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Inverse Problems in Science and
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The body force in a three-dimensional
Lamé system: identification and
regularization
a

b

Dang Duc Trong , Phan Thanh Nam & 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;
Department of Mathematical Sciences, University of Copenhagen, Copenhagen, Denmark

b

Downloaded by [University of Guelph] at 05:20 13 June 2012

(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) Â (0, 1) Â (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])
@2 u
À Du À ð þ ÞrðdivðuÞÞ ¼ F, x 2 , t 2 ð0, T Þ,
ð1Þ
@t2
where F(x, t) ¼ (F1(x, t), F2(x, t), F3(x, t)) is the body force and div(u) ¼ r Á u ¼@u1/@x1 þ @u2/
@x2 þ @u3/@x3. The Lame´ constants  and  satisfy  > 0 and  þ 2 > 0. The system (1) is
associated with the initial condition
8
< ðu
1 ðxÞ, g2 ðxÞ, g3 ðxÞÞ, x 2 ,
 1 ðx, 0Þ, u2 ðx, 0Þ, u3 ðx, 0ÞÞ ¼ ð g

@u1
@u2
@u3
ð2Þ
ðx, 0Þ,
ðx, 0Þ,
ðx, 0Þ ¼ ðh1 ðxÞ, h2 ðxÞ, h3 ðxÞÞ, x 2 ,
:
@t
@t
@t
and the Dirichlet boundary condition
ðu1 ðx, tÞ, u2 ðx, tÞ, u3 ðx, tÞÞ ¼ ð0, 0, 0Þ,

x 2 @, t 2 ð0, T Þ,

namely the boundary of the elastic body is clamped.
*Corresponding author. Email:
ISSN 1741–5977 print/ISSN 1741–5985 online
ß 2012 Taylor & Francis
/>

ð3Þ


518

D.D. Trong et al.

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.
0
10 1 0 1
1 12 13
n1
X1
@ 21 2 23 A@ n2 A ¼ @ X2 A, x 2 @, t 2 ð0, T Þ,
ð4Þ
31 32 3
n3
X3

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


@uj @uk
@uj
þ
, j, k 2 f1, 2, 3g:
, j ¼  divðuÞ þ 2
jk ¼ 
@xk @xj
@xj
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 b
f ðÞ
with jj large. The idea is that because b
f ðÞ is an analytic function (since f has compact
support), we can use some interpolation process to recover b
f ðÞ with 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 f1 ðxÞ, f2 ðxÞ, f3 ðxÞÞ,

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,
À
Á
I 2 L1 ð0, T Þ, ðL1 ð0, T, L1 ð@ÞÞÞ3 , ðL2 ðÞÞ3 , ðL2 ðÞÞ3 :


519

Inverse Problems in Science and Engineering
Forpffiffiffiffiffiffiffiffiffiffi
¼ (ffi 1, 2, 3),  ¼ (1, 2, 3) 2 C3, we set  Á  ¼ 11 þ 22 þ 33, jj ¼
jj0 ¼ j Á j. For  2 C3 and x 2 R3, denote

pffiffiffiffiffiffiffiffi
 Á  and

Gð, xÞ ¼ G11 ð, xÞ ¼ G22 ð, xÞ ¼ G33 ð, xÞ
¼ cosð1 x1 Þ cosð2 x2 Þ cosð3 x3 Þ,
G12 ð, xÞ
G13 ð, xÞ
G23 ð, xÞ

¼ G21 ð, xÞ ¼ Àsinð1 x1 Þ sinð2 x2 Þ cosð3 x3 Þ,
¼ G31 ð, xÞ ¼ Àsinð1 x1 Þ cosð2 x2 Þ sinð3 x3 Þ,
¼ G32 ð, xÞ ¼ Àcosð1 x1 Þ sinð2 x2 Þ sinð3 x3 Þ:

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Sometimes we shall write Gjk instead of Gjk ð, 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
E1j ðIÞðÞ þ EÃ1j ðÞ E2j ðIÞðÞ þ EÃ2j ðÞ
þ
, j 2 f1, 2, 3g
fj G dx ¼
D1 ðIÞðÞ
D2 ðIÞðÞ

for all  ¼ (1, 2, 3) 2 C3 such that
jj20 ¼ Àð21 þ 22 þ 23 Þ ! 0 and

D1 ðIÞðÞ 6¼ 0, D2 ðIÞðÞ 6¼ 0,

where
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð þ 2Þ jj0 ðT À tÞ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 dt,
D1 ðI ÞðÞ ¼ Àjj20
’ðtÞ
0
cosh ð þ 2Þ jj0 T
pffiffiffiffi


ZT
sinh  jj0 ðT À tÞ
pffiffiffiffi
 dt,
D2 ðI ÞðÞ ¼ Àjj20
’ðtÞ
0
cosh  jj0 T
Z
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
À
Á
E1j ðIÞðÞ ¼ À  þ 2 jj0 j
1 g1 Gj1 þ 2 g2 Gj2 þ 3 g3 Gj3 dx

