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INVERSE PROBLEM FOR THE HEAT EQUATION AND THE ¨ SCHRODINGER EQUATION ON A TREE LIVIU I. IGNAT, ADEMIR F. PAZOTO, AND LIONEL ROSIER Abstract. In this paper we establish global Carleman estimates for the heat and Schr¨ odinger equations on a network. The heat equation is considered on a general tree and the Schr¨ odinger equation on a star-shaped tree. The Carleman inequalities are used to prove the Lipschitz stability for an inverse problem consisting in retrieving a stationary potential in the heat (resp. Schr¨ odinger) equation from boundary measurements.

1. Introduction In this paper we consider two inverse problems on a network formed by the edges of a tree. The problems we address here enter in the framework of quantum graphs. The name quantum graph is used for a graph considered as a one-dimensional singular variety and equipped with a differential operator. Those quantum graphs are metric spaces which can be written as the union of finitely many intervals, which are compact or [0, ∞) and any two of these intervals are either disjoint or intersect only at one of their endpoints. Quantum graphs arise as simplified models in mathematics, physics, chemistry, and engineering (e.g., nanotechnology and microelectronics), when one considers propagation of waves through a quasi-one-dimensional system that looks like a thin neighborhood of a graph. We can mention in particular the quantum wires and thin waveguides. Differential operators on metric graphs arise in a variety of applications, to quote a few: carbon nano-structures [31], photonic crystals [19], high-temperature granular superconductors [1], quantum waveguides [15], freeelectron theory of conjugated molecules in chemistry, quantum chaos, etc. For more details we refer the reader to the review papers [28], [30], [29], [18] and the references therein for more informations on this topic. To be more precise we consider the heat equation on a 1-D network Γ given by the edges of a general tree and the Schr¨ odinger equation on a star-shaped tree. The first system we consider is the following one  ut − ∆Γ u + pu = 0, in Γ × (0, T ),     u = h, on ∂Γ × (0, T ), (1.1)     u(·, 0) = u0 , in Γ, where ∆Γ is the Laplace operator on the network Γ. The system is closed with the coupling conditions at the internal nodes of the tree, namely the continuity and the Kirchhoff’s law on the Key words and phrases. Carleman estimate; tree; network; heat equation; Schr¨ odinger equation; inverse problem. 1

2

LIVIU I. IGNAT, ADEMIR F. PAZOTO, AND LIONEL ROSIER

flux at all internal vertices of Γ. Here, u is a collection of functions uα each of them satisfying a heat equation on some edge of the network. Simultaneously with problem (1.1) we consider the following problem  iut + ∆Γ u + pu = 0, in Γ × (0, T ),     u = h, on ∂Γ × (0, T ), (1.2)     u(·, 0) = u0 , in Γ, under similar coupling conditions as in the previous model. In both cases we are interested in determining the potential p, a collection of functions defined on the edges of Γ, from boundary measurements. In [2, 3], it was proved that the connectivity, the lengths of the edges and the potentials on them can be retrieved from the knowledge of the dynamic Dirichlet-to-Neumann map for the wave equation. Here, we determine the potentials, but only from a finite number of measurements (more explicitly, the measurements at k ≤ N external nodes of the normal derivatives for the solution of the Cauchy problem corresponding to one initial data.) In the case of the first system, we are able to prove that we can recover p using only N − 1 measurements, where N is the total number of exterior nodes of the network Γ. However, in the case of the second system, besides of the fact that we need to deal with a star-shaped network, we only can recover the potential p from measurements performed at all the exterior nodes of Γ. The use of Carleman estimates to achieve uniqueness and stability results in inverse problems is well known. Some authors use local Carleman inequalities and deduce uniqueness and H¨ older estimates. Others make use of global Carleman inequalities and deduce Lipschitz stability results and hence uniqueness results. We shall follow that second approach. Inverse problems with a finite number of measurements have been widely studied by Bukhgeim and Klibanov (see [11], [24], and [25]) by means of Carleman estimates (see also the book [23] and the references therein). For a wide class of partial differential equations, their method provides the stability in the inverse problem, whenever a suitable Carleman estimate is available. Since [11], there have been many works based upon their methodology. The theory of global Carleman estimates for parabolic operator has been largely developed since the work by Fursikov-Imanuvilov [20] and it has been applied to many situations (e.g. to prove the controllability along the trajectories or the stability in inverse problems). Since a complete list of references is too long we refer the reader to [37] for a quite complete review of the state of art. Concerning the Schr¨ odinger equation we refer to [6, 8, 10, 12, 13, 33, 38] where Carleman estimates are proved and used to establish the stability for some inverse problems (see also [22, 32, 39] for some other Carleman estimates for Schrödinger equation). The same approach has given many results for the wave equation. Since a complete list is too long we quote only some of them, related to the same inverse problem consisting in retrieving a stationary potential in wave equation: [34] and [36] for Dirichlet boundary data and a Neumann measurement and [21] for Neumann boundary data and a Dirichlet measurement. These references are based on the use of local or global Carleman estimates. In the framework of

¨ INVERSE PROBLEM FOR THE HEAT EQUATION AND THE SCHRODINGER EQUATION ON A TREE

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Carleman inequalities on networks we mention the recent paper [5] where the authors establish a global Carleman estimate for the wave equation on a star-shaped tree and used it to derive the Lipschitz stability in an inverse problem. The Carleman estimate in [5] involves some positive definite matrix introduced in [9] to derive a Carleman estimate for the one-dimensional heat equation with discontinuous coefficients. As far as we know, the determination of a time-independent potential for the heat or Schrödinger equation in a network-like structure has not been addressed in the literature yet. This type of problems has been studied for example for membranes or elastic strings (see for instance [4] and the references therein). Let us now state the main results of the paper. For a given initial data u0 and a given boundary data h, we denote by u(p) the solution of the above systems associated with the potential p ∈ L∞ (Γ, R). We introduce the space H 2,1 (Γ × (0, T )) := L2 (0, T ; H 2 (Γ)) ∩ H 1 (0, T ; L2 (Γ)). (See below Section 2 for the definition of H 2 (Γ).) We also introduce the ball Bm (0) := {q ∈ L∞ (Γ, R); ||q||L∞ (Γ) ≤ m}. Then the following stability results hold. Theorem 1.1. Assume that p ∈ L∞ (Γ), u0 = u0 (x), h = h(x, t) and r > 0 are such that the solution u(p) of (1.1) fulfills u(p) ∈ H 2,1 (Γ × (0, T )), ∂t u(p) ∈ H 2,1 (Γ × (0, T )), and such that for some t0 ∈ (0, T ) it holds |u(p)(·, t0 )| ≥ r a.e. on Γ. Then, for any m > 0 there exists a constant C = C(m, ||∂t u(p)||L∞ (Γ×(0,T )) , r) such that for any q ∈ Bm (0) satisfying ∂x [u(p) − u(q)](v, .) ∈ H 1 (0, T ) for all exterior nodes v, we have kp−qkL2 (Γ) X ≤ C k[u(p) − u(q)](·, t0 )kH 2 (Γ) + k∂x [u(p) − u(q)](v, ·)kH 1 (0,T ) , v∈E

where E denotes the set of all the exterior vertices of Γ except one. For the second system, under the assumption that the network is a star-shaped tree, we can prove a similar stability result. Theorem 1.2. Assume that p ∈ L∞ (Γ; R), u0 = u0 (x), h = h(x, t) and r > 0 are such that the solution of (1.2) satisfies • u0 (x) ∈ R or iu0 (x) ∈ R a.e. in Γ, • |u0 (x)| ≥ r > 0 a.e. in Γ, and • ∂t u(p) ∈ H 2,1 (Γ × (0, T )). Then, for any m ≥ 0, there exists a constant C = C(m, ||∂t u(p)||H 2,1 (Γ×(0,T )) , r) > 0 such that for any q ∈ Bm (0) satisfying ∂t u(q) ∈ H 2,1 (Γ × (0, T )),

4


we have ||p − q||L2 (Γ) ≤ C

X

||∂x [u(p) − u(q)](v, .)||H 1 (0,T ) ·

v∈∂Γ

The above theorems extend to networks classical results on inverse problems. To prove those results, we need to establish (new) global Carleman estimates for the heat (resp. the Schrödinger) equation on trees. Note that if we impose Kirchhoff-type conditions to the weight function at the internal vertices, the Carleman estimate cannot be derived. In our Carleman estimates, the weight function has to fulfill some nonlinear flux condition at each internal vertex. Note that we shall write Carleman estimates with two parameters (namely s and λ), although a single parameter (s) is usually sufficient in dimension one. This is just to simplify the construction of the weight function. On the other hand, for the Schrödinger equation posed on a star-shaped tree with N external vertices, we consider a combination of N weight functions in order to cancel some “bad” terms at the internal vertices involving time derivatives (unfortunately, the extension of our result to a tree with two internal nodes cannot be done by that approach, see below). That strategy was used in [7], with two different weight functions, in order to improve the observation region for the wave equation. The article is organized as follows. In Section 2 we introduce the notations and some classical facts about the heat and Schr¨ odinger equations on trees. Section 3 presents the analysis in the case of the heat equation. The Schrödinger equation is considered in Section 4. Finally we discuss some open problems in Section 5. 2. Notations and Preliminaries Let Γ = (V, E) be a graph where V is the set of vertices and E the set of edges. The edges are assumed to be of finite length and their ends are the vertices of V . For each v ∈ V we denote Ev = {e ∈ E : v ∈ e}. The multiplicity of a vertex of Γ is equal to the number of edges that branch out from it. If the multiplicity is equal to one, the vertex is said to be exterior, otherwise it is said to be interior. We assume that Γ does not contain vertices with multiplicity two, since they are irrelevant for our models. From now on, we assume that Γ is a tree, that is, Γ is a planar finite connected graph without circuit (closed path). We fix an orientation of Γ and for each oriented edge e, we denote by I(e) its initial vertex and by T (e) its terminal one. We identify every edge e of Γ with an interval Ie , where Ie = [0, le ], le being the length of e. This identification introduces a coordinate xe along the edge e. Let v be a vertex of V and e be an edge in Ev . We set ( 0 if v = I(e), i(v, e) = le if v = T (e). We identify any function u on Γ with a collection {ue }e∈E of functions ue defined on the edges e of Γ. Each ue can be considered as a function on the interval Ie . In fact, we use the same notation ue for both the function on the edge e and the function on the interval Ie identified with e. For a function u : Γ → C, u = {ue }e∈E , we denote by f (u) : Γ → C the family {f (ue )}e∈E , where f (ue ) : e → C.


