Research ArticleResolving Power of Algorithm for Solving the CoefficientInverse Problem for the Geoelectric Equation

K T Iskakov and Zh O Oralbekova

LN Gumilyov Eurasian National University Faculty of Information Technologies Astana 010008 Kazakhstan

Correspondence should be addressed to Zh O Oralbekova oralbekovabkru

Received 9 April 2014 Revised 27 June 2014 Accepted 13 July 2014 Published 6 August 2014

Academic Editor Valery G Yakhno

Copyright copy 2014 K T Iskakov and Zh O Oralbekova This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We considered the inverse coefficient problem for the geoelectric equation For the purpose of research of the conditional stability ofthe inverse problem solution we used integral formulation of the inverse geoelectric problem By implementing the relevant normsand using the close system of Volterra integral equations we managed to estimate the conditional stability of the solution of inverseproblem or rather lower changes in input data imply lower changes in the solution (of the numerical method) When determiningthe additional information the device errors are possibleThat is why this research is important for experimental studies with usageof ground penetrating radars

1 Introduction

Inverse problems for hyperbolic equations in particular forthe acoustics and geoelectrics were investigated by manyauthors notably a detailed bibliography is given in themonography of Kabanikhin [1] We will present the mainscientific results on this problem Blagoveshchenskii appliedGelfand-Levitan method for proving the uniqueness of thesolution of the inverse acoustic problem [2] Romanov proveda comparable theorem for the following equation [3]

119908119905119905(119909 119905) = 119908

119909119909(119909 119905) minus 119902 (119909)119908 (119909 119905) (1)

which is consolidated from the acoustic equation with well-known transformation (see [4])

119908 (119909 119905) = 119906 (119909 119905) exp minus12

ln120590 (119909)

119902 (119909) = minus

1

2

[ln120590 (119909)]10158401015840 + 1

4

[

1205901015840(119909)

120590 (119909)

]

2

(2)

Romanov and Yamamoto [5] obtained the estimationof conditional stability in 119871

2for getting a multidimension

analog of the inverse problem (1)

Numerical algorithm of inverse acoustic problem solvingin the discrete case was given in work [6] for the first time

Bamberger and his coauthors used a conjugate gradientmethod to define the acoustic impedance [7 8]

He and Kabanikhin used the optimization method tosolve the inverse problem for three-dimension acoustic equa-tion [9]

Azamatov and Kabanikhin studied the conditional stabil-ity of the solution to Volterra operator equation in 119871

2[10]

Problems of uniqueness of the inverse problem solutionand set of numerical methods for solving the geoelectricequation were given in the monograph of Romanov andKabanikhin [11]

For solving inverse acoustic problem in integral caseformulation the estimation of the conditional stability in1198671was obtained in the work of Kabanikhin et al [12]

Further in works [13 14] for minimizing purposes theybuilt and investigated a special form of the compositefunctional that allowed proving the following theorems inthe space 119871

2 the local correctness theorem the correctness

theorem of the inverse problem for small amount of data andthe correctness theorem in the envelope of the exact solutionin 1198712

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 545689 9 pageshttpdxdoiorg1011552014545689

2 Mathematical Problems in Engineering

Bukhgeim and Klibanov suggested using the method ofCarleman estimates when proving uniqueness theorems ofthe coefficient inverse problems [15] A broad overview on theuse of Carleman estimates in the theory of multidimensioncoefficient inverse problems is given in the work [16]

The problem of uniqueness of inverse problem solutionfor determination of the coefficients of the permittivity andconductivity for Maxwellrsquos equation system is considered inthe work [17]

Approbation of the globally convergent numerical algo-rithm with the use of experimental radar data for determina-tion of the permittivity is given in work [18] They presentedan analysis of convergence of the method and it has beenshown that the computed and real values of permittivity werein enough agreement A wide range of globally convergentalgorithms of solving a class of problems is described in work[19]

Comparative analysis of the classical equation methodsand globally convergent numerical method of solving thecoefficient inverse problems was given in work [20] Thesecomparisons were performed for both computationally sim-ulated and experimental data

In the work [21] continuation problem from the time-like surface for the 2D Maxwellrsquos equation was consideredThe gradient method for the continuation and coefficientinverse problem was explained The results of computationalexperiment were presented

In this research following the methods which weredescribed in the work [12] we obtained the estimation resultsof the conditional stability of the geoelectric equation in1198671

Herein after the second paragraph there is the conclusionof the main equations which were derived from the system ofMaxwellrsquos equations [11]

In the third paragraph we had amplified the inverse prob-lem for the geoelectric equation with data on characteristicsIt allows us to obtain a close system of integral equations

Finally in the fourth paragraph the implementation ofthe relevant class of input data functions and the class ofsolutions of the inverse problem allowed us to estimate theconditional stability of the inverse problem solution for thegeoelectric equation

2 Statement of the Problems

The propagation process of electromagnetic waves in amedium is described by Maxwellrsquos equations [11]

120576

120597

120597119905

119864 minus rot119867 + 120590119864 + 119895cm

= 0

1199093

= 0 (1199091 1199092 1199093) isin 1198773

120583

120597

120597119905

119867 + rot119864 = 0 119905 gt 0

(3)