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 Z À
Á
1 h1 Gj1 þ 2 h2 Gj2 þ 3 h3 Gj3 dx
À tanh  þ 2 jj0 T j



ffi
Z T Z sinh pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 þ 2 jj0 ðT À tÞ À
Á
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 j 1 X1 Gj1 þ 2 X2 Gj2 þ 3 X3 Gj3 d! dt,
À

0
@
cosh  þ 2 jj0 T
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
 þ 2 jj
Àpffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 Á j
EÃ1j ðÞ ¼
cosh  þ 2 jj0 T
Z

Â
1 u1 ðx, TÞGj1 þ 2 u2 ðx, TÞGj2 þ 3 u3 ðx, TÞGj3 dx,
Z

sinh

T



pffiffiffiffi
E2j ðIÞðÞ ¼ À  jj0

Z
Á
Àjj20 gj Gjj À j ð1 g1 Gj1 þ 2 g2 Gj2 þ 3 g3 Gj3 dx


Àpffiffiffiffi
Á

À tanh  jj0 T

Z
Á
Àjj20 hj Gjj À j ð1 h1 Gj1 þ 2 h2 Gj2 þ 3 h3 Gj3 dx



520

D.D. Trong et al.
ZT Z
þ
0 @

Á
pffiffiffiffi
sinh ð  jj0 ðT À tÞ
Á
pffiffiffiffi
cosh ð  jj0 T


Á
 jj20 Xj Gjj þ j ð1 X1 Gj1 þ 2 X2 Gj2 þ 3 X3 Gj3 d! dt,
Z
pffiffiffiffi
 jj
Àpffiffiffiffi 0 Á ðÀjj20 uj ðx, TÞGjj
EÃ2j ðÞ ¼

cosh  jj0 T


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À j 1 u1 ðx, TÞGj1 þ 2 u2 ðx, TÞGj2 þ 3 u3 ðx, TÞGj3 Þdx:
Note that EÃ1j and EÃ2j in Lemma 1 depend on u(x, T ) instead of the data I ¼ (’, X, g, h).
Therefore, in general these terms are unknown. However, our observation is that with jj
large, EÃ1j and EÃ2j are relativelyRsmall in comparison with E1j and E2j, and can be relaxed
when computing the integrals  fj G dx which are the Fourier coefficients of fj’s. So we
introduce some convenient notations.
Definition 1 (Information from data) For I ¼ (’, X, g, h) and  2 C3 such that  Á  < 0,
denote (using notations of Lemma 1).
8
< E1j ðIÞ E2j ðIÞ
þ
, if D1 ðIÞðÞ Á D2 ðIÞðÞ 6¼ 0,
Hj ðIÞðÞ ¼ D1 ðIÞ D2 ðIÞ
:
0,
if D1 ðIÞðÞ Á D2 ðIÞðÞ ¼ 0:
Definition 2 (Fourier co-efficients)

For  ¼ (1, 2, 3) 2 C3, w 2 L2(), denote

Z

Z

F ðwÞðÞ ¼

wðxÞ cosð1 x1 Þ cosð2 x2 Þ cosð3 x3 Þdx:

wðxÞGð, xÞdx ¼




Note that any function w 2 L2() admits the representation
wðxÞ ¼

X

ðm, n, pÞF ðwÞðm, n, pÞ cosðmx1 Þ cosðnx2 Þ cosð px3 Þ,

ð5Þ

m,n, p!0

where (m, n, p) :¼ (1 þ 1{m6¼0})(1 þ 1{n6¼0})(1 þ 1{p6¼0}).
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 ’ and T.
(W1) There exist Ã(’) 2 (0, T ) and C(’) > 0 such that either ’(t) ! C(’) for a.e
t 2 (0, Ã(’)) or ’(t) ÀC(’) for a.e t 2 (0, Ã(’)). pffiffiffi
1 ffi
1 ffi
g, or (W20 ) T 4 12 5 Â maxfp1ffiffiffi , pffiffiffiffiffiffiffiffi

g.
(W2) T 4 2 maxfp1ffiffiffi , pffiffiffiffiffiffiffiffi
þ2

þ2

Remark 1 If ’ is continuous at t ¼ 0 then the condition (W1) is equivalent to ’(0) 6¼ 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)–(4)
has at most one solution
À
Á
ðu, f Þ 2 ðC2 ð½0, T Š; L2 ðÞÞ \ L2 ð0, T; H2 ðÞÞÞ3 , ðL2 ðÞÞ3 :


521

Inverse Problems in Science and Engineering

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
!
n

Y z À xk
X
À Á
w xj :
L½A; wŠðzÞ ¼
x À xk
j¼1
k6¼j j

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THEOREM 2 (Regularization) Assume that (u0, f 0) is the exact solution to the system
(1)–(4) with the exact data I0 ¼ (’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

