Research ArticleA Numerical Approach to Solving an Inverse Heat ConductionProblem Using the Levenberg-Marquardt Algorithm

Tao Min12 Xing Chen1 Yao Sun1 and Qiang Huang2

1 School of Science Xirsquoan University of Technology Xirsquoan Shaanxi 710054 China2 State Key Laboratory of Eco-Hydraulic Engineering in Shaanxi Xirsquoan University of Technology Xirsquoan Shaanxi 710048 China

Correspondence should be addressed to Tao Min mintaoxauteducn

Received 14 April 2014 Accepted 8 June 2014 Published 24 June 2014

Academic Editor Hua Fan

Copyright copy 2014 Tao Min et alThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This paper is intended to provide a numerical algorithm involving the combined use of the Levenberg-Marquardt algorithm and theGalerkin finite elementmethod for estimating the diffusion coefficient in an inverse heat conduction problem (IHCP) In the presentstudy the functional form of the diffusion coefficient is unknown a priori The unknown diffusion coefficient is approximated bythe polynomial form and the present numerical algorithm is employed to find the solution Numerical experiments are presentedto show the efficiency of the proposed method

1 Introduction

The numerical solution of the inverse heat conduction prob-lem (IHCP) requires determining diffusion coefficient fromadditional information Inverse heat conduction problemshave many applications in various branches of science andengineering mechanical and chemical engineers mathe-maticians and specialists in many other science branches areinterested in inverse problems eachwith different applicationin mind [1ndash15]

In this work we propose an algorithm for numerical solv-ing of an inverse heat conduction problem The algorithm isbased on the Galerkin finite element method and Levenberg-Marquardt algorithm [16 17] in conjunction with the least-squares scheme It is assumed that no prior information isavailable on the functional form of the unknown diffusioncoefficient in the present study thus it is classified as thefunction estimation in inverse calculation Run the numericalalgorithm to solve the unknown diffusion coefficient whichis approximated by the polynomial form The Levenberg-Marquardt optimization is adopted to modify the estimatedvalues

The plan of this paper is as follows In Section 2 we for-mulate a one-dimensional IHCP In Section 3 the numericalalgorithm is derived Calculation of sensitivity coefficients

will be discussed in Section 4 In order to discuss somenumerical aspects two examples are given in Section 5Section 6 ends this paper with a brief discussion on somenumerical aspects

2 Description of the Problem

The mathematical formulation of a one-dimensional heatconduction problem is given as follows

120597119906

120597119905=120597

120597119909[119902 (119909)

120597119906

120597119909] + 119891 (119909 119905)

(119909 119905) isin (0 119871) times (0 119879]

(1)

with the initial condition

119906 (119909 0) = 1199060(119909) 0 le 119909 le 119871 (2)

and Dirichlet boundary conditions

119906 (0 119905) = 1198921(119905) 0 le 119905 le 119879

119906 (1 119905) = 1198922(119905) 0 le 119905 le 119879

(3)

where 119891(119909 119905) 1199060(119909) 119892

1(119905) 1198922(119905) and 119902(119909) are continuous

known functionsWe consider the problem (1)ndash(3) as a direct

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 626037 11 pageshttpdxdoiorg1011552014626037

2 Mathematical Problems in Engineering

problem As we all know if 1199060(119909) 119892

1(119905) 1198922(119905) are continuous

functions and 119902(119909) is known the problem (1)ndash(3) has a uniquesolution

For the inverse problem the diffusion coefficient 119902(119909)is regarded as being unknown In addition an overspeci-fied condition is also considered available To estimate theunknown coefficient 119902(119909) the additional information on theboundary 119909 = 119909

0 0 lt 119909

0lt 119871 is required Let the 119906(119909 119905)

taken at 119909 = 1199090over the time period [0 119879] be denoted by

119906 (1199090 119905) = 119892 (119905) 0 le 119905 le 119879 (4)

It is evident that for an unknown function 119902(119909) the problem(1)ndash(3) is underdetermined and we are forced to imposeadditional information (4) to provide a unique solution pair(119906(119909 119905) 119902(119909)) to the inverse problem (1)ndash(4)

We note that the measured overspecified condition119906(1199090 119905) = 119892(119905) should contain measurement errors There-

fore the inverse problem can be stated as follows by utilizingthe above-mentioned measured data estimate the unknownfunction 119902(119909)

In this work the polynomial form is proposed forthe unknown function 119902(119909) before performing the inversecalculation Therefore 119902(119909) is approximated as

119902 (119909) asympand

119902 (119909)

= 1199011+ 1199012119909 + 11990131199092+ sdot sdot sdot + 119901

119898+1119909119898

(5)

where 1199011 1199012 119901

119898+1are constants which remain to be

determined simultaneously The unknown coefficients1199011 1199012 119901

119898+1can be determined by using least-squares

method The error in the estimate

119865 (1199011 1199012 119901

119898+1)

=

119899

sum

119894=1

[119906 (1199090 119905119894 1199011 1199012 119901

119898+1) minus 119892 (119905

119894)]2

(6)

is to be minimized Here 119906(1199090 119905119894 1199011 1199012 119901

119898+1) are the

calculated results These quantities are determined from thesolution of the direct problem which is given previously by

using an approximatedand

119902(119909) for the exact 119902(119909) The estimatedvalues of 119901

119895 119895 = 1 2 119898 + 1 are determined until the

value of 119865(1199011 1199012 119901

119898+1) is minimum Such a norm can be

written as

119865 (P) = [U (P) minus G]119879 [U (P) minus G] (7)

whereP119879 = [1199011 1199012 119901

119898+1] denotes the vector of unknown

parameters and the superscript 119879 above denotes transposeThe vector [U(P) minus G]119879 is given by

[U (P) minus G]119879

= [119906 (1199090 1199051P) minus 119892 (119905

1) 119906 (119909

0 1199052P)

minus119892 (1199052) 119906 (119909

0 119905119899P) minus 119892 (119905

119899)]

(8)

119865(P) is real-valued bounded function defined on a closedbounded domain 119863 sub 119877

119898+1 The function 119865(P) may

have many local minima in 119863 but it has only one globalminimumWhen119865(P) and119863have some attractive propertiesfor instance 119865(P) being a differentiable concave function and119863 being a convex region then a local maximum problem canbe solved explicitly by mathematical programming methods

3 Overview of theLevenberg-Marquardt Method

The Levenberg-Marquardt method originally devised forapplication to nonlinear parameter estimation problems hasalso been successfully applied to the solution of linear ill-conditioned problems Such a method was first derived byLevenberg (1944) by modifying the ordinary least-squaresnorm Later Marquardt (1963) derived basically the sametechnique by using a different approach Marquardtrsquos inten-tion was to obtain a method that would tend to the Gaussmethod in the neighborhood of theminimumof the ordinaryleast-squares norm and would tend to the steepest descentmethod in the neighborhood of the initial guess used for theiterative procedure

To minimize the least-squares norm (7) we need toequate to zero the derivatives of 119865(P) with respect to each ofthe unknown parameters [119901

1 1199012 119901

119898+1] that is

120597119865 (P)1205971199011

=120597119865 (P)1205971199012

= sdot sdot sdot =120597119865 (P)120597119901119898+1

= 0 (9)

Let us introduce the sensitivity or Jacobianmatrix as follows

J (P) = [120597U119879(P)

120597P]

119879

=

[[[[

[

1199061199011

(1199090 1199051P) 119906

1199012

(1199090 1199051P) sdot sdot sdot 119906

119901119898+1

(1199090 1199051P)

1199061199011

(1199090 1199052P) 119906

1199012

(1199090 1199052P) sdot sdot sdot 119906

119901119898+1

(1199090 1199052P)

sdot sdot sdot

1199061199011

(1199090 119905119899P) 119906

1199012

(1199090 119905119899P) sdot sdot sdot 119906

119901119898+1

(1199090 119905119899P)

]]]]

]

(10)

or

119869119894119895= 119906119901119895

(1199090 119905119894P) =

120597119906 (1199090 119905119894P)

120597119901119895

119894 = 1 2 119899 119895 = 1 2 119898 + 1

(11)

The elements of the sensitivitymatrix are called the sensitivitycoefficients and the results of differentiation (9) can bewritten down as follows

minus2J119879 (P) [U (P) minus G] = 0 (12)