Here 119864 = (1198641 1198642 1198643)lowast and 119867 = (119867

1 1198672 1198673)lowast are the

electric and magnetic fields intensity vectors 120576 is dielectricpermittivity of themedium 120583 is magnetic permeability of themedium 120590 is conductivity of the medium 119895cm is source ofexternal currents

Consider geophysical model of the medium consisting oftwo half spaces 1198773

minus= 119909 isin 119877

3 1199093lt 0mdashair 1198773

+= 119909 isin

1198773 1199093gt 0mdashearth

Let the external current source take the following form

119895cm

= (0 1 0)lowast119892 (1199091) 120575 (1199093) 120579 (119905) (4)

where 119892(1199091) is the function which describes the transversal

dimension of the source 120575(1199093) is Dirac delta function and

120579(119905) is Heaviside functionSetting the external current in the form (4) makes it an

instantaneous inclusion current parallel to the axis 1199092at time

scales of 10ndash50 ns (nanoseconds)Using the definition of the curl we get finally from

Maxwellrsquos equations

120576

120597

120597119905

1198641+ 1205901198641=

120597

1205971199092

1198673minus

120597

1205971199093

1198672

120576

120597

120597119905

1198642+ 1205901198642= minus

120597

1205971199091

1198673+

120597

1205971199093

1198671+ 1205742

120576

120597

120597119905

1198643+ 1205901198643=

120597

1205971199091

1198672minus

120597

1205971199092

1198671

120583

120597

120597119905

1198671= minus

120597

1205971199092

1198643+

120597

1205971199093

1198642

120583

120597

120597119905

1198672= minus

120597

1205971199091

1198643minus

120597

1205971199093

1198641

120583

120597

120597119905

1198673= minus

120597

1205971199091

1198642minus

120597

1205971199092

1198641

(5)

Assuming that the coefficients of Maxwellrsquos equations donot depend on the variable 119909

2and are of the special choice of

the source in the form (4) the system will retain only threenonzero components 119864

2 1198671 and 119867

3[11] Excluding the last

two components the final equations are written such that

120576

1205972

12059711990521198642+ 120590

120597

120597119905

1198642

=

120597

1205971199091

(

1

120583

120597

1205971199091

1198642) +

120597

1205971199093

(

1

120583

120597

1205971199093

1198642)

+ 119892 (1199091) 120578 (1199093) 1205791015840(119905) 119909

3gt 0 119905 gt 0

(6)

1198642

1003816100381610038161003816119905gt0

= 0 (7)

1198642

10038161003816100381610038161199093=+0

= 120593(1)(1199091 119905) (8)

(

1

120583

120597

1205971199093

1198642)

100381610038161003816100381610038161003816100381610038161199093=+0

=

120597

120597119905

120593(2)(1199093 119905) (9)

Particular attention has aggravated conditions (8) and (9)Condition (8) is taken as additional information (the

response of the medium)Condition (9) is unknown but it is necessary for solving

direct and inverse problems in a half space 1199093gt 0 (earth)

Mathematical Problems in Engineering 3

In this situation we proceed as shown in [11] in the halfspace 119909

3le 0 where 120590 = 0 we solve the direct problem by

the known data 120576 120583

120576

1205972

12059711990521198642=

120597

1205971199091

(

1

120583

120597

1205971199091

1198642) +

120597

1205971199093

(

1

120583

120597

1205971199093

1198642)

+ 119892 (1199091) 120578 (1199093) 1205791015840(119905) 119909

3lt 0 119905 gt 0

(10)

1198642

1003816100381610038161003816119905lt0

= 0 (11)

1198642

10038161003816100381610038161199093=+0

= 120593(1)(1199091 119905) (12)

In the last system we consider known additional infor-mation (8) as a boundary condition for solving the directproblem in the area 119909

3lt 0 (air) This fact enables us to

restrict the numerical solution of the inverse problem for theminimum possible size of the area in the plane 119909

3gt 0

If the coefficients of (10) do not depend on the variable1199091[11] then applying the Fourier transform 119865

1199091[sdot] to (10)ndash

(12) and similar to (6)ndash(9) we write the final statement of theproblem

In the air domain 1199093

lt 0 we have the followingstatement of the direct problem

120576120592119905119905=

1

120583

12059211990931199093

minus

1205822

120583

120592 + 119892120582120578 (1199093) 1205791015840(119905) 119909

3lt 0

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0

120592 (0 119905) = 119891(1)(119905)

(13)

In the earth domain 1199093gt 0 we have the following

statement of the direct problem

120592119905119905+

120590

120576

120592119905=

1

120583120576

12059211990931199093

minus

1205822

120583120576

119892120582120575 (1199093 119905) 119909

3gt 0 119909

3isin 1198771

(14)

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0 (15)

1

120583

1205921199093(0 119905) = 119891

(2)(119905) (16)

120592 (0 119905) = 119891(1)(119905) (17)

Here 120582 is a Fourier parameter and 120592(119909 119905) = 1198651199091[1198642(1199091

0 1199093 119905)] 119891

(1)(119905) = 119865

1199091[120593(1)(1199091 119905)] and 119891

(2)(119905) = 119865

1199091[(120597

120597119905)120593(2)(1199091 119905)] are Fourier images

Direct Problem By the known values of 120576 120583 and 120590 find120592(1199093 119905) as the solution of the mixed problem (14)ndash(16)