’ À ’"  1
L

",




 "
Xj À X0j 

L1

",




 "
gj À g0j 

L2

",




 "
hj À h0j 

L2

":

Construct the regularized solution f"j from I" ¼ (’", X", g", h") by
X
ðm, n, pÞF " ðm, n, pÞGðm, n, pÞ,
f"j ¼
0 m, n, p r"

where

!

lnð"À1 Þ lnð"À1 Þ
r" ¼ Z \
,
þ1 ,
60
60
È
É
Br" ¼ Æð5r" þ jÞ : j ¼ 1, 2, . . . , 24r" ,
Â
Ã
F " ðm, n, pÞ ¼ L Br" , Hj ðI" ÞðÀi:, n, pÞ ðimÞ:
Then we have, for all j 2 {1, 2, 3},
(i) (Convergence) f"j 2 C1 ðR3 Þ and f"j ! fj0 in L2() as " ! 0.
(ii) (L2-estimate) If fj0 2 H1 ðÞ 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),


 À ÁÀ12

 "
C0 ln "À1
:
fj À fj0  2
L ð Þ

1

(iii) (H -estimate) If fj0 2 H2 ðÞ then for all " 2 (0, "0),



 À ÁÀ14

 "
C0 ln "À1
:
fj À fj0  1
H ðÞ

Remark 2 In our construction, the convergence in H2() is not expected even if
fj0 2 C1 ðÞ 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].


522

D.D. Trong et al.

Proof of Lemma 1 Fix  ¼ (1, 2, 3) 2 C3 with jj20 ¼ Àð21 þ 22 þ 23 Þ ! 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
Z
Z
À 2
Á
d2

j
2
2
u
G
dx
þ


þ

þ

uk Gjk dx
k k
1
2
3
dt2 

Z
À
Á
1 u1 Gj1 þ 2 u2 Gj2 þ 3 u3 Gj3 dx
þ ð þ Þk

Z
Z
j
Xk Gk d! À ’ðtÞ fk Gjk dx:

ðAk Þ
¼


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@

Multiplying (Ak) with jk, and then getting the sum of k ¼ 1, 2, 3, we have
Z
À
Á
d2
j 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx
2
dt 
Z
À
Á
2
j 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx
À ð þ 2Þjj0

Z
À
Á
j
j 1 X1 G1 þ j 2 X2 Gj2 þ j 3 X3 Gj3 d!
¼
@

Z
À
Á
j 1 f1 Gj1 þ j 2 f2 Gj2 þ j 3 f3 Gj3 dx:
þ ’ðtÞ

ð6Þ



Choosing k ¼ j in (Ak), then multiplying the result by jj20 , and adding to (6), we obtain
Z 

d2
j
j
jÁ
2
u
G
þ
ð

u
G
þ


u
G

þ


u
G
dx
jj
j
j
1
1
j
2
2
j
3
3
0
1
2
3
dt2 
Z 
Á
þ jj20
Àjj20 uj Gjj À ðj 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx

Z 
Á
jj20 Xj Gjj þ ðj 1 X1 Gj1 þ j 2 X2 Gj2 þ j 3 X3 Gj3 d!

¼
@
Z 
Á
jj20 fj Gjj þ ðj 1 f1 Gj1 þ j 2 f2 Gj2 þ j 3 f3 Gj3 dx:
ð7Þ
þ ’ðtÞ


We can consider (6) and (7) as the differential equations of the form
y00 ðtÞ À 2 yðtÞ ¼ hðtÞ,

ð8Þ

where  > 0 is independent of t. Getting the inner product (in L2(0, T )) of (8) with
sinhððTÀtÞÞ
coshðTÞ , we have
ZT

sinhððT À tÞÞ
yðTÞ ¼
dt:
ð9Þ
hðtÞ
coshðTÞ
coshðTÞ
0
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffi
Applying (9) to (6) and (7) with  ¼  þ 2 jj0 and  ¼  jj0 , respectively, we get

Z
À
Á
1
Ã
E1j ðIÞðÞ þ E1j ðÞ ¼
D ðI ÞðÞ j 1 f1 Gj1 þ 2 f2 Gj2 þ 3 f3 Gj3 dx,
2 1
Àjj0

Z 

1
j
j
j
jÁ
2
j
j
D
ð
I
Þ
ð

Þ
f
G
À


ð
f
G
þ

f
G
þ

f
G
E2j ðI ÞðÞ þ EÃ2j ðÞ ¼
À

2
j
j
1
1
2
2
3
3
j
0
1
2
3 dx:
Àjj20


Ày0 ð0Þ tanhðT Þ À yð0Þ þ


Inverse Problems in Science and Engineering

523

It follows from the latter equations that
Z
fj G dx ¼


E1j ðIÞðÞ þ E1jà ðÞ E2j ðIÞðÞ þ E2jà ðÞ
þ
:
D1 ðIÞðÞ
D2 ðIÞðÞ
g

This is the desired result.

