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Applied Mathematical Modelling 39 (2015) 5480–5494

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Applied Mathematical Modelling

journal homepage: www.elsevier .com/locate /apm

A one-dimensional inverse problem in composite materials:Regularization and error estimates

http://dx.doi.org/10.1016/j.apm.2015.01.0040307-904X/� 2015 Elsevier Inc. All rights reserved.

⇑ Corresponding author.E-mail address: [email protected] (Y.C. Hon).

X.T. Xiong a, W.X. Shi a, Y.C. Hon b,⇑a Department of Mathematics, Northwest Normal University, Lanzhou, Gansu, Chinab Department of Mathematics, City University of Hong Kong, Hong Kong, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 29 November 2012Received in revised form 1 March 2014Accepted 5 January 2015Available online 14 January 2015

Keywords:Inverse problemIll-posednessError estimateRegularizationComposite materials

In this paper we investigate an inverse one-dimensional heat conduction problem in multi-layer medium. The inverse problem is first formulated in the frequency domain via Fouriertransform technique. An effective regularization method for the stable reconstruction ofsolution is given with proven error estimates. Several numerical examples are constructedto demonstrate the effectiveness of the proposed method.

� 2015 Elsevier Inc. All rights reserved.

1. Introduction

The transient temperature distribution in a composite medium consisting of several layers in contact has numerous appli-cations in engineering. For instance, it is important in the aerospace engineering to reconstruct the surface temperature andheat flux in high temperature composite materials. One of these works evolved as a boundary identification problem [1] inseveral-layer domain arises from steel-making industry in which the iron was heated and melted in one container composedof different materials. The result of this work is important to detect any corrosion inside the inner-surface of the containerand prevent any disaster damage due to the leakage of the steel fluid. Mathematically, these kinds of problems can be treatedas boundary identification problems. Another example is the design problem of shielded thermocouple which is a measure-ment device used for monitoring the temperature in the hostile environments [2]. The shielded thermocouple deviceconsists of composite materials each with different thermal property. In order to simulate the concrete measurement situ-ations, a mathematical model of the shielded thermocouple is usually formulated as an Inverse Heat Conduction Problem(IHCP) [3] in several-layers domain.

The IHCP arises in thermal manufacturing processes of solids and has attracted much attention [4–7]. In this inverse IHCPproblem, it is required to recover the surface temperature and heat flux on an inaccessible boundary from the measurementon an accessible boundary, which is also called as a non-characteristic Cauchy problem of heat equation. It is well knownthat non-characteristic Cauchy problem of heat equation is ill-posed [8] in the Hadamard sense that any ‘‘small’’ measure-ment error in the data can induce extremely ‘‘large’’ error in the solution. Under an additional condition, a continuousdependence of the solution on the Cauchy data can be obtained. This is called conditional stability [9]. In other words, in

X.T. Xiong et al. / Applied Mathematical Modelling 39 (2015) 5480–5494 5481

theory we can stably reconstruct the solution of the ill-posed problem under a priori condition which is called the ‘‘sourcecondition’’. Due to the severe ill-posedness of the problem, most classical numerical methods failed to produce satisfactorystable approximation to the solution of the Cauchy problem of heat conduction equation. Some kinds of regularizationstrategies [10] needs to be employed. For the IHCP in a single-layer domain, theoretical investigation and computationalimplementation have been well studied, e.g., [11–27]. In alternative to the single-layer IHCP, the multi-layer IHCP is muchmore difficult in both numerical and theoretical studies. In general, the IHCP in multi-layer domain will be dissolved into anIHCP in each layer and the solution the multi-layer IHCP will be obtained by solving the IHCP in each single layer. Thisapproach is computational inefficient but the author can refer to the recent works [28–30] in which the authors used themethod of fundamental solutions combined with regularization to solve each IHCP layer by layer.

This paper aims at establishing a mathematical framework to recover the surface temperature and heat flux on the inac-cessible boundary in a two-layer domain [31] as follow.

Consider a two-layer body that consists of the first layer in 0 6 x 6 l1 and the second layer in l1 6 x 6 l2. The two layersare in perfect thermal contact at x ¼ l1 as displayed in Fig. 1.

Let k1; k2 > 0 be the thermal conductivities and a1;a2 > 0 be the thermal diffusivities for the first and second layer,respectively. The temperature distributions in the first and the second layers are denoted by u1ðx; tÞ and u2ðx; tÞ respectively.These temperature distributions satisfy the following partial differential equations in the two domains D1 :¼ fxj0 6 x 6 l1gand D2 :¼ fxjl1 6 x 6 l2g:

@u1

@t� a1

@2u1

@x2 ¼ 0; 0 < x < l1; t > 0; ð1:1Þ

@u2

@t� a2

@2u2

@x2 ¼ 0; l1 < x < l2; t > 0; ð1:2Þ

subject to the initial and boundary conditions

u1ðx;0Þ ¼ u2ðx;0Þ ¼ 0; 0 < x < l2; ð1:3Þu2ðl2; tÞ ¼ gðtÞ; t > 0; ð1:4Þ@u2

@xðl2; tÞ ¼ 0; t > 0; ð1:5Þ

u1ðl1; tÞ ¼ u2ðl1; tÞ; t > 0; ð1:6Þ

k1@u1

@xðl1; tÞ ¼ k2

@u2

@xðl1; tÞ; t > 0: ð1:7Þ

For the similar IHCP, Shcheglov [32,33] analyzed the convergence of the problem by using a hyperbolic equation perturba-tion method. For 2D IHCP in two-layer Cartesian bodies, some numerical approaches have been proposed [34,35]. To theknowledge of the authors, the convergence rate on the IHCP in multi-layer domain has not yet been given.