For linear inverse problem the sensitivity matrix is not afunction of the unknown parameters Equation (12) can besolved then in explicit form as follows

P = (J119879J)minus1

J119879G (13)

In the case of a nonlinear inverse problem the matrix J hassome functional dependence on the vector P The solution

Mathematical Problems in Engineering 3

of (12) requires an iterative procedure which is obtained bylinearizing the vector U(P) with a Taylor series expansionaround the current solution at iteration 119896 Such a linearizationis given by

U (P) = U (P119896) + J119896 (P minus P119896) (14)

where U(P119896) and J119896 are the estimated temperatures and thesensitivity matrix evaluated at iteration 119896 respectively Equa-tion (14) is substituted into (13) and the resulting expression isrearranged to yield the following iterative procedure to obtainthe vector of unknown parameters P

P119896+1 = P119896 + [(J119896)119879

J119896]minus1

(J119896)119879

[G minus U (P119896)] (15)

The iterative procedure given by (15) is called the Gaussmethod Such method is actually an approximation for theNewton (orNewton-Raphson)methodWe note that (13) andthe implementation of the iterative procedure given by (15)require the matrix J119879J to be nonsingular or

10038161003816100381610038161003816J119879J10038161003816100381610038161003816 = 0 (16)

where | sdot | is the determinantFormula (16) gives the so-called identifiability condition

that is if the determinant of J119879J is zero or even very small theparameters 119901

119895 for 119895 = 1 2 119898 + 1 cannot be determined

by using the iterative procedure of (15)Problems satisfying |J119879J| asymp 0 are denoted as ill-

conditioned Inverse heat transfer problems are generallyvery ill-conditioned especially near the initial guess usedfor the unknown parameters creating difficulties in theapplication of (13) or (15)The Levenberg-Marquardtmethodalleviates such difficulties by utilizing an iterative procedurein the form

P119896+1 = P119896 + [(J119896)119879

J119896 + 120583119896Ω119896]minus1

(J119896)119879

[G minus U (P119896)] (17)

where 120583119896 is a positive scalar named damping parameter andΩ119896 is a diagonal matrixThe purpose of the matrix term 120583

119896Ω119896 is to damp oscilla-

tions and instabilities due to the ill-conditioned character ofthe problem by making its components large as compared tothose of J119879J if necessary 120583119896 is made large in the beginning ofthe iterations since the problem is generally ill-conditionedin the region around the initial guess used for iterativeprocedure which can be quite far from the exact parametersWith such an approach the matrix J119879J is not required to benonsingular in the beginning of iterations and the Levenberg-Marquardt method tends to the steepest descent methodthat is a very small step is taken in the negative gradientdirection The parameter 120583119896 is then gradually reduced asthe iteration procedure advances to the solution of theparameter estimation problem and then the Levenberg-Marquardt method tends to the Gauss method given by (15)The following criteria were suggested in literature [13] to stop

the iterative procedure of the Levenberg-Marquardt methodgiven by (17)

119865 (P119896+1) lt 1205761 (18)

10038171003817100381710038171003817(J119896) [G minus U (P119896)]10038171003817100381710038171003817 lt 1205762 (19)10038171003817100381710038171003817P119896+1 minus P11989610038171003817100381710038171003817 lt 1205763 (20)

where 1205761 1205762 and 120576

3are user prescribed tolerances and sdot

denotes the Euclidean norm The criterion given by (18)tests if the least-squares norm is sufficiently small which isexpected in the neighborhood of the solution for the problemSimilarly (19) checks if the norm of the gradient of 119865(P) issufficiently small since it is expected to vanish at the pointwhere 119865(P) is minimum The last criterion given by (20)results from the fact that changes in the vector of parametersare very small when the method has converged Generallythese three stopping criteria need to be tested and the iterativeprocedure of the Levenberg-Marquardt method is stopped ifany of them is satisfied

Different versions of the Levenberg-Marquardt methodcan be found in the literature depending on the choice of thediagonal matrixΩ119896 and on the form chosen for the variationof the damping parameter 120583119896 In this paper we chooseΩ119896 as

Ω119896= diag [(J119896)

119879

J119896] (21)

Suppose that the vectors of temperature measurements G =

[119892(1199051) 119892(1199052) 119892(119905

119899)] are given at times 119905

119894 119894 = 1 2 119899

and an initial guess P0 is available for the vector of unknownparameters P Choose a value for 1205830 say 1205830 = 0001 and 119896 =0 Then consider the following

Step 1 Solve the direct problem (1)ndash(3) with the avail-able estimate P119896 in order to obtain the vector U(P119896) =

[119906(1199090 1199051P119896) 119906(119909

0 1199052P119896) 119906(119909

0 119905119899P119896)]

Step 2 Compute 119865(P119896) from (7)

Step 3 Compute the sensitivity matrix J119896 from (11) and thenthe matrixΩ119896 from (21) by using the current value of P119896

Step 4 Solve the following linear system of algebraic equa-tions obtained from (17)

[(J119896)119879J119896 + 120583119896Ω119896]ΔP119896 = (J119896)119879[G minus U(P119896)] in order tocompute ΔP119896 = P119896+1 minus P119896

Step 5 Compute the new estimate P119896+1 as P119896+1 = P119896 + ΔP119896

Step 6 Solve the exact problem (1)ndash(3) with the new estimateP119896+1 in order to find 119880(P119896+1) Then compute 119865(P119896+1)

Step 7 If 119865(P119896+1) ge 119865(P119896) replace 120583119896 by 10120583119896 and return toStep 4

Step 8 If 119865(P119896+1) le 119865(P119896) accept the new estimate P119896+1 andemplace 120583119896 by 01120583119896

4 Mathematical Problems in Engineering

Table 1 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 05 05 05 1 1 1 10 10 10 50 50 50

Iteration 120999729028233135 0999729028233183 0999729028233194 09997290283072610499885876453067 0499885876453056 0499885876453057 04998858764541690252009862457275 0252009862457315 0252009862457325 025200986249336

Error 119865 87564944405 times 10minus14

87564944427 times 10minus14

87564944420 times 10minus14

87564944420 times 10minus14

0 2 4 6 8 10 120

1

2

3

4

5

6 Convergence plot

Number of iterations

Res

idua

l F(p)

times10minus4

Figure 1 All the initial values for the parameters are set to 05

Step 9 Check the stopping criteria given by (18) Stop theiterative procedure if any of them is satisfied otherwisereplace 119896 by 119896 + 1 and return to Step 3

4 Calculation of Sensitivity Coefficients

Generally there have been two approaches for determin-ing the gradient the first is a discretize-then-differentiateapproach and the second is a differentiate-then-discretizeapproach

The first approach is to approximate the gradient of thefunctional by a finite difference quotient approximation butin general we cannot determine the sensitivities exactly sothis method may lead to larger error

Here we intend to use differentiate-then-discretizeapproach which we refer to as the sensitivity equationmethod This method can be determined more efficientlywith the help of the sensitivities

119906119896=120597119906

120597119901119896

119896 = 1 2 119898 + 1 (22)

We first differentiate the flow system (1)ndash(3) with respectto each of the design parameters [119901

1 1199012 119901

119898+1] to obtain

the119898+1 continuous sensitivity systems for 119896 = 1 2 119898+1

120597119906119896

120597119905=120597

120597119909[(1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119896

120597119909+ 119909119896minus1 120597119906

120597119909]

119906119896(119909 0) = 0

Res

idua

l F(p)

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9Convergence plot

Number of iterations

times10minus5

Figure 2 All the initial values for the parameters are set to 1

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25Convergence plot

Number of iterations

times10minus3

Figure 3 All the initial values for the parameters are set to 10

119906119896(0 119905) = 0

119906119896(119871 119905) = 0

(23)

There are (119898 + 2) equations we can make them in onesystem equation and use the finite element methods to solve

Mathematical Problems in Engineering 5

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25

3Convergence plot

Number of iterations

times10minus3

Figure 4 All the initial values for the parameters are set to 50

the system of equation Here we give the vector form of theequation as follows

(P1)

120597U120597119905

+ nabla sdot Γ = FU (119909 0) = U

0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(24)

where

U =

[[[[[[

[

119906

1199061

1199062

119906119898+1

]]]]]]

]