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The following lemma gives a lower bound for jDj(I )()j (defined in Lemma 1) when ’
satisfies the condition (W1).
LEMMA 2 Let ’ 2 L1(0, T ) satisfy the condition (W1), then there exists a constant R(’) > 0
such that for all  ¼ (1, 2, 3) 2 C3,  Á  < 0 and jj0 ! R(’),
&
'



1
Dj ðI ÞðÞ ! 1 jj0 Cð’Þ min p1ffiffiffiffi , pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
for j ¼ 1, 2:
4

 þ 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffi
Proof Denote k1 ¼  þ 2 and k2 ¼ . By the triangle inequality, one finds that
 


D ðI ÞðÞ  Z T
sinhðkj jj0 ðT À tÞÞ 
 
 j
À
Á dt
’ðtÞ
 ¼ À


 jj20   0
cosh kj jj0 T

Z
 Ãð’Þ

sinhðkj jj0 ðT À tÞÞ 

À
Á dt
!
’ðtÞ

 0
cosh kj jj0 T

Z
 T
sinhðkj jj0 ðT À tÞÞ 

À
Á dt:
À
’ðtÞ

 Ãð’Þ
cosh kj jj0 T
We have, with jj0 large,

Z
 T
sinhðkj jj0 ðT À tÞÞ 

dt
’ðtÞ



 Ãð’Þ
coshðkj jj0 TÞ

  sinhðkj jj0 ðT À Ãð’ÞÞÞ
’ 1
L
coshðkj jj0 TÞ
Cð’Þ
:
4 maxfk1 , k2 gjj0

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

Z
Z Ã ð’ Þ
 Ãð’Þ
sinhðkj jj0 ðT À tÞÞ 
sinhðkj jj0 ðT À tÞÞ

À
Á dt ! Cð’Þ
À
Á dt
’ðtÞ


 0
cosh kj jj0 T
cosh kj jj0 T

0
!
Cð’Þ
cosh ðkj jj0 ðT À Ãð’ÞÞÞ
Cð’Þ
¼
1À
!
kj jj0
coshðkj jj0 TÞ
2kj jj0
with jj0 large. 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]).


524

D.D. Trong et al.

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


lnðrÞ
! À1:
lim sup

r
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 (u1 À u2, f1 À f 2) 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 fj 6¼ 0 for some j 2 {1, 2, 3}. For any n, p 2 N [ {0}, let us
consider the entire function
Z
z ° n,p ðzÞ ¼
fj ðxÞ cosðizx1 Þ cosðnx2 Þ cosð px3 Þdx:

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Because
d

n,p

dz

Z
ix1 fj ðxÞ sinðmx1 Þ cosðnx2 Þ cosðpx3 Þdx

ðimÞ ¼


and {sin(mx1)cos(nx2)cos( px3)}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
Z
EÃ1j ðr Þ
EÃ2j ðr Þ
þ
,
ð10Þ
ðrÞ
¼
f
ðxÞGð
,
xÞdx
¼
n0 ,p0
j
r
D1 ðI Þðr Þ D2 ðI Þðr Þ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi
where r ¼ (ir, n0, p0). Fix " > 0 such that minfT  þ 2, T g 4 2 þ ". Then it is
straightforward to see, from the explicit formulas in Lemma 1 and the lower bound in
Lemma 2, that
jEÃej ðr Þj

C1 eÀð1þ"Þr ,

jDe ðIÞðr Þj ! C2 r, 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

n0 ,p0 ðrÞj

Àð1þ"Þr À1
2C1 CÀ1
r
2 e

for r > 0 large. This yields
lim sup
r!1


ln




n 0 , p 0 ð rÞ

r

Àð1 þ "Þ,

which is a contradiction to Lemma 3. Thus f  0.
Now following the proof of Lemma 1 up to Equations (6) and (7) (2 :¼ 21 þ 32 þ 33
need not be negative at that time) we obtain
Z

À
Á
d2
j 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx
2
dt 
Z
À
Á
j 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx ¼ 0,
ð11Þ
þ ð þ 2Þ2



Inverse Problems in Science and Engineering
d2
dt2

525

Z 


2 uj G À ðj 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 Þ dx

Z 

þ 2
2 uj Gjj À ðj 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 Þ dx ¼ 0

ð12Þ



for all  2 C3. Note that y  0 is the unique solution to the differential equation
8
00
2
>
< y ðtÞ þ  yðtÞ ¼ 0,
yð0Þ ¼ 0,
>
: 0
y ð0Þ ¼ 0,
where  ! 0 is independent of t. Applying this to (11) and (12) with  2 R3, we get
Z
À
Á
j 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 dx ¼ 0

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and
Z 

2 uj G À ðj 1 u1 Gj1 þ j 2 u2 Gj2 þ j 3 u3 Gj3 Þ dx ¼ 0:


Adding the latter equations, we obtain
Z
uj ðx, :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
Z
d2
uj ðx, tÞdx ¼ 0:
dt2 
Since
Z

!

d
uj ðx, tÞdx
¼0¼
dt

t¼0

!