Inverse Problem Find 120590(1199093) and 120592(119909

3 119905) from (14) to (16) for

given 119891(1)(119905) with fixed 120582 = 120582

0

To study conditional stability of the inverse geoelectricproblem it is convenient to use the integral formulation

Now we introduce the following notations 119887(1199093) =

1120583120576(1199093) and 119886(119909

3) = 120590(119909

3)120576(1199093) and change the variables

and functions

119911 = 119911 (1199093) = int

1199093

0

radic120583120576 (120585)119889120585 1199093= 120596 (119911)

119886 (119911) =

120590 (120596 (119911))

120576 (120596 (119911))

119887 (119911) =

1

120583120576 (120596 (119911))

119906 (119911 119905) = 120592 (120596 (119911) 119905) 1198950= minus119892 (120582)radic

120583

120576 (0)

(18)

Then (14)ndash(16) can be written in the form

119906119905119905(119911 119905) = 119906

119911119911(119911 119905) minus 119886 (119911) 119906

119905(119911 119905)

minus

1198871015840(119911)

119887 (119911)

119906119911(119911 119905) minus (120582119887 (119911))

2119906 (119911 119905)

(19)

119906|119905lt0

= 0 119906119905|119905lt0

= 0 (20)

1

120583

119906119911(0 119905) = 119891

(2)(119905) (21)

119906 (0 119905) = 119891(1)(119905) (22)

In the future we will get (19) without the derivative 119906119911 for

this we assume that

119906 (119911 119905) = 119866 (119911) V (119911 119905) (23)

Now we calculate derivatives as follows

119906119911= 1198661015840V + 119866V

119911

119906119911119911= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911

119906119905= 119866V119905 119906

119905119905= 119866V119905119905

(24)

Substituting (24) into (19) we obtain

119866V119905119905= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911minus 119886 (119911) 119866V

119905

minus

1198871015840

119887

(1198661015840V + 119866V

119911) minus (120582119887)

2119866V

(25)

Grouped together we obtain

V119905119905= V119911119911+ (2

1198661015840

119866

minus

1198871015840

119887

) V119911

minus 119886 (119911) V119905+ (

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2) V

(26)

Put that

2

1198661015840

119866

minus

1198871015840

119887

= 0 (27)

119892 (119911) =

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2 (28)

4 Mathematical Problems in Engineering

Finally we have

V119905119905= V119911119911minus 119886 (119911) V

119905+ 119892 (119911) V

V|119905lt0

= 0 V119905|119905lt0

= 0

1

120583

V119911(0 119905) = 119891

(2)(119905)

V (0 119905) = 119891(1)(119905)

(29)

From (27) we have

1198661015840

119866

=

1198871015840

2119887

(ln119866)1015840 = (lnradic119887)1015840

ln119866 = lnradic119887 + ln 119878 (0) 119878 (0) = 1

119866 (119911) = radic119887 (119911)

(30)

Thus the function 119892(119911) is uniquely determined from (28)by the formula (30)

3 Statement of the Problem with the Data onthe Characteristics

In the domain Δ(119897) = (119911 119905)|0 lt |119911| lt 119905 lt 119897 we consider theinverse problem with data on the characteristics [11]

V119905119905(119911 119905) = V

119911119911(119911 119905) minus 119875V (119911 119905) (119911 119905) isin Δ119897 (31)

V (119911 119911) = 119878 (119911) 0 le 119911 le 119897 (32)

V (0 119905) = 119891 (119905) 0 le 119905 le 2119897 (33)

V119911(0 119905) = 120593 (119905) 0 le 119905 le 2119897 (34)

Here

119875V (119911 119905) = 119886 (119911) V119905(119911 119905) + 119892 (119911) V (119911 119905)

119891 (119905) = 119891(1)(119905) 120593 (119905) = 120583119891

(2)(119905)

(35)

We deem that 119886(119911) is an unknown function and thefunction 119892(119911) is to be known

Function 119878(119911) is a solution to Volterra integral equationof the second kind

119878 (119911) =

1

2

1205740minus

1

2

int

119911

0

119886 (120585) 119878 (120585) 119889120585 119911 isin (0 119897) (36)

Inverting the operator (12059721205971199052) minus (12059721205971199112) in (31) and

taking into account (33) and (34) we obtain

V (119911 119905) = Φ (119911 119905) + 119860119905119911[119875V] (119911 119905) isin Δ (119897) (37)

Here we use the following notations

Φ (119911 119905) =

1

2

[119891 (119905 + 119911) + 119891 (119905 minus 119911)] +

1

2

int

119905+119911

119905minus119911

120593 (120585) 119889120585

119860119905119911[V] =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

V (120585 120591) 119889120591 119889120585

(38)

Differentiating (37) with respect to 119905 we obtain

V119905(119911 119905)

= Φ119905(119911 119905) +

1

2

int

119911

0

[119875V (120585 119905 + 119911 minus 120585) minus 119875V (120585 119905 minus 119911 + 120585)] 119889120585

(39)

Put 119905 = 119911 + 0 in (37) and use condition (32) then we have

119878 (119911) = Φ (119911 119911 + 0) + 119860119911+0119911

[119875V] (40)