Γ =

[[[[[[[[[[[[[

[

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909minus (

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909minus 119909(

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909minus (119909119898 120597119906

120597119909)

]]]]]]]]]]]]]

]

F =[[[[[[

[

119891 (119909 119905)

0

0

0

]]]]]]

]

U0(119909) = [119906

0(119909) 0 0 0]

119879

G1(119905) = [119892

1(119905) 0 0 0]

119879

G2(119905) = [119892

2(119905) 0 0 0]

119879

(25)

We use the Galerkin finite element method approxima-tion for discretizing problem (24) For this we multiply (24)by a test function V [0 119871] rarr 119877 V isin 119881

0= 119867

1

0(0 119871)

and integrate the obtained equation in space form 0 to 119871 Weobtain the following equation

int

119871

0

120597U (119909 119905)120597119905

sdot V (119909) 119889119909 minus int119871

0

nablaΓ sdot V (119909) 119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(26)

integrating by parts gives

int

119871

0

nablaΓ sdot V (119909) 119889119909 = (Γ sdot V (119909))|1198710minus int

119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 (27)

We can change the first derivative in time and the integralWe have V(0) = 0 = V(119871) because V isin 119881

0 This leads to an

equivalent problem (P1) forall119905 gt 0 find U(119909 119905) satisfying

119889

119889119905int

119871

0

U (119909 119905) sdot V (119909) 119889119909 + int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(28)

for all V isin 1198810= 1198671

0(0 119871) To simplify the notation we use the

scalar product in 1198712(0 119871)

(119891 119892) = int

119871

0

119891 (119909) sdot 119892 (119909) 119889119909 (29)

We also can define the following bilinear form

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

problem As we all know if 1199060(119909) 119892

1(119905) 1198922(119905) are continuous

functions and 119902(119909) is known the problem (1)ndash(3) has a uniquesolution

For the inverse problem the diffusion coefficient 119902(119909)is regarded as being unknown In addition an overspeci-fied condition is also considered available To estimate theunknown coefficient 119902(119909) the additional information on theboundary 119909 = 119909

0 0 lt 119909

0lt 119871 is required Let the 119906(119909 119905)

taken at 119909 = 1199090over the time period [0 119879] be denoted by

119906 (1199090 119905) = 119892 (119905) 0 le 119905 le 119879 (4)

It is evident that for an unknown function 119902(119909) the problem(1)ndash(3) is underdetermined and we are forced to imposeadditional information (4) to provide a unique solution pair(119906(119909 119905) 119902(119909)) to the inverse problem (1)ndash(4)

We note that the measured overspecified condition119906(1199090 119905) = 119892(119905) should contain measurement errors There-

fore the inverse problem can be stated as follows by utilizingthe above-mentioned measured data estimate the unknownfunction 119902(119909)

In this work the polynomial form is proposed forthe unknown function 119902(119909) before performing the inversecalculation Therefore 119902(119909) is approximated as

119902 (119909) asympand

119902 (119909)

= 1199011+ 1199012119909 + 11990131199092+ sdot sdot sdot + 119901

119898+1119909119898

(5)

where 1199011 1199012 119901

119898+1are constants which remain to be

determined simultaneously The unknown coefficients1199011 1199012 119901

119898+1can be determined by using least-squares

method The error in the estimate

119865 (1199011 1199012 119901

119898+1)

=

119899

sum

119894=1

[119906 (1199090 119905119894 1199011 1199012 119901

119898+1) minus 119892 (119905

119894)]2

(6)

is to be minimized Here 119906(1199090 119905119894 1199011 1199012 119901

119898+1) are the

calculated results These quantities are determined from thesolution of the direct problem which is given previously by

using an approximatedand

119902(119909) for the exact 119902(119909) The estimatedvalues of 119901

119895 119895 = 1 2 119898 + 1 are determined until the

value of 119865(1199011 1199012 119901

119898+1) is minimum Such a norm can be

written as

119865 (P) = [U (P) minus G]119879 [U (P) minus G] (7)

whereP119879 = [1199011 1199012 119901

119898+1] denotes the vector of unknown

parameters and the superscript 119879 above denotes transposeThe vector [U(P) minus G]119879 is given by

[U (P) minus G]119879

= [119906 (1199090 1199051P) minus 119892 (119905

1) 119906 (119909

0 1199052P)

minus119892 (1199052) 119906 (119909

0 119905119899P) minus 119892 (119905

119899)]

(8)

119865(P) is real-valued bounded function defined on a closedbounded domain 119863 sub 119877

119898+1 The function 119865(P) may

have many local minima in 119863 but it has only one globalminimumWhen119865(P) and119863have some attractive propertiesfor instance 119865(P) being a differentiable concave function and119863 being a convex region then a local maximum problem canbe solved explicitly by mathematical programming methods

3 Overview of theLevenberg-Marquardt Method

The Levenberg-Marquardt method originally devised forapplication to nonlinear parameter estimation problems hasalso been successfully applied to the solution of linear ill-conditioned problems Such a method was first derived byLevenberg (1944) by modifying the ordinary least-squaresnorm Later Marquardt (1963) derived basically the sametechnique by using a different approach Marquardtrsquos inten-tion was to obtain a method that would tend to the Gaussmethod in the neighborhood of theminimumof the ordinaryleast-squares norm and would tend to the steepest descentmethod in the neighborhood of the initial guess used for theiterative procedure

To minimize the least-squares norm (7) we need toequate to zero the derivatives of 119865(P) with respect to each ofthe unknown parameters [119901

1 1199012 119901

119898+1] that is

120597119865 (P)1205971199011

=120597119865 (P)1205971199012

= sdot sdot sdot =120597119865 (P)120597119901119898+1

= 0 (9)

Let us introduce the sensitivity or Jacobianmatrix as follows

J (P) = [120597U119879(P)

120597P]

119879

=

[[[[

[

1199061199011

(1199090 1199051P) 119906

1199012

(1199090 1199051P) sdot sdot sdot 119906

119901119898+1

(1199090 1199051P)

1199061199011

(1199090 1199052P) 119906

1199012

(1199090 1199052P) sdot sdot sdot 119906

119901119898+1

(1199090 1199052P)

sdot sdot sdot

1199061199011

(1199090 119905119899P) 119906

1199012

(1199090 119905119899P) sdot sdot sdot 119906

119901119898+1

(1199090 119905119899P)

]]]]

]

(10)

or

119869119894119895= 119906119901119895

(1199090 119905119894P) =

120597119906 (1199090 119905119894P)

120597119901119895

119894 = 1 2 119899 119895 = 1 2 119898 + 1

(11)

The elements of the sensitivitymatrix are called the sensitivitycoefficients and the results of differentiation (9) can bewritten down as follows

minus2J119879 (P) [U (P) minus G] = 0 (12)

For linear inverse problem the sensitivity matrix is not afunction of the unknown parameters Equation (12) can besolved then in explicit form as follows

P = (J119879J)minus1

J119879G (13)

In the case of a nonlinear inverse problem the matrix J hassome functional dependence on the vector P The solution

Mathematical Problems in Engineering 3

of (12) requires an iterative procedure which is obtained bylinearizing the vector U(P) with a Taylor series expansionaround the current solution at iteration 119896 Such a linearizationis given by

U (P) = U (P119896) + J119896 (P minus P119896) (14)

where U(P119896) and J119896 are the estimated temperatures and thesensitivity matrix evaluated at iteration 119896 respectively Equa-tion (14) is substituted into (13) and the resulting expression isrearranged to yield the following iterative procedure to obtainthe vector of unknown parameters P

P119896+1 = P119896 + [(J119896)119879

J119896]minus1

(J119896)119879

[G minus U (P119896)] (15)

The iterative procedure given by (15) is called the Gaussmethod Such method is actually an approximation for theNewton (orNewton-Raphson)methodWe note that (13) andthe implementation of the iterative procedure given by (15)require the matrix J119879J to be nonsingular or

10038161003816100381610038161003816J119879J10038161003816100381610038161003816 = 0 (16)

where | sdot | is the determinantFormula (16) gives the so-called identifiability condition

that is if the determinant of J119879J is zero or even very small theparameters 119901