Z

,

uj ðx, tÞdx


t¼0

we have
Z
uj ðx, ÁÞdx ¼ 0:

ð14Þ



Putting (13) and (14) together, we arrive at
Z
Z
uj G ¼
uj ðx, :Þ cosð1 x1 Þ cosð2 x2 Þ cosð3 x3 Þdx ¼ 0 for all  2 R3 :




This implies that u  0 because {cos(mx1)cos(mx2)cos(nx3)}m,n,p!0 forms an orthogg
onal basis of L2().

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


526

D.D. Trong et al.

adaption of Lemma 4 in [6]. The only difference is that we are working on a 3-dim
problem, and hence we require more interpolation points.
LEMMA 4 (Interpolation inequality) Let Br ¼ {Æzjjzj ¼ 5r þ j, j ¼ 1, 2, . . . , 24r} for some
integer r ! 50. Let !: C ! C be an entire function such that j!(z) j Aejzj} for some constant
A. Then for any function g: Br ! C, one has




sup !ðzÞ À L½Br , gŠðzÞ AeÀr þ 48re30r sup!ðzÞ À gðzÞ:
jzj r

z2Br

Proof Fix z 2 C with jzj
{z} [ Br,

r. We have the following residue formula at 49 simple poles

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Z

24r z2 À z2


!ðxÞ Y
j
dx
¼
2i
!
ð
z
Þ
À
L
½
B
,
!
Š
ð
z
Þ
:
r
2
2
¼fx2C:jxj¼50rg x À z j¼1 x À zj

This gives the simple bound



!ðzÞ À L½Br , !ŠðzÞ

(

)
24r jz2 À z2 j
j!ðxÞj Y
j
50r sup
:
2
2
x2 jx À zj j¼1 jx À zj j

Now we bound the right-hand side of (15). For x 2 C, jxj ¼ 50r, we have j!(x) j
jx À zj ! (50 À )r and
24r jz2 À z2 j
Y
j
j¼1

jx2 À z2j j

24r jzj2 þ z2
Y
j
j¼1

jxj2 À z2j

24r ðrÞ2 þ z2
Y
j
j¼1

ð50rÞ2 À z2j

:

ð15Þ
Ae50r,

ð16Þ

We shall show that
24r ðrÞ2 þ z2
Y
j

50 À  À51r
e
,
50

2
2
j¼1 ð50rÞ À zj

8r ! 50:

2

ð17Þ

ðrÞ þx
Note that the function x ° vðxÞ ¼ lnðð50rÞ
Þ. v(x) is increasing and concave in [0, (29r)2].
2
Àx
Using Jensen’s inequality, we get
!
!
24r  
6
4hr
6
4hr
 
X
X
X
X
X
1
2
2
2
v zj ¼
v zj

v
z
4r
4r j¼4ðhÀ1Þrþ1 j
j¼1
h¼1 j¼4ðhÀ1Þrþ1
h¼1

¼ 4r




6
X
31 2
1
v 24h þ 16h2 þ
r þ ð3 þ 4hÞr þ
3
6
h¼1




6
X
31 2
r2

r2
2
4r
v 24h þ 16h þ
r þ ð3 þ 4hÞ þ
3
50 6 Â 502
h¼1
!
3þ4h
1
6
X
2 þ 24h þ 16h2 þ 31
3 þ 50 þ 6Â502
¼ 4r
ln
3þ4h
1
502 À ð24h þ 16h2 þ 31
h¼1
3 þ 50 þ 6Â502 Þ


50 À 
5 À51r þ ln
:
50

Inverse Problems in Science and Engineering
Thus (17) holds. Replacing (16) and (17) into (15), we obtain


!ðzÞ À L½Br , !ŠðzÞ AeÀr , 8z 2 C, jzj

r:

527

ð18Þ

Next, we observe that
!
24r
Y z2 À z2 z þ zj  À Á
X
À Á
k
L½Br , !ŠðzÞ À L½Br , gŠðzÞ ¼
À
g
zj
!
z
j
z2 À z2k 2zj
j¼1
k6¼j j
!

24r
Y z2 À z2 z À zj  À Á
X
À Á
k
þ
! Àzj À g Àzj :
2
2
z À zk À2zj
j¼1
k6¼j j

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Therefore


jL½Br , !ŠðzÞ À L½Br , gŠðzÞ




24r Y 2
2
X
z
À
z
k


2
 z2 À z2  ,
k
j¼1  k6¼j j

ð19Þ

where  ¼ supz2Br j!ðzÞ À gðzÞj.
Now we bound the right-hand side of (19). We have



Y 2
 z À z2k  Y jzj2 þ z2k Y jzj2 þ z2k
z


À
Á Á k 
¼
 z2 À z2 
2 À z2 j

jz
z
À
zk 
z
þ

z
z
j
j
k
k
 k6¼j j
k
k
k6¼j j
k6¼j
Y
k6¼j

z
 k 
 zj À zk 

ð5r þ 2Þð5r þ 3Þ . . . ð29rÞ
ðj À 1Þ!ð24r À jÞ!