Differentiating both sides of the resulting equality withrespect to 119911 gives

1198781015840(119911) = Φ

1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585 (41)

It is not difficult to see that the function 119902(119911) = [119878(119911)]minus1

satisfies Volterra integral equation of the second kind

119902 (119911) = 120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585 120574 =

1205740

2

(42)

Taking this into account and the relation 119886(119911) =

21198781015840(119911)119878(119911) we get

119886 (119911) = 2 [Φ1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585]

sdot [120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585]

(43)

Thus we obtain a closed system of integral equations (37)(39) (42) and (43)

We write this system in vector form as follows

Υ = 119865 + 119870 (Υ) (44)

where

Υ (119911 119905) = (Υ1 Υ2 Υ3 Υ4)119879

119865 (119911 119905) = (1198651 1198652 1198653 1198654)119879

119870 (Υ) = (1198701(Υ) 119870

2(Υ) 119870

3(Υ) 119870

4(Υ))119879

Υ1(119911 119905) = V (119911 119905) Υ

2(119911 119905) = V

119905(119911 119905)

Υ3(119911) = 119902 (119911) Υ

4(119911) = 119886 (119911)

1198651(119911 119905) = Φ (119911 119905) 119865

2(119911 119905) = Φ

119905(119911 119905)

1198653= 1205740

minus1 119865

4(119911) = 120594 (119911)

(45)

Mathematical Problems in Engineering 5

where 120594(119911) = 2120574minus1Φ1015840(119911 119911 + 0)

1198701(Υ) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ (120585 120591) 119889120591 119889120585 (119911 119905) isin Δ (119897)

1198702(Υ) =

1

2

int

119911

0

[119875Υ (120585 119905 + 119911 minus 120585) minus 119875Υ (120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ) =

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585

1198704(Υ) = Φ

1015840(119911 119911 + 0) int

119911

0

Υ4(120585) Υ3(120585) 119889120585

+ 2int

119911

0

119875Υ (120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585)

(46)

Here

119875Υ (119911 119905) = Υ4(119911) sdot Υ

2(119911 119905) minus 119892 (119911) Υ

1(119911 119905) (47)

We deem that Υ = (Υ1 Υ2 Υ3 Υ4) isin 1198712(119897) if

Υ119895(119911 119905) isin 119871

2(Δ (119897)) 119895 = 1 2

Υ119895(119911) isin 119871

2(0 119897) 119895 = 3 4

(48)

Let Υ(119895) = (Υ(119895)1(119911 119905) Υ

(119895)

2(119911 119905) Υ

(119895)

3(119911) Υ

(119895)

4(119911))

119879

119895 = 1 2We define the scalar product and the norm as follows

⟨Υ(1) Υ(2)⟩

=

2

sum

119896=1

int

119897

0

int

2119897minus119911

119911

Υ(1)

119896(119911 119905) Υ

(2)

119896(119911 119905) 119889119905 119889119911

+

4

sum

119896=3

int

119897

0

Υ(1)

119896(119911) Υ(2)

119896(119911) 119889119911

Υ2= ⟨Υ Υ⟩

(49)

Inverse Problem Find vector Υ isin 1198712(119897) from (44) for given

119865 isin 1198712(119897)

4 Conditional Stability

Studying 1198671 conditional stability is similar to that in [12]

where it was done for the inverse acoustic problemWe suppose 1198862

1198712(0119897)= 1198721 11989121198712(0119897)

+ 1198911015840

2

1198712(0119897)= 1198722

1205932

1198712(0119897)= 1198723 and 1198922

1198712(0119897)le 1198724to be known

We define sum(1198971198721 119886lowast) as the class of possible solutions

of the inverse problem namely 119886(119911) isin sum(1198971198721 119886lowast) if 119886(119911)

satisfies the following conditions

(1) 119886(119911) isin 1198671(0 119897) cap 119862

1(0 119897)

(2) 1198861198671(0119897)

le 1198721

(3) 0 lt 119886lowastle 119886(119911) 119909 isin (0 119897)

We also define119865(119897119872211987231198724 1198960) as the class of possible

initial data namely 119891 isin 119865(119897 119876 1198960) if 119891 satisfies the following

conditions

(1) 119891 isin 1198671(0 2119897)

(2) 1198911198671(02119897)

le 1198722

(3) 119891(+0) = 1198960 1205931198671(02119897)

le 1198723

Suppose that for 119891(1) 119891(2) isin 119865(119897119872211987231198724 1198960) there

exist 119886(1) and 119886(2) from sum(1198971198721 119886lowast) which solve the inverse

problem

V(119895)119905119905(119911 119905) = V(119895)

119911119911(119911 119905) minus 119875V(119895) (119911 119905) (119911 119905) isin Δ (119897)

V(119895) (119911 119911) = 119878(119895) (119911) 0 le 119911 le 119897

V(119895) (0 119905) = 119891(119895) (119905) V119911(0 119905) = 120593

(119895)(119905) 0 le 119905 le 2119897

(50)

for 119895 = 1 2 respectivelyHere

119875V(119895) (119911 119905) = 119886(119895) (119911) V(119895)119905(119911 119905) + 119892 (119911) V(119895) (119911 119905)

