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On: 13 June 2012, At: 05:20Publisher: Taylor & Francis Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, Lo

On: 13 June 2012, At: 05:20

Publisher: Taylor & Francis

Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Inverse Problems in Science and Engineering

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The body force in a three-dimensional Lamé system: identification and

regularization

Dang Duc Trong a , Phan Thanh Nam b & Phung Trong Thuc a a

Department of Mathematics, Vietnam National University, Ho Chi Minh City, Vietnam

b Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark

Available online: 14 Dec 2011

To cite this article: Dang Duc Trong, Phan Thanh Nam & Phung Trong Thuc (2012): The body force

in a three-dimensional Lamé system: identification and regularization, Inverse Problems in Science and Engineering, 20:4, 517-532

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Inverse Problems in Science and Engineering

Vol 20, No 4, June 2012, 517–532

The body force in a three-dimensional Lame´ system:

identification and regularization

Dang Duc Tronga, Phan Thanh Namb and Phung Trong Thuca* a

Department of Mathematics, Vietnam National University, Ho Chi Minh City, Vietnam;

b

Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark

(Received 14 June 2011; final version received 23 September 2011) Let a three-dimensional isotropic elastic body be described by the Lame´ system with the body force of the form F(x, t) ¼ ’(t)f (x), where ’ is known We consider the problem of determining the unknown spatial term f (x) of the body force when the surface stress history is given as the overdetermination This inverse problem

is ill-posed Using the interpolation method and truncated Fourier series,

we construct a regularized solution from approximate data and provide explicit error estimates

Keywords: body force; elastic; ill-posed problem; interpolation; Fourier series AMS Subject Classifications: 35L20; 35R30

1 Introduction

Let  ¼ (0, 1)  (0, 1)  (0, 1) represent a three-dimensional isotropic elastic body and let

T >0 be the length of the observation time For each x :¼ (x1, x2, x3) 2 , we denote by u(x, t) ¼ (u1(x, t), u2(x, t), u3(x, t)) the displacement, where uj is the displacement in the

xj-direction

As known, u satisfies the Lame´ system (see, e.g [1–3])

@2u

@t2Du   þ ð ÞrðdivðuÞÞ ¼F, x 2, t 2 ð0, T Þ, ð1Þ where F(x, t) ¼ (F1(x, t), F2(x, t), F3(x, t)) is the body force and div(u) ¼ r  u ¼@u1/@x1þ@u2/

@x2þ@u3/@x3 The Lame´ constants  and  satisfy  > 0 and  þ 2 > 0 The system (1) is associated with the initial condition

ðu1ðx, 0Þ, u2ðx, 0Þ, u3ðx, 0ÞÞ ¼ ð g1ðxÞ, g2ðxÞ, g3ðxÞÞ, x 2 ,

@u1

@t ðx, 0Þ,

@u2

@t ðx, 0Þ,

@u3

@t ðx, 0Þ

¼ ðh1ðxÞ, h2ðxÞ, h3ðxÞÞ, x 2 ,

8

<

and the Dirichlet boundary condition

ðu1ðx, tÞ, u2ðx, tÞ, u3ðx, tÞÞ ¼ ð0, 0, 0Þ, x 2 @, t 2 ð0, T Þ, ð3Þ namely the boundary of the elastic body is clamped

*Corresponding author Email: phungtrongthuc@vanlanguni.edu.vn

ISSN 1741–5977 print/ISSN 1741–5985 online

ß 2012 Taylor & Francis

http://dx.doi.org/10.1080/17415977.2011.638068

The direct problem is to determine u from u(0, x), ut(0, x) and F We are, however, interested in the inverse problem of determining both (u, F ) Of course, to ensure the uniqueness of the solution we shall require some additional information (the overdetermination) Similar to [4,5], we shall assume that the surface stress is given

on the boundary of the body, i.e

1 12 13

21 2 23

31 32 3

0

@

1 A

n1

n2

n3

0

@

1

A ¼ XX12

X3

0

@

1

A, x 2 @, t 2 ð0, T Þ, ð4Þ

where n ¼ (n1, n2, n3) is the outward unit normal vector of @ and the stresses  and  are defined by

jk¼ @uj

@xk

þ@uk

@xj

, j¼divðuÞ þ 2@uj

@xj , j, k 2 f1, 2, 3g:

