tree. If d = 1,then there is only one edge (I = {1}) and s1 solves

s1,t � c1s1,x = 0, t > 0, 0 < x < 1, (97)

s1(1, t) = 0, t > 0, (98)

s1(., 0) = s01 := v01 + c1u01,x. (99)

Then it is easily seen that

s1(x, t) =

⇢s01(x+ c1t) if x+ c1t 1,0 if x+ c1t � 1.

(100)

Thuss1(x, t) = 0 8x 2 [0, 1], 8t � c�1

1

and (96) is established for d = 1.Assume now Lemma 4.2 established for any tree of depth at most d � 1, where

d � 2. Pick a tree T of depth d, and a sequence (↵n

)n2NM satisfying (P). Denote

by R0 the node of index n = 1, and by Ti

, for i = 2, ..., k1, the subtree of T of rootR0 and of first edge the edge of T of index i. Since T

i

is of depth at most d� 1, weinfer from the induction hypothesis that for i > 1

si

(x, t) = 0 8x 2 [0, 1], 8t � ti

. (101)

It remains to prove (96) for i = 1. Since the condition (92) is satisfied for n = 1,we infer from (34) that

s1(1, t) =k1X

i=2

si

(0, t), 8t � 0.

It follows then from (101) that

s1(1, t) = 0 8t � maxi2I1

ti

.

Finally, using (97), we infer that

s1(x, t) = 0 8x 2 [0, 1], 8t � c�11 +max

i2I1

ti

= t1.

The proof of Lemma 4.2 is complete.Let us go back to the proof of Theorem 4.1.

FINITE-TIME STABILIZATION IN A NETWORK 737

(i) Assume first that U0 2 H0, and let (ui

)i2I denote the solution of (12)-(16)

and (17). From Lemma 4.2, we have that for all i 2 Isi

(x, t) = 0 8x 2 [0, 1], 8t � T (R). (102)

From (17), we infer that d1(0, t) + s1(0, t) = 0 for all t � 0, and hence

d1(0, t) = 0, 8t � T (R).

Using (37), we infer that

d1(x, t) = 0, 8x 2 [0, 1], 8t � c�11 + T (R).

Combined with (43)-(45) (with k = k1) and (102), this yields

d2(0, t) = · · · = dk1(0, t) = 0, 8t � c�1

1 +maxi2I1

c�1i

+ T (R).

Using the second definition of T (R) and proceeding inductively, we arrive to

di

(x, t) = 0 8i 2 I, 8x 2 [0, 1], 8t � 2T (R). (103)

Gathering together (102) and (103), we infer the existence of some constant C 2 Rsuch that

ui

(x, t) = C, 8i 2 I, 8x 2 [0, 1], 8t � 2T (R).

Using (17), we see that C = 0. This proves that solutions of (12)-(16) and (17) arenull for t � T (R). Combined with the strong continuity of the semigroup (etAD )

t�0

in H0, this yields the finite-time stability.(ii) Assume now that u0 2 H and let (u

i

)i2I denote the solution of (12)-(16) and

(18). From (17), we infer that d1(0, t) � s1(0, t) = 0 for all t � 0. The same proofas in (i) then yields

si

(x, t) = di

(x, t) = 0, 8i 2 I, 8x 2 [0, 1], 8t � 2T (R).

Thus there exists a constant C 2 R such that

ui

(x, t) = C, 8i 2 I, 8x 2 [0, 1], 8t � 2T (R).

(iii) Pick a solution (ui

)i2I of (12)-(16) and (19). Then it follows from Lemma

4.2 that for all i 2 Isi

(x, t) = 0 8x 2 [0, 1], 8t � T (R). (104)

For any given i 2 I, we pick a sequence i1 < i2 < · · · < ip

such that i1 = 1, i = iq

for some q 2 [1, p], and the final point of the edge of index ip

is an external point,that we call R. If we exchange R and R, we notice that d

i

is linked to the sj

’s(associated with the new root R) by:

di

(x, t) = sip�i+1(1 � x, t).

We infer that

di

(x, t) = 0 8x 2 [0, 1], 8t � T (T ). (105)

Therefore, there exists a constant C 2 R such that

ui

(x, t) = C, 8i 2 I, 8x 2 [0, 1], 8t � T (T ).

The proof of Theorem 4.1 is complete.

738 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

5. Sharpness of the condition (92). The condition (92), which is su�cient toyield the finite-time stability, is expected to be also necessary in some situations.(However, we already noticed that it was not necessary for a tree with two internalnodes.)

Here we consider a star-shaped tree, with the homogeneous Dirichlet boundarycondition at one external node and the transparent boundary conditions at theother external nodes. We consider any possible value of the coe�cient ↵ at theinternal node, and exhibit an eigenvalue of the underlying operator when (28) holdsand (92) fails.