5

A function u = {ue }e∈E is continuous if and only if ue is continuous on Ie for every e ∈ E, and u is continuous at the vertices of Γ: 0

ue (i(v, e)) = ue (i(v, e0 )),

∀ e, e0 ∈ Ev .

The space Lp (Γ), 1 ≤ p < ∞ consists of all the functions u = {ue }e∈E on Γ that belong to e ) for each edge e ∈ E. That space is endowed with the norm X kukpLp (Γ) = kue kpLp (Ie ) < ∞.

Lp (I

e∈E

Similarly, the space L∞ (Γ) consists of all the functions u = {ue }e∈E that belong to L∞ (Ie ) for each edge e ∈ E. The corresponding norm is kukL∞ (Γ) = sup kue kL∞ (Ie ) < ∞. e∈E

N∗ ,

H m (Γ),

The Sobolev space with m ∈ consists of all the continuous functions on Γ (viewed 2 m as a closed subset of R ) that belong to H (Ie ) for each e ∈ E. It is endowed with the norm X kuk2H m (Γ) = kue k2H m (e) < ∞. e∈E

The spaces L2 (Γ) and H m (Γ) are Hilbert spaces when endowed with the inner products X XZ (u, v)L2 (Γ) = (ue , v e )L2 (Ie ) = ue (x)v e (x)dx e∈E

e∈E

Ie

and (u, v)H m (Γ) =

X

e

e

(u , v )H m (Ie ) =

e∈E

m Z XX e∈E k=0

Ie

dk ue dk v e dx. dxk dxk

H01 (Γ) denotes the set of functions in H 1 (Γ) that vanish at the exterior vertices. We now introduce the Laplace operator ∆Γ on the tree Γ. Even if it is a standard procedure, we prefer to recall it following [14], for the sake of completeness. Consider the sesquilinear continuous form ϕ on H01 (Γ) defined by XZ ϕ(u, v) = (ux , vx )L2 (Γ) = uex (x)vxe (x)dx. e∈E

Ie

We denote by D(∆Γ ) the set of all the functions u ∈ H01 (Γ) such that the linear map v ∈ H01 (Γ) → ϕu (v) := ϕ(u, v) satisfies |ϕu (v)| ≤ CkvkL2 (Γ)

for all v ∈ H01 (Γ).

For u ∈ D(∆Γ ), we can extend ϕu to a linear continuous mapping on L2 (Γ). There is a unique element in L2 (Γ), denoted by ∆Γ u, such that ϕ(u, v) = −(∆Γ u, v)L2 (Γ)

for all v ∈ H01 (Γ).

6


We now define the normal exterior derivative of a function u = {ue }e∈E at the endpoints of the edges. For each e ∈ E and v an endpoint of e we consider the normal derivative of the restriction of u to the edge e of Ev evaluated at i(v, e) to be defined by: ( −uex (0+ ) if i(v, e) = 0, ∂ue (i(v, e)) = ∂ne ue (l− ) if i(v, e) = l . x

e

e

With this notation it is easy to characterize D(∆Γ ) (see [14]): n X ∂ue D(∆Γ ) = u = {ue }e∈E ∈ H 2 (Γ) ∩ H01 (Γ); (i(v, e)) = 0 ∂ne

for any interior vertex v

o

e∈Ev

and (∆Γ u)e = (ue )xx

for all e ∈ E, u ∈ D(∆Γ ).

In other words D(∆Γ ) is the space of all the continuous functions u = {ue }e∈E on Γ, such that for each edge e ∈ E, ue ∈ H 2 (Ie ), and which vanish at each exterior node and fulfill the following Kirchhoff-type condition X X uex (le− ) − uex (0+ ) = 0 e∈E; T (e)=v

e∈E; I(e)=v

at each interior node v. It is easy to verify that (∆Γ , D(∆Γ )) is a linear, unbounded, self-adjoint, dissipative operator on L2 (Γ), i.e. Re (∆Γ u, u)L2 (Γ) ≤ 0 for all u ∈ D(∆Γ ). 3. The heat equation 3.1. Preliminaries and notations. In this section we introduce the notations for the elements of the considered tree. We mainly follow the notations of [17]. We first describe the procedure to index the edges and vertices of the tree. We first choose an exterior vertex, called the root of the tree and denoted by R. The remaining edges and vertices will be denotedSby eα and Oα , respectively, where α = (α1 , . . . , αk ) is a multi-index (taking value in {1} ∪ k≥2 Nk ). The multi-indices are defined by induction in the following way. For the edge containing the root R we choose the index 1. That edge is denoted by e1 and its second end is denoted by O1 . Assume now that the interior vertex Oα , which is the end of the edge eα , has multiplicity equal to mα + 1. The mα edges, different from eα , that branch out from Oα are denoted by eαβ with β ∈ {1, . . . , mα }. (See Figure 1.) Let now I be the set of the interior vertices of Γ and E be the set of the exterior vertices of Γ, R being excepted. We denote by II = {α, Oα ∈ I},

IE = {α, Oα ∈ E}.

the sets of the indices for the interior and exterior vertices (except the root R). With these notations I = II ∪ IE is the set of the indices of all the vertices except the root R. The length of the edge eα will be denoted by lα . Each eα is parameterized by the interval [0, lα ], so that the end Oα of eα corresponds to x = lα while the origin of eα corresponds to x = 0.


7

R

e1

O1 e13 e11

O13

e12

e132

O12

O11

e131 e121

e122

O121

O131

e1211 O1211

O132

e1212

O122

O1212

Measurement No measurement

Figure 1. A tree with 10 edges.

3.2. Carleman estimate for the heat equation. In this section we derive a Carleman estimate for the heat equation on a tree. The following properties for a function u = {uα }α∈I : Γ → R will be relevant for our work. (C1) Continuity condition at the internal vertices: uα (lα ) = uαβ (0) for all α ∈ II and β ∈ [[1, mα ]]. m Pα αβ (C2) Flux condition at the internal vertices: uαx (lα ) = ux (0) for all α ∈ II . β=1

(C3) Vanishing condition at the root R and at the external vertices: u(v) = 0 for all v ∈ {R}∪E. Throughout the paper we shall use the notation [[1, m]] := [1, m] ∩ N = {1, ..., m}. We introduce the set Z = {u = {uα }α∈I : Γ × [0, T ] → R; uα ∈ C 2,1 ([0, lα ] × [0, T ]), u(·, t) satisfies (C1)-(C3)}. Note that u(·, t) ∈ D(∆Γ ) for u ∈ Z and t ∈ [0, T ]. The aim of this section is to define a continuous weight function ψ = {ψ α }α∈I : Γ → (0, ∞) and a constant Cψ > 0 such that if we set θ(x, t) =

eλψ(x) , t(T − t)

ϕ(x, t) =

we have the following Carleman estimate.

eλCψ − eλψ(x) , x ∈ Γ, t ∈ (0, T ), t(T − t)

8


Proposition 3.1. There exist a continuous function ψ : Γ → (0, +∞) and some positive constants λ0 , s0 , C such that for all λ ≥ λ0 , s ≥ s0 and q ∈ Z, it holds Z TZ (sθ)−1 (|qt |2 + |∆Γ q|2 ) + λ2 (sθ)|qx |2 + λ4 (sθ)3 |q|2 e−2sϕ dxdt 0

Γ T

Z +

λ(sθ)(|qx |2 e−2sϕ )(R, t)dt

0

≤C

Z 0

T

Z

2 −2sϕ

|qt + ∆Γ q| e

dxdt +

Γ

XZ v∈E

T

λ(sθ)(|qx |2 e−2sϕ )(v, t)dt .

(3.3)

0

In the above proposition we have used the following notations |q|2 = {|q α |2 }α∈I , |qt |2 = {|qtα |2 }α∈I , |qx |2 = {|qxα |2 }α∈I , etc. and Z XZ u dx = uα dx. Γ

In the sequel, ϕt = ∂ϕ/∂t, ϕx = ∂ϕ/∂x, ϕnx

α∈I Iα = ∂ n ϕ/∂xn ,

etc.

Remark 3.2. (1) The same inequality holds for the operator ∂t − ∆Γ instead of ∂t + ∆Γ just by changing t into T − t. (2) In the definition of Z, we can replace C 2,1 by H 2,1 , as well. (3) We note that the result is false with only N − 2 measurements for a star-shaped tree with N edges of length 1. Indeed, a non-trivial solution of the heat equation that vanishes on N − 2 edges, at the internal vertex and all the external vertices is as follows. Consider q = {q α }α∈I with q 1 ≡ 0, 2

q 11 (x, t) = e−π t sin(πx), 12

q (x, t) = −e q

1j

≡ 0,

−π 2 t

x ∈ (0, 1), t ≥ 0,

sin(πx),

x ∈ (0, 1), t ≥ 0,

3 ≤ j ≤ N − 1.

Then q is a non trivial solution of the heat equation on the star-shaped tree for which the measurements at the external nodes R and O1j for 3 ≤ j ≤ N − 1 vanish. Proof. Let us consider the operator P = ∂t + ∆Γ . Set u = e−sϕ q and w = e−sϕ P (esϕ u). Following [35] we obtain w = M u = ut + sϕt u + (∆Γ u + 2sϕx ux + s(∆Γ ϕ)u + s2 |ϕx |2 u) = M1 u + M2 u, where M1 u = ∆Γ u + sϕt u + s2 |ϕx |2 u

(3.4)

and M2 u = ut + 2sϕx ux + s(∆Γ ϕ)u are the self-adjoint and skew-adjoint parts of M , respectively. Then

(3.5)

kwk2 = kM1 u + M2 uk2 = kM1 uk2 + kM2 uk2 + 2(M1 u, M2 u), where k · k and (·, ·) denote the norm and the inner product of L2 (Γ × (0, T )), respectively.


Step 1. Exact computation of (M1 u, M2 u). Recall that XZ T Z (M1 u, M2 u) = α∈I

0

lα

9

(M1 u)α (M2 u)α dxdt.

0

We compute the integral term in the r.h.s. of the above identity only for one (arbitrary) edge eα , that we denote by e for simplicity. We assume that e is parameterized by x ∈ [0, l]. Also, RR R RT Rl RT where there is no confusion, we use the symbols and to denote 0 0 and 0 , respectively. We write Z Z T

l

(M1 u)(M2 u) dxdt = I1 + I2 + I3 + I4 0

0

with Z

T

l

Z

uxx ut ,

I1 = 0

Z

0 T

l

Z

I2 =

uxx (2sϕx ux + sϕxx u), 0

Z

0 T

l

Z

(sϕt u + s2 ϕ2x u)(ut + 2sϕx ux ),

I3 = 0

Z

0 T

l

Z

(sϕt u + s2 ϕ2x u)(sϕxx u).