Grasselli et al [4] showed that the body force of the form F(x, t) ¼ ’(t)f (x) is uniquely determined from (1)–(4) provided that ’ 2 C1([0, T ]) is given such that ’(0) 6¼ 0 and the time of observation T > 0 is large enough In spite of the uniqueness, the problem of determining the spatial term f is still ill-posed, i.e a small error of data may cause a large error of solutions Therefore, it is important in practice to find a regularization process, namely to construct an approximate solution using approximate data

Recently, the regularization problem was solved partially in [5], where a regularized solution for the time-independent term f is produced using further information on the final condition u(x, T ) The final condition plays an essential role in [5] since it enables the authors to find an explicit formula for the Fourier transform of f, and then use this information to recover f

It was left as an open problem in [5] (see their Conclusion) to find a regularization process without using this technical condition The aim of this article is to solve this problem completely, i.e to find a regularization process of f using only the data in (1)–(4)

We follow the interpolation method introduced in [6] where the authors constructed a regularized solution for the heat source of a heat equation More precisely, lacking the final condition, we are only able to find an approximation for the Fourier transform bf ðÞ with jj large The idea is that because bf ðÞis an analytic function (since f has compact support), we can use some interpolation process to recover bf ðÞwith jj small

This article is organized as follows In Section 2, we shall set some notations and state our main results Then we prove the uniqueness in Section 3 and the regularization in Section 4 Finally, in Section 5 we test our regularization process on an explicit numerical example

2 Main results

Recall that our aim is to recover the spatial term

f ðxÞ ¼ ð f1ðxÞ, f2ðxÞ, f3ðxÞÞ, x 2,

of the body force F(x, t) ¼ ’(t)f (x) from the system (1)–(4) The Lame´ constants always satisfy  > 0 and  þ 2 > 0, and the data I ¼ (’, X, g, h) is allowed to be non-smooth,

I 2 L 1ð0, T Þ, ðL1ð0, T, L1ð@ÞÞÞ3, ðL2ðÞÞ3, ðL2ðÞÞ3

:

For  ¼ (1, 2, 3),  ¼ (1, 2, 3) 2 C3, we set    ¼ 11þ22þ33, jj ¼ ffiffiffiffiffiffiffiffi

  

p and jj0¼ ffiffiffiffiffiffiffiffiffiffiffi

j  j

p For  2 C3and x 2 R3, denote Gð, xÞ ¼ G11ð, xÞ ¼G22ð, xÞ ¼G33ð, xÞ

¼cos ð 1x1Þcos ð 2x2Þcos ð 3x3Þ,

G12ð, xÞ ¼G21ð, xÞ ¼ sin ð 1x1Þsin ð 2x2Þcos ð 3x3Þ,

G13ð, xÞ ¼G31ð, xÞ ¼ sin ð 1x1Þcos ð 2x2Þsin ð 3x3Þ,

G23ð, xÞ ¼G32ð, xÞ ¼ cos ð 1x1Þsin ð 2x2Þsin ð 3x3Þ:

Sometimes we shall write Gjkinstead of Gjkð, xÞ if there is no confusion

We start with the following lemma

LEMMA 1 If u 2(C2([0, T ]; L2()) \ L2(0, T; H2()))3 and f 2(L2())3 satisfy the system(1)–(4) with data I ¼ (’, X, g, h), then

Z



fjGdx ¼E1jðIÞ ð Þ þE

 1jðÞ

D1ðIÞ ð Þ þ

E2jðIÞ ð Þ þE

2jðÞ

D2ðIÞ ð Þ , j 2 f1, 2, 3g for all  ¼(1, 2, 3) 2 C3such that

jj20 ¼ ð21þ22þ23Þ 0 and D1ð ÞIð Þ 6¼ 0, D2ð ÞIð Þ 6¼ 0, where

D1ð ÞI ð Þ ¼ jj 20

ZT 0

’ tð Þ sinh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0ðT  tÞ

cosh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0T

D2ð ÞI ð Þ ¼ jj 20

ZT 0

’ tð Þ sinh ffiffiffiffi

 p



j j0ðT  tÞ

 cosh ffiffiffiffi

 p



j j0T

 dt,

E1jðIÞðÞ ¼  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0j Z



1g1Gj1þ2g2Gj2þ3g3Gj3

dx

tanh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0T

j Z



1h1Gj1þ2h2Gj2þ3h3Gj3

dx



Z T 0

Z

@

sinh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0ðT  tÞ cosh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0T j1X1Gj1þ2X2Gj2þ3X3Gj3