119895 for 119895 = 1 2 119898 + 1 cannot be determined

by using the iterative procedure of (15)Problems satisfying |J119879J| asymp 0 are denoted as ill-

conditioned Inverse heat transfer problems are generallyvery ill-conditioned especially near the initial guess usedfor the unknown parameters creating difficulties in theapplication of (13) or (15)The Levenberg-Marquardtmethodalleviates such difficulties by utilizing an iterative procedurein the form

P119896+1 = P119896 + [(J119896)119879

J119896 + 120583119896Ω119896]minus1

(J119896)119879

[G minus U (P119896)] (17)

where 120583119896 is a positive scalar named damping parameter andΩ119896 is a diagonal matrixThe purpose of the matrix term 120583

119896Ω119896 is to damp oscilla-

tions and instabilities due to the ill-conditioned character ofthe problem by making its components large as compared tothose of J119879J if necessary 120583119896 is made large in the beginning ofthe iterations since the problem is generally ill-conditionedin the region around the initial guess used for iterativeprocedure which can be quite far from the exact parametersWith such an approach the matrix J119879J is not required to benonsingular in the beginning of iterations and the Levenberg-Marquardt method tends to the steepest descent methodthat is a very small step is taken in the negative gradientdirection The parameter 120583119896 is then gradually reduced asthe iteration procedure advances to the solution of theparameter estimation problem and then the Levenberg-Marquardt method tends to the Gauss method given by (15)The following criteria were suggested in literature [13] to stop

the iterative procedure of the Levenberg-Marquardt methodgiven by (17)

119865 (P119896+1) lt 1205761 (18)

10038171003817100381710038171003817(J119896) [G minus U (P119896)]10038171003817100381710038171003817 lt 1205762 (19)10038171003817100381710038171003817P119896+1 minus P11989610038171003817100381710038171003817 lt 1205763 (20)

where 1205761 1205762 and 120576

3are user prescribed tolerances and sdot

denotes the Euclidean norm The criterion given by (18)tests if the least-squares norm is sufficiently small which isexpected in the neighborhood of the solution for the problemSimilarly (19) checks if the norm of the gradient of 119865(P) issufficiently small since it is expected to vanish at the pointwhere 119865(P) is minimum The last criterion given by (20)results from the fact that changes in the vector of parametersare very small when the method has converged Generallythese three stopping criteria need to be tested and the iterativeprocedure of the Levenberg-Marquardt method is stopped ifany of them is satisfied

Different versions of the Levenberg-Marquardt methodcan be found in the literature depending on the choice of thediagonal matrixΩ119896 and on the form chosen for the variationof the damping parameter 120583119896 In this paper we chooseΩ119896 as

Ω119896= diag [(J119896)

119879

J119896] (21)

Suppose that the vectors of temperature measurements G =

[119892(1199051) 119892(1199052) 119892(119905

119899)] are given at times 119905

119894 119894 = 1 2 119899

and an initial guess P0 is available for the vector of unknownparameters P Choose a value for 1205830 say 1205830 = 0001 and 119896 =0 Then consider the following

Step 1 Solve the direct problem (1)ndash(3) with the avail-able estimate P119896 in order to obtain the vector U(P119896) =

[119906(1199090 1199051P119896) 119906(119909

0 1199052P119896) 119906(119909

0 119905119899P119896)]

Step 2 Compute 119865(P119896) from (7)

Step 3 Compute the sensitivity matrix J119896 from (11) and thenthe matrixΩ119896 from (21) by using the current value of P119896

Step 4 Solve the following linear system of algebraic equa-tions obtained from (17)

[(J119896)119879J119896 + 120583119896Ω119896]ΔP119896 = (J119896)119879[G minus U(P119896)] in order tocompute ΔP119896 = P119896+1 minus P119896

Step 5 Compute the new estimate P119896+1 as P119896+1 = P119896 + ΔP119896

Step 6 Solve the exact problem (1)ndash(3) with the new estimateP119896+1 in order to find 119880(P119896+1) Then compute 119865(P119896+1)

Step 7 If 119865(P119896+1) ge 119865(P119896) replace 120583119896 by 10120583119896 and return toStep 4

Step 8 If 119865(P119896+1) le 119865(P119896) accept the new estimate P119896+1 andemplace 120583119896 by 01120583119896

4 Mathematical Problems in Engineering

Table 1 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 05 05 05 1 1 1 10 10 10 50 50 50

Iteration 120999729028233135 0999729028233183 0999729028233194 09997290283072610499885876453067 0499885876453056 0499885876453057 04998858764541690252009862457275 0252009862457315 0252009862457325 025200986249336

Error 119865 87564944405 times 10minus14

87564944427 times 10minus14

87564944420 times 10minus14

87564944420 times 10minus14

0 2 4 6 8 10 120

1

2

3

4

5

6 Convergence plot

Number of iterations

Res

idua

l F(p)

times10minus4

Figure 1 All the initial values for the parameters are set to 05

Step 9 Check the stopping criteria given by (18) Stop theiterative procedure if any of them is satisfied otherwisereplace 119896 by 119896 + 1 and return to Step 3

4 Calculation of Sensitivity Coefficients

Generally there have been two approaches for determin-ing the gradient the first is a discretize-then-differentiateapproach and the second is a differentiate-then-discretizeapproach

The first approach is to approximate the gradient of thefunctional by a finite difference quotient approximation butin general we cannot determine the sensitivities exactly sothis method may lead to larger error

Here we intend to use differentiate-then-discretizeapproach which we refer to as the sensitivity equationmethod This method can be determined more efficientlywith the help of the sensitivities

119906119896=120597119906

120597119901119896

119896 = 1 2 119898 + 1 (22)

We first differentiate the flow system (1)ndash(3) with respectto each of the design parameters [119901

1 1199012 119901

119898+1] to obtain

the119898+1 continuous sensitivity systems for 119896 = 1 2 119898+1

120597119906119896

120597119905=120597

120597119909[(1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119896

120597119909+ 119909119896minus1 120597119906

120597119909]

119906119896(119909 0) = 0

Res

idua

l F(p)

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9Convergence plot

Number of iterations

times10minus5

Figure 2 All the initial values for the parameters are set to 1

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25Convergence plot

Number of iterations

times10minus3

Figure 3 All the initial values for the parameters are set to 10

119906119896(0 119905) = 0

119906119896(119871 119905) = 0

(23)

There are (119898 + 2) equations we can make them in onesystem equation and use the finite element methods to solve

Mathematical Problems in Engineering 5

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25

3Convergence plot

Number of iterations

times10minus3

Figure 4 All the initial values for the parameters are set to 50

the system of equation Here we give the vector form of theequation as follows

(P1)

120597U120597119905

+ nabla sdot Γ = FU (119909 0) = U

0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(24)

where

U =

[[[[[[

[

119906

1199061

1199062

119906119898+1

]]]]]]

]

Γ =

[[[[[[[[[[[[[

[

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909minus (

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909minus 119909(

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909minus (119909119898 120597119906

120597119909)

]]]]]]]]]]]]]

]

F =[[[[[[

[

119891 (119909 119905)

0

0

0

]]]]]]

]

U0(119909) = [119906

0(119909) 0 0 0]

119879

G1(119905) = [119892

1(119905) 0 0 0]

119879

G2(119905) = [119892

2(119905) 0 0 0]

119879

(25)

We use the Galerkin finite element method approxima-tion for discretizing problem (24) For this we multiply (24)by a test function V [0 119871] rarr 119877 V isin 119881

0= 119867

1

0(0 119871)

and integrate the obtained equation in space form 0 to 119871 Weobtain the following equation

int

119871

0

120597U (119909 119905)120597119905

sdot V (119909) 119889119909 minus int119871

0

nablaΓ sdot V (119909) 119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(26)

integrating by parts gives

int

119871

0

nablaΓ sdot V (119909) 119889119909 = (Γ sdot V (119909))|1198710minus int

119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 (27)

We can change the first derivative in time and the integralWe have V(0) = 0 = V(119871) because V isin 119881

0 This leads to an

equivalent problem (P1) forall119905 gt 0 find U(119909 119905) satisfying

119889

119889119905int

119871

0

U (119909 119905) sdot V (119909) 119889119909 + int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(28)

for all V isin 1198810= 1198671

0(0 119871) To simplify the notation we use the

scalar product in 1198712(0 119871)

(119891 119892) = int

119871

0

119891 (119909) sdot 119892 (119909) 119889119909 (29)