I4 = 0

0

For I1 we have that ZZ

Z

I1 = −

ux uxt + I21

Z l l ux ut = ux ut . 0

0

I22

We write the second term as I2 = + where ZZ ZZ I21 = 2s uxx ϕx ux and I22 = s uxx ϕxx u. Thus I21

ZZ = −s

2

ϕxx |ux | + s

Z

l ϕx |ux |2

0

and I22

ZZ

Z

l = −s ux (ϕ3x u + ϕxx ux ) + s ux ϕxx u 0 ZZ Z ZZ Z l 2 l u s 2 2 ϕxx |ux | + s ϕxx uux . ϕ4x u − s ϕ3x − s = 2 2 0 0

I3 is decomposed as I3 = I31 + I32 where I31 I32

ZZ =

ZZ =

(sϕt u + s2 |ϕx |2 u)ut

(sϕt u + s2 |ϕx |2 u)(2sϕx ux ).

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Then I31 I32

ZZ =−

(sϕtt + 2s2 ϕx ϕxt )

|u|2 , 2

ZZ

sϕx (sϕt + s2 |ϕx |2 )∂x (|u|2 ) ZZ Z l 2 3 2 = −s (ϕx ϕt )x + s(ϕx )x |u| + sϕx (sϕt + s2 |ϕx |2 )|u|2 . =

0

Finally, ZZ I4 =

(s2 ϕt + s3 |ϕx |2 )ϕxx |u|2 .

We conclude that for the edge e, Z TZ l Z TZ l Z 2 M1 uM2 u dxdt = −2s ϕxx |ux | + 0

0 T

Z + 0

T

Z

l

|u|2

hs

(ϕ4x − ϕtt ) − s2 (|ϕx |2 )t − s3 ϕx (|ϕx |2 )x

2 0 0 0 0 h i l s ux ut + sϕxx uux + s|ux |2 ϕx + |u|2 − ϕ3x + s2 ϕx ϕt + s3 (ϕx )3 . 2 0

(3.6)

Summing now the above identity over all the edges {eα }α∈I we obtain the exact expression of the scalar product (M1 u, M2 u): Z TZ Z TZ hs i 2 (M1 u, M2 u) = −2s (∆Γ ϕ)|ux | + |u|2 (ϕ4x − ϕtt ) − s2 (|ϕx |2 )t − s3 ϕx (|ϕx |2 )x 2 0 Γ 0 Γ XZ T h i lα s + uαx uαt + sϕαxx uα uαx + s|uαx |2 ϕαx + |uα |2 − ϕα3x + s2 ϕαx ϕαt + s3 (ϕαx )3 . (3.7) 2 0 0 α∈I

Step 2. Terms in the inner product related to the internal nodes. Let us pick an internal node Oα . Using our previous notations, its parent edge is eα and its children edges are denoted by eαβ with β ∈ [[1, mα ]]. Let us denote by X α the sum of the boundary terms involving this internal node Oα in the right hand side of (3.7). Thus Z Th i s α X = uαx uαt + sϕαxx uα uαx + s|uαx |2 ϕαx + |uα |2 − ϕα3x + s2 ϕαx ϕαt + s3 (ϕαx )3 (lα , t)dt 2 0 Z T X h αβ αβ αβ αβ αβ 2 αβ − uαβ x ut + sϕxx u ux + s|ux | ϕx 0

β∈[[1,mα ]]

i s 2 αβ αβ 3 αβ 3 + |uαβ |2 − ϕαβ + s ϕ ϕ + s (ϕ ) (0, t)dt. x x t 2 3x Moreover, in (3.7) we also have contributions from the exterior nodes in E and from the root R. These contributions are given by Z T XZ T Y = −s |u1x |2 ϕ1x (0, t)dt + s |uαx |2 ϕαx (lα , t)dt. (3.8) 0

α∈IE

0

i


11

Let us now define the weight function ψ = {ψ α }α∈I on the tree as follows. The components : [0, lα ] → R are chosen in such a way that ψ α ∈ C ∞ ([0, lα ]) and

ψα

α (x) ≥ 0 on [0, l ], (B1) |ψxα (x)|2 + ψxx α

(B2) ψxα > 0 on [0, lα ], (B3)

3 4 Cψ

≥ ψ α > 32 Cψ on [0, lα ], for some positive constant Cψ ,

α | ≤ Kψ α on [0, l ] for some positive constant K, (B4) |ψxx α x

(B5) ψ α (lα ) = ψ αβ (0) for all α ∈ II , β ∈ [[1, mα ]], (B6) ψxαβ (0) − (mα + 1)ψxα (lα ) > 0 for all α ∈ II , β ∈ [[1, mα ]], P P αβ 3 α 3 α (B7) (ψx (0)) −(ψx (lα )) −(mα +1)ψx (lα ) ψxα (lα )− β∈[[1,mα ]]

β∈[[1,mα ]]

2 αβ ψx (0)

> 0 for all α ∈ II .

Finding a set of functions as above is easy. We can even take ψ α to be affine, ψ α (x) = aα x+bα . The coefficients aα and bα are positive numbers that satisfy

(P1)

3 4 Cψ

≥ aα lα + bα > bα > 23 Cψ for all α ∈ I,

(P2) aα lα + bα = bαβ for all α ∈ II and β ∈ [[1, mα ]], (P3) aαβ − (mα + 1)aα > 0 for all α ∈ II and β ∈ [[1, mα ]], 2 P P (P4) (aαβ )3 − (aα )3 − (mα + 1)aα aα − aαβ > 0 for all α ∈ II . β∈[[1,mα ]]

β∈[[1,mα ]]

Let us first deal with the conditions (P2)-(P4). We define the constants corresponding to the edge e1 by a1 = 2 and b1 = 1. Assuming that we have already constructed aα and bα for some multi-index α, then bαβ is given by (P2). Next, we have to find aαβ large enough to satisfy (P3)(P4). Let us choose aαβ = rα aα . Obviously, for large enough rα , depending on mα , conditions (P3) and (P4) are satisfied. Finally, assume that all the coefficients aα and bα have been defined to satisfy (P2)-(P4). Adding 23 Cψ to all the bαβ , we see that (P1) is fulfilled for Cψ large enough, while (P2)-(P4) still hold true.

12


Let us split X α into X α = X1α + X2α + X3α + X4α , where X1α :=

Z

X2α :=

Z

T

h [uαx uαt ](lα , t) −

0

β∈[[1,mα ]] T

h [sϕαxx uα uαx ](lα , t) −

0

X3α :=

Z

X4α :=

Z

i αβ αβ [sϕαβ xx u ux ](0, t) dt,

X β∈[[1,mα ]]

T

h [s|uαx |2 ϕαx ](lα , t) −

0

0

i αβ [uαβ u ](0, t) dt, x t

X

i 2 αβ [s|uαβ | ϕ ](0, t) dt, x x

X β∈[[1,mα ]]

T

nh i s |uα |2 − ϕα3x + s2 ϕαx ϕαt + s3 (ϕαx )3 (lα , t) 2 o X h i s 2 αβ αβ 3 αβ 3 − (0, t) dt. |uαβ |2 − ϕαβ 3x + s ϕx ϕt + s (ϕx ) 2 β∈[[1,mα ]]

We now estimate each term Xiα , i = 1, ..., 4. Using the definition of the function u we have for any index α ∈ I the following identities α

uαx = e−sϕ (−sϕαx q α + qxα ),

α

uαt = e−sϕ (−sϕαt q α + qtα ).

Let us set u(Oα , t) = uα (lα , t) = uαβ (0, t) and ϕ(Oα , t) = ϕα (lα , t) = ϕαβ (0, t) for any α ∈ II and β ∈ [[1, mα ]]. With these notations we have X1α =

Z

T

uαx uαt (lα , t) −

0

Z

αβ (0, t) dt u uαβ x t

X β∈[[1,mα ]]

T

0

Z

αβ α ϕαβ x q (0) + qx (lα ) −

β∈[[1,mα ]] T

X

ut (Oα , t)e−sϕ(Oα ,t) − sϕαx q α (lα ) + s

= 0

Z

X

ut (Oα , t)e−sϕ(Oα ,t) − sϕαx q α (lα ) + s

=

X

qxαβ (0) dt

β∈[[1,mα ]]

αβ ϕαβ x q (0) dt

β∈[[1,mα ]] T

ut (Oα , t)u(Oα , t) − sϕαx (lα ) + s

= 0

X

ϕαβ x (0) dt

β∈[[1,mα ]]

Z = −

T

− sϕαxt (lα , t) + s

0

X β∈[[1,mα ]]

|u(Oα , t)|2 ϕαβ dt. xt (0, t) 2

Let us estimate X2α . Using property (B4) we infer that α

|ϕαxx |

λeλψ T2 α 2 α = |λ(ψxα )2 + ψxx |≤ |ϕ | + K|ϕαx |. t(T − t) 4 x

(3.9)


13

This gives that

|X2α |

Z =

T

s2 2

Z

0

≤

X

sϕαxx uα uαx (lα , t) − s

αβ αβ u u (0, t) dt ϕαβ xx x

β∈[[1,mα ]] T

Z T2 α 1 T α α2 2 α |u | ( |ϕx | + K) |ϕx |(lα , t)dt + |ϕx ||ux | (lα , t)dt 4 2 0 Z o X n s2 Z T T2 1 T αβ αβ 2 2 αβ |uαβ |2 ( |ϕαβ |ϕ | + K) |ϕ |(0, t)dt + ||u | (0, t)dt x x x 2 0 4 x 2 0 α 2

0

+

β∈[[1,mα ]]

=

s2

T

h T2 i X T2 2 αβ ( |ϕαβ | + K) |ϕ |(0, t) dt |u(Oα , t)|2 ( |ϕαx | + K)2 |ϕαx |(lα , t) + x 2 0 4 4 x β∈[[1,mα ]] Z i X 1 Th α α2 αβ 2 + |ϕx ||ux | (lα , t)dt + ||u | (0, t) dt. (3.10) |ϕαβ x x 2 0 Z

β∈[[1,mα ]]

To estimate the term

RT 0

X

uαx (lα , t) −

|ϕαx ||uαx |2 (lα , t)dt which occurs in (3.10) and in X3α , we notice that

−sϕ(Oα ,t) uαβ (−s) ϕαx (lα , t) − x (0, t) = e

β∈[[1,mα ]]

X

ϕαβ x (0, t) q(Oα , t),

β∈[[1,mα ]]

hence

|uαx (lα , t)|2 ≤ (mα + 1)

2 α 2 2 |uαβ x (0, t)| + s |u(Oα , t)| ϕx (lα , t) −

X β∈[[1,mα ]]

2 ϕαβ x (0, t) .