119891(119895)(119905) = 119891

(119895)

(1)(119905) 120593

(119895)(119905) = 120583119891

(119895)

(2)(119905)

(51)

We deem that the function 119892(119911) is known and 119886(119895)(119911) isunknown 119895 = 1 2 We write the early resulting closed systemin the vector form as follows

Υ(119895)= 119865(119895)+ 119870 (Υ

(119895)) 119895 = 1 2 (52)

where

Υ(119895)= (Υ(119895)

1 Υ(119895)

2 Υ(119895)

3 Υ(119895)

4)

119879

Υ(119895)

1(119911 119905) = V(119895) (119911 119905) Υ

(119895)

2(119911 119905) = V(119895)

119905(119911 119905)

(53)

Υ(119895)

3(119911) = 119902 (119911) Υ

(119895)

4(119911) = 119886

(119895)(119911)

119865(119895)= (119865(119895)

1 119865(119895)

2 119865(119895)

3 119865(119895)

4)

119879

119895 = 1 2

119865(119895)

1(119911 119905) = Φ

(119895)(119911 119905) 119865

(119895)

2(119911 119905) = Φ

(119895)

119905(119911 119905)

1198653= 119895minus1 119865

4(119911) = 120594

(119895)(119911) 119895 = 1 2

(54)

1198701(Υ(119895)) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ(119895)(120585 120591) 119889120591 119889120585 (119911 119897) isin Δ (119897)

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

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2 Mathematical Problems in Engineering

Bukhgeim and Klibanov suggested using the method ofCarleman estimates when proving uniqueness theorems ofthe coefficient inverse problems [15] A broad overview on theuse of Carleman estimates in the theory of multidimensioncoefficient inverse problems is given in the work [16]

The problem of uniqueness of inverse problem solutionfor determination of the coefficients of the permittivity andconductivity for Maxwellrsquos equation system is considered inthe work [17]

Approbation of the globally convergent numerical algo-rithm with the use of experimental radar data for determina-tion of the permittivity is given in work [18] They presentedan analysis of convergence of the method and it has beenshown that the computed and real values of permittivity werein enough agreement A wide range of globally convergentalgorithms of solving a class of problems is described in work[19]

Comparative analysis of the classical equation methodsand globally convergent numerical method of solving thecoefficient inverse problems was given in work [20] Thesecomparisons were performed for both computationally sim-ulated and experimental data

In the work [21] continuation problem from the time-like surface for the 2D Maxwellrsquos equation was consideredThe gradient method for the continuation and coefficientinverse problem was explained The results of computationalexperiment were presented

In this research following the methods which weredescribed in the work [12] we obtained the estimation resultsof the conditional stability of the geoelectric equation in1198671

Herein after the second paragraph there is the conclusionof the main equations which were derived from the system ofMaxwellrsquos equations [11]

In the third paragraph we had amplified the inverse prob-lem for the geoelectric equation with data on characteristicsIt allows us to obtain a close system of integral equations

Finally in the fourth paragraph the implementation ofthe relevant class of input data functions and the class ofsolutions of the inverse problem allowed us to estimate theconditional stability of the inverse problem solution for thegeoelectric equation

2 Statement of the Problems

The propagation process of electromagnetic waves in amedium is described by Maxwellrsquos equations [11]

120576

120597

120597119905

119864 minus rot119867 + 120590119864 + 119895cm

= 0

1199093

= 0 (1199091 1199092 1199093) isin 1198773

120583

120597

120597119905

119867 + rot119864 = 0 119905 gt 0

(3)

Here 119864 = (1198641 1198642 1198643)lowast and 119867 = (119867

1 1198672 1198673)lowast are the

electric and magnetic fields intensity vectors 120576 is dielectricpermittivity of themedium 120583 is magnetic permeability of themedium 120590 is conductivity of the medium 119895cm is source ofexternal currents

Consider geophysical model of the medium consisting oftwo half spaces 1198773

minus= 119909 isin 119877

3 1199093lt 0mdashair 1198773

+= 119909 isin

1198773 1199093gt 0mdashearth

Let the external current source take the following form

119895cm

= (0 1 0)lowast119892 (1199091) 120575 (1199093) 120579 (119905) (4)

where 119892(1199091) is the function which describes the transversal

dimension of the source 120575(1199093) is Dirac delta function and

120579(119905) is Heaviside functionSetting the external current in the form (4) makes it an

instantaneous inclusion current parallel to the axis 1199092at time

scales of 10ndash50 ns (nanoseconds)Using the definition of the curl we get finally from

Maxwellrsquos equations

120576

120597

120597119905

1198641+ 1205901198641=

120597

1205971199092

1198673minus

120597

1205971199093

1198672

120576

120597

120597119905

1198642+ 1205901198642= minus

120597

1205971199091

1198673+

120597

1205971199093

1198671+ 1205742

120576

120597

120597119905

1198643+ 1205901198643=

120597

1205971199091

1198672minus

120597

1205971199092

1198671

120583

120597

120597119905

1198671= minus

120597

1205971199092

1198643+

120597

1205971199093

1198642

120583

120597

120597119905

1198672= minus

120597

1205971199091

1198643minus

120597

1205971199093

1198641

120583

120597

120597119905

1198673= minus

120597

1205971199091

1198642minus

120597

1205971199092

1198641

(5)