ð5r þ 2Þð5r þ 3Þ . . . ð29rÞ
¼: JðrÞ:
ð12r À 1Þ!ð12rÞ!
Since J(1)

e30 and, by direct expansion,

Jðr þ 1Þ
ð29r þ 1Þð29r þ 2Þ . . . ð29r þ 29Þ

¼
JðrÞ
ð5r þ 2Þ . . . ð5r þ 6Þ Á ½ð12r þ 1Þ . . . ð12r þ 11ފ2 Á ð12rÞ Á ð12r þ 12Þ
2929
5 e30 ,
55 Á 1224
we conclude that J(r)

e30r for all r ! 1. Thus (19) reduces to


jL½Br , !ŠðzÞ À L½Br , gŠðzÞ 48re30r :

The desired result follows from (18), (20) and the triangle inequality.

ð20Þ
g

The last of our preparation to prove the regularization result is the following useful
lemma on the truncated Fourier series. The proof is elementary and we refer to Lemma 5
in [7] for details (Trong et al. [8] the authors dealt with the two-dimensional case, but the
proof for three-dimensional case is essentially the same).
For each w 2 L2() and r > 0, we define
X
ðm, n, pÞF ðwÞðm, n, pÞ cosðmx1 Þ cosðnx2 Þ cosð px3 Þ:
Àr ðwÞðxÞ ¼

LEMMA 5

0 m,n,p r

528

D.D. Trong et al.

Then
(i) Àr(w) ! w in L2() as r ! 1.
(ii) If w 2 H1() then Àr(w) ! w in H1() and


Àr ðwÞ À w 2
L ðÞ

1
pffiffi kwkH1 ðÞ :
 r



Àr ðwÞ À w 1
H ðÞ

4
pffi kwkH2 ðÞ :
½4Šr

(iii) If w 2 H2() then

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We are now ready to prove the main result.
Proof of Theorem 2 We shall first estimate the error fj" À Àr" fj0 , where Àr is defined in
Lemma 5. Then we compare Àr" fj0 and f0j by employing Lemma 5. The conclusion follows
from the triangle inequality. In the following, C, C1, C2 always stand for constants
depending only on the exact data, and saying that " > 0 small enough means that " 2 (0, "0)
for some constant "0 > 0 depending only on the exact data.
Step 1 Bound on jF"j ðm, n, pÞ À F ð fj0 Þðm, n, pÞj for m, n, p 2 [{0, r"].
Applying Lemma 4 to gðzÞ ¼ H"j ðz, n, pÞ and
Z
!ðzÞ ¼


fj0 ðx1 , x2 , x3 Þ cosðÀizx1 Þ cosðnx2 Þ cosðpx3 Þdx1 dx2 dx3 ,

we find that for m, n, p 2 [0, r"] and " > 0 small enough



 
Â
Ã


 "
Fj ðm, n, pÞ À F fj0 ðm, n, pÞ ¼ jL Br" , g ðimÞ À !ðimÞ



kfj0 kL1 ðÞ eÀr" þ 48 r" e30r" sup !ðzÞ À gðzÞ: ð21Þ

z2Br"

The error j!(z) À g(z)j can be bound by direct computation. Indeed, for z 2 Br" ,
we have


   
!ðzÞ À gðzÞ ¼ F f 0 ðÀiz, n, pÞ À H" ðz, n, pÞ
j
j

  


F fj0 ðÀiz, n, pÞ À H0j ðz, n, pÞ




þ H0j ðz, n, pÞ À H"j ðz, n, pÞ

 Ã
 E1j ðÀiz, n, pÞ
EÃ2j ðÀiz, n, pÞ 


D ðI ÞðÀiz, n, pÞ þ D ðI ÞðÀiz, n, pÞ
1 0
2 0



E1j ðI0 ÞðÀiz, n, pÞ E1j ðI" ÞðÀiz, n, pÞ

þ 
À
D1 ðI0 ÞðÀiz, n, pÞ D1 ðI" ÞðÀiz, n, pÞ 


E2j ðI0 ÞðÀiz, n, pÞ E2j ðI" ÞðÀiz, n, pÞ
:
þ 
À
D2 ðI0 ÞðÀiz, n, pÞ D2 ðI" ÞðÀiz, n, pÞ 

ð22Þ


Inverse Problems in Science and Engineering

529

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It is straightforward to see from the explicit formulas of E1j, E2j, EÃ1j , EÃ2j that


E1j ðI" ÞðÀiz, n, pÞ À E1j ðI0 ÞðÀiz, n, pÞ C1 r3 e29r" ",
"