We also can define the following bilinear form

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

of (12) requires an iterative procedure which is obtained bylinearizing the vector U(P) with a Taylor series expansionaround the current solution at iteration 119896 Such a linearizationis given by

U (P) = U (P119896) + J119896 (P minus P119896) (14)

where U(P119896) and J119896 are the estimated temperatures and thesensitivity matrix evaluated at iteration 119896 respectively Equa-tion (14) is substituted into (13) and the resulting expression isrearranged to yield the following iterative procedure to obtainthe vector of unknown parameters P

P119896+1 = P119896 + [(J119896)119879

J119896]minus1

(J119896)119879

[G minus U (P119896)] (15)

The iterative procedure given by (15) is called the Gaussmethod Such method is actually an approximation for theNewton (orNewton-Raphson)methodWe note that (13) andthe implementation of the iterative procedure given by (15)require the matrix J119879J to be nonsingular or

10038161003816100381610038161003816J119879J10038161003816100381610038161003816 = 0 (16)

where | sdot | is the determinantFormula (16) gives the so-called identifiability condition

that is if the determinant of J119879J is zero or even very small theparameters 119901

119895 for 119895 = 1 2 119898 + 1 cannot be determined

by using the iterative procedure of (15)Problems satisfying |J119879J| asymp 0 are denoted as ill-

conditioned Inverse heat transfer problems are generallyvery ill-conditioned especially near the initial guess usedfor the unknown parameters creating difficulties in theapplication of (13) or (15)The Levenberg-Marquardtmethodalleviates such difficulties by utilizing an iterative procedurein the form

P119896+1 = P119896 + [(J119896)119879

J119896 + 120583119896Ω119896]minus1

(J119896)119879

[G minus U (P119896)] (17)

where 120583119896 is a positive scalar named damping parameter andΩ119896 is a diagonal matrixThe purpose of the matrix term 120583

119896Ω119896 is to damp oscilla-

tions and instabilities due to the ill-conditioned character ofthe problem by making its components large as compared tothose of J119879J if necessary 120583119896 is made large in the beginning ofthe iterations since the problem is generally ill-conditionedin the region around the initial guess used for iterativeprocedure which can be quite far from the exact parametersWith such an approach the matrix J119879J is not required to benonsingular in the beginning of iterations and the Levenberg-Marquardt method tends to the steepest descent methodthat is a very small step is taken in the negative gradientdirection The parameter 120583119896 is then gradually reduced asthe iteration procedure advances to the solution of theparameter estimation problem and then the Levenberg-Marquardt method tends to the Gauss method given by (15)The following criteria were suggested in literature [13] to stop

the iterative procedure of the Levenberg-Marquardt methodgiven by (17)

119865 (P119896+1) lt 1205761 (18)

10038171003817100381710038171003817(J119896) [G minus U (P119896)]10038171003817100381710038171003817 lt 1205762 (19)10038171003817100381710038171003817P119896+1 minus P11989610038171003817100381710038171003817 lt 1205763 (20)

where 1205761 1205762 and 120576

3are user prescribed tolerances and sdot

denotes the Euclidean norm The criterion given by (18)tests if the least-squares norm is sufficiently small which isexpected in the neighborhood of the solution for the problemSimilarly (19) checks if the norm of the gradient of 119865(P) issufficiently small since it is expected to vanish at the pointwhere 119865(P) is minimum The last criterion given by (20)results from the fact that changes in the vector of parametersare very small when the method has converged Generallythese three stopping criteria need to be tested and the iterativeprocedure of the Levenberg-Marquardt method is stopped ifany of them is satisfied

Different versions of the Levenberg-Marquardt methodcan be found in the literature depending on the choice of thediagonal matrixΩ119896 and on the form chosen for the variationof the damping parameter 120583119896 In this paper we chooseΩ119896 as

Ω119896= diag [(J119896)

119879

J119896] (21)

Suppose that the vectors of temperature measurements G =

[119892(1199051) 119892(1199052) 119892(119905

119899)] are given at times 119905

119894 119894 = 1 2 119899

and an initial guess P0 is available for the vector of unknownparameters P Choose a value for 1205830 say 1205830 = 0001 and 119896 =0 Then consider the following

Step 1 Solve the direct problem (1)ndash(3) with the avail-able estimate P119896 in order to obtain the vector U(P119896) =

[119906(1199090 1199051P119896) 119906(119909

0 1199052P119896) 119906(119909

0 119905119899P119896)]

Step 2 Compute 119865(P119896) from (7)

Step 3 Compute the sensitivity matrix J119896 from (11) and thenthe matrixΩ119896 from (21) by using the current value of P119896

Step 4 Solve the following linear system of algebraic equa-tions obtained from (17)

[(J119896)119879J119896 + 120583119896Ω119896]ΔP119896 = (J119896)119879[G minus U(P119896)] in order tocompute ΔP119896 = P119896+1 minus P119896

Step 5 Compute the new estimate P119896+1 as P119896+1 = P119896 + ΔP119896

Step 6 Solve the exact problem (1)ndash(3) with the new estimateP119896+1 in order to find 119880(P119896+1) Then compute 119865(P119896+1)

Step 7 If 119865(P119896+1) ge 119865(P119896) replace 120583119896 by 10120583119896 and return toStep 4

Step 8 If 119865(P119896+1) le 119865(P119896) accept the new estimate P119896+1 andemplace 120583119896 by 01120583119896

4 Mathematical Problems in Engineering

Table 1 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 05 05 05 1 1 1 10 10 10 50 50 50

Iteration 120999729028233135 0999729028233183 0999729028233194 09997290283072610499885876453067 0499885876453056 0499885876453057 04998858764541690252009862457275 0252009862457315 0252009862457325 025200986249336

Error 119865 87564944405 times 10minus14

87564944427 times 10minus14

87564944420 times 10minus14

87564944420 times 10minus14

0 2 4 6 8 10 120

1

2

3

4

5

6 Convergence plot

Number of iterations

Res

idua

l F(p)

times10minus4

Figure 1 All the initial values for the parameters are set to 05

Step 9 Check the stopping criteria given by (18) Stop theiterative procedure if any of them is satisfied otherwisereplace 119896 by 119896 + 1 and return to Step 3

4 Calculation of Sensitivity Coefficients

Generally there have been two approaches for determin-ing the gradient the first is a discretize-then-differentiateapproach and the second is a differentiate-then-discretizeapproach

The first approach is to approximate the gradient of thefunctional by a finite difference quotient approximation butin general we cannot determine the sensitivities exactly sothis method may lead to larger error

Here we intend to use differentiate-then-discretizeapproach which we refer to as the sensitivity equationmethod This method can be determined more efficientlywith the help of the sensitivities

119906119896=120597119906

120597119901119896

119896 = 1 2 119898 + 1 (22)

We first differentiate the flow system (1)ndash(3) with respectto each of the design parameters [119901

1 1199012 119901

119898+1] to obtain

the119898+1 continuous sensitivity systems for 119896 = 1 2 119898+1

120597119906119896

120597119905=120597

120597119909[(1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119896

120597119909+ 119909119896minus1 120597119906

120597119909]

119906119896(119909 0) = 0

Res

idua

l F(p)

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9Convergence plot

Number of iterations

times10minus5

Figure 2 All the initial values for the parameters are set to 1

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25Convergence plot

Number of iterations

times10minus3

Figure 3 All the initial values for the parameters are set to 10

119906119896(0 119905) = 0

119906119896(119871 119905) = 0

(23)

There are (119898 + 2) equations we can make them in onesystem equation and use the finite element methods to solve

Mathematical Problems in Engineering 5

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25

3Convergence plot

Number of iterations

times10minus3

Figure 4 All the initial values for the parameters are set to 50

the system of equation Here we give the vector form of theequation as follows

(P1)

120597U120597119905

+ nabla sdot Γ = FU (119909 0) = U

0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(24)

where

U =

[[[[[[

[

119906

1199061

1199062

119906119898+1

]]]]]]

]

Γ =

[[[[[[[[[[[[[

[

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909minus (

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909minus 119909(

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909minus (119909119898 120597119906

120597119909)

]]]]]]]]]]]]]

]

F =[[[[[[

[

119891 (119909 119905)

0

0

0

]]]]]]

]

U0(119909) = [119906

0(119909) 0 0 0]