Assuming that the coefficients of Maxwellrsquos equations donot depend on the variable 119909

2and are of the special choice of

the source in the form (4) the system will retain only threenonzero components 119864

2 1198671 and 119867

3[11] Excluding the last

two components the final equations are written such that

120576

1205972

12059711990521198642+ 120590

120597

120597119905

1198642

=

120597

1205971199091

(

1

120583

120597

1205971199091

1198642) +

120597

1205971199093

(

1

120583

120597

1205971199093

1198642)

+ 119892 (1199091) 120578 (1199093) 1205791015840(119905) 119909

3gt 0 119905 gt 0

(6)

1198642

1003816100381610038161003816119905gt0

= 0 (7)

1198642

10038161003816100381610038161199093=+0

= 120593(1)(1199091 119905) (8)

(

1

120583

120597

1205971199093

1198642)

100381610038161003816100381610038161003816100381610038161199093=+0

=

120597

120597119905

120593(2)(1199093 119905) (9)

Particular attention has aggravated conditions (8) and (9)Condition (8) is taken as additional information (the

response of the medium)Condition (9) is unknown but it is necessary for solving

direct and inverse problems in a half space 1199093gt 0 (earth)

Mathematical Problems in Engineering 3

In this situation we proceed as shown in [11] in the halfspace 119909

3le 0 where 120590 = 0 we solve the direct problem by

the known data 120576 120583

120576

1205972

12059711990521198642=

120597

1205971199091

(

1

120583

120597

1205971199091

1198642) +

120597

1205971199093

(

1

120583

120597

1205971199093

1198642)

+ 119892 (1199091) 120578 (1199093) 1205791015840(119905) 119909

3lt 0 119905 gt 0

(10)

1198642

1003816100381610038161003816119905lt0

= 0 (11)

1198642

10038161003816100381610038161199093=+0

= 120593(1)(1199091 119905) (12)

In the last system we consider known additional infor-mation (8) as a boundary condition for solving the directproblem in the area 119909

3lt 0 (air) This fact enables us to

restrict the numerical solution of the inverse problem for theminimum possible size of the area in the plane 119909

3gt 0

If the coefficients of (10) do not depend on the variable1199091[11] then applying the Fourier transform 119865

1199091[sdot] to (10)ndash

(12) and similar to (6)ndash(9) we write the final statement of theproblem

In the air domain 1199093

lt 0 we have the followingstatement of the direct problem

120576120592119905119905=

1

120583

12059211990931199093

minus

1205822

120583

120592 + 119892120582120578 (1199093) 1205791015840(119905) 119909

3lt 0

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0

120592 (0 119905) = 119891(1)(119905)

(13)

In the earth domain 1199093gt 0 we have the following

statement of the direct problem

120592119905119905+

120590

120576

120592119905=

1

120583120576

12059211990931199093

minus

1205822

120583120576

119892120582120575 (1199093 119905) 119909

3gt 0 119909

3isin 1198771

(14)

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0 (15)

1

120583

1205921199093(0 119905) = 119891

(2)(119905) (16)

120592 (0 119905) = 119891(1)(119905) (17)

Here 120582 is a Fourier parameter and 120592(119909 119905) = 1198651199091[1198642(1199091

0 1199093 119905)] 119891

(1)(119905) = 119865

1199091[120593(1)(1199091 119905)] and 119891

(2)(119905) = 119865

1199091[(120597

120597119905)120593(2)(1199091 119905)] are Fourier images

Direct Problem By the known values of 120576 120583 and 120590 find120592(1199093 119905) as the solution of the mixed problem (14)ndash(16)

Inverse Problem Find 120590(1199093) and 120592(119909

3 119905) from (14) to (16) for

given 119891(1)(119905) with fixed 120582 = 120582

0

To study conditional stability of the inverse geoelectricproblem it is convenient to use the integral formulation

Now we introduce the following notations 119887(1199093) =

1120583120576(1199093) and 119886(119909

3) = 120590(119909

3)120576(1199093) and change the variables

and functions

119911 = 119911 (1199093) = int

1199093

0

radic120583120576 (120585)119889120585 1199093= 120596 (119911)

119886 (119911) =

120590 (120596 (119911))

120576 (120596 (119911))

119887 (119911) =

1

120583120576 (120596 (119911))

119906 (119911 119905) = 120592 (120596 (119911) 119905) 1198950= minus119892 (120582)radic

120583

120576 (0)

(18)

Then (14)ndash(16) can be written in the form

119906119905119905(119911 119905) = 119906

119911119911(119911 119905) minus 119886 (119911) 119906

119905(119911 119905)

minus

1198871015840(119911)

119887 (119911)

119906119911(119911 119905) minus (120582119887 (119911))

2119906 (119911 119905)

(19)

119906|119905lt0

= 0 119906119905|119905lt0

= 0 (20)

1

120583

119906119911(0 119905) = 119891

(2)(119905) (21)

119906 (0 119905) = 119891(1)(119905) (22)

In the future we will get (19) without the derivative 119906119911 for

this we assume that

119906 (119911 119905) = 119866 (119911) V (119911 119905) (23)