d! dt,

E

1jð Þ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0 cosh ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p



j j0T

 Z





1u1ðx, TÞGj1þ2u2ðx, TÞGj2þ3u3ðx, TÞGj3

 dx,

E2jð ÞIð Þ ¼  ffiffiffiffi



p



j j0 Z





j j20gjGjjjð1g1Gj1þ2g2Gj2þ3g3Gj3

dx

tanh ffiffiffiffi



p



j j0T

j j20hjGjjjð1h1Gj1þ2h2Gj2þ3h3Gj3

dx Inverse Problems in Science and Engineering 519

ZT 0

Z

@

sinh đ ffiffiffiffipj j0đT  tỡ cosh đ ffiffiffiffipj j0T







j j20XjGjjợjđ1X1Gj1ợ2X2Gj2ợ3X3Gj3

d! dt,

E2jđ ỡ Ử

ffiffiffiffi

 p



j j0 cosh pffiffiffiffi



j j0T

Z

 đj j20ujđx, TỡGjj

j

1u1đx, TỡGj1ợ2u2đx, TỡGj2ợ3u3đx, TỡGj3

ỡdx:

Note that E

1jand E

2jin Lemma 1 depend on u(x, T ) instead of the data I Ử (Ỗ, X, g, h) Therefore, in general these terms are unknown However, our observation is that with jj large, E

1j and E

2jare relatively small in comparison with E1jand E2j, and can be relaxed when computing the integrals R

fjGdx which are the Fourier coefficients of fjỖs So we introduce some convenient notations

Definition 1 (Information from data) For I Ử (Ỗ, X, g, h) and  2 C3 such that    < 0, denote (using notations of Lemma 1)

HjđIỡđỡ Ử

E1jđIỡ

D1đIỡợ

E2jđIỡ

D2đIỡ, if D1đIỡđỡ  D2đIỡđỡ 6Ử0,

0, if D1đIỡđỡ  D2đIỡđỡ Ử0:

8

<

: Definition 2(Fourier co-efficients) For  Ử (1, 2, 3) 2 C3, w 2 L2(), denote

F đwỡ đ ỡ Ử

Z

 wđxỡGđ, xỡdx Ử

Z

 wđxỡcos đ 1x1ỡcos đ 2x2ỡcos đ 3x3ỡdx:

Note that any function w 2 L2() admits the representation

wđxỡ Ử X m,n, p0

where (m, n, p) :Ử (1 ợ 1{m6Ử0})(1 ợ 1{n6Ử0})(1 ợ 1{p6Ử0})

As we have explained above, we hope to approximate F ( fj)() by Hj(I )() with jj large To do this, we need some lower bounds on jD1(I )()j and jD2(I )()j We shall require the following assumptions on Ỗ and T

(W1) There exist (Ỗ) 2 (0, T ) and C(Ỗ) > 0 such that either Ỗ(t)  C(Ỗ) for a.e

t 2(0, (Ỗ)) or Ỗ(t)  C(Ỗ) for a.e t 2 (0, (Ỗ))

(W2) T 4 2 maxfp 1ffiffiffi, ffiffiffiffiffiffiffiffiffi1

ợ2

p g, or (W20) T 4 12 ffiffiffi

5

p

maxfp 1ffiffiffi, ffiffiffiffiffiffiffiffiffi1

ợ2

p g

Remark 1 If Ỗ is continuous at t Ử 0 then the condition (W1) is equivalent to Ỗ(0) 6Ử 0 The conditions (W2) and (W20) mean that the observation time must be long enough The condition (W2) is enough for the uniqueness, while the stronger condition (W20) is required in our regularization These conditions should be compared to similar conditions (2.7) and (2.8) in [4]