119879

G1(119905) = [119892

1(119905) 0 0 0]

119879

G2(119905) = [119892

2(119905) 0 0 0]

119879

(25)

We use the Galerkin finite element method approxima-tion for discretizing problem (24) For this we multiply (24)by a test function V [0 119871] rarr 119877 V isin 119881

0= 119867

1

0(0 119871)

and integrate the obtained equation in space form 0 to 119871 Weobtain the following equation

int

119871

0

120597U (119909 119905)120597119905

sdot V (119909) 119889119909 minus int119871

0

nablaΓ sdot V (119909) 119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(26)

integrating by parts gives

int

119871

0

nablaΓ sdot V (119909) 119889119909 = (Γ sdot V (119909))|1198710minus int

119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 (27)

We can change the first derivative in time and the integralWe have V(0) = 0 = V(119871) because V isin 119881

0 This leads to an

equivalent problem (P1) forall119905 gt 0 find U(119909 119905) satisfying

119889

119889119905int

119871

0

U (119909 119905) sdot V (119909) 119889119909 + int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(28)

for all V isin 1198810= 1198671

0(0 119871) To simplify the notation we use the

scalar product in 1198712(0 119871)

(119891 119892) = int

119871

0

119891 (119909) sdot 119892 (119909) 119889119909 (29)

We also can define the following bilinear form

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

Table 1 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 05 05 05 1 1 1 10 10 10 50 50 50

Iteration 120999729028233135 0999729028233183 0999729028233194 09997290283072610499885876453067 0499885876453056 0499885876453057 04998858764541690252009862457275 0252009862457315 0252009862457325 025200986249336

Error 119865 87564944405 times 10minus14

87564944427 times 10minus14

87564944420 times 10minus14

87564944420 times 10minus14

0 2 4 6 8 10 120

1

2

3

4

5

6 Convergence plot

Number of iterations

Res

idua

l F(p)

times10minus4

Figure 1 All the initial values for the parameters are set to 05

Step 9 Check the stopping criteria given by (18) Stop theiterative procedure if any of them is satisfied otherwisereplace 119896 by 119896 + 1 and return to Step 3

4 Calculation of Sensitivity Coefficients

Generally there have been two approaches for determin-ing the gradient the first is a discretize-then-differentiateapproach and the second is a differentiate-then-discretizeapproach

The first approach is to approximate the gradient of thefunctional by a finite difference quotient approximation butin general we cannot determine the sensitivities exactly sothis method may lead to larger error

Here we intend to use differentiate-then-discretizeapproach which we refer to as the sensitivity equationmethod This method can be determined more efficientlywith the help of the sensitivities

119906119896=120597119906

120597119901119896

119896 = 1 2 119898 + 1 (22)

We first differentiate the flow system (1)ndash(3) with respectto each of the design parameters [119901

1 1199012 119901

119898+1] to obtain

the119898+1 continuous sensitivity systems for 119896 = 1 2 119898+1

120597119906119896

120597119905=120597

120597119909[(1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119896

120597119909+ 119909119896minus1 120597119906

120597119909]

119906119896(119909 0) = 0

Res

idua

l F(p)

0 2 4 6 8 10 120

1

2

3

4

5

6

7

8

9Convergence plot

Number of iterations

times10minus5

Figure 2 All the initial values for the parameters are set to 1

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25Convergence plot

Number of iterations

times10minus3

Figure 3 All the initial values for the parameters are set to 10

119906119896(0 119905) = 0

119906119896(119871 119905) = 0

(23)

There are (119898 + 2) equations we can make them in onesystem equation and use the finite element methods to solve

Mathematical Problems in Engineering 5

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25

3Convergence plot

Number of iterations

times10minus3

Figure 4 All the initial values for the parameters are set to 50

the system of equation Here we give the vector form of theequation as follows

(P1)

120597U120597119905

+ nabla sdot Γ = FU (119909 0) = U

0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(24)

where

U =

[[[[[[

[

119906

1199061

1199062

119906119898+1

]]]]]]

]

Γ =

[[[[[[[[[[[[[

[

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909minus (

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909minus 119909(

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909minus (119909119898 120597119906

120597119909)

]]]]]]]]]]]]]

]

F =[[[[[[

[

119891 (119909 119905)

0

0

0

]]]]]]

]

U0(119909) = [119906

0(119909) 0 0 0]

119879

G1(119905) = [119892

1(119905) 0 0 0]

119879

G2(119905) = [119892

2(119905) 0 0 0]

119879

(25)

We use the Galerkin finite element method approxima-tion for discretizing problem (24) For this we multiply (24)by a test function V [0 119871] rarr 119877 V isin 119881

0= 119867

1

0(0 119871)

and integrate the obtained equation in space form 0 to 119871 Weobtain the following equation

int

119871

0

120597U (119909 119905)120597119905

sdot V (119909) 119889119909 minus int119871

0

nablaΓ sdot V (119909) 119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(26)

integrating by parts gives

int

119871

0

nablaΓ sdot V (119909) 119889119909 = (Γ sdot V (119909))|1198710minus int

119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 (27)

We can change the first derivative in time and the integralWe have V(0) = 0 = V(119871) because V isin 119881

0 This leads to an

equivalent problem (P1) forall119905 gt 0 find U(119909 119905) satisfying

119889

119889119905int

119871

0

U (119909 119905) sdot V (119909) 119889119909 + int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(28)

for all V isin 1198810= 1198671

0(0 119871) To simplify the notation we use the

scalar product in 1198712(0 119871)

(119891 119892) = int

119871

0

119891 (119909) sdot 119892 (119909) 119889119909 (29)

We also can define the following bilinear form

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

Res

idua

l F(p)

0 2 4 6 8 10 120

05

1

15

2

25

3Convergence plot

Number of iterations

times10minus3

Figure 4 All the initial values for the parameters are set to 50

the system of equation Here we give the vector form of theequation as follows

(P1)

120597U120597119905

+ nabla sdot Γ = FU (119909 0) = U

0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(24)

where

U =

[[[[[[

[

119906

1199061

1199062

119906119898+1

]]]]]]

]

Γ =

[[[[[[[[[[[[[

[

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909minus (

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909minus 119909(

120597119906

120597119909)

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909minus (119909119898 120597119906

120597119909)

]]]]]]]]]]]]]

]

F =[[[[[[

[

119891 (119909 119905)

0

0

0

]]]]]]

]

U0(119909) = [119906

0(119909) 0 0 0]

119879

G1(119905) = [119892

1(119905) 0 0 0]

119879

G2(119905) = [119892

2(119905) 0 0 0]

119879

(25)

We use the Galerkin finite element method approxima-tion for discretizing problem (24) For this we multiply (24)by a test function V [0 119871] rarr 119877 V isin 119881

0= 119867

1

0(0 119871)

and integrate the obtained equation in space form 0 to 119871 Weobtain the following equation

int

119871

0

120597U (119909 119905)120597119905

sdot V (119909) 119889119909 minus int119871

0

nablaΓ sdot V (119909) 119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(26)

integrating by parts gives

int

119871

0

nablaΓ sdot V (119909) 119889119909 = (Γ sdot V (119909))|1198710minus int

119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 (27)

We can change the first derivative in time and the integralWe have V(0) = 0 = V(119871) because V isin 119881

0 This leads to an

equivalent problem (P1) forall119905 gt 0 find U(119909 119905) satisfying

119889

119889119905int

119871

0

U (119909 119905) sdot V (119909) 119889119909 + int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909

= int

119871

0

F (119909 119905) sdot V (119909) 119889119909(28)

for all V isin 1198810= 1198671

0(0 119871) To simplify the notation we use the

scalar product in 1198712(0 119871)

(119891 119892) = int

119871

0

119891 (119909) sdot 119892 (119909) 119889119909 (29)

We also can define the following bilinear form

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Numerical simulationTrue solution

Figure 5 The comparison chart with all the initial values for theparameters being set to 05

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 6 The comparison chart with all the initial values for theparameters being set to 1

119886 (U V) = int119871

0

Γ sdot120597V (119909 119905)120597119909

119889119909 =

int

119871

0

minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906

120597119909

120597V120597119909119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199061