Now we calculate derivatives as follows

119906119911= 1198661015840V + 119866V

119911

119906119911119911= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911

119906119905= 119866V119905 119906

119905119905= 119866V119905119905

(24)

Substituting (24) into (19) we obtain

119866V119905119905= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911minus 119886 (119911) 119866V

119905

minus

1198871015840

119887

(1198661015840V + 119866V

119911) minus (120582119887)

2119866V

(25)

Grouped together we obtain

V119905119905= V119911119911+ (2

1198661015840

119866

minus

1198871015840

119887

) V119911

minus 119886 (119911) V119905+ (

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2) V

(26)

Put that

2

1198661015840

119866

minus

1198871015840

119887

= 0 (27)

119892 (119911) =

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2 (28)

4 Mathematical Problems in Engineering

Finally we have

V119905119905= V119911119911minus 119886 (119911) V

119905+ 119892 (119911) V

V|119905lt0

= 0 V119905|119905lt0

= 0

1

120583

V119911(0 119905) = 119891

(2)(119905)

V (0 119905) = 119891(1)(119905)

(29)

From (27) we have

1198661015840

119866

=

1198871015840

2119887

(ln119866)1015840 = (lnradic119887)1015840

ln119866 = lnradic119887 + ln 119878 (0) 119878 (0) = 1

119866 (119911) = radic119887 (119911)

(30)

Thus the function 119892(119911) is uniquely determined from (28)by the formula (30)

3 Statement of the Problem with the Data onthe Characteristics

In the domain Δ(119897) = (119911 119905)|0 lt |119911| lt 119905 lt 119897 we consider theinverse problem with data on the characteristics [11]

V119905119905(119911 119905) = V

119911119911(119911 119905) minus 119875V (119911 119905) (119911 119905) isin Δ119897 (31)

V (119911 119911) = 119878 (119911) 0 le 119911 le 119897 (32)

V (0 119905) = 119891 (119905) 0 le 119905 le 2119897 (33)

V119911(0 119905) = 120593 (119905) 0 le 119905 le 2119897 (34)

Here

119875V (119911 119905) = 119886 (119911) V119905(119911 119905) + 119892 (119911) V (119911 119905)

119891 (119905) = 119891(1)(119905) 120593 (119905) = 120583119891

(2)(119905)

(35)

We deem that 119886(119911) is an unknown function and thefunction 119892(119911) is to be known

Function 119878(119911) is a solution to Volterra integral equationof the second kind

119878 (119911) =

1

2

1205740minus

1

2

int

119911

0

119886 (120585) 119878 (120585) 119889120585 119911 isin (0 119897) (36)

Inverting the operator (12059721205971199052) minus (12059721205971199112) in (31) and

taking into account (33) and (34) we obtain

V (119911 119905) = Φ (119911 119905) + 119860119905119911[119875V] (119911 119905) isin Δ (119897) (37)

Here we use the following notations

Φ (119911 119905) =

1

2

[119891 (119905 + 119911) + 119891 (119905 minus 119911)] +

1

2

int

119905+119911

119905minus119911

120593 (120585) 119889120585

119860119905119911[V] =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

V (120585 120591) 119889120591 119889120585

(38)

Differentiating (37) with respect to 119905 we obtain

V119905(119911 119905)

= Φ119905(119911 119905) +

1

2

int

119911

0

[119875V (120585 119905 + 119911 minus 120585) minus 119875V (120585 119905 minus 119911 + 120585)] 119889120585

(39)

Put 119905 = 119911 + 0 in (37) and use condition (32) then we have

119878 (119911) = Φ (119911 119911 + 0) + 119860119911+0119911

[119875V] (40)

Differentiating both sides of the resulting equality withrespect to 119911 gives

1198781015840(119911) = Φ

1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585 (41)

It is not difficult to see that the function 119902(119911) = [119878(119911)]minus1

satisfies Volterra integral equation of the second kind

119902 (119911) = 120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585 120574 =

1205740

2

(42)

Taking this into account and the relation 119886(119911) =

21198781015840(119911)119878(119911) we get

119886 (119911) = 2 [Φ1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585]

sdot [120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585]

(43)

Thus we obtain a closed system of integral equations (37)(39) (42) and (43)

We write this system in vector form as follows

Υ = 119865 + 119870 (Υ) (44)

where

Υ (119911 119905) = (Υ1 Υ2 Υ3 Υ4)119879

119865 (119911 119905) = (1198651 1198652 1198653 1198654)119879

119870 (Υ) = (1198701(Υ) 119870

2(Υ) 119870

3(Υ) 119870

4(Υ))119879

Υ1(119911 119905) = V (119911 119905) Υ

2(119911 119905) = V

119905(119911 119905)

Υ3(119911) = 119902 (119911) Υ

4(119911) = 119886 (119911)

1198651(119911 119905) = Φ (119911 119905) 119865

2(119911 119905) = Φ

119905(119911 119905)

1198653= 1205740

minus1 119865

4(119911) = 120594 (119911)

(45)

Mathematical Problems in Engineering 5

where 120594(119911) = 2120574minus1Φ1015840(119911 119911 + 0)

1198701(Υ) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ (120585 120591) 119889120591 119889120585 (119911 119905) isin Δ (119897)