THEOREM 1 (Uniqueness) Assume that (W1) and (W2) hold true Then the system (1)Ờ(4) has at most one solution

đu, f ỡ 2 đC 2đơ0, T ; L2đỡỡ \ L2đ0, T; H2đỡỡỡ3, đL2đỡỡ3

:

The main point in our regularization is to recover F ( f )() with jj small from approximate values of F ( f )() with jj large As in [6] we shall use the Lagrange interpolation polynomial

Definition 3 (Lagrange interpolation polynomial) Let A ¼ {x1, x2, , xn} be a set of n mutually distinct complex numbers and let w be a complex function Then the Lagrange interpolation polynomial L[A; w] is

L A; w½ ð Þ ¼z Xn

j¼1

Y k6¼j

z  xk

xjxk

!

w xj

  :

THEOREM 2 (Regularization) Assume that (u0, f0) is the exact solution to the system (1)–(4) with the exact data I0¼(’0, X0, g0, h0), where the conditions (W1) and (W20) are satisfied Let " > 0, consider the inexact data I"¼(’", X", g", h") such that, for all j 2 {1, 2, 3},

’0’"

L 1", X"jX0j

L 1", g"j g0j

L 2", h"j h0j

L 2 ":

Construct the regularized solution f"

j from I"¼(’", X", g", h") by

f"j ¼ X 0m, n, pr "

m, n, pð ÞF"ðm, n, pÞ ð Þ, where

r"¼Z \ lnð"

1Þ

60 ,

lnð"1Þ

60 þ1



,

Br " ¼ ð5r"þjÞ: j ¼ 1, 2, , 24r"

,

F"ðm, n, pÞ ¼L Br ", HjðI"Þ i:ð Þ

ð Þ:

Then we have, for all j 2 {1, 2, 3},

(i) (Convergence) f"

j 2C1ðR3Þand f"

j !fj0in L2() as " ! 0

(ii) (L2-estimate) If fj02H1ðÞ then f"

j !fj0 in H1() and there exist constants "0>0 and C0>0 depending only on the exact data such that for all " 2 (0, "0),

fj"fj0

L 2 ð Þ  C0



ln " 1 1

: (iii) (H1-estimate) If f0

j 2H2ðÞ then for all " 2 (0, "0),

fj"fj0

H 1 ð Þ  C0



ln " 11

:

Remark 2 In our construction, the convergence in H2() is not expected even if

f0

j 2C1ðÞ since @f"

j=@n ¼ 0 on @

3 Uniqueness

In this section we shall prove Theorem 1 We start with the proof of Lemma 1 by using the argument in [5]

Inverse Problems in Science and Engineering 521

Proof of Lemma 1 Fix  ¼ (1, 2, 3) 2 C3 with jj20¼ ð2þ2þ2Þ 0 For any

j 2{1, 2, 3}, get the inner product (in L2()) of the k-th equation of the system (1) with Gjk (k ¼ 1, 2, 3), then using the integral by part and the boundary conditions (3) and (4),

we have

d2

dt2

Z



ukGjkdx þ   21þ22þ23 Z



ukGjkdx

þð þ Þk

Z



1u1Gj1þ2u2Gj2þ3u3Gj3

dx

¼ Z

@

XkGjkd!  ’ tð Þ

Z



Multiplying (Ak) with jk, and then getting the sum of k ¼ 1, 2, 3, we have

d2

dt2

Z



j1u1Gj1þj2u2Gj2þj3u3Gj3

dx

ð þ2Þjj20

Z



j1u1Gj1þj2u2Gj2þj3u3Gj3

dx

¼ Z

@

j1X1Gj1þj2X2Gj2þj3X3Gj3

d!

þ’ tð Þ Z



j1f1Gj1þj2f2Gj2þj3f3Gj3

Choosing k ¼ j in (Ak), then multiplying the result by jj2

0, and adding to (6), we obtain

d2

dt2

Z



 jj20ujG þ ðj1u1Gj1þj2u2Gj2þj3u3Gj3

dx

þjj20 Z





jj20ujGjj ðj1u1Gj1þj2u2Gj2þj3u3Gj3

dx

¼ Z

@

 jj20XjGjjþ ðj1X1Gj1þj2X2Gj2þj3X3Gj3

d!