X β∈[[1,mα ]]

(3.11) We infer that

Z

T

|ϕαx ||uαx |2 (lα , t)dt

0

Z ≤ (mα + 1) 0

T

|ϕαx (lα , t)|

X β∈[[1,mα ]]

2 |uαβ x (0, t)|

+s

2

|u(Oα , t)|2 ϕαx (lα , t)

−

X

2 αβ ϕx (0, t)

β∈[[1,mα ]]

(3.12)

14


Combined to (3.10) and to the fact that ϕαβ x (0, t) ≤ 0 by (B2), this yields X2α + X3α Z h T2 i X s2 T T2 2 αβ ≥− |u(Oα , t)|2 ( |ϕαx | + K)2 |ϕαx |(lα , t) + ( |ϕαβ | + K) |ϕ |(0, t) dt x 2 0 4 4 x β∈[[1,mα ]] Z T X 1 αβ 2 |ϕαβ + (s − ) x ||ux | (0, t)dt 2 0 β∈[[1,mα ]] Z T X X 2 1 2 2 αβ 2 α . |uαβ (0, t)| + s |u(O (l ϕ (0, t) − (s + )(mα + 1) |ϕαx (lα , t)| , t)| ϕ , t) − α α x x x 2 0 β∈[[1,mα ]]

β∈[[1,mα ]]

(3.13) Using the definition of X α and estimates (3.9), (3.13) we obtain that X α ≥ Z1α + Z2α where Z1α

Z

T

− sϕαxt (lα , t) + s

=− 0

X

ϕαβ xt (0, t)

β∈[[1,mα ]]

|u(Oα , t)|2 dt 2

T

h T2 i X T2 2 αβ |u(Oα , t)|2 ( |ϕαx | + K)2 |ϕαx |(lα , t) + ( |ϕαβ x | + K) |ϕx |(0, t) dt 4 4 0 β∈[[1,mα ]] Z T 2 X 1 2 3 ϕαβ (0, t) |u(Oα , t)|2 |ϕαx (lα , t)| ϕαx (lα , t) − − (s + s )(mα + 1) dt x 2 0 β∈[[1,mα ]] Z T i s + |u(Oα , t)|2 | − ϕα3x + s2 ϕαx ϕαt + s3 (ϕαx )3 (lα , t)dt 2 0 Z T X s 2 αβ αβ 3 αβ 3 − |u(Oα , t)|2 − ϕαβ 3x + s ϕx ϕt + s (ϕx ) (0, t)dt 2 0 −

s2 2

Z

β∈[[1,mα ]]

and 1 Z2α = (s− ) 2

X

Z

β∈[[1,mα ]] 0

T

αβ αβ 2 1 |ϕx ||ux | (0, t)dt−(s+ )(mα +1) 2

Z

T

h

|ϕαx (lα , t)|

0

We notice that |ϕαx |3 ≥ c(λθ)3 while, with (B3), |ϕαx | + |ϕαx |2 + |ϕαxt | + |ϕαx ϕαt | + |ϕα3x | ≤ c(λθ)3 .

X β∈[[1,mα ]]

i 2 |uαβ (0, t)| dt. x


15

It follows that Z T α Z1 = |u(Oα , t)|2 × 0

h × s3 (ϕαx )3 (lα , t) −

α 3 α (ϕαβ + 1)|ϕ , t)| ) (0, t) − (m (l ϕx (lα , t) − α x x α

X β∈[[1,mα ]]

X

2 ϕαβ (0, t) x

β∈[[1,mα ]]

i

+ ... Z T |u(Oα , t)|2 × = 0

h

× (sλθ)3 − (ψxα )3 (lα ) +

X

(ψxαβ )3 (0) − (mα + 1)|ψxα (lα )| ψxα (lα ) −

β∈[[1,mα ]]

X

2 ψxαβ (0)

β∈[[1,mα ]]

i + O(s2 λ3 θ3 )

and Z2α

Z

T

X

= 0

h i 2 αβ α |uαβ x (0, t)| sλθ ψx (0) − (mα + 1)|ψx (lα )| + O(λθ) dt.

β∈[[1,mα ]]

Looking at the coefficient of s3 in Z1α and of s in Z2α and using (B6), (B7), and (3.12), we obtain that for s ≥ s0 and λ ≥ λ0 (with s0 , λ0 large enough) X α ≥ Z1α + Z2α Z Z T ≥ C s3 λ3 θ3 |u(Oα , t)|2 dt + C 0

T

sλθ |uαx (lα , t)|2 +

0

X

2 dt. (3.14) (0, t)| |uαβ x

β∈[[1,mα ]]

In particular, X α > 0. Step 3. Estimation of the integrals along the edges. We need the following lemma. Lemma 3.3. [35, Claim 1] There exist λ1 ≥ λ0 , s1 ≥ s0 and A > 0 such that for all λ ≥ λ1 , s ≥ s1 , it holds Z TZ Z TZ i h 2 s 2 2 3 2 3 |u| (ϕ4x − ϕtt ) − s (|ϕx | )t − s ϕx (|ϕx | )x dxdt ≥ Aλs |u|2 |ϕx |3 dxdt. 2 0 Γ 0 Γ (3.15) As the proof of [35, Claim 1] does not involve any integration by parts in x, it is still valid in our context. The following lemma is inspired by [35, Claim 2].

16


Lemma 3.4. There exist s2 ≥ s1 , λ2 ≥ λ1 , and a positive constant C such that for all λ ≥ λ2 and s ≥ s2 Z

T

Z

|ϕx ||ux |2 + λs−1

λs 0

Z

T

Z

T

Z

0

Γ

−1

≤C s

2

kM1 uk + λs

3

Z

Γ

+ λs 0

X

3

Z

2

T

|ϕx | |u| + λs 0

T

Z

|ϕx |−1 |∆Γ u|2

0

Γ

X

|ϕαx uα uαx |(0) + |ϕαx uα uαx |(lα )

α∈I

|(|ϕαx |)x ||uα |2 (0) + |(|ϕαx |)x ||uα |2 (lα ) .

(3.16)

α∈I

Proof. Let us pick any edge eα . To simplify the writing, we remove the index α in our computations. Using the definition of M1 u (see (3.4)) we have that s

−1

Z

T

Z

l

|ϕx | 0

−1

−1

2

T

Z

Z

|uxx | = s

0

0

≤ Cs−1

T

Z

Z

0

l

l

|ϕx |−1 |M1 u − sϕt u − s2 ϕ2x u|2

0

i |ϕx |−1 |M1 u|2 + s2 |ϕt |2 u2 + s4 |ϕx |4 u2 . h

0

Using property (B3) we get that |ϕt | ≤ C|ϕx |2 . Therefore, we have for some constant A > 0 s−1

Z

T

l

Z

0

|ϕx |−1 |uxx |2 ≤ A

0

kM1 uk2 + s3 λs

Z 0

T

Z

l

|ϕx |3 u2 .

(3.17)

0

The first term in the left hand side of (3.16) satisfies ZZ λs

2

|ϕx ||ux | = λs

Z Z

ZZ |ϕx |(−uxx )u −

Z (|ϕx |)x ux u + 0

T

l |ϕx |uux 0

Z T λ λs3 λs |u|2 l −1 2 3 2 2 ≤ |ϕx | |uxx | + |ϕx | |u| + (|ϕx |)xx |u| + λs |ϕx |uux − (|ϕx |)x 2s 2 2 2 0 0 ZZ 2 λ kM1 uk + s3 |ϕx |3 u2 ≤ A 2 λs ZZ ZZ Z T λs3 λs |u|2 l 3 2 2 + |ϕx | |u| + (|ϕx |)xx |u| + λs |ϕx |uux − (|ϕx |)x 2 2 2 0 0 ZZ ZZ ZZ 3 A −1 λs λs ≤ s kM1 uk2 + s3 λ |ϕx |3 u2 + |ϕx |3 |u|2 + (|ϕx |)xx |u|2 2 2 2 Z T Z T |u|2 |u|2 + λs |ϕx ||u||ux | + |(|ϕx |)x | (0) + λs |ϕx ||u||ux | + |(|ϕx |)x | (l). 2 2 0 0 (3.18) ZZ

ZZ

ZZ

The claim follows by summing (3.17) and (3.18) over all the edges.


17

Step 4. Conclusion. By (3.7), (3.15) and (B1), we get for λ ≥ 1 kwk2 = kM1 u + M2 uk2 = kM1 uk2 + kM2 uk2 + 2(M1 u, M2 u) nX = kM1 uk2 + kM2 uk2 + 2 Xα + Y α∈II T

Z

Z

2

T

Z

Z

ϕxx |ux | +

− 2s

|u|2

hs

2 n 0X Γ α ≥ kM1 uk + kM2 uk + 2 X +Y 0

(ϕ4x − ϕtt ) − s2 (|ϕ2x |)t − s3 ϕx (|ϕx |2 )x

io

Γ

2

2

α∈II T

Z

Z

+ 2s

(λ 0

2

ψ 2x

2

+ λψ xx )θ|ux | + Aλs

Γ 2

T

Z

3

Z

0 2

≥ kM1 uk + kM2 uk +

X

α

X + Y + Aλs

3

|u|2 |ϕx |3

Γ

Z

T

Z

0

α∈II

o

|u|2 |ϕx |3 .

(3.19)

Γ

Multiplying (3.16) by A/2C and adding it to (3.19) we get ZZ ZZ ZZ A Aλs3 Aλs Aλ 3 2 2 kM2 uk + kM1 uk (1 − ) + |u| |ϕx | + |ϕx ||ux | + |ϕx |−1 |∆Γ u|2 2s 2 2C 2sC X A + X α + Y ≤ kwk2 + B, (3.20) 2 2

2

α∈II

where B = B1 + B2 Z TX Z α α α α α α |ϕx u ux |(0) + |ϕx u ux |(lα ) + λs = λs 0

T

0

α∈I

X

|(|ϕαx |)x ||uα |2 (0) + |(|ϕαx |)x ||uα |2 (lα ) .

