LI Yaxin, XIE Weisong

(School of Mathematics, Tianjin University, Tianjin 300350, China)

Abstract: The problem of identifying unknown damping coefficient of hyperbolic equations on Riemannian manifolds is studied. For an initial boundary value problem of hyperbolic equation x∈M,0

Keywords: inverse problem; stability; Carleman estimate; hyperbolic equation

0 Introduction

One of the major topics to be investigated in inverse problems is the coefficient identification problem, the problem typically involves the estimation of certain coefficients based on inexact measurements of other measurable quantities[1-3]. Hussein considers the inverse problem of the hyperbolic equation which consists of determining an unknown space-dependent force function acting on a vibrating structure from Cauchy boundary data[4]. Lesnic investigates a couple of inverse problems of simultaneously determining time and space dependent coefficients in the parabolic heat equation with initial and boundary conditions of the direct problem and determination conditions[5].

In this paper, we’ll consider an equation with damping coefficient over the time interval (0,T), and extend the function to (-T,0) in order to apply the Carleman estimate, and we’ll establish the Lipschitz stability estimate ‖p1-p2‖L2(M)≤C‖?v?tup1,0-?v?tup2,0‖L2(Σ0). The rest of this paper is organized as follows. In Section 1, some preliminary results are given. In Section 2, the Carleman estimate is introduced. In Section 3, a stability estimate for the wave equation is established, the proof is based on Carleman estimate.

1 Preliminary results

Definition1Ann-dimensional differentiable manifoldMis defined as a Hausdorff spaceMwith a complete atlas.

Definition2LetMbe aCM-manifold. A Riemannian metric g onMis defined as a map which associates to anyCM-vector fieldsXandYonM, and aCM-functiong(X,Y) onMsatisfies

g(X1+X2,Y)=g(X1,Y)+g(X2,Y),g(X,Y1+Y2)=g(X,Y1)+g(X,Y2)

g(fX,Y)=fg(X,Y)=g(X,fY),g(X,Y)=g(Y,X)

for all real-valuedCM-functionsfand vector fieldsX,X1,X2,Y,Y1,Y2. Andg(X,X)>0 ifX≠0.

A Riemannian manifold (M,g) be a manifoldMwith metricg.

Lemma 4[9]LetT>0 andp,q∈L∞(M) be given. Suppose that

Consider the equations

(1)

‖u‖C([0,T];H1(M))+‖?tu‖C([0,T];L2(M))≤C(‖u0‖H1(M)+‖u1‖L2(M)+‖F(xiàn)‖L1(0,T;L2(M)))

Lemma5Ifp∈L∞(Ω),F∈L1(0,T;L2(M)),z1∈L2(M),zsatisfies

(2)

ProofMultiplying the first equation in (2) by 2?tzand integrating over (0,T), we have

By Green’s formula andz|Σ=0, we have

sincez(·,0)=z(·,T)=?tz(·,T)=0, we obtain

Lemma6Ifφ∈L2(-T,T;L∞(M)), then

The lemma means that there exists a functionkdefined inR, such that

2 Carleman estimate

This section gives a Carleman estimate on a Riemannian manifoldM. Let’s give some notes.

The weight function is defined byφ:M×R→R,φ(x,t)=eγψ(x,t), whereγ>0 is a fixed sufficiently large constant.

(M,g) is a Riemannian compact manifold with boundary ?M, g>0 onMand g∈C2(M), Γ0??Mis a subboundary. Suppose that there exists a positive and smooth functionψ0onMwhich satisfies the following assumptions:

(3)

(b)ψ0has no critical points onM, in namely minx∈M|?gψ0(x)|>0.

(4)

(c)Γ0??Msatisfies {x∈?M;?vψ0≥0}?Γ0.

(5)

ψ(x,t)=ψ0(x)-βt2+β0, 0<β

(6)

where the parameterβ0>0 is chosen such that the functionψis positive in the domain.

Proposition1If (3 ～6) hold, there is a constantγ*>0 such that for anyγ>γ*, there exist constantss*=s*(γ) andC>0 satisfying

for alls≥s*， And the solutionu∈H2(Q±) fits

u(·,±T)=?tu(·,±T)=0,u(·,0)=?tu(·,0)=0,u=0 on Σ

Since there exists a constantC>0 such that

C-1|?u(x)|≤|?gu(x)|≤C|?u(x)|,x∈M

for allu∈C1(M), we can replace the term |?gu|2by |?u|2.