þ’ tð Þ Z



 jj20fjGjjþ ðj1f1Gj1þj2f2Gj2þj3f3Gj3

We can consider (6) and (7) as the differential equations of the form

y00ð Þ t 2y tð Þ ¼h tð Þ, ð8Þ where > 0 is independent of t Getting the inner product (in L2(0, T )) of (8) with sinhððTtÞÞ

coshðTÞ , we have

y0ð Þ0 tanh Tð Þ yð Þ þ0

cosh Tð Þy Tð Þ ¼

ZT 0

h tð Þsinh T  tð ð ÞÞ cosh Tð Þ dt: ð9Þ Applying (9) to (6) and (7) with ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p

jj0 and ¼pffiffiffiffi

jj0, respectively, we get

E1jð ÞIð Þ þ E1jð Þ ¼ 1

j j20D1ð ÞI ð Þ

Z



j1f1Gj1þ2f2Gj2þ3f3Gj3

dx,

E2jð ÞI ð Þ þ E2jð Þ ¼ 1

j j2D2ð ÞI ð Þ

Z





j j20fjGjjjð1f1Gj1þ2f2Gj2þ3f3Gj3

dx:

It follows from the latter equations that

Z



fjGdx ¼E1jðIÞ ð Þ þE

 1jðÞ

D1ðIÞ ð Þ þ

E2jðIÞ ð Þ þE

2jðÞ

D2ðIÞ ð Þ :

The following lemma gives a lower bound for jDj(I )()j (defined in Lemma 1) when ’ satisfies the condition (W1)

LEMMA 2 Let ’ 2L1(0, T ) satisfy the condition (W1), then there exists a constant R(’) > 0 such that for all  ¼(1, 2, 3) 2 C3,    < 0 and jj0R(’),

Djð ÞI ð Þ

 1

4j j0C ’ð Þmin

1 ffiffiffiffi



ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p

for j ¼1, 2:

Proof Denote k1¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p

and k2 ¼pffiffiffiffi

By the triangle inequality, one finds that

Djð ÞI ð Þ jj20











 ¼ 

Z T 0

’ tð Þsinhðkjj j0ðT  tÞÞ cosh k jj j0T dt















Z ’ ð Þ 0

’ tð Þsinhðkjj j0ðT  tÞÞ cosh k jj j0T dt















ZT

 ’ ð Þ

’ tð Þsinhðkjj j0ðT  tÞÞ cosh k jj j0T dt











:

We have, with jj0large,

ZT

 ’ ð Þ

’ tð Þsinhðkjj j0ðT  tÞÞ coshðkjj j0TÞ dt











  ’ L 1

sinhðkjj j0ðT ð’ÞÞÞ coshðkjj j0TÞ

4 maxfk1, k2gjj0:

On the other hand, the condition (W1) implies that

Z ’ ð Þ

0

’ tð Þsinhðkjj j0ðT  tÞÞ cosh k jj j0T dt











  C ’ð Þ

Z ’ ð Þ 0

sinhðkjj j0ðT  tÞÞ cosh k jj j0T dt

¼ Cð’Þ

kjj j0 1 cosh ðkjj j0ðT ð’ÞÞÞ

coshðkjjj0TÞ

!

 Cð’Þ 2kjjj0 with jj0large The desired result follows immediately from the above inequalities g The proof of the uniqueness below follows the argument in [6] We shall need a useful result of entire functions (see, e.g [7, Section 6.1])

Inverse Problems in Science and Engineering 523

LEMMA 3 (Beurling) Let be a non-constant entire function satisfying the condition: there exists a constant k >0 such that M(r)  ker, for all r > 0, where M(r) ¼ j(z)j: jzj ¼ r Then

lim sup r!1

ln r ð Þ

r  1:

Proof of Theorem 1 Suppose that (u1, f1) and (u2, f2) are two solutions to the system (1)–(4) with the same the data I ¼ (’, X, g, h) Then (u, f ) :¼ (u1u2, f1f2) is a solution to (1)–(4) with data (’, 0, 0, 0) We shall show that (u, f) ¼ 0