In this paper, we aim at obtaining an analytical solution to the above IHCP in multi-layer domain via Laplace and Fouriertransform techniques. Due to the severe ill-posedness of the problem, a regularization strategy is derived for the stablereconstruction of the solution. For illustration, we present the regularization method based on the analytical solution forthe IHCP in a two-layer domain in which we adapt a spectral regularization method and derive the error estimate withproven optimal order. The proposed method is simple and effective [14]. Furthermore, the proposed method can easily begeneralized to solve IHCP in the multi-layer domain.

The paper is organized as follows. We first give a theoretical analysis on the IHCP problem in Section 2. In Section 3, thespectral regularization technique is adopted for the reconstruction of stable solution with proven error estimates. Finally,several numerical examples are constructed in the last section to demonstrate the validity of the proposed regularizationmethod although the so-called ‘‘inverse crime’’ [36] has been used.

0

D2D1

l1xl2

Lateral Interface Lateral

Fig. 1. Schematic illustration of a one-dimensional inverse heat conduction problem in a two-layer domain.

5482 X.T. Xiong et al. / Applied Mathematical Modelling 39 (2015) 5480–5494

2. Mathematical analysis of the problem

Throughout this paper, we suppose that the exact data g 2 L2ð0;1Þ. As a solution of the problem (1.1)–(1.7), the functionsu1ðx; tÞ;u2ðx; tÞ satisfy (1.1)–(1.7) in the classical sense. Assume that, for a fixed x 2 ½0; l2�, the functions u1ðx; �Þ;u2ðx; �Þ andtheir derivatives @u1ðx;�Þ

@x ; @u2ðx;�Þ@x belong to L2ð0;1Þ.

The inverse problem is to seek the solutions @u1@x ðx; tÞ; u1ðx; tÞ for 0 6 x 6 l1 in the space L2ð0;1Þ from the given data

gðtÞ 2 L2ð0;1Þ and the insulated condition at the accessible boundary x ¼ l2. In practice, gðtÞ can only be measured withmeasurement errors to give some data functions gdðtÞ 2 L2ð0;1Þ, for which

kgdð�Þ � gð�Þk 6 d; ð2:1Þ

where the constant d > 0 represents a bound on the measurement error, k � k denotes the L2-norm, and there exists a con-stant E > 0, such that the following a priori bound exist for the problem

ku1ð0; �Þkp 6 E; ð2:2Þ

with p P 0, where k � kp denotes the norm of Sobolev space HpðRÞ (when p ¼ 0;H0ð�Þ ¼ L2ð�Þ).In fact, in the domain D2, we have

@u2

@t� a2

@2u2

@x2 ¼ 0; l1 < x < l2; t > 0 ð2:3Þ

u2ðx;0Þ ¼ 0; l1 < x < l2; ð2:4Þu2ðl2; tÞ ¼ gðtÞ; t > 0; ð2:5Þ@u2

@xðl2; tÞ ¼ 0; t > 0: ð2:6Þ

Applying the Laplace transform to both sides of (2.3) with respect to t and according to the property of Laplace transform,we get

sU2ðx; sÞ � u2ðx;0Þ � a2@2U2

@x2 ðx; sÞ ¼ 0; ð2:7Þ

where s denotes the variable of Laplace transform on t. Applying the homogenous initial condition, we have

a2@2U2

@x2 ðx; sÞ ¼ sU2ðx; sÞ; ð2:8Þ

which is a second-order ordinary differential equation. From the boundary conditions, we obtain

U2ðx; sÞ ¼ coshðffiffiffiffiffiffiffiffiffiffis=a2

pðl2 � xÞÞGðsÞ; ð2:9Þ

whereffiffiffiffiffiffiffiffiffiffis=a2

pis taken to be the principal square root and GðsÞ is the Laplace transform of gðtÞ.

Throughout this paper, we extend all the functions to the whole line �1 < t <1. Let

f ðnÞ ¼ 1ffiffiffiffiffiffiffi2pp

Z 1

�1f ðtÞe�intdt

be the Fourier transform of the function f ðtÞ 2 L2ðRÞ.For the function f ðtÞ which vanishes on the negative t axis, the Fourier and Laplace transforms are related via

FðinÞ ¼ffiffiffiffiffiffiffi2pp

f ðnÞ:

Therefore, from (2.9), on setting s ¼ in, the solution of problem (2.3)–(2.6) can be formulated in the frequency domain forl1 6 x 6 l2:

u2ðx; nÞ ¼ coshððl2 � xÞ

ffiffiffiffiffiina2

sÞgðnÞ; ð2:10Þ

where
ffiffiffiffiffiina2
s¼ jn=ð2a2Þj

12ð1þ isgnðnÞÞ; the symbol sgn is the usual sign function: ð2:11Þ

Inverse Fourier transform on u2ðx; nÞ gives u2ðx; tÞ. From (2.10), we have for l1 6 x 6 l2:

@u2

@xðx; nÞ ¼ �

ffiffiffiffiffiina2

ssinhððl2 � xÞ

ffiffiffiffiffiina2

sÞgðnÞ: ð2:12Þ


In the domain D1;u1 satisfies

@u1

@t� a1

@2u1

@x2 ¼ 0; 0 < x < l1; t > 0 ð2:13Þ

u1ðx;0Þ ¼ 0; 0 < x < l1; ð2:14Þu1ðl1; tÞ ¼ u2ðl1; tÞ; t > 0; ð2:15Þ

k1@u1

@xðl1; tÞ ¼ k2

@u2

@xðl1; tÞ; t > 0: ð2:16Þ

Similarly, from Laplace transform we have

sU1ðx; sÞ � u1ðx;0Þ � a1@2U1

@x2 ðx; sÞ ¼ 0; 0 < x < l1; ð2:17Þ

U1ðl1; sÞ ¼ U2ðl1; sÞ; ð2:18Þ

k1@U1

@xðl1; sÞ ¼ k2

@U2

@xðl1; sÞ: ð2:19Þ

From the relation between the Fourier transform and the Laplace transform, we have for 0 6 x 6 l1

u1ðx; nÞ ¼ cosh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A cosh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1AgðnÞ þ k2ffiffiffiffiffia1p

k1ffiffiffiffiffia2p sinh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A sinh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1AgðnÞ;

ð2:20Þ

which yields

@u1ðx; nÞ@x

¼ �

ffiffiffiffiffiina1

ssinh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A cosh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1AgðnÞ

� k2ffiffiffiffiffia1p

k1ffiffiffiffiffia2p

ffiffiffiffiffiina1

scosh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A sinh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1AgðnÞ: ð2:21Þ

Taking inverse Fourier transform on (2.20) and (2.21), we have u1ðx; tÞ and @u1ðx;tÞ@x . In fact, we need to prove that the inverse

Fourier transforms u1ðx; tÞ and u2ðx; tÞ satisfy (1.3). In fact, if bgðnÞ decays rapidly in the frequency domain, we can prove thatu1ðx; tÞ is a unique solution to the problem. Refer [37] for the detailed proof. However. since the given data usually containmeasurement error, we cannot expect the error to have the same decay rate in frequency domain as the exact data bgðnÞ. Inother words, the corresponding perturbed solution will in general not lie in L2ðRÞ. Therefore, the problem considered in thispaper is an ill-posed problem.

Denote

k :¼ k2ffiffiffiffiffia1p

k1ffiffiffiffiffia2p ; ð2:22Þ

bAðx; nÞ :¼ cosh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A cosh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1Aþ k sinh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A sinh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1A; ð2:23Þ

bBðx; nÞ :¼ �

ffiffiffiffiffiina1

ssinh

ffiffiffiffiffiina1

sðl1 � xÞ

0@ 1A cosh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1A� k

ffiffiffiffiffiina1

scosh

ffiffiffiffiffiina1

sðl1 � xÞÞ sinhð

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1A: ð2:24Þ

Formulating (2.20) and (2.21) as operator equations, we have

u1ðx; nÞ ¼ bAðx; nÞgðnÞ; ð2:25Þ@u1ðx; nÞ

@x¼ bBðx; nÞgðnÞ: ð2:26Þ

From above two equations, we can see that bAðx; nÞ; bBðx; nÞ : L2ðRÞ ! L2ðRÞ are unbounded multiplication operators for0 6 x 6 l1.

In order to obtain the error estimates, first we need two inequalities on bAðx; nÞ and bBðx; nÞ.


Proposition 2.1 [31]. Let 0 < x < l1, then

ð1Þ jbAðx; nÞj 6 94ð1þ kÞe

ðl1�xÞffiffiffiffiffijnj

2a1

qeðl2�l1Þ

ffiffiffiffiffijnj

2a2

q; ð2:27Þ

ð2Þ jbBðx; nÞj 6 94ð1þ kÞ

ffiffiffiffiffiffiffiffijnj

2a1

seðl1�xÞ

ffiffiffiffiffijnj

2a1

qeðl2�l1Þ

ffiffiffiffiffijnj

2a2

q: ð2:28Þ

3. Regularization method and error estimates

In this section, we prove the error estimates for the regularized solutions between the solution u1ðx; tÞ and the heat fluxsolution @u1

@x ðx; tÞ, respectively.

3.1. On temperature u1ðx; tÞ

In the frequency domain, the spectral method for solving the problem can be given as:

unmax1 ðx; nÞ ¼ u1ðx; nÞvmax;

@u1

@x

nmax

ðx; nÞ ¼ @u1

@xðx; nÞvmax;

where nmax is the regularization parameter, vmax denotes the characteristic function on the interval ½�nmax; nmax�, i.e.,

vmaxðnÞ ¼1; �nmax 6 n 6 nmax;

0; jnj > nmax:

�
Let us recall dfðtÞ ¼ Fðf ðtÞÞ ¼ bf ðnÞ ¼ 1ffiffiffiffiffiffiffi
2pp

Z 1

�1f ðtÞe�intdt;

represents the Fourier transform of f ðtÞ 2 L2ðRÞ, and

f ðtÞ ¼ F�1ðbf ðnÞÞ ¼ 1ffiffiffiffiffiffiffi2pp

Z 1

�1

bf ðnÞeintdn;

represents the corresponding inverse Fourier transform.Denote

Rðu1ðx; tÞÞ ¼ F�1ðunmax1 ðx; nÞÞ ¼ 1ffiffiffiffiffiffiffi

2pp

Z 1

�1u1ðx; nÞvmaxeinxdn; ð3:1Þ

R@u1

@xðx; tÞ

� �¼ F�1ð@u1

@x

nmax

ðx; nÞÞ ¼ 1ffiffiffiffiffiffiffi2pp

Z 1

�1

@u1

@xðx; nÞvmaxeinxdn: ð3:2Þ

In order to derive the error estimates, we need a lower bound on bAð0; nÞ.Proposition 3.1. If jnj > ð