Assume that T is a star-shaped tree with N edges (d = 2, k1 = N), and considerthe boundary conditions (14)-(16) and (17). (See Figure 3.)

3

4

5

4

5

3

2

2

110

Dirichlet boundary condition

Transparent boundary condition

Figure 3. A star-shaped tree.

We assume that ↵1 6= N , so that the system (12)-(16) and (17) is well-posedin H0 according to Theorem 2.1. According to Theorem 4.1, there is a finite-timestabilization when ↵1 = N �2. We shall show that this condition is sharp, i.e. thata finite-time stabilization cannot hold if ↵1 62 {N � 2, N}.

Let ↵1 2 R be given. The operator AD

reads

AD

�(u

i

, vi

)i2I�= (v

i

, c2i

u00i

)i2I

with

D(AD

) = {(ui

, vi

)i2I 2 H0; (v

i

, c2i

u00i

)i2I 2 H0,

ci

u0i

(1) = �vi

(1) for 2 i N,X

2iN

ci

u0i

(0) � c1u01(1) = �↵1v1(1),

and (ui

(0), vi

(0)) = (u1(1), v1(1)) for 2 i N},where 0 = d/dx, 00 = d2/dx2, etc. Setting U := (u

i

, vi

)i2I , we see that (12)-(16)

and (17) may be written as

Ut

= AD

U (106)

U(0) = U0 = (u0i

, u1i

)i2I (107)

If AD

U0 = �U0 with U0 6= 0, then the solution U of (106)-(107) reads U(t) = e�tU0

(exponential solution), and hence ||U(t)||H = e(Re�)t||U0||H > 0 for all t � 0. Thusif the operator A

D

has at least one eigenvalue, then the finite-time stabilizationcannot hold.

FINITE-TIME STABILIZATION IN A NETWORK 739

Proposition 2. Let T denote a star-shaped tree with N edges, and assume that

↵1 6= N . Then the operator AD

has at least one eigenvalue if, and only if,

↵1 6= N � 2. (108)

Furthermore, if (108) holds, then the discrete spectrum of AD

is �d

(AD

) = {�k

; k 2Z} where

�k

=c12log�⇡

2(N � 2 � ↵1

N � ↵1) + ic1k⇡ (109)

In particular, if (108) holds, then the finite-time stabilization of (12)-(16) and (17)in H0 fails.

Remark 3. (1) It follows from the proof of Proposition 2 that the operator AD

has no eigenvalue when ↵1 = N . However, as the well-posedness of the system failsby Theorem 2.1, that value of ↵1 is not considered for the analysis of the finite-timestability of the system. Actually, there may exist solutions that do not reach aconstant value in finite time (pick h(t) = t in Remark 1 (2)).(2) If we replace the Dirichlet boundary condition u1(0, t) = 0 by the transparentboundary condition u1,t(0, t) = c1u1,x(0, t) and take any value ↵1 6= N , then sinced1(0, t) = s2(1, t) = · · · = s

N

(1, t) = 0 for all t � 0, we infer from (43)-(45) and(46) that s1(1, t) = d2(0, t) = · · · = d

N

(0, t) = 0 for all t � max1iN

c�1i

, so thatfor some constant C 2 R

ui

(x, t) = C, 8i 2 [1, N ], 8x 2 [0, 1], 8t � 2 max1iN

c�1i

.

Proof. Let � 2 C and U = (ui

, vi

)i2I 2 D(A

D

). Then the equation AD

U = �U isequivalent to the following system

(vi

, c2i

u00i

) = �(ui

, vi

), 1 i N, (110)

u1(0) = 0, (111)

ci

u0i

(1) = �vi

(1), 2 i N, (112)X

2iN

ci

u0i

(0) � c1u01(1) = �↵1v1(1), (113)

ui

(0) = u1(1), 2 i N. (114)

Note that the conditions v1(0) = 0 and vi

(0) = v1(1) for 2 i N are satisfiedwhenever (110)-(111) and (114) hold. (110) is easily solved as

ui

(x) = ai

e�x/ci + bi

e��x/ci , vi

(x) = �ui

(x), 1 i N, (115)

where ai

, bi

2 C are constants to be determined. Substituting the above expressionof u

i

(x) in (111)-(114) yields the system

a1 + b1 = 0, (116)

�ai

= 0, 2 i N, (117)

�X

2iN

(ai

� bi

) � �(a1e�/c1 � b1e

��/c1) = �↵1�(a1e�/c1 + b1e

��/c1), (118)

ai

+ bi

= a1e�/c1 + b1e

��/c1 , 2 i N. (119)