Assume that f 6¼ 0, namely fj6¼0 for some j 2 {1, 2, 3} For any n, p 2 N [ {0}, let us consider the entire function

z ° n,pð Þ ¼z

Z



fjðxÞcos izxð 1Þ ð 2Þ ð 3Þdx:

Because

d n,p

dz ð

Z



ix1fjðxÞ ð 1Þ ð 2Þ ð 3Þdx

1 2 3)}m2N,n,p2N[{0} is an orthogonal basis on L2(), there exist some (n0, p0) such that n0, p0 is non-constant

On the other hand, recall from Lemma 1 that

n 0 ,p 0ðrÞ ¼

Z



fjðxÞGðr, xÞdx ¼ E

 1jðrÞ

D1ð ÞI ðrÞþ

E 2jðrÞ

D2ð ÞI ðrÞ, ð10Þ where r¼(ir, n0 , p0 ) Fix " > 0 such that minfT ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 þ2

p

, T ffiffiffiffipg4 2 þ " Then it is straightforward to see, from the explicit formulas in Lemma 1 and the lower bound in Lemma 2, that

jEejðrÞj C1eð1þ"Þr, jDeðIÞðrÞj C2r, e ¼ 1, 2, with r > 0 large, where C1>0 and C2>0 are independent of r Therefore, it follows from (10) that

j n 0 ,p 0ðrÞj 2C1C12 eð1þ"Þrr1 for r > 0 large This yields

lim sup r!1

ln  n0, p0ð Þr

r  ð1 þ "Þ,

Now following the proof of Lemma 1 up to Equations (6) and (7) (2:¼ 21þ32þ33 need not be negative at that time) we obtain

d2

dt2

Z



j1u1Gj1þj2u2Gj2þj3u3Gj3

dx

þð þ2Þ2

Z



j1u1Gj1þj2u2Gj2þj3u3Gj3

dx ¼ 0, ð11Þ

dt2

Z





2ujG  ðj1u1Gj1þj2u2Gj2þj3u3Gj3Þ

 dx

þ2 Z





2ujGjj ðj1u1Gj1þj2u2Gj2þj3u3Gj3Þ



dx ¼ 0 ð12Þ for all  2 C3

y00ð Þ þt 2y tð Þ ¼0,

yð Þ ¼0 0,

y0ð Þ ¼0 0,

8

>

>

where  0 is independent of t Applying this to (11) and (12) with  2 R3, we get

Z



j1u1Gj1þj2u2Gj2þj3u3Gj3

dx ¼ 0 and

Z





2ujG  ðj1u1Gj1þj2u2Gj2þj3u3Gj3Þ

dx ¼ 0:

Adding the latter equations, we obtain

Z



ujðx, :ÞGð, xÞdx ¼ 0 for all  2 R3,  6¼ 0: ð13Þ

To get the same equation to (13) with  ¼ 0, we can simply use identity (Ak) with  ¼ 0

to get

d2

dt2

Z



ujðx, tÞdx ¼ 0:

Since

Z



ujðx, tÞdx



t¼0

¼0 ¼ d

dt

Z



ujðx, tÞdx



t¼0 ,

we have

Z



Putting (13) and (14) together, we arrive at

Z



ujG ¼ Z



ujðx, :Þ cosð1x1Þcosð2x2Þcosð3x3Þdx ¼ 0 for all  2 R3:

1 2 3)}m,n,p0 forms an

4 Regularization

Lemmas 1 and 2 allow us to compute F ( f )() approximately with jj large To recover F( f)() with jj small, we shall need the following interpolation inequality This is an

Inverse Problems in Science and Engineering 525

...

j1u1Gj1ỵj2u2Gj2ỵj3u3Gj3

dx ẳ 0, ð11Þ

dt2... j1u1Gj1ỵj2u2Gj2ỵj3u3Gj3ị

dx ẳ 0:

Adding the latter equations, we obtain

Z... cosð1x1Þcosð2x2Þcosð3x3Þdx ¼ for all  R3:

1 3)}m,n,p0 forms an

4 Regularization

Lemmas and allow us to compute F ( f )() approximately with jj large To