ffiffiffiffiffiffi2a2

pl2�l1

ln 2Þ2

, it holds

jbAð0; nÞjP 1þ k8

el1

ffiffiffiffiffijnj

2a1

qeðl2�l1Þ

ffiffiffiffiffijnj

2a2

q: ð3:3Þ

Proof. Let a :¼ l1

ffiffiffiffiffiffijnj

2a1

q; b :¼ ðl2 � l1Þ

ffiffiffiffiffiffijnj

2a2

q; a1 :¼ asgnðnÞ; b1 :¼ bsgnðnÞ, we have

jbAð0;nÞj ¼ cosh

ffiffiffiffiffiina1

sl1

0@ 1A cosh

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1Aþ k sinh

ffiffiffiffiffiina1

sl1Þ sinhð

ffiffiffiffiffiina2

sðl2 � l1Þ

0@ 1A��

¼ cosh l1


2a1

s !ð1þ isgnðnÞÞ

!cosh ðl2 � l1Þ


2a2


!þ k sinh l1


2a1


!�� sinh ðl2 � l1Þ


2a2


!��¼ j coshðaþ ia1Þ coshðbþ ib1Þ þ k sinhðaþ ia1Þ sinhðbþ ib1Þj¼ jA1 þ kA2j;


where

A1 ¼ coshðaþ ia1Þ coshðbþ ib1Þ ¼eaþia1 þ e�a�ia1

2ebþib1 þ e�b�ib1

2

¼ 12½cosh ðaþ bÞ þ iða1 þ b1Þð Þ þ cosh ða� bÞ þ iða1 � b1Þð Þ�;

A2 ¼ sinhðaþ ia1Þ sinhðbþ ib1Þ ¼eaþia1 � e�a�ia1

2ebþib1 � e�b�ib1

2

¼ 12½cosh ðaþ bÞ þ iða1 þ b1Þð Þ � cosh ða� bÞ þ iða1 � b1Þð Þ�:

Indeed, first we have

j coshððaþ bÞ þ iða1 þ b1ÞÞj � j coshðða� bÞ þ iða1 � b1ÞÞj ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficosh2ðaþ bÞ cos2ða1 þ b1Þ þ sinh2ðaþ bÞ sin2ða1 þ b1Þ

q�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficosh2ða� bÞ cos2ða1 � b1Þ þ sinh2ða� bÞ sin2ða1 � b1Þ

qP

12ðeaþb � e�ðaþbÞÞ � 1

2ea�b ¼ 1

2eaþb � 1

2e�ðaþbÞ þ ea�b� �

P12

eaþb � ea�b:

If jnj > ðffiffiffiffiffiffi2a2

pl2�l1

ln 2Þ2

, i.e., b > ln 2, then 14 eaþb � ea�b P 0, therefore

jbAð0; nÞj ¼ 12jð1þ kÞ coshððaþ bÞ þ iða1 þ b1ÞÞ þ ð1� kÞ coshðða� bÞ þ iða1 � b1ÞÞj

P12jð1þ kÞ coshððaþ bÞ þ iða1 þ b1ÞÞj � jð1� kÞ coshðða� bÞ þ iða1 � b1ÞÞjj j

P1þ k

2j coshððaþ bÞ þ iða1 þ b1ÞÞj � j coshðða� bÞ þ iða1 � b1ÞÞjj jP 1þ k

212

eaþb � ea�b

� �P

1þ k8

eaþb:

By the definitions of a; b, we have the inequality holds. h

For readability, we stress that the following assumptions: the noisy data gdðtÞ satisfies kgdðtÞ � gðtÞk 6 d and the ‘‘sourcecondition’’ ku1ð0; �Þkp 6 E; ðp P 0Þ holds. Meanwhile, we define a symbol:

‘ :¼ l1ffiffiffiffiffiffiffiffi2a1p þ ðl2 � l1Þffiffiffiffiffiffiffiffi

2a2p : ð3:4Þ

The following lemma shows the stability of the regularization solution.

Lemma 3.1. If Rðu1ðx; tÞÞ;Rðud1ðx; tÞÞ are the regularized solutions with data given by gðtÞ and gdðtÞ, respectively, and

kgdðtÞ � gðtÞk 6 d, if we choose the regularization parameter
ffiffiffiffiffiffiffiffi2a2p
l2 � l1ln 2

� �2

< nmax ¼1‘

lnEd

lnEd

� ��2p !" #2

; ð3:5Þ

then

kRðud1ðx; tÞÞ � Rðu1ðx; tÞÞk 6 c1 ln

Ed

� ��2p !1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ ; ð3:6Þ

where c1 is a positive constant which is independent on d and E.

Proof. Using Parseval’s equality, we have

kRðud1ðx; �ÞÞ � Rðu1ðx; �ÞÞk2 ¼ k dRðud

1ðx; �ÞÞ � dRðu1ðx; �ÞÞk2 ¼Z nmax

�nmax

jbAðx; nÞj2jbgdð�Þ � bgð�Þj2dn

6 supjnj<nmaxjbAðx; nÞj2 Z nmax

�nmax

jbgdð�Þ � bgð�Þj2dn:


From kgdð�Þ � gð�Þk 6 d, we have

kRðud1ðx; �ÞÞ � Rðu1ðx; �ÞÞk2

68116ð1þ kÞ2e

2ðl1�xÞffiffiffiffiffiffiffiffijnmax j

2a1

qe

2ðl2�l1Þffiffiffiffiffiffiffiffijnmax j

2a2

q Z nmax

�nmax

jbgdð�Þ � bgð�Þj2dn

68116ð1þ kÞ2e

2l1�xffiffiffiffiffi

2a1

p þ l2�l1ffiffiffiffiffi2a2

p� � ffiffiffiffiffiffiffi

nmax

pd2

By nmax ¼ ½1‘ lnðEd ðln EdÞ�2pÞ�

2, we have

xxffiffiffiffiffip

kRðud1ðx; �ÞÞ � Rðu1ðx; �ÞÞk2

68116ð1þ kÞ2d2eð2�2

xffiffiffiffiffi2a1

p‘Þ ln E

d lnEdð Þ�2p

� �6

8116ð1þ kÞ2d2 E

d

� �2�2

ffiffiffiffiffi2a1

p‘

lnEd

� ��2p !2�2

2a1‘

6 c21 ln

Ed

� ��2p !2�2

xffiffiffiffiffi2a1

p‘

E2�2

xffiffiffiffiffi2a1

p‘ d2

xffiffiffiffiffi2a1

p‘ ;

i.e., The proof of (3.6) is completed. h

The following lemma shows the convergence of the regularized solution.