120597119909

120597V120597119909

minus120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)1205971199062

120597119909

120597V120597119909

minus 119909120597119906

120597119909

120597V120597119909)119889119909

int

119871

0

(minus (1199011+ 1199012119909 + sdot sdot sdot + 119901

119898+1119909119898)120597119906119898+1

120597119909

120597V120597119909

minus 119909119898+1 120597119906

120597119909

120597V120597119909)119889119909

(30)

Finally we obtain with this notation the weak problem of(P1)

(P2)

119889

119889119905(U V)1198712 + 119886 (U V) = (F V)1198712

U (119909 0) = U0 (119909)

U (0 119905) = G1(119905)

U (119871 119905) = G2(119905)

(31)

41 Space Discretization with the Galerkin Method In thissection we search a semidiscrete approximation of the weakproblem (P2) using the Galerkin finite elementmethodThisleads to a first-order Cauchy problem in time

Let119881ℎbe an119873

119909+1-dimensional subspace of119881 and119881

0ℎ=

119881ℎcap 1198810 Then the following problem is an approximation

of the weak problem find 119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎisin 119881ℎthat

satisfies

119889

119889119905(Uℎ Vℎ) + 119886 (U

ℎ Vℎ) = (F V

ℎ)

Uℎ(119909 0) = U

0ℎ(119909)

Uℎ(0 119905) = G

1(119905)

Uℎ (119871 119905) = G

2 (119905)

(32)

for all Vℎisin 1198810ℎ where U

ℎ= [119906ℎ 1199061ℎ 1199062ℎ 119906

119898+1sdotℎ]119879

The choice of119881ℎis completely arbitrary So we can choose

it theway that for later treatment it will be as easy as possibleFor example we subdivide the interval [0 119871] into partitionsof equal distances ℎ

0 = 1198861lt 1198862lt sdot sdot sdot lt 119886

119873119909

lt 119886119873119909+1= 119871

119886119894= (119894 minus 1) sdot ℎ

119881ℎ= Vℎisin 1198620[0 119871] Vℎ

10038161003816100381610038161003816[119886119894119886119894+1]isin Ρ1 forall119894 = 1 119873

119909

1198810ℎ= Vℎisin 119881ℎ Vℎ(0) = V

ℎ(119871) = 0

(33)

Note that the finite dimension allows us to build a finite basefor the corresponding space In the case of 119881

0ℎ we have

120593119894119873119909

119894=2 where forall119894 = 2 119873

119909

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Discrete Dynamics in Nature and Society

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Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

Table 2 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 0999729028233183 1 0 01 105223771450306 10525 00605593190239173 006055202806016692 110978659802209 111 0120511797611786 01204990402717963 117237567879026 11725 0179257059078521 01792420659047164 124000495680758 124 0236207449080344 02361941640646665 131267443207404 13125 0290793943250869 02907862882126926 139038410458965 139 0342471828361625 03424729718900647 147313397435441 14725 0390726114897089 03907377788388248 156092404136831 156 0435076630410587 04350984630621639 165375430563136 16525 0475082717530532 047511178726701610 175162476714355 175 0510347406713368 0510377951544573

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 7 The comparison chart with all the initial values for theparameters being set to 10

Consider

120593119894(119909) =

0 119909 isin [1198860 119886119894minus1]

119909

ℎminus (119894 minus 2) 119909 isin [119886

119894minus1 119886119894]

119894 minus119909

ℎ119909 isin [119886

119894 119886119894+1]

0 119909 isin [119886119894+1 119886119873119909+1]

(34)

while we add for 119881ℎthe two functions 120593

1and 120593

119873119909+1

definedas

1205931(119909) =

1 minus119909

ℎ if 119909 isin [119886

1 1198862]

0 if 119909 isin [1198862 119886119873119909+1]

120593119873119909+1(119909) =

0 if 119909 isin [1198861 119886119873119909

]119909

ℎminus 119873119909+ 1 if 119909 isin [119886

119873119909

119886119873119909+1]

(35)

Numerical simulationTrue solution

0 01 02 03 04 05 06 07 08 09 109

1

11

12

13

14

15

16

17

18

x

q(x)

Figure 8 The comparison chart with all the initial values for theparameters being set to 50

so that we can write Uℎas a linear combination of the basic

elements

Uℎ(119909 119905) =

119873119909+1

sum

119895=1

U119895(119905) sdot 120593

119895(119909)

U0ℎ (119909 119905) =

119873119909+1

sum

119895=1

U0(119909119895) sdot 120593119895 (119909)

(36)

where U1(119905) = G

0(119905) and U

119873119909+1(119905) = G

1(119905) Knowing that

119886(sdot sdot) is bilinear form and that (32) is valid for each elementof the base 120593

119894119873119909

119894=2 we obtain

119873119909+1

sum

119895=1

119889

119889119905U119895(119905) sdot (120593

119895 120593119894)

+

119873119909+1

sum

119895=1

U119895 (119905) sdot 119886 (120593119895 120593119894) = (F 120593119894)

forall119894 = 2 119873119909

(37)

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

0 2 4 6 8 10 12 14 16 18 200

0005

001

0015

002

0025

003

0035

004

0045Convergence plot

Number of iterations

Res

idua

l F(p)

Figure 9 All the initial values for the parameters are set to 01

This equation can be written in a vector form For this wedefine the vectors u u

0 and F with components

F119894(119905) = (F 120593

119894)1198712 u

119895(119905) = u

119895(119905) u

0119895= u0(x119895) (38)

and matricesM and A as

119898119894119895= (120593119894 120593119895)1198712 119886

119894119895= 119886 (120593

119894 120593119895) (39)

Note thatMA isin R119873119909minus1times119873119909+1 u isin R119873119909+1 and F isin R119873119909minus1 Sothat (37) is equal to the Cauchy problem

M 119889

119889119909u (119905) + A sdot u (119905) = F (119905)

u (1199050) = u0

(40)

the Crank-Nicolsonmethod can be applied to (40) at time 119905119896

resulting in

M sdot (u119896+1minusu119896

Δ119905) +

1

2A sdot u119896+1

+1

2A sdot u119896=1

2(F119896+F119896+1) (41)

where u119896= u(119905119896) F119896= F(119905119896) 119896 = 0 1

Equation (41) can be written in simple form as

(M +Δ119905

2A) sdot u

119896+1=(MminusΔ119905

2A) sdot u

119896+Δ119905

2(F119896+F119896+1) (42)

The algebraic system (42) is solved by Gauss eliminationmethod

5 Numerical Experiment

In this section we are going to demonstrate some numericalresults for (119906(119909 119905) 119902(119909)) in the inverse problem (1)ndash(4)Therefore the following examples are considered and thesolutions are obtained

0 2 4 6 8 10 12 14 16 18 200

1

2

Convergence plot

Number of iterations

times10minus4

Res

idua

l F(p)

Figure 10 All the initial values for the parameters are set to 05

Example 1 Consider (1)ndash(3) with

119906 (119909 0) = sin119909 0 le 119909 le 1

119906 (0 119905) = 0 0 le 119905 le 1

119906 (1 119905) = sin (1) 119890minus119905 0 le 119905 le 1

119891 (119909 119905) = (sin119909(1199092

4+119909

2+ 1)) 119890

minus119905

minus(119909 + 1)

2cos (119909) 119890minus119905 minus sin119909119890minus119905

0 le 119909 le 1 0 le 119905 le 1

(43)

We obtain the unique exact solution

119902 (119909) = 1 + 05119909 + 0251199092

119906 (119909 119905) = sin (119909) 119890minus119905(44)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = sin (05) 119890minus119905 0 le 119905 le 1 (45)

The unknown function 119902(119909) is defined as the following form

and

119902(119909)= 1199011+ 1199012119909 + 11990131199092 (46)

where 1199011 1199012 and 119901

3are unknown coefficients

Table 1 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 12 iterations when it isinitialized in four different points

Figures 1 2 3 and 4 show the fitness of the estimatedparameters and the rate of convergence

Figures 5 6 7 and 8 show the comparison between the

inversion resultsand

119902(119909) and the exact value 119902(119909)Table 2 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

Table 3 Performance of the algorithm when it is run to solve the model using four different parameters guesses

Starting point 01 01 01 01 05 05 05 05 1 1 1 1 2 2 2 201 01 01 01 05 05 05 05 1 1 1 1 2 2 2 2