α∈I

P We now prove that for s large enough, the term B is small compared to α∈II X α , so that B can be absorbed by the left hand side of (3.20). Using (3.14) and the fact that u vanishes at the vertices of E ∪ R = ∂Γ, we see that B2 ≤ Cλ3 s

Z 0

T

X α∈I

θ|u(Oα , t)|2 dt = Cλ3 s

Z 0

T

X α∈II

θ|u(Oα , t)|2 dt ≤

C X α X . s2 α∈II

(3.21)

18


Using again the fact that u vanishes at the vertices of ∂Γ, we obtain with (3.14) that T

Z B1 ≤ Cλs 0 T

Z

|ϕαx uα |(Oα , t) |uαx (lα , t)| +

α∈II

X

≤C 0

X

|uαβ (0, t)| x

X β∈[[1,mα ]]

α∈II

2 |uαβ x (0, t)| ) dt

X

(sλ)2 |ϕαx ||uα |2 (Oα , t) + |ϕαx (Oα , t)|(|uαx (lα , t)|2 +

β∈[[1,mα ]]

C X α ≤ X . s

(3.22)

α∈II

Gathering together (3.20), (3.21) and (3.22), we obtain Z Z ZZ Z Z Aλs Aλ T Aλs3 T A 3 2 2 |u| |ϕx | + |ϕx ||ux | + |ϕx |−1 |∆Γ u|2 kM2 uk + kM1 uk (1 − ) + 2s 2 2C 2sC 0 Γ 0 Γ C X α + (1 − ) X + Y ≤ kwk2 . s 2

2

α∈II

Writing explicitly the term Y and tacking into account the sign of the functions ψxα occuring in Y , we get for s and λ large enough 2

2

kM1 uk + kM2 uk + λs +

X

3

Z

Z

Xα +

T

Z

Z

3

2

T

Z

|u| |ϕx | + λs 0 T

2

|ϕx ||ux | + λs 0

Γ

Z

T

Z

0

Γ

Z

λsθ1 |ux |2 (R, t)dt ≤ C kwk2 +

0

α∈II

−1

0

T

X

|ϕx |−1 |∆Γ u|2

Γ

λsθα |uαx |2 (lα , t)dt . (3.23)

α∈IE

Finally, using the definition of M2 we get λs

−1

Z

T

Z

−1

|ϕx | 0

−1

2

T

Z

Z

|ut | dxdt ≤ Cλs

0

Γ

Z

T

Z

≤ C 0

|ϕx |−1 |M2 u|2 + s2 |ϕx |2 |ux |2 + s2 |ϕxx |2 |u|2

Γ

(s−1 |M2 u|2 + λs|ϕx ||ux |2 + λs|ϕx |−1 |ϕxx |2 |u|2 ).

(3.24)

Γ

From (3.23) and (3.24), we infer that for s ≥ s3 and λ ≥ λ3 (with s3 , λ3 large enough) we have that Z TZ Z TZ Z TZ 2 2 3 2 3 2 −1 kM1 uk + kM2 uk + λs |u| |ϕx | + λs |ϕx ||ux | + λs |ϕx |−1 (|∆Γ u|2 + |ut |2 ) +

X α∈II

Xα +

Z 0

0 T

Γ

0

Γ

λsθ1 |ux |2 (R, t)dt ≤ C kwk2 +

0

Z 0

T

X

Γ

λsθα |uαx |2 (lα , t)dt .

α∈IE

Replacing u by e−sϕ q in the last inequality, we readily obtain (3.3).


19

3.3. Inverse problem. Before proving the stability result in Theorem 1.1 we need to analyze the following system:  uαt (x, t) = uαxx (x, t) + bα (x)uα (x, t) + Rα (x, t)f α (x), (x, t) ∈ (0, lα ) × (0, T ), α ∈ I,        t ∈ (0, T ), α ∈ IE , uα (lα , t) = 0,        u1 (0, t) = 0, t ∈ (0, T ),     uα (lα , t) = uαβ (0, t),      m  Pα αβ  α   ux (0, t),  ux (lα , t) =

t ∈ (0, T ), α ∈ II , β ∈ [[1, mα ]], t ∈ (0, T ), α ∈ II ,

β=1

(3.25) where b = {bα }α∈I ∈ L∞ (Γ). Proposition 3.5. Assume that u = {uα }α∈I is a solution of (3.25) which satisfies ut ∈ H 2,1 (Γ× (0, T )). If R = {Rα (x, t)}α∈I is such that Rt ∈ L∞ (Γ × (0, T )) and that |R(x, t0 )| ≥ r > 0,

for a.e. x ∈ Γ and some t0 ∈ (0, T ),

then there exists a positive constant C = C(||Rt ||L∞ (Γ×(0,T )) , ||b||L∞ (Γ) , r) such that ! X kf kL2 (Γ) ≤ C ku(·, t0 )kH 2 (Γ) + k∂xt u(v, ·)kL2 (0,T )

(3.26)

(3.27)

v∈E

for any f ∈ L2 (Γ). Proof. We proceed as in [37]. Set z = ∂t u. Then z = {z α }α∈I satisfies  α (x, t) + bα (x)z α (x, t) + Rα (x, t)f α (x), (x, t) ∈ (0, l ) × (0, T ), α ∈ I, ztα (x, t) = zxx  α t       z α (lα , t) = 0, t ∈ (0, T ), α ∈ IE ,        z 1 (0, t) = 0, t ∈ (0, T ),     z α (lα , t) = z αβ (0, t),      m  Pα αβ  α (l , t) =   z zx (0, t), α x  β=1

t ∈ (0, T ), α ∈ II , β ∈ [[1, mα ]], t ∈ (0, T ), α ∈ II . (3.28)

On the other hand, R(x, t)f (x) = ∂t u(x, t) − (∆Γ u)(x, t) − bu(x, t). Using a change of variables it is sufficient to prove (3.27) when t0 = T /2.

(3.29)

20


We now apply the Carleman estimate (3.3) with q = z = ∂t u (and some fixed λ > 0) T

Z 0

Z h Z i −1 2 2 3 2 −2sϕ (sθ) |∂tt u| + sθ|∂xt u| + (sθ) |∂t u| e +

T

(sθ)(|∂xt u|2 e−2sϕ )(R, t)dt

0

Γ

≤C

Z

T

Z

|∂tt u − ∆Γ ∂t u| e

0

≤C

Z

2 −2sϕ

+

Γ T

XZ v∈E

Z

2

2

−2sϕ

|(∂t R)f | + |∂t u| e

0

Γ

+

T

sθ(|∂xt u|2 e−2sϕ )(v, t)dt

0

XZ

T

sθ(|∂xt u|2 e−2sϕ )(v, t)dt ,

0

v∈E

which gives, for s large enough, T

Z 0

Z h Z i −1 2 2 3 2 −2sϕ (sθ) |∂tt u| + sθ|∂xt u| + (sθ) |∂t u| e + Γ

T

(sθ)(|∂xt u|2 e−2sϕ )(R, t)dt

0

≤C

Z

T

Z

2 −2sϕ

|(∂t R)f | e

0

Γ

+

T

XZ v∈E

sθ(|∂xt u|2 e−2sϕ )(v, t)dt .

0

Since limt→0 e−2sϕ(x,t) = 0 for x ∈ Γ and |ϕt (x, t)| ≤ Cθ 2 (x, t) for all x ∈ Γ and t > 0, we get Z T /2 Z T 2 −2sϕ(x, T ) ∂ 2 |∂t u(x, )| e |∂t u(x, t)|2 e−2sϕ(x,t) dx dt dx = 2 ∂t Γ Γ 0 Z T /2 Z = 2∂t u∂tt u − 2s∂t ϕ|∂t u|2 e−2sϕ(x,t) dxdt

Z

0

Z

Γ T /2 Z

2|∂t u||∂tt u| + Csθ 2 |∂t u|2 e−2sϕ(x,t) dxdt

≤ 0

Γ T /2 Z

Z

≤C 0

C ≤ s

Z

(s2 θ)−1 |∂tt u|2 + (sθ)2 |∂t u|2 e−2sϕ(x,t) dxdt

Γ T

Z

2 −2sϕ

|(∂t R)f | e 0

−Cs

dxdt + Ce

Γ

XZ v∈E

T

|∂xt u|2 (v, t)dt.

(3.30)

0

Using (3.26), (3.29) and (3.30) we obtain that Z

Z |f (x)|2 e−2sϕ(x,T /2) dx ≤ C |R(x, T /2)f (x)|2 e−2sϕ(x,T /2) dx Γ Γ Z T 2 T T ≤C |∂t u(x, )| + |∆Γ u(x, )|2 + |u(x, )|2 e−2sϕ(x,T /2) dx 2 2 2 Γ Z TZ Z T X T C ≤ |(∂t R)f |2 e−2sϕ dxdt + Ce−Cs |∂xt u|2 (v, t)dt + Cku(·, )k2H 2 (Γ) . s 0 Γ 2 0 v∈E


21

Since ∂t R ∈ L∞ (Γ × (0, T )), it holds Z |f (x)|2 e−2sϕ(x,T /2) dx Γ

≤

T

Z

C s

0

Z

|f |2 e−2sϕ dxdt + Ce−Cs

Γ

T

XZ

|∂xt u|2 (v, t)dt + Cku(·,

0

v∈E

T 2 )k 2 . 2 H (Γ)

It follows from the definition of ϕ that ϕ(x,

T ) ≤ ϕ(x, t) 2

for all (x, t) ∈ Γ × (0, T ),

so that Z

T

Z

2 −2sϕ(x,t)

|f (x)| e 0

Z

|f (x)|2 e−2sϕ(x,T /2) dt.

dxdt ≤ T

Γ

Γ

Therefore CT ) (1 − s

Z

2 −2sϕ(x,T /2)

|f (x)| e Γ

dt ≤ C

XZ v∈E

T

|∂xt u|2 (v, t)dt + Cku(·,

0

The desired inequality follows for s large enough.

T 2 )k 2 . 2 H (Γ)

We are now able to prove the stability result for system (1.1). Proof of Theorem 1.1. Let us denote w = u(p) − u(q). It satisfies the following system wt = ∆Γ w − qw + Rf w(x, t) = 0,

in Γ × (0, T ), on ∂Γ × (0, T ),

where f = q − p, R = u(p). Note that R ∈ C([0, T ]; H 1 (Γ)) ⊂ C(Γ × [0, T ]), for u ∈ L2 (0, T ; H 2 (Γ)) and ut ∈ L2 (0, T ; L2 (Γ)). Using our hypothesis, we see that |R(·, t0 )| ≥ r > 0 on Γ, hence we can apply Proposition 3.5 to obtain X kp − qkL2 (Γ) ≤ C k[u(p) − u(q)](·, t0 )kH 2 (Γ)) + k∂x [u(p) − u(q)](v, ·)kH 1 (0,T ) , v∈E

where C = C(k∂t u(p)kL∞ (Γ×(0,T )) , kqkL∞ (Γ) , r). The proof is now completed.

¨ dinger equation on a star-shaped tree 4. Schro In this section, we consider a network Γ which is a star-shaped tree constituted by N edges ej (with N ≥ 3) connected at the internal node O. Here, the parameterization of the edge ej is chosen so that the origin O of ej corresponds to x = 0, while the endpoint Oj of ej corresponds to x = lj , for all j ∈ [[1, N ]].