Lemma 3.2. Suppose Rðu1ðx; tÞÞ; u1ðx; tÞ are the regularized solution and the exact solution with the exact data gðtÞ respectively,and nmax is chosen as (3.5), it holds error estimate for d! 0

kRðu1ðx; tÞÞ � u1ðx; tÞk 6 c2 lnEd

� ��2p" #1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ ð1þ oð1ÞÞ; ð3:7Þ


Proof. Using the Parseval’s equality, we have the follow relation

kRðu1ðx; �ÞÞ � u1ðx; �Þk2 ¼ k dRðu1ðx; �ÞÞ � du1ðx; �Þk2 ¼Zjnj>nmax

jbAðx; nÞj2jbgðnÞj2dn:

Since bu1ð0; nÞ ¼ bAð0; nÞbgðnÞ, then � �
kRðu1ðx; �ÞÞ � u1ðx; �Þk2 ¼ k dRðu1ðx; �ÞÞ � du1ðx; �Þk2 ¼
Zjnj>nmax

bAðx; nÞbAð0; nÞ��

2

jbu1ð0; nÞj2dn:

According to ku1ð0; �Þkp 6 E, it yields

kRðu1ðx; �ÞÞ � u1ðx; �Þk2 ¼ k dRðu1ðx; �ÞÞ � du1ðx; �Þk2 ¼Zjnj>nmax


��ð1þ n2Þ�p2

" #2

ð1þ n2Þpjbu1ð0; nÞj2dn

6 supjnj>nmax


��2

n�2p

24 35Zjnj>nmax

ð1þ n2Þpjbu1ð0; nÞj2dn 6 E2supjnj>nmax


��2

n�2p

8<:9=;:

Using Proposition 2.1 and 3.1, then

supjnj>nmax


�� 6 ce�x

ffiffiffiffiffijnj

2a1

q; c is a constant:

By (3.5), we have 0 1 0 1
kRðu1ðx; �ÞÞ � u1ðx; �Þk2
6 c1‘

lnEdðln E

dÞ�2p

!" #�2p@ A2

E2 exp �2

xffiffiffiffiffiffi2a1

p

‘ln

Edðln E

dÞ�2p

!@ A

¼ c22 ln

Edðln E

dÞ�2p

! !�2p24 352

lnEd

� ��2p" #�2

xffiffiffiffiffi2a1

p‘

E2�2

xffiffiffiffiffi2a1

p‘ d2

xffiffiffiffiffi2a1

p‘ :

The proof has been completed. h


Now we can claim the main conclusion of this subsection. For the temperature u1ðx; tÞ, we have the following errorestimate:

Theorem 3.1. Suppose that Rðud1ðx; tÞÞ; u1ðx; tÞ are the regularized solution with the noisy data and the exact solution with the

exact data respectively, and the a priori condition ku1ð0; �Þkp 6 E with p > 0 holds, ifffiffiffiffiffiffi2a2

pl2�l1

ln 2� �2

< nmax ¼ 1‘ ln E

d ln Ed

� ��2p h i2

,

then for d! 0 it holds the error estimate

kRðud1ðx; tÞÞ � u1ðx; tÞk 6 c3 ln

Ed

� ��2p !1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ ð1þ oð1ÞÞ: ð3:8Þ


Proof. Due to Parseval’s equation and triangle inequality, it yields

kRðud1ðx; tÞÞ � u1ðx; tÞk

6 kRðud1ðx; tÞÞ � Rðu1ðx; tÞÞk þ kRðu1ðx; tÞÞ � u1ðx; tÞk:

According to Lemma 3.1 and 3.2 (see (3.6)), we have

kRðud1ðx; tÞÞ � u1ðx; tÞk 6 c1 ln

Ed

� ��2p !1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ þ c2 ln

Edðln E

dÞ�2p

!�2p24 35 ln

Ed

� ��2p" #� xffiffiffiffiffi

2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘

¼ c3 lnEd

� ��2p !1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ ð1þ oð1ÞÞ; d! 0:

Thus, we have the conclusion (3.8). h

Remark 3.1. When p ¼ 0, i.e., ku1ð0; �Þk 6 E holds, for 0 < x < l1 we have the following error estimate

kRðud1ðx; tÞÞ � u1ðx; tÞk 6 c3E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1

p‘ :

It is a Hölder-type error estimate.

Remark 3.2. If x ¼ 0 and kuð0; �Þkp 6 E with p > 0 holds, we have the following error estimate

kRðud1ð0; tÞÞ � u1ð0; tÞk 6 c3E ln

Ed

� ��2p

ð1þ oð1ÞÞ; d! 0:

It is a logarithm-type error estimate.Obviously, above two results in the Remarks are consistent with the results obtained in single-layer case [13].In the forthcoming subsection, we will deal with the problem on estimating the regularized heat flux. Usually, this prob-

lem is more difficult than the problem of estimating the regularized temperature (see [27]). Using the proposed method forsingle-layer case [27], fortunately, we can obtain the corresponding results for the multi-layer case.