3 Inverse problem

Consider the initial boundary value problem

(7)

For givenp,qand arbitrarily fixedM0>0, there is a solutionu=up,qof (7) such that

(8)

?V?tup,q∈L2(Σ)

(9)

Let Γ0??MandT>0 is given,q(x) is called the potential coefficient,p(x) is called the damping coefficient, if the damping coefficientp(x) is unknown, consider the following inverse problem:

Determine the coefficientp(x) from the data ?vup,0on Γ0×(0,T). For any fixedM0>0, denote the admissible set for the coefficientpas

D(M0)={p∈W1,∞(M);‖p‖W1,∞(M)≤M0}

Theorem1Set

(10)

there is a constantC>0 such that

‖p1-p2‖L2(M)≤C‖?v?tup1,0-?v?tup2,0‖L2(Σ0),p1,p2∈D(M0)

ifT>T0, (u0,u1)∈H(M), (3)～(6), (8) and (9) hold, there exists a constantm0>0 satisfying

|u1(x)|≥m0,x∈M

(11)

Theorem2Consider the problem

(12)

there exists a constantC>0 such that

‖f‖L2(M)≤C‖?V?tu‖L2(Σ0)

if (3)～(6) are satisfied,T>T0, and

R,?tR∈L2(0,T;L∞(M))

|R(x,0)|≥m0>0 almost everywhere onM

p0∈L∞(M) andusatisfies the regularity conditions of (8) and (9).

In fact, setu=up1,0-up2,0,p0=p1,f=p2-p1,R=?tup2,0, the problem (12) can be obtained, thus Theorem 1 is equal to Theorem 2.

ProofThe proof of Theorem 2 can be divided into four steps.

Step1Extension of solution

Set

Q±=M×(-T,T), Σ±=?M×(-T,T), Σ0,±=Γ0×(-T,T)

By Lemma 4, consider?tu, the unique solution of (12) satisfies

u∈C2([0,T];L2(M))∩C1([0,T];H1(M))∩C((0,T);H2(M))

?v?tu∈L2(Σ)

First, let’s extenduto (-T,0) keeping the hyperbolicity of the equation (12).

Set

(13)

(14)

(15)

sinceu(x,0)=?tu(x,0)=0 forx∈M, we can verify that for -T

(16)

(17)

therefore

u∈C2([-T,T];L2(M))∩C1([-T,T];H1(M))∩C((-T,T);H2(M))

R∈H1(-T,T;L∞(M))

Thus the equation (12) can be extended to

(18)

Now we can verify that

(19)

Since?tu∈C2([0,T];L2(M)), integration by parts

Thus

Differentiating with respect totin the hyperbolic equation (18), and setv=?tu, we obtain

(20)

Step2Cut off function

Let’s introduce a cut off functionη∈C∞(R) satisfying

(21)

Setω(x,t)=η(t)?tu(x,t)=η(t)v(x,t),(x,t)∈Q±. We have

?tω=?tη·v+η·?tv

ω(x,0)=η(0)v(x,0)=0

?tω(x,0)=?tη(0)v(x,0)+η(0)?tv(x,0)=f(x)R(x,0)

Thuswsatisfies

(22)

Furthermore, we have

ω(x,±T)=?tw(x,±T)=0 for allx∈M

Hence, applying the Carleman estimate to the functionw, we can obtain

for anys≥s*.

Notice that the second integral on the right side does not vanish int∈(-T+ε,-T+2ε)∪(T-2ε,T-ε).

Set

φ(x,t)=eγψ(x,t)=eγ(ψ0(x)+β0)·e(-γβt2)=φ0(x)μ0(t)

(23)

whereφ0(x)≥1 andμ0(t)≤1 are defined by

(24)

Chooseδ>0andβ>0 such that

So we have

and there is a smallε>0 such that

ψ(x,t)≤β0-2δfor allx∈MandT-2ε≤t≤T

Therefore

φ(x,t)=eγψ(x,t)≤eγ(β0-2δ)=d1

(25)

By (25) and Lemma 4, we have

Hence, there is a constantC>0such that

(26)

Step3Energy Estimate

Letz=esφω, we have

And

-s2esφ(?gφ)2ω-sesφΔgφ·ω-2sesφ?gφ·?gω-esφΔgω

+psesφ?tφ·ω+pesφ?tω

=s2esφ(|?tφ|2-|?gφ|2)ω+2sesφ(?tφ·?tw-?gφ·?gω)

Let

Asω=s2esφ(|?tφ|2-|?gφ|2)ω+2sesφ(?tφ·?tw-?gφ·?gω)

(27)

(28)

Applying Lemma 5 to (28) and noticing thatR(x,0)≥m0>0 andz(·,±T)=?tz(·,±T)=0 inM, we obtain

(29)

Byz=esφωand (27), we have

+‖Asω?tz‖L1(Q±)

(30)

Since|Asω?tz|≤Ce2sφ(s2|ω|+s|?ω|+s|?tω|)(s|ω|+|?tω|), by the Cauchy-Schwarz inequality, we have

(31)

Substituting (26) and (31) into the right side of (30), we obtain

Step4Absorption of the norm offon the right side

Using Lemma 6 andω=η·?tu, 0≤η≤1, we obtain

(32)

Notice that the first term on the right side of (32) can be absorbed into the left side if we takes>0 large enough.

On the other hand, since

φ0(x)≥d0>d1for allx∈M

we have

for sufficiently larges>0.

The proof of Theorem 2 is completed.