E2j ðI" ÞðÀiz, n, pÞ À E2j ðI0 ÞðÀiz, n, pÞ C1 r3 e29r" ",
"


E1j ðI0 ÞðÀiz, n, pÞ C1 r3 e29r" ,
"


E2j ðI0 ÞðÀiz, n, pÞ C1 r3 e29r" ,
"


C1 r3" e29r"

 Ã
,
E1j ðÀiz, n, pÞ
pffiffipffiffiffiffiffiffiffiffiffi
e 5 þ2 T r"
 C r3 e29r"


 Ã
1 "
pffiffipffiffiffi
:
E2j ðÀiz, n, pÞ
e 5  T r"
On the other hand, Lemma 2 ensures that C2 r" jDj ðI0 ÞðÀiz, n, pÞj C1 r2" and


 

Dj ðI" ÞðÀiz, n, pÞ ! Dj ðI0 ÞðÀiz, n, pÞ


À Dj ðI" ÞðÀiz, n, pÞ À Dj ðI0 ÞðÀiz, n, pÞ
! C2 r" À C1 r2" ",
for all z 2 Br" and " > 0 small enough. Therefore, from (21) and (22) we get


 

 "
Fj ðm,n,pÞ À F fj0 ðm, n,pÞ
"
#
1
3
3
4
 
1
1
1
C1 ð30
lnð"À1 ÞÞ
C1 ð30
lnð"À1 ÞÞ
4C21 ð30
lnð"À1 ÞÞ e59 "60

1
 0
pffiffipffiffiffi
þ
:
þ
fj  1 "60 þ 48
pffiffipffiffiffiffiffiffiffiffiffi
1
1
1
À1
L ð Þ
C "À60ð 5 þ2 TÀ59Þ C2 "À60ð 5  TÀ59Þ C2 ðC2 À 30 C1 lnð" Þ"Þ
2

Moreover, the condition (W20 ) gives

1
1 pffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5
5  þ 2T À 59 ,
60
60


1
1 pffiffiffipffiffiffiffi
5
5 T À 59 :

60
60

Thus we obtain


 

 "
Fj ðm, n, pÞ À F fj0 ðm, n, pÞ

À Á 1
Cðln "À1 Þ4 "60

ð23Þ

for all " > 0 small enough.
Step 2

Conclusion.

(i) Using the Parseval equality and the estimate (23), one gets
2


 "
fj À Àr" fj0  2

L ð Þ

¼

X
0 m, n, p r"

2

 


ðm, n, pÞF"j ðm, n, pÞ À F fj0 ðm, n, pÞ :
1

Cðlnð"À1 ÞÞ11 "30 :

ð24Þ

Since kfj" À Àr" fj0 kL2 ðÞ ! 0 by (24) and kÀr" fj0 À fj0 kL2 ðÞ ! 0 by Lemma 5, we obtain
k fj" À fj0 kL2 ðÞ ! 0 as " ! 0 by the triangle inequality.
(ii) Assume that fj0 2 H1 ðÞ. Then by Lemma 5 one has kÀr" fj0 À fj0 kH1 ðÞ ! 0 and
1
kÀr" fj0 À fj0 kL2 ðÞ Cðlnð"À1 ÞÞÀ2 : Combining the latter estimate with (24), we conclude


530

D.D. Trong et al.

using the triangle inequality that




 "
 fj À fj0 

L 2 ð Þ

À Á 1
Cðln "À1 ÞÀ2

for " > 0 small enough.
Moreover, using (23) and the Parseval equality for H1, we have
2
2


 
X


 "

ðm,n,pÞð1 þ 2 ðm2 þ n2 þ p2 ÞÞF"j ðm,n,pÞ À F fj0 ðm,n,pÞ :
fj À Àr" fj0  1 ¼
H ð Þ

0 m, n, p r"

1

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Cðlnð"À1 ÞÞ13 "30 ! 0:

ð25Þ

Thus k fj" À fj0 kH1 ðÞ ! 0 by the triangle inequality (in H1-norm).
(iii) Assume that fj0 2 H2 ðÞ. Then by Lemma 5 one has




Cðlnð"À1 ÞÞÀ1=4 :
Àr" fj0 À fj0  1
H ðÞ

This estimate together with (25) and the triangle inequality yield



 "
Cðlnð"À1 ÞÞÀ1=4
fj À fj0  1
H ð Þ

for " > 0 small enough.
g

5. Numerical example
In this section, we test our regularization process in an explicit example. Choose  ¼ 1,

 ¼ À1, T ¼ 30 and consider the system (1)–(4) with the exact data I0 ¼ (’0, X0, g0, h0)
given by
’0 ðtÞ ¼232 cosðtÞ,