3.2. On heat flux @u1@x ðx; tÞ

In this subsection, a stronger a priori condition ku1ð0; �Þkp 6 E for p > 1=2 is assumed. Here, we need to estimate @u1@x ðx; tÞ.

Similarly, we first give the stability result.

Lemma 3.3. Suppose that R @ud1ðx;tÞ@x

; R @u1ðx;tÞ

@x

are the regularized solutions with the noisy data and the exact data respectively,

and kgdðtÞ � gðtÞk 6 d, then if the nmax satisfiesffiffiffiffiffiffi2a2

pl2�l1

ln 2� �2

< nmax ¼ 1‘ ln E

d ln Ed

� ��2p h i2

, which is the same as (3.5), it holds for

d! 0x

!
R
@ud1

@xðx; tÞ

� �� R

@u1

@xðx; tÞ

� �� 6 c4 lnEd

� �1�2p 1�

ffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1




Proof. From Parseval’s equality, it yields

R@ud

1

@xðx; �Þ

� �� R

@u1

@xðx; �Þ

� �� 2

¼Z nmax

�nmax

jbBðx; nÞj2jbgdðnÞ � bgðnÞj2dn 6 supjnj<nmaxjbBðx; nÞj2 Z nmax

�nmax

jbgdðnÞ � bgðnÞj2dn

6

8116 ð1þ kÞ2

2a1jnmaxje

2ðl1�xÞffiffiffiffiffiffiffiffijnmax j

2a1

qe

2ðl2�l1Þffiffiffiffiffiffiffiffijnmax j

2a2

q Z nmax

�nmax

jbgdðnÞ � bgðnÞj2dn:

Noting that nmax ¼ 1‘

ln Ed ln E

d

� ��2p h i2

, we have

R@ud

1

@xðx; �Þ

� �� R

@u1

@xðx; �Þ

� �� 2

ð3:10Þ

6

8116 ð1þ kÞ2

2a1d2jnmaxj exp 2ðl1 � xÞ

ffiffiffiffiffiffiffiffiffiffiffiffijnmaxj2a1

s !exp 2ðl2 � l1Þ


s !

¼ 9ð1þ kÞ4ffiffiffiffiffiffiffiffi2a1p

� �2 1‘

lnEd

lnEd

� ��2p !" #2

d2 exp 2ðl1 � xÞ




s !

¼ c24 ln

Edðln E

dÞ�2p

!" #2

d2 exp 2ðl1 � xÞ




s !¼ c2

4 lnEdðln E

dÞ�2p

!" #2

E2�2

xffiffiffiffiffi2a1

p‘ d2

xffiffiffiffiffi2a1

p‘

¼ c24 ln

Ed

� �� 2�4p 1�

xffiffiffiffiffi2a1

p‘

!E2�2

xffiffiffiffiffi2a1

p‘ d2

xffiffiffiffiffi2a1


Thus (3.9) holds. h

Secondly, we have the following convergence result.

Lemma 3.4. Suppose that R @u1@x ðx; tÞ� �

; @u1@x ðx; tÞ are the regularized solution and the exact solution with the exact data respectively,

and the a priori condition kuð0; tÞkp 6 E with p > 12 holds, if the regularization parameter nmax is chosen as (3.5), then for d! 0 it

holds xffiffiffiffiffi2ap

xffiffiffiffiffip xffiffiffiffiffip
kRð@u1
@xðx; tÞÞ � @u1

@xðx; tÞk 6 c5ðln

EdÞ

1�2pð1� 1‘ Þ

E1�2a1‘ d

2a1‘ ð1þ oð1ÞÞ; ð3:11Þ


Proof. As before, we have

R@u1

@xðx; tÞ

� �� @u1

@xðx; tÞ

�� 2

¼Zjnj>nmax

jbBðx; nÞbgðnÞj2dn; ð3:12Þ

According to bu1ðx; nÞ ¼ bAðx; nÞbgðnÞ, bgðnÞ ¼ bu1ð0;nÞbAð0;nÞ , we have

R@u1

@xðx; tÞ

� �� @u1

@xðx; tÞ

�� 2

¼Zjnj>nmax

bBðx; nÞbAð0; nÞ��

��2

jcu1ð0; nÞj2dn;

Due toffiffi2p

a2l2�l1

ln 2 2

< nmax ¼ 1‘

ln Ed ðln E

dÞ�2p

h i2, and ku1ð0; �Þkp 6 E, it yields

R@u1

@xðx; tÞ

� �� @u1

@xðx; tÞ

�� 2

6 c2Zjnj>nmax

ffiffiffiffiffiffiffiffin

2a1

se�x

ffiffiffiffiffin

2a1

q��

2

ð1þ n2Þ�pð1þ n2Þpjbu1ð0; nÞj2dn

61

2a1c2Zjnj>nmax

jnj1�2pe�2x

ffiffiffiffiffijnj

2a1

qð1þ n2Þpjbu1ð0; nÞj2dn 6

12a1

c2supjnj>nmaxjnj1�2pe

�2x

ffiffiffiffiffijnj

2a1

q Zjnj>nmax

ð1þ n2Þpjcu1ð0; nÞj2dn

61

2a1c2E2n1�2p

max e�2x

ffiffiffiffiffiffiffinmax2a1

q6 c2

5½lnðEdÞ�

2�4pð1�

xffiffiffiffiffi2a1

p‘ÞE2�2

xffiffiffiffiffi2a1

p‘ d2

xffiffiffiffiffi2a1


Thus, we have (3.11). h

Now we have the main conclusion of this subsection.