If � = 0, we infer from (115)-(116) and (119) that U = 0, which is excluded. Assumefrom now on that � 6= 0. Then the system (116)-(119) is found to be equivalent to

740 F. ALABAU-BOUSSOUIRA, V. PERROLLAZ AND L. ROSIER

the system

b1 = �a1, (120)

ai

= 0, 2 i N, (121)

�(N � 1)a1(e�/c1 � e��/c1) � a1(e

�/c1 + e��/c1)

= �↵1a1(e�/c1 � e��/c1), (122)

bi

= a1(e�/c1 � e��/c1), 2 i N. (123)

The existence of a nontrivial solution (a1 6= 0) holds if, and only if, the coe�cientabove a1 in (122) vanishes, i.e.

(�N + ↵1)e�/c1 + (N � 2 � ↵1)e

��/c1 = 0. (124)

For ↵1 6= N , (124) is equivalent to

e2�c1 =

N � 2 � ↵1

N � ↵1·

(5) has a solution � 2 C if and only if ↵1 6= N � 2, and in that case the solutionsof (5) read

�k

=c12log�⇡

2(N � 2 � ↵1

N � ↵1) + ic1k⇡, k 2 Z. (125)

Remark 4. For k 2 Z and �k

as in (125), we introduce the sequence of eigenfunc-tions U

k

= ((ui,k

, vi,k

))1iN,k2Z where

u1,k(x) = e�kx/c1 � e��kx/c1 , v1,k(x) = �k

u1,k(x),

ui,k

(x) = (e�k/c1 � e��k/c1)e��kx/ci , vi,k

(x) = �k

ui,k

(x), for 2 i N.

Then the family (ak

Uk

)k2Z may fail to be a Riesz basis in H0 for any choice of

the sequence of numbers (ak

)k2Z. Consider e.g. N = 2 and c2 = c1/2. Then, for

N � 2 < ↵1 < N ,

u2,k(x) = (e�k/c1 � e��k/c1)e� log |N�2�↵1N�↵1

|x�i⇡xe�i2k⇡x.

Let U = (ui

, vi

)i=1,2 2 H0 be given. If (a

k

Uk

)k2Z is a Riesz basis in H0, then U

can be expended in terms of the Uk

’s in H0 as

(ui

, vi

) =X

k2Zdk

ak

(ui,k

, vi,k

), i = 1, 2

for some sequence (dk

)k2Z 2 L2(Z). Writing

elog |N�2�↵1N�↵1

|x+i⇡xu2(x) =X

k2Zck

e�i2k⇡x

we have, by harmonicity, that

ck

= dk

ak

(e�k/c1 � e��k/c1), k 2 Z,and hence

u1(x) =X

k2Z

ck

e�k/c1 � e��k/c1(e�kx/c1 � e��kx/c1)

in L2(0, 1). Therefore, u1 is uniquely determined by the ck

’s, and hence by u2,which is a property much stronger than the conditions u1(0) = 0 and u1(1) = u2(0)present in the definition of H0. This shows that the family (a

k

Uk

)k2Z is not total

in H0.

FINITE-TIME STABILIZATION IN A NETWORK 741

It is natural to conjecture a decay of all the trajectories like

||U(t)||H0 C(↵1)ec12 log

���N�2�↵1N�↵1

���t||U(0)||H0 , t � 0, (126)

for N � 2 < ↵1 < N . (Note that lim↵1&N�2 log |N�2�↵1

N�↵1| = �1.) Without a Riesz

basis of eigenvectors in the full space H0, the validity of (126) seems hard to check.

6. Conclusion. In this paper, we addressed the issue of the finite-time stabilizationof a network of strings. We showed in particular that, when using the homogeneousDirichlet boundary condition at the root and transparent boundary conditions atthe other external nodes, and incorporating in Kirchho↵ law the damping term↵(n)u

t

with ↵(n) = k(n) � 2 at each internal node n, we obtained the finite-timestability of the system. The condition on the coe�cients ↵(n) proved also to besharp for a star-shaped tree.

Several questions remain open:

1. Can we obtain a sharp result for a general network? Same question when (12)is replaced by (8)?

2. Can we obtain finite-time stability results for networks containing circuits?

Acknowledgments. The authors wish to acknowledge the contribution of anony-mous Referees in o↵ering valuable suggestions and in bringing the papers [18, 21]to their attention.

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Received October 2014; revised January 2015.

E-mail address: fatiha.alabau@univ-lorraine.frE-mail address: Vincent.Perrollaz@lmpt.univ-tours.frE-mail address: Lionel.Rosier@mines-paristech.fr