22


O

e4 e1

e3 e2

O4 O3

O1 O2

Figure 2. A star-shaped tree with 4 edges

We consider the following Cauchy problem

iyj,t + yj,xx + pj (x)yj = fj (x, t), yj (0, t) = yl (0, t), X yj,x (0, t) = 0,

x ∈ (0, lj ), j ∈ [[1, N ]], t ∈ (0, T ), t ∈ (0, T ), j, l ∈ [[1, N ]],

(4.31) (4.32)

t ∈ (0, T ),

(4.33)

j ∈ [[1, N ]], t ∈ (0, T ), x ∈ Γ,

(4.34) (4.35)

1≤j≤N

y(lj , t) = 0, y(x, 0) = y0 (x),

where p = {pj }j=1,N ∈ L∞ (Γ) is some given potential function. Our main aim is to prove the stability for the inverse problem consisting in retrieving the potential p from the measurement of yx (lj , t) for j ∈ [[1, N ]]. This is done thanks to some Carleman estimate in following the classical Bukhgeim-Klibanov method. The first step will be the proof of a Carleman inequality on Γ. The key point is that choosing only one weight function ψ = {ψj }j=1,N : Γ → R as in the case of the heat equation is not convenient since we fail to control some boundary terms. Instead, we consider a family of weights {ψ k }k=1,N allowing us to get rid of some bad boundary terms at the internal node.


23

4.1. Carleman estimate. Assume given a family (ψjk )1≤j,k≤N of functions fulfilling the following properties ψjk : [0, lj ] → R is of class C 2 , ψjk11 (0) = ψjk22 (0), |(ψjk )0 (x)|2 + (ψjk )00 (x) ≥ 0, (ψjk )0 (x)

6= 0, C C ≥ ψjk (x) > , 2 3

∀j, k ∈ [[1, N ]],

(4.36)

∀j1 , k1 , j2 , k2 ∈ [[1, N ]],

(4.37)

∀x ∈ [0, lj ], ∀j, k ∈ [[1, N ]],

(4.38)

∀x ∈ [0, lj ], ∀j, k ∈ [[1, N ]],

(4.39)

∀x ∈ [0, lj ], ∀j, k ∈ [[1, N ]],

(4.40)

where C > 0 is some positive constant. We also assume that the following flux conditions at x = 0 are satisfied: X (ψjk )0 (0) = 0, ∀k ∈ [[1, N ]], (4.41) 1≤j≤N

X

(ψjk )0 (0) = 0,

∀j ∈ [[1, N ]],

(4.42)

|(ψjk )0 (0)|2 = C1 ,

∀j ∈ [[1, N ]],

(4.43)

(ψjk )00 (0) = C2 ,

∀j ∈ [[1, N ]],

(4.44)

[(ψjk )0 (0)]3 > 0,

∀j ∈ [[1, N ]],

(4.45)

1≤k≤N

X 1≤k≤N

X 1≤k≤N

X 1≤k≤N

for some constants C1 > 0 and C2 ∈ R. Such a family of weights functions (ψjk )1≤j,k≤N exists. 5 It is sufficient to pick (affine) functions of the form ψjk (x) = akj x + 12 C with C 1 and akj :=

N −1 −1

if j = k, if j = 6 k.

Note that if another internal node would be present, the condition (4.45) (with the opposite sign due to the orientation) would not be satisfied. Let us introduce the families of weights k

k

θjk (x, t)

eλψj (x) , = t(T − t)

ϕkj (x, t)

eλC − eλψj (x) = , t(T − t)

and the class of functions Z = {q = (qj )j=1,N ∈ C(Γ×[0, T ]); qj ∈ C 2,1 ([0, lj ]×[0, T ]) ∀j ∈ [[1, N ]], and (4.32)−(4.34) hold}. Proposition 4.1. Assume that the family of weights (ψkj ) fulfills (4.36)-(4.45). Then there exist some constants λ0 ≥ 1, s0 ≥ 1 and C0 > 0 such that for all λ ≥ λ0 , all s ≥ s0 , and all q ∈ Z, it

24


holds Z TZ

X 1≤j,k≤N

0

1≤j,k≤N

where i =

√

f1k q)j |2 + |(M f2k q)j |2 ]e−2sϕkj dxdt [λ2 sθjk |qj,x |2 + λ4 (sθjk )3 |qj |2 + |(M

0

X

≤ C0

lj

Z TZ

lj

2 −2sϕkj

|qj,t + iqj,xx | e 0

Z

T

dxdt +

k λsθjk (lj )|qj,x (lj )|2 e−2sϕj dt

, (4.46)

0

0

fk and M fk denote the operators −1 and M 1 2 f1k q)j := [s(ϕkj,t + iϕkj,xx ) − 2is2 |ϕkj,x |2 ]qj + 2isϕkj,x qj,x , (M

(4.47)

f2k q)j := [−s(ϕkj,t + iϕkj,xx ) + 2is2 |ϕkj,x |2 ]qj + qj,t − 2isϕkj,x qj,x + iqj,xx . (M

(4.48)

Proof. In what follows, the letter c will denote a constant (independent of s, λ, q, j, k) which may vary from line to line. Let q ∈ Z be given, and for j, k ∈ [[1, N ]], let k

ukj = e−sϕj qj ,

k

k

wjk = e−sϕj L(esϕj ukj )

where L denotes the operator L = ∂t + i∂x2 . Straightforward computations show that wk = M k uk with wjk = (M k uk )j := ukj,t + sϕkj,t ukj + i(ukj,xx + 2sϕkj,x ukj,x + sϕkj,xx ukj + s2 |ϕkj,x |2 ukj ), the operator M k acting simply on the components of uk along the different edges. Let M1k and M2k denote respectively the (formal) adjoint and skew-adjoint parts of the operator M k . We readily obtain that (M1k uk )j (M2k uk )j

= i(2sϕkj,x ukj,x + sϕkj,xx ukj ) + sϕkj,t ukj =

ukj,t

+

i(ukj,xx

2

+s

|ϕkj,x |2 ukj ).

(4.49) (4.50)

fk q)j := esϕkj (M k uk )j and (M fk q)j := esϕkj (M k uk )j , we easily check that (4.47) and Letting (M 1 1 2 2 (4.48) hold. On the other hand, ||wk ||2 = ||M1k uk + M2k uk ||2 = ||M1k uk ||2 + ||M2k uk ||2 + 2 Re (M1k uk , M2k uk ) R TR l P where (u, v) := 1≤j≤N 0 0 j uj (x, t)vj (x, t)dxdt and ||w||2 = (w, w). The proof of the Carleman estimate is inspired by those of [33, Proposition 2.1]. In the first step, we compute precisely Re (M1k uk , M2k uk ). In the second step, we check that the boundary terms related to the internal node O give positive contributions. The third step is completely similar to the second step in the proof of [33, Proposition 2.1]. Step 1. Exact computation of Re (M1k uk , M2k uk ). We intend to compute X Z TZ Re (M1k uk , M2k uk ) = Re 1≤j≤N

0

0

lj

(M1k uk )j (M2k uk )j dxdt.


25

R TR l Let us fix any pair (j, k) of indices in [[1, N ]] and let us compute 0 0 j (M1k uk )j (M2k uk )j dxdt. For the sake of simplicity, we shall drop the indices j and k during the computations, and we RR RTRl R RT shall write u for 0 0 j u(x, t)dxdt, and h for 0 h(t)dt. Then Z TZ 2Re 0

0

lj

(M1k uk )j (M2k uk )j

ZZ dxdt = 2Re

[i(2sϕx ux + sϕxx u) + sϕt u][ut − i(uxx + s2 |ϕx |2 u)]

= I1 + I2 + I3

where ZZ I1 = 2Re

i(2sϕx ux + sϕxx u)(ut − i(uxx + s2 |ϕx |2 u)),

ZZ sϕt u(ut − iuxx ),

I2 = 2Re ZZ I3 = 2Re

sϕt u(−is2 |ϕx |2 u).

Obviously, I3 = 0. Let us begin with the computation of I1 . ZZ I1 = 2Re (2sϕx ux + sϕxx u)(uxx + s2 |ϕx |2 u) ZZ +2Re i(2sϕx ux + sϕxx u)ut = I11 + I12 . To calculate I11 , we need to evaluate the real part of the integral term J := grating by part yields Z ZZ l ux (ϕxx ux + ϕx uxx ) + ϕx |ux |2 , J =−

RR

uxx ϕx ux . Inte-

0

where l stands for lj . On the other hand ZZ

ZZ 2Re

ZZ

ϕx (ux uxx + ux uxx ) = ϕx ∂x |ux |2 ZZ Z l 2 = − ϕxx |ux | + ϕx |ux |2 .

ux ϕx uxx =

0

Therefore ZZ 2Re J

2

ZZ

2

= −2 ϕxx |ux | + ϕxx |ux | − ZZ Z l = − ϕxx |ux |2 + ϕx |ux |2 . 0

Z

Z l l ϕx |ux | + 2 ϕx |ux |2 2

0

0

26


It follows that

I11

ZZ

3

ZZ

3

3

ZZ

2

2

ϕxx uuxx + 2s (ϕx ) ux u + s ϕxx |ϕx | |u| ZZ Z ZZ l 3 (ϕx )3 ∂x |u|2 = 4s Re J − 2s Re (ϕ3x u + ϕxx ux )ux + 2s Re ϕxx uux + 2s 0 ZZ 3 +2s ϕxx |ϕx |2 |u|2 Z Z Z ZZ ZZ Z l ϕ4x |u|2 − [ϕ3x |u|2 ]l0 − 2 ϕxx |ux |2 = 2s{− ϕxx |ux |2 + ϕx |ux |2 } + s 0 Z ZZ Z ZZ l l 3 2 2 3 3 2 3 +2s Re ϕxx uux − 6s (ϕx ) ϕxx |u| + 2s (ϕx ) |u| + 2s ϕxx |ϕx |2 |u|2 0 0 ZZ ZZ ZZ 2 2 3 = −4s ϕxx |ux | + s ϕ4x |u| − 4s (ϕx )2 ϕxx |u|2 Z l + [2sϕx |ux |2 + (−sϕ3x + 2s3 (ϕx )3 )|u|2 + 2sϕxx Re (uux )] . = 2 Re 2sJ + s

0

On the other hand,

I12

ZZ

ZZ

(2sϕx ux + sϕxx u)ut − i (2sϕx ux + sϕxx u)ut ZZ −i (2sϕxt ux + 2sϕx uxt + sϕxxt u + sϕxx ut )u ZZ Z ZZ l +i 2s(ϕxx ut + ϕx uxt )u − i 2sϕx uut − i sϕxx uut 0 ZZ Z l −i (sϕxxt |u|2 + 2sϕxt ux u) − i 2sϕx uut 0 ZZ ZZ Z Z l l ϕxt ux u − i 2sϕx uut i sϕxt (uux + ux u) − i sϕxt |u|2 − 2is 0 0 ZZ Z l i sϕxt (uux − ux u) + i [sϕx (uut − ut u)] .