Theorem 3.2. Suppose R @ud1ðx;tÞ@x

; @u1ðx;tÞ

@x are the regularized solution with the noisy data and the exact solution with the exact data

respectively, and kgdðtÞ � gðtÞk 6 d; ku1ð0; �Þkp 6 E with p > 12 hold, if

ffiffiffiffiffiffi2a2

pl2�l1

ln 2� �2

< nmax ¼ 1‘ ln E

d ðln EdÞ

� ��2ph i2

, we have the

following error estimate for d! 0 !
R
@ud1

@xðx; tÞ

� �� @u1

@xðx; tÞ

�� 6 c6 lnEd

� �1�2p 1�

xffiffiffiffiffi2a1

p‘

E1�

xffiffiffiffiffi2a1

p‘ d

xffiffiffiffiffi2a1



Proof. Similar to Theorem 3.1, according to the triangle inequality and Lemma 3.3 and 3.4, we have (3.13). h

In practice, we are more interesting in the convergence rate at x ¼ 0. By Theorem 3.2, we have the following remark.

Remark 3.3. Suppose R @ud1ð0;tÞ@x

; @u1ð0;tÞ

@x are the regularized solution with the noisy data and the exact solution with the exact

data respectively, and kgdðtÞ � gðtÞk 6 d; ku1ð0; �Þkp 6 E with p > 12 hold, if nmax is chosen as (3.5), we have the following error

estimate for d! 0

R@ud

1

@xð0; tÞ

� �� @u1

@xð0; tÞ

�� 6 c6 lnEd

� �1�2p

ð1þ oð1ÞÞ; d! 0: ð3:14Þ

4. Numerical examples

In general, for an ill-posed problem, we can only obtain the worst-case error for regularized methods. However, in prac-tical computation, the errors in numerical computation of regularization methods are far less than the worst-case errors. Thisphenomenon has been observed in many literatures, e.g. [6]. In this paper, the formula (3.5) is difficult for computationexcept that one can compute the exact E. In the following examples, we found that the regularization parameter nmax withina large range is available for good numerical reconstruction. Moreover, in practice the test of an inversion process avoidingthe ‘‘inverse crime’’ can be done using a model for the numerically simulated data and a different one to invert the data. Inthis paper, we can use a simple traditional numerical method (e.g. finite difference method), which is different from Fouriermethod, to solve the direct problem. However, this is a time-consuming task [38]. Based on the theoretical analysis derivedin the above section, we construct in the following several numerical examples to verify the convergence of the proposedmethod.

The following numerical implementation is performed by using Matlab. The regularized solutions are computed by usingthe discrete Fast Fourier Transform (FFT) and inverse discrete Fast Fourier Transform according to the formulas (3.1) and(3.2) in Section 3.

Firstly we solve the corresponding direct problem

@u1

@t� a1

@2u1

@x2 ¼ 0; 0 < x < l1; t > 0; ð4:1Þ

@u2

@t� a2

@2u2

@x2 ¼ 0; l1 < x < l2; t > 0; ð4:2Þ

subject to the initial and boundary conditions

u1ðx;0Þ ¼ u2ðx;0Þ ¼ 0; 0 < x < l2; ð4:3Þu1ð0; tÞ ¼ f ðtÞ; t > 0; ð4:4Þ@u2

@xðl2; tÞ ¼ 0; t > 0; ð4:5Þ

u1ðl1; tÞ ¼ u2ðl1; tÞ; t > 0; ð4:6Þ

k1@u1

@xðl1; tÞ ¼ k2

@u2

@xðl1; tÞ; t > 0: ð4:7Þ

The solution of (4.1)–(4.7) in the frequency domain is given by

u2ðl2; nÞ :¼ gðnÞ ¼ 1bAð0; nÞ f ðnÞ: ð4:8Þ

Inverse Fourier transform gives the data gðtÞ. Then we generate the noisy data gd:

gd ¼ g þ gmax � r randðsizeðgÞÞ; ð4:9Þ


r indicates the error level, gmax is the maximum value of sampled data g, RMS denotes the root mean square for a sampledfunction W which is defined by

RMSðWÞ ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1

N þ 1

XNþ1

j¼1

ðWðtjÞÞ2vuut ; ð4:10Þ

where N þ 1 is the total number of test points. According to (4.10) we can define the root mean square error (RMSE) for eachpair of the computed data and exact data. The symbol randðsizeð�ÞÞ is a random number uniformly distributed on ½0;1�.

In the numerical test, we take a1 ¼ 9; k1 ¼ 6; a2 ¼ 4; k2 ¼ 3; l1 ¼ 0:5; l2 ¼ 1; x ¼ 0; N ¼ 100. The numerical simulationis produced by the following steps:

� Step 1. Use discrete FFT to obtain the data in the frequency domain according (4.8), then perform inverse FFT to obtain thedata gðtÞ.� Step 2. According to (4.9), generate the noisy data gdðtÞ.� Step 3. Use discrete FFT to obtain the regularized solution with the noisy data in the frequency domain, according to (3.1)

and (3.2), then perform inverse FFT to obtain the data ud1;að0; tÞ and

@ud1;a@x ð0; tÞ.