h01 ¼ h02 ¼ h03 ¼ 0,

X01 ðx1 , x2 , x3 , tÞ ¼ cosðtÞðÀ4 sinð2x2 Þ sinð2x3 Þn1
À 2 sinð4x1 Þ sinð2x3 Þn2 À 2 sinð4x1 Þ sinð2x2 Þn3 Þ,
X02 ðx1 , x2 , x3 , tÞ ¼ cosðtÞðÀ2 sinð4x2 Þ sinð2x3 Þn1
À 4 sinð2x1 Þ sinð2x3 Þn2 À 2 sinð2x1 Þ sinð4x2 Þn3 Þ,
X03 ðx1 , x2 , x3 , tÞ ¼ cosðtÞðÀ2 sinð2x2 Þ sinð4x3 Þn1
À 2 sinð2x1 Þ sinð4x3 Þn2 À 4 sinð2x1 Þ sinð2x2 Þn3 Þ,
g01 ðx1 , x2 , x3 Þ ¼ sinð4x1 Þ sinð2x2 Þ sinð2x3 Þ,
g02 ðx1 , x2 , x3 Þ ¼ sinð2x1 Þ sinð4x2 Þ sinð2x3 Þ,
g03 ðx1 , x2 , x3 Þ ¼ sinð2x1 Þ sinð2x2 Þ sinð4x3 Þ:


531

Inverse Problems in Science and Engineering

Note that in this example the conditions (W1) and (W20 ) are satisfied. The exact
solution of the system (1)–(4) can be computed explicitly
u01 ðx1 , x2 , x3 , tÞ ¼ cosðtÞ sinð4x1 Þ sinð2x2 Þ sinð2x3 Þ,
u02 ðx1 , x2 , x3 , tÞ ¼ cosðtÞ sinð2x1 Þ sinð4x2 Þ sinð2x3 Þ,
u03 ðx1 , x2 , x3 , tÞ ¼ cosðtÞ sinð2x1 Þ sinð2x2 Þ sinð4x3 Þ,
f10 ðx1 , x2 , x3 Þ ¼ sinð4x1 Þ sinð2x2 Þ sinð2x3 Þ,
f20 ðx1 , x2 , x3 Þ ¼ sinð2x1 Þ sinð4x2 Þ sinð2x3 Þ,
f30 ðx1 , x2 , x3 Þ ¼ sinð2x1 Þ sinð2x2 Þ sinð4x3 Þ:

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Now, we consider the disturbed data, for n 2 N, j 2 {1, 2, 3},
’n ¼ ’,

hnj ¼ 0,

À2 cosðtÞ
pffiffiffi
ðsinð2nx2 Þ sinð2nx3 Þn1
n
þ sinð2nx1 Þ sinð2nx3 Þn2 þ sinð2nx1 Þ sinð2nx2 Þn3 Þ,
sinð2nx1 Þ sinð2nx2 Þ sinð2nx3 Þ
gnj ¼ g0j þ
:
3
n2

Xnj ¼ X0j þ

The disturbed solution of the system (1)–(4) with the disturbed data is
cosðtÞ sinð2nx1 Þ sinð2nx2 Þ sinð2nx3 Þ
pffiffiffiffiffi
,
n3
ðÀ1 þ 12n2 Þ sinð2nx1 Þ sinð2nx2 Þ sinð2nx3 Þ
pffiffiffiffiffi
:
fnj ¼ fj0 þ
23 n3

unj ¼ u0j þ

We can see that the error of the data is small
pffiffiffi
2
¼ pffiffiffiffiffi ,
4 n3




 n
gj À g0j 

L 2 ð Þ




 n
Xj À X0j 

L1 ð0, T, L1 ð@ÞÞ

360
pffiffiffi
n

(in fact, they even converge in the uniform norm). However, the error of the solution is

large since



n
fj À fj0 

L 2 ð Þ

pffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 144n4 À 24n2 þ 1
!1
¼
4
529n3

as n ! 1. Thus the problem is ill-posed and a regularization is necessary.
Now, we apply our regularization procedure for " ¼ 0.01 and the disturbed data with
n ¼ 10. The resulting regularized solution is
f"1 ðx1 , x2 , x3 Þ % 0:035 À 0:063 cosðx1 Þ À 0:156 cosðx2 Þ À 0:455 cosðx3 Þ
þ 0:033 cosðx1 Þ cosðx2 Þ þ 0:027 cosðx1 Þ cosðx3 Þ
þ 0:15 cosðx2 Þ cosðx3 Þ À 0:005 cosðx1 Þ cosðx2 Þ cosðx3 Þ:


532

D.D. Trong et al.

The error between the regularized solution and the exact solution is
"


f À f 0 2 2 % 0:273:
1

1

L ð Þ

To see the effect of our regularization, note that the corresponding disturbed solution
0 2
5
(with data error " ¼ 0.01) causes an extremely large error kf10
j À fj kL2 ðÞ % 3:4 Â 10 .
Acknowledgements
The work was done when P.T. Thuc was a Master student in Ho Chi Minh City University of
Education.

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References
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