= i =

= = =

0

It remains to estimate I2 .

ZZ I2 =

ZZ sϕt (uut + uut ) +

sϕt (−iuuxx + iuuxx ) =: I21 + I22 .


27

We find that

I21

ZZ = −

sϕtt |u|2 ,

ZZ

Z l (ϕxt u + ϕt ux )ux − is ϕt uux 0 ZZ Z l −is (ϕtx u + ϕt ux )ux + is ϕt uux 0 Z ZZ l = 2 Re (is)ϕxt uux + 2 Re (−is)ϕt uux .

I22 = is

0

Thus

I1 + I2 + I3 ZZ ZZ ZZ 2 2 3 = −4s ϕxx |ux | + s ϕ4x |u| − 4s |ϕx |2 ϕxx |u|2 ZZ ZZ ZZ 2 +i sϕxt (uux − ux u) − s ϕtt |u| + 2 Re (is)ϕtx uux Z + [2sϕx |ux |2 + (−sϕ3x + 2s3 (ϕx )3 )|u|2 + 2sϕxx Re (uux ) l +isϕx (uut − ut u) + 2 Re {(−is)ϕt uux }] . 0

We conclude that (with the indices written again)

X 1≤k≤N

||wk ||2 =

X h

||M1k uk ||2 + ||M2k uk ||2

i

1≤k≤N

Z TZ −4s

X

+

1≤j,k≤N

Z TZ + 0

Z +

lj

0

0

lj

ϕkj,xx |ukj,x |2

Z TZ − 4s Im 0

0

lj

ϕkj,xt ukj ukj,x

|ukj |2 [s(ϕkj,4x − ϕkj,tt ) − 4s3 (ϕkj,x )2 ϕkj,xx ]

0 T

[2sϕkj,x |ukj,x |2 + (−sϕkj,3x + 2s3 (ϕkj,x )3 )|ukj |2 + 2sϕkj,xx Re (ukj ukj,x ) 0 l k k k k k k k k +2sϕj,t Re (−iuj uj,x ) + isϕj,x (uj uj,t − uj,t uj )] . (4.51) 0

28


Step 2. Estimation of the boundary terms at the internal node O. We estimate each term in Z T XZ T X k k 2 (sϕkj,3x (0) − 2s3 (ϕkj,x (0))3 )|u(0)|2 ϕj,x (0)|uj,x (0)| + (−2s) 0

j,k

+

X (−2s)

j,k

+

X

Z

T

ϕkj,t (0) Re (−iu(0)ukj,x (0))

(−2s) 0

j,k

T

XZ j,k

ϕkj,xx (0) Re (u(0)ukj,x (0)) +

0

j,k

0

T

Z

(−is)ϕkj,x (0)(u(0)ut (0) − ut (0)u(0)) =: J1 + J2 + J3 + J4 + J5 .

0

In the above equation and in the following ones, we write merely u(0) := ukj (0, t),

ϕ(0) = ϕkj (0, t),

etc.

Using (4.42), (4.43), and (4.33) (for the qj ’s) we see that XZ T X ϕkj,x (0)|e−sϕ(0) (−sϕkj,x qj + qj,x )|2 J1 = (−2s) 0

j

= −2s3

Z

T

0

k

X XZ (ϕkj,x (0))3 |u(0)|2 − 2s j

j,k

XZ

+4s2 Re

j

= −2s3

Z

T

0

T

0

T

X ( ϕkj,x (0))e−2sϕ(0) |qj,x (0)|2 k

X ( [ϕkj,x (0)]2 )u(0)qj,x (0)

0

k

X (ϕkj,x (0))3 |u(0)|2 . j,k

Therefore J1 + J2 =

XZ j,k

T

(sϕkj,3x (0) − 4s3 (ϕkj,x (0))3 )|u(0)|2 .

(4.52)

0

On the other hand, using (4.43), (4.44) and (4.33), we have that XZ T J3 = −2s ϕkj,xx (0) Re [u(0)e−sϕ(0) (−sϕkj,x (0)qj (0) + qj,x (0))] 0

j,k

= 2s2 Re

Z

T



 X

 0

j,k

Z

T

+2sλ Re 0

= 2s2 Re

Z

T

 X 

0

ϕkj,xx (0)ϕkj,x (0) |u(0)|2

j,k

X u(0)e−sϕ(0) eλψ(0) X X k 00 (ψj ) (0) + λ [(ψjk )0 (0)]2 t(T − t) j k k 

ϕkj,xx (0)ϕkj,x (0) |u(0)|2 .

! qj,x (0)

(4.53)


Combining (4.52), (4.53) and (4.45), we obtain that for s ≥ s1 and λ ≥ λ1 , !3 X Z T eλψjk (0) J1 + J2 + J3 ≥ cs3 λ3 |u(0)|2 . t(T − t) 0

29

(4.54)

j,k

Finally, we claim that J4 = J5 = 0. Indeed, using (4.33), we obtain that Z T X ϕt (0)u(0) J4 = −2s Im ukj,x (0)dt 0

j,k

T

Z = −2s Im

ϕt (0)u(0) 0

X

(−sϕkj,x (0)q(0) + qj,x (0))e−sϕ(0) dt

j,k

= 0, while J5 = 0 by (4.41). Thus we conclude that J1 + J2 + J3 + J4 + J5 ≥ cs3 λ3

Z

T

eλψ(0) t(T − t)

0

!3 |u(0)|2 .

(4.55)

for s ≥ s1 , λ ≥ λ1 . Step 3. Estimation of the integrals along the edges. Direct estimations as in [33, Proposition 2.1] (without any integration by parts) yield that for some constant A > 0 Z T Z lj Z T Z lj X k k 2 ϕkj,xt ukj ukj,x ϕj,xx |uj,x | − 4s Im {(−4s) 0

0

j,k

Z

T

lj

Z

+ 0

≥A

X j,k

0

0

|ukj |2 [s(ϕkj,4x − ϕkj,tt ) − 4s3 (ϕkj,x )2 ϕkj,xx ]}

0 2

Z

T

Z

{λ s 0

0

lj

k

eλψj |(ψ k )0 uk |2 + λs3 t(T − t) j j,x

Z

T

0

Z

lj

|ϕkj,x |3 |ukj |2 }

(4.56)

0

provided that s ≥ s2 , λ ≥ λ2 . Combining (4.51), (4.55) and (4.56), we infer that ( k Z TZ lj Z TZ lj X Z TZ lj eλψj k k 2 k k 2 2 k 0 k 2 3 [|(M1 u )j | + |(M2 u )j | ] + λ s |(ψj ) uj,x | + λs |ϕkj,x |3 |ukj |2 t(T − t) 0 0 0 0 0 0 j,k  !3 Z T Z T  X Z TZ lj eλψ(0) 3 3 2 k 2 k k 2 |u(0)| ≤c +cs λ |wj | + s |ϕj,x (lj )| |uj,x (lj )| dt . (4.57)  t(T − t) 0 0 0 0 j,k

−sϕkj

Replacing ukj by e

qj in (4.57) gives (4.46).

Remark 4.2. Note that (4.46) is still valid if, in the definition of Z, one replaces qj ∈ C 2,1 ([0, lj ] × [0, T ])

∀j ∈ [[1, N ]]

30


by q ∈ H 2,1 (Γ × (0, T )). 4.2. The boundary problem. We consider the following boundary initial-value problem  iuj,t + uj,xx + pj (x)uj = 0, x ∈ (0, lj ), j ∈ [[1, N ]], t ∈ (0, T ),     j, k ∈ [[1, N ]], t ∈ (0, T ),  u Pj (0, t) = ul (0, t), u (0, t) = 0, t ∈ (0, T ), (4.58) j,x 1≤j≤N   u (l , t) = h (t), j ∈ [[1, N ]], t ∈ (0, T ),  j   j j u(x, 0) = u0 (x), x ∈ Γ. In what follows we fix the initial data u0 and the boundary data h = {hj }j=1,N , and we denote by u(p) the solution of the system (4.58) associated with the potential p ∈ L∞ (Γ). Theorem 4.3. Assume that p ∈ L∞ (Γ; R), u0 ∈ L∞ (Γ) and r > 0 are such that • u0 (x) ∈ R or iu0 (x) ∈ R a.e. in Γ, • |u0 (x)| ≥ r > 0 a.e. in Γ, and • ∂t u(p) ∈ H 2,1 (Γ × (0, T )). Then, for any m ≥ 0, there exists a constant C = C(m, ||∂t u(p)||H 2,1 (Γ×(0,T )) , r) > 0 such that for any q ∈ Bm (0) ⊂ L∞ (Γ; R) satisfying ∂t u(q) ∈ H 2,1 (Γ × (0, T )), we have that ||p − q||L2 (Γ) ≤ C

X

||∂x [u(p) − u(q)]j (lj , .)||H 1 (0,T ) ·

1≤j≤N

Proof. Pick any p, q as in the statement of the theorem, and introduce the difference y := u(p) − u(q) of the corresponding solutions of (4.58). Then y fulfills the system  iyj,t + yj,xx + qj (x)yj = fj (x)Rj (x, t), x ∈ (0, lj ), j ∈ [[1, N ]], t ∈ (0, T ),     j, k ∈ [[1, N ]], t ∈ (0, T ),  yP j (0, t) = yl (0, t), t ∈ (0, T ), (4.59) 1≤j≤N yj,x (0, t) = 0,   y (l , t) = 0, j ∈ [[1, N ]], t ∈ (0, T ),    j j y(x, 0) = 0, x ∈ Γ, with fj = qj − pj (real valued) and Rj := (u(p))j . To complete the proof of Theorem 4.3, we need the following result. Proposition 4.4. Suppose that R = {Rj }j=1,N satisfies • R(x, 0) ∈ R or iR(x, 0) ∈ R a.e. in Γ, • |R(x, 0)| ≥ r > 0 a.e. in Γ, • R ∈ H 1 (0, T ; L∞ (Γ)), and • ∂t y ∈ H 2,1 (Γ × (0, T )). Then for any m ≥ 0 there exists a constant C = C(m, ||Rt ||L2 (0,T ;L∞ (Γ)) , r) such that for any q ∈ L∞ (Γ, R) with ||q||L∞ (Γ) ≤ m and for all f ∈ L2 (Γ, R), the solution y of (4.59) satisfies X ||f ||L2 (Γ) ≤ C ||yj,x (lj , .)||H 1 (0,T ) · (4.60) 1≤j≤N