Example 1. Let the exact data for the corresponding direct problem be given by

f ðtÞ ¼1; if 0:1 6 t 6 0:4 and 0:5 6 t 6 0:9;0; else:

�
This function belongs to HpðRÞ with p < 1=2.
First, we can computed the exact input data gðtÞ according to (4.8). The result is shown in Fig. 2.To show the effect of the proposed method, we compute the solution f ðtÞ :¼ u1ð0; tÞ according to the formula (2.20)

directly without any regularization. The result is displayed in Fig. 3 where r ¼ 1%.Now in order to show the convergence of the proposed regularization method, we computed the regularization solutions

by tuning the regularization parameter nmax gradually. The results are shown in Fig. 4 where r ¼ 1%.From Fig. 4, we can see that the important role of regularization parameter nmax is obvious for accurate

approximation. If the parameter nmax is too large or too small, the numerical approximation to the solution willcompletely fail.

We give two computation results to close this example.Fig. 5 shows the result for the reconstruction of temperature with r ¼ 1% and nmax ¼ 80.Fig. 6 shows the result for the reconstruction of flux with r ¼ 1% and nmax ¼ 200.Despite the discontinuities of the temperature in the form of square-wave step function and the heat flux in the form of

sharp corner, the numerical approximations are in reasonably match with the exact solutions of the temperature and theheat flux.

From the above figures, we can see that the proposed method works well too.

0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the

inpu

t dat

a g(

t)

Fig. 2. The exact input data gðtÞ.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the

solu

tion

u 1(0,t)

and

its

appr

oxim

atio

n

exact solutionno regularization

Fig. 3. The result without any regularization.

0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the s

olu

tion u

1(0

,t)

and it

s appro

xim

atio

n

exact solutionFourier regularization

0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the s

olu

tion u

1(0

,t)

and it

s appro

xim

atio

n


0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the s

olu

tion u

1(0

,t)

and it

s appro

xim

atio

n


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

t

the

solu

tion

u 1(0,t)

and

its

appr

oxim

atio

n


Fig. 4. (a): nmax ¼ 10 , (b): nmax ¼ 40, (c): nmax ¼ 70, (d) nmax ¼ 300.


0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the

solu

tion

u 1(0,t)

and

its

appr

oxim

atio

n


Fig. 5. Reconstruction of temperature with RMSE = 0.13.

0 0.2 0.4 0.6 0.8 1−4

−3

−2

−1

0

1

2

3

4

t

the

solu

tion

(u1) x(0

,t) a

nd it

s ap

prox

imat

ion


Fig. 6. Reconstruction of gradient of temperature with RMSE = 0.40.

0 0.2 0.4 0.6 0.8 1−0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t

the

inpu

t dat

a g(

t)

Fig. 7. the exact input data gðtÞ.


0 0.2 0.4 0.6 0.8 1−0.2

0

0.2

0.4

0.6

0.8

1

1.2

t

the

solu

tion

u 1(0,t)

and

its

appr

oxim

atio

n


Fig. 8. Reconstruction of temperature with RMSE = 0.02.

0 0.2 0.4 0.6 0.8 1−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

t

the

solu

tion

(u1) x(0

,t) a

nd it

s ap

prox

imat

ion


Fig. 9. Reconstruction of flux (gradient) of temperature of with RMSE = 0.04.


Example 2. Let the exact data be given by

f ðtÞ ¼

23 ð5t � 1Þ; if 0:2 6 t 6 0:5;� 2

3 ð5t � 4Þ; if 0:5 6 t 6 0:8;0; else:

8><>:
This function belongs to HpðRÞ with p < 3=2.
Without more explanation, we give the numerical results as follows:Fig. 7 shows the exact input data gðtÞ for the inverse problem.Fig. 8 shows the result for the reconstruction of temperature with r ¼ 3% and nmax ¼ 80.Fig. 9 shows the result for the flux(gradient) of temperature with r ¼ 3% and nmax ¼ 80.From these numerical results we conclude that the proposed method works well. In these examples we managed to solve

the inverse problem with acceptable accuracy. In the numerical tests, we found that the quality of the numerical solution isnot very sensitive to variations of the cutoff level within a suitable range. It is relatively easy to find an appropriate regular-ization parameter nmax, which is very important for practical computation.

5. Concluding Remarks

Although there are several regularization methods for stabilizing the inverse heat conduction problem in a single-layerbody by using an a priori information on the exact solution, the regularization error estimates for the inverse heat conduc-tion problem in a multi-layer body are still very rare. This is due to the complexity of the forward operators as shown in(2.23) and (2.24). Therefore, the direct extension of the existing methods for solving the IHCP in single-layer domain is


unavailable. However, the idea for stabilizing the IHCP in single-layer domain can be used. In this paper, we found that thespectral regularization method is efficient in solving the IHCP in two layer domain.

Spectral method is a simple and effective method and has been investigated by many authors for IHCP, e.g.[15][35]. Themethod can be implemented by the mollification method with some special mollifier in the physical domain (see [35]).Relative to the IHCP in single-layer domain, the IHCP in multi-layer domain is much more difficult. As we see that thecorresponding estimates on the forward operators are more difficult to be established. Fortunately, we can successfullyestimate the forward operators in the frequency domain(see Proposition 2.1 of this paper). Furthermore, we obtain the errorestimates for the spectral regularization method for solving the IHCP in two-layer domain. In theoretical aspect, the order ofthe error estimates is ’optimal’. The constructed numerical examples also verify that the proposed regularization method iseffective for solving the IHCP in the two-layer domain.

Acknowledgement

The work described in this paper was partially supported by a grant from the Research Council of the Hong Kong SpecialAdministrative Region, China (Project No. CityU 101112). The research of X.T. Xiong was partially supported by a grant fromthe National Natural Science Foundation of China (No. 11001223), the Research Fund for the Doctoral Program of HigherEducation of China (No. 20106203120001). The authors would like to thank the reviewers for their valuable commentsand suggestions, in particularly the reminder of the concept of ‘‘inverse crime’’.

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