Iteration 20

101536263526644 101536263500695 101536263525763 1015362635259050896348846894057 0896348850692318 0896348847022403 08963488469997360954285303464511 0954285278486704 0954285302637587 0954285302790922minus0890298938193057 minus0890298849338373 minus0890298935334171 minus0890298935876618140131927153131 140131909032315 140131926588117 140131926696099

minus0871276408882294 minus0871276197318301 minus0871276402487896 minus0871276403706480183785623507722 0183785492186491 018378561965322 018378562038081400359103726343979 00359104061952515 00359103735933848 00359103734148875

Error 119865 789749200363512 times 10minus11

789749200363504 times 10minus11

789749200353888 times 10minus11

78974920035389 times 10minus11

0 2 4 6 8 10 12 14 16 18 200

02

04

06

08

1

12

14

16

18Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 11 All the initial values for the parameters are set to 1

Example 2 Consider (1)ndash(3) with

119906 (119909 0) = 119909119890minus119909 0 le 119909 le 1

119906 (0 119905) = 119905119890minus119905 0 le 119905 le 1

119906 (1 119905) = (1 + 119905) 119890minus1minus119905

0 le 119905 le 1

119891 (119909 119905) = 119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

minus 119890119909(119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

+ 119890119909(2119890minus(119905+119909)

minus (119905 + 119909) 119890minus(119905+119909)

)

0 le 119909 le 1 0 le 119905 le 1

(47)

We obtain the unique exact solution

119902 (119909) = 119890119909

119906 (119909 119905) = (119909 + 119905) 119890minus119909minus119905

(48)

We take the observed data 119892 as

119892 (119905) = 119906 (05 119905) = (05 + 119905) 119890(minus05minus119905)

0 le 119905 le 1 (49)

0 2 4 6 8 10 12 14 16 18 200

05

1

15

2

25

3

35

4 Convergence plot

Number of iterations

times10minus3

Res

idua

l F(p)

Figure 12 All the initial values for the parameters are set to 2

The unknown function 119902(119909) is defined as the following formand

119902(119909)= 1199011+1199012119909+11990131199092+11990141199093+11990151199094+11990161199095+11990171199096 where

1199011 1199012 119901

7are unknown coefficients

Table 3 shows how the Levenberg-Marquardt algorithmcan find the best parameters after 20 iterations when it isinitialized in four different points

Figures 9 10 11 and 12 show the fitness of the estimatedparameters and the rate of convergence

Figures 13 14 15 and 16 show the comparison between

the inversion resultsand

119902(119909) and the exact value 119902(119909)Table 4 shows the values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 =

119895Δ119909 with all the initial values being set to 1

6 Conclusions

A numerical method to estimate the temperature 119906(119909 119905) andthe coefficient 119902(119909) is proposed for an IHCP and the followingresults are obtained

(1) The present study successfully applies the numericalmethod involving the Levenberg-Marquardt algo-rithm in conjunction with the Galerkin finite elementmethod to an IHCP

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Table 4 The values of 119902(119895Δ119909) and 119906(119895Δ119909 05) in 119909 = 119895Δ119909 with all the initial values being set to 1

119895Numerical Exact Numerical Exact119902(119895Δ119909) 119902(119895Δ119909) 119906(119895Δ119909 05) 119906(119895Δ119909 05)

0 101536263525763 1 0303342962644088 03032653298563171 111378168059013 110517091807565 0329201964677126 03292869816564162 122765694959399 122140275816017 0347492882224886 03476097126539873 135549021305874 1349858807576 0359347568702678 03594631712937774 14973722149265 149182469764127 0365808711321159 03659126937665395 165432828415206 164872127070013 0367792378208857 03678794411714426 182785056869393 182211880039051 0366091218454182 03661581920678877 201963499046464 201375270747048 0361387761473796 03614330542946438 223154102006879 222554092849247 0354267869273063 03542913309442169 246579237015687 245960311115695 0345233023618059 034523574951824910 272543670622333 271828182845905 0334712604803175 0334695240222645

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

Numerical simulationTrue solution

x

q(x)

Figure 13 The comparison chart with all the initial values for theparameters being set to 01

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 14 The comparison chart with all the initial values for theparameters being set to 05

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 15 The comparison chart with all the initial values for theparameters being set to 1

0 01 02 03 04 05 06 07 08 09 11

12

14

16

18

2

22

24

26

28

x

q(x)

Numerical simulationTrue solution

Figure 16 The comparison chart with all the initial values for theparameters being set to 2

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

(2) From the illustrated example it can be seen that theproposed numerical method is efficient and accurateto estimate the temperature 119906(119909 119905) and the coefficient119902(119909)

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work of the author is supported by the Special Fundsof the National Natural Science Foundation of China (nos51190093 and 51179151) The author would like to thank thereferees for constructive suggestions and comments

References

[1] A Shidfar and G R Karamali ldquoNumerical solution of inverseheat conduction problem with nonstationary measurementsrdquoApplied Mathematics and Computation vol 168 no 1 pp 540ndash548 2005

[2] A Shidfar G R Karamali and J Damirchi ldquoAn inverse heatconduction problem with a nonlinear source termrdquo NonlinearAnalysisTheory Methods amp Applications vol 65 no 3 pp 615ndash621 2006

[3] A Shidfar and R Pourgholi ldquoNumerical approximation ofsolution of an inverse heat conduction problem based onLegendre polynomialsrdquoAppliedMathematics and Computationvol 175 no 2 pp 1366ndash1374 2006

[4] A Shidfar and H Azary ldquoAn inverse problem for a nonlineardiffusion equationrdquo Nonlinear Analysis Theory Methods ampApplications vol 28 no 4 pp 589ndash593 1997

[5] K Kurpisz and A J Nowak ldquoBEM approach to inverse heatconduction problemsrdquo Engineering Analysis with BoundaryElements vol 10 no 4 pp 291ndash297 1992

[6] H Han D B Ingham and Y Yuan ldquoThe boundary elementmethod for the solution of the backward heat conductionequationrdquo Journal of Computational Physics vol 116 no 2 pp292ndash299 1995

[7] J Skorek ldquoApplying the least squares adjustment technique forsolving inverse heat conduction problemsrdquo in Proceedings of the8th Conference onNumericalMethods in Laminar and TurbulentFlow C Taylor Ed pp 189ndash198 Pinerdage Press Swansea UKAugust 1993

[8] R Pasquetti and C le Niliot ldquoBoundary element approachfor inverse heat conduction problems application to a bidi-mensional transient numerical experimentrdquo Numerical HeatTransfer B Fundamentals vol 20 no 2 pp 169ndash189 1991

[9] D B Ingham and Y Yuan ldquoThe solution of a nonlinear inverseproblem in heat transferrdquo IMA Journal of Applied Mathematicsvol 50 no 2 pp 113ndash132 1993

[10] E Hensel Inverse Theory and Applications for Engineers Pren-tice Hall New Jersey NJ USA 1991

[11] M N Ozisik Inverse Heat Transfer Fundamentals and Applica-tions Taylor amp Francis New York NY USA 2000

[12] R Pourgholi N Azizi Y S Gasimov F Aliev and H KKhalafi ldquoRemoval of numerical instability in the solutionof an inverse heat conduction problemrdquo Communications in

Nonlinear Science and Numerical Simulation vol 14 no 6 pp2664ndash2669 2009

[13] A Shidfar R Pourgholi and M Ebrahimi ldquoA numericalmethod for solving of a nonlinear inverse diffusion problemrdquoComputers ampMathematics with Applications vol 52 no 6-7 pp1021ndash1030 2006

[14] J Wang and N Zabaras ldquoA Bayesian inference approach to theinverse heat conduction problemrdquo International Journal of Heatand Mass Transfer vol 47 no 17-18 pp 3927ndash3941 2004

[15] M Dehghan ldquoDetermination of an unknown parameter ina semi-linear parabolic equationrdquo Mathematical Problems inEngineering vol 8 no 2 pp 111ndash122 2002

[16] J Lihua and M Changfeng ldquoOn L-M method for nonlinearequationsrdquo Journal of Mathematics vol 29 no 3 pp 253ndash2592009

[17] P Chen ldquoWhy not use the Levenberg-Marquardt method forfundamental matrix estimationrdquoThe Institution of Engineeringand Technology vol 4 no 4 pp 286ndash288 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of