31

Proof of Proposition 4.4. Let f ∈ L2 (Γ; R) and R ∈ H 1 (0, T ; L∞ (Γ)) be such that R(x, 0) ∈ R a.e. in Γ, and let y be the solution of (4.59). We take the even-conjugate extensions of y and R to the interval (−T, T ); i.e., we set y(x, t) = y(x, −t) for t ∈ (−T, 0) and similarly for R. Since R(x, 0) ∈ R a.e. in Γ, we have that R ∈ H 1 (−T, T ; L∞ (Γ)), and y satisfies the system (4.59) in Γ × (−T, T ). In the case when R(x, 0) ∈ iR a.e. in Γ, the proof is still valid by taking odd-conjugate extensions. Let z(x, t) = yt (x, −t). Then z satisfies the following system:  zj,t + izj,xx + iqj (x)zj = ifj (x)Rj,t (x, t), x ∈ (0, lj ), j ∈ [[1, N ]], t ∈ (−T, T ),     j, k ∈ [[1, N ]], t ∈ (−T, T ),  zP j (0, t) = zl (0, t), z (0, t) = 0, t ∈ (−T, T ), (4.61) j,x 1≤j≤N   z (l , t) = 0, j ∈ [[1, N ]], t ∈ (−T, T ),    j j z(x, 0) = −i f (x)R(x, 0), x ∈ Γ. We apply Proposition 4.1, but on the time interval (−T, T ) instead of (0, T ). Therefore, here we consider k

k

θjk (x, t) =

eλψj (x) , (T + t)(T − t)

ϕkj (x, t) =

eλC − eλψj (x) , (T + t)(T − t)

∀(x, t) ∈ Γ × (−T, T ).

k gk z) = esϕkj (M k wk ) and As in the proof of Proposition 4.1, we introduce wjk = e−sϕj zj , (M j 2 2 j k k k k 2 k 2 k (M2 w )j = wj,t + i(wj,xx + s |ϕj,x | wj ). Next, we set X Z 0 Z lj −2sϕk gk z) z dxdt. j (M e J= 2 j j

1≤j,k≤N

Then we have XZ 0Z J = j,k

=

−T 0

X Z { j,k

lj

0

(M2k wk )j wjk dxdt

0 Z lj

−T 0

−T

k wj,t wjk dxdt + i

Z

0 Z lj

−T 0

k 2 (−|wj,x | + s2 |ϕkj,x |2 |wjk |2 )dxdt + i

Z

0

−T

lj k wj,x wjk dt}. 0

Note that, by (4.61) and (4.41), X X k wj,x (0)wjk (0) = (zj,x (0) − sϕkj,x (0)z(0))e−2sϕ(0) z(0) = 0. j

j

Therefore 1X Re (J) = 2 j,k

Z

lj

|wjk (x, 0)|2 dxdt

0

1X = 2 j,k

Z

lj

k

e−2sϕj (x,0) |f (x)|2 |R(x, 0)|2 dx.

0

Using the hypothesis on R(x, 0), we infer that Z r2 X lj −2sϕkj (x,0) Re (J) ≥ e |f (x)|2 dx. 2 0 j,k

(4.62)

32


On the other hand, we have that   12 Z 0 Z lj 21  X Z 0 Z lj −2sϕk k gk z) )|2 dxdt j |(M |J| ≤ e e−2sϕj |zj |2 dxdt 2 j  −T 0  −T 0 j,k Z 0 Z lj Z 0 Z lj 1X −2sϕkj g −2sϕkj 2 2 23 2 −2 − 32 k ≤ e |(M2 z)j | dxdt + λ s e |zj | dxdt λ s 2 −T 0 −T 0 j,k Z 0 Z lj X Z 0 Z lj −2sϕk g k 3 −2sϕkj 2 −2 − 23 2 4 3 k j ≤ cλ s (θj ) e |zj | dxdt , e |(M2 z)j | dxdt + λ s

(4.63)

−T 0

−T 0

j,k

where we used the fact that θjk ≥ T −2 . From (4.63), the Carleman estimate (4.46) (applied on the interval (−T, T ) instead of (0, T )), k and the fact that ϕkj (x, 0) ≤ ϕkj (x, t) for all (x, t) ∈ (0, lj ) × (−T, T ), that θjk e−2sϕj is bounded from above in (0, lj ) × (−T, T ), that q ∈ L∞ (Γ), and that Rt ∈ L2 (−T, T ; L∞ (Γ)), we infer that for s and λ large enough Z T X Z T Z lj −2sϕk −2 − 23 2 k −2sϕkj 2 j e |f Rt | dxdt + λs θj e |zj,x (lj , t)| dt |J| ≤ cλ s 3

≤ cλ−2 s− 2

XZ j,k

−T

−T 0

j,k

lj

1

k

e−2sϕj (x,0) |f |2 dx + cλ−1 s− 2

XZ

0

j

T

|zj,x (lj , t)|2 dt.

(4.64)

−T

It follows from (4.62), (4.64), and the fact that z(x, t) = −z(x, −t) for (x, t) ∈ Γ × (−T, 0), that for s and λ large enough XZ 0 X Z lj −2sϕk (x,0) 2 j e |f (x)| dx ≤ c |zj,x (lj , t)|2 dt. (4.65) j,k

0

j

−T

Then (4.60) follows from (4.65) since k

e−2sϕj (x,0) ≥ e−2sT

−2 (eλC −1)

.

This completes the proof of Proposition 4.4 and of Theorem 4.3.

5. Open problems We now mention a few open problems related to our work. One of them is whether it is possible to reduce the number of measurements at the boundaries. It could be interesting to combine the ideas of the paper with those appearing in [16], [17] where less measurements on the boundary are needed but some rationality assumptions on the lengths of the edges have to be made. For the Schr¨ odinger equation, the question whether a Carleman estimate on a tree with N exterior vertices can be written with only one weight function and N − 1 boundary observations seems to be challenging.


33

The extension of the present work to more general graphs with other kind of coupling is also an open problem. We recall here the works of Kostrykin and Schrader [26, 27] where self-adjoint Laplace operators with general coupling conditions are introduced. Acknowledgements The authors wish to thank Institut Henri Poincaré (Paris, France) for providing a very stimulating environment during the “Control of Partial Differential Equations and Applications” program in the Fall 2010. LI was supported by a grant of the Romanian National Authority for Scientific Research, CNCS – UEFISCDI, project number PN-II-ID-PCE-2011-3-0075 and by the grants MTM2011-29306-C02-00 (Spanish MEC), FP7-246775 NUMERIWAVES (European Research Council). LR was partially supported by the Agence Nationale de la Recherche, Project CISIFS, grant ANR-09-BLAN-0213-02. AP was partially supported by CNPq, Math-Amsud project CIP-PDE, and French-Brasil agreement. References [1] S. Alexander. Superconductivity of networks. A percolation approach to the effects of disorder. Phys. Rev. B (3), 27(3):1541–1557, 1983. [2] S. Avdonin and P. Kurasov. Inverse problems for quantum trees. Inverse Probl. Imaging, Vol. 2, no. 1, 121, 2008. [3] S. Avdonin, P. Kurasov, and M. Nowaczyk. Inverse problems for quantum trees II: recovering matching conditions for star graphs. Inverse Probl. Imaging Vol. 4, no. 4, 579598, 2010. [4] S. Avdonin, G. Leugering, and V. Mikhaylov. On an inverse problem for tree-like networks of elastic strings. ZAMM Z. Angew. Math. Mech., 90(2):136–150, 2010. [5] L. Baudouin, E. Crépeau and J. Valein. Global Carleman estimate on a network for the wave equation and application to an inverse problem, preprint (hal-00576296), 2011. [6] L. Baudouin, A. Mercado. An inverse problem for Schr¨ odinger equations with discontinuous main coefficient. Appl. Anal. Vol. 87, no. 10-11, 11451165, 2008. [7] L. Baudouin, A. Mercado, A. Osses. A global Carleman estimate in a transmission wave equation and application to a one-measurement inverse problem. Inverse Problems Vol. 23, no. 1, 257278, 2007. [8] L. Baudouin and J.-P. Puel. Uniqueness and stability in an inverse problem for the Schr¨ odinger equation. Inverse Problems, 18(6):1537–1554, 2002. [9] A. Benabdallah, Y. Dermenjian and J. Le Rousseau. Carleman estimates for the one-dimensional heat equation with a discontinuous coefficient and applications to controllability and an inverse problem. J. Math. Anal. Appl. Vol. 336, no. 2, 865887, 2007. [10] M. Bellassoued, M. Choulli. Logarithmic stability in the dynamical inverse problem for the Schr¨ odinger equation by arbitrary boundary observation. J. Math. Pures Appl. Vol. 9, no. 91, 233-255, 2009. [11] A. L. Bukhge˘ım and M. V. Klibanov. Uniqueness in the large of a class of multidimensional inverse problems. Dokl. Akad. Nauk SSSR, 260(2):269–272, 1981. [12] L. Cardoulis, M. Cristofol, P. Gaitan. Inverse problem for the Schr¨ odinger operator in an unbounded strip. J. Inverse Ill-Posed Probl. vol. 16, no. 2, 127146, 2008. [13] L. Cardoulis, P. Gaitan. Simultaneous identification of the diffusion coefficient and the potential for the Schr¨ odinger operator with only one observation. Inverse Problems Vol. 26, no. 3, 035012, 10 pp., 2010. [14] C. Cattaneo. The spectrum of the continuous Laplacian on a graph. Monatsh. Math., 124(3):215–235, 1997. [15] J. P. Carini, J. T. Londergan, D. P. Murdock, D. Trinkle and C. S. Young. Bound states in waveguides and bent quantum wires. I. Applications to waveguide systems, Phys. Rev. B, 55:9842–9851, 1997. [16] R. D´ ager. Observation and control of vibrations in tree-shaped networks of strings. SIAM J. Control Optim., 43(2):590–623 (electronic), 2004.

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(L. I. Ignat) Institute of Mathematics “Simion Stoilow” of the Romanian Academy, 21 Calea Grivitei Street, 010702 Bucharest, Romania and BCAM - Basque Center for Applied Mathematics, Bizkaia Technology Park, Building 500 Derio, Basque Country, Spain. E-mail address: [email protected] ´ tica, Universidade Federal do Rio de Janeiro, P.O. Box 68530, (A. F. Pazoto) Instituto de Matema CEP 21945-970, Rio de Janeiro, RJ, Brasil E-mail address: [email protected] `s(L. Rosier) Institut Elie Cartan, UMR 7502 UHP/CNRS/INRIA, B.P. 70239, 54506 Vandœuvre-le Nancy Cedex, France E-mail address: [email protected]