1198702(Υ) =

1

2

int

119911

0

[119875Υ (120585 119905 + 119911 minus 120585) minus 119875Υ (120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ) =

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585

1198704(Υ) = Φ

1015840(119911 119911 + 0) int

119911

0

Υ4(120585) Υ3(120585) 119889120585

+ 2int

119911

0

119875Υ (120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585)

(46)

Here

119875Υ (119911 119905) = Υ4(119911) sdot Υ

2(119911 119905) minus 119892 (119911) Υ

1(119911 119905) (47)

We deem that Υ = (Υ1 Υ2 Υ3 Υ4) isin 1198712(119897) if

Υ119895(119911 119905) isin 119871

2(Δ (119897)) 119895 = 1 2

Υ119895(119911) isin 119871

2(0 119897) 119895 = 3 4

(48)

Let Υ(119895) = (Υ(119895)1(119911 119905) Υ

(119895)

2(119911 119905) Υ

(119895)

3(119911) Υ

(119895)

4(119911))

119879

119895 = 1 2We define the scalar product and the norm as follows

⟨Υ(1) Υ(2)⟩

=

2

sum

119896=1

int

119897

0

int

2119897minus119911

119911

Υ(1)

119896(119911 119905) Υ

(2)

119896(119911 119905) 119889119905 119889119911

+

4

sum

119896=3

int

119897

0

Υ(1)

119896(119911) Υ(2)

119896(119911) 119889119911

Υ2= ⟨Υ Υ⟩

(49)

Inverse Problem Find vector Υ isin 1198712(119897) from (44) for given

119865 isin 1198712(119897)

4 Conditional Stability

Studying 1198671 conditional stability is similar to that in [12]

where it was done for the inverse acoustic problemWe suppose 1198862

1198712(0119897)= 1198721 11989121198712(0119897)

+ 1198911015840

2

1198712(0119897)= 1198722

1205932

1198712(0119897)= 1198723 and 1198922

1198712(0119897)le 1198724to be known

We define sum(1198971198721 119886lowast) as the class of possible solutions

of the inverse problem namely 119886(119911) isin sum(1198971198721 119886lowast) if 119886(119911)

satisfies the following conditions

(1) 119886(119911) isin 1198671(0 119897) cap 119862

1(0 119897)

(2) 1198861198671(0119897)

le 1198721

(3) 0 lt 119886lowastle 119886(119911) 119909 isin (0 119897)

We also define119865(119897119872211987231198724 1198960) as the class of possible

initial data namely 119891 isin 119865(119897 119876 1198960) if 119891 satisfies the following

conditions

(1) 119891 isin 1198671(0 2119897)

(2) 1198911198671(02119897)

le 1198722

(3) 119891(+0) = 1198960 1205931198671(02119897)

le 1198723

Suppose that for 119891(1) 119891(2) isin 119865(119897119872211987231198724 1198960) there

exist 119886(1) and 119886(2) from sum(1198971198721 119886lowast) which solve the inverse

problem

V(119895)119905119905(119911 119905) = V(119895)

119911119911(119911 119905) minus 119875V(119895) (119911 119905) (119911 119905) isin Δ (119897)

V(119895) (119911 119911) = 119878(119895) (119911) 0 le 119911 le 119897

V(119895) (0 119905) = 119891(119895) (119905) V119911(0 119905) = 120593

(119895)(119905) 0 le 119905 le 2119897

(50)

for 119895 = 1 2 respectivelyHere

119875V(119895) (119911 119905) = 119886(119895) (119911) V(119895)119905(119911 119905) + 119892 (119911) V(119895) (119911 119905)

119891(119895)(119905) = 119891

(119895)

(1)(119905) 120593

(119895)(119905) = 120583119891

(119895)

(2)(119905)

(51)

We deem that the function 119892(119911) is known and 119886(119895)(119911) isunknown 119895 = 1 2 We write the early resulting closed systemin the vector form as follows

Υ(119895)= 119865(119895)+ 119870 (Υ

(119895)) 119895 = 1 2 (52)

where

Υ(119895)= (Υ(119895)

1 Υ(119895)

2 Υ(119895)

3 Υ(119895)

4)

119879

Υ(119895)

1(119911 119905) = V(119895) (119911 119905) Υ

(119895)

2(119911 119905) = V(119895)

119905(119911 119905)

(53)

Υ(119895)

3(119911) = 119902 (119911) Υ

(119895)

4(119911) = 119886

(119895)(119911)

119865(119895)= (119865(119895)

1 119865(119895)

2 119865(119895)

3 119865(119895)

4)

119879

119895 = 1 2

119865(119895)

1(119911 119905) = Φ

(119895)(119911 119905) 119865

(119895)

2(119911 119905) = Φ

(119895)

119905(119911 119905)

1198653= 119895minus1 119865

4(119911) = 120594

(119895)(119911) 119895 = 1 2

(54)

1198701(Υ(119895)) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ(119895)(120585 120591) 119889120591 119889120585 (119911 119897) isin Δ (119897)

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

In this situation we proceed as shown in [11] in the halfspace 119909

3le 0 where 120590 = 0 we solve the direct problem by

the known data 120576 120583

120576

1205972

12059711990521198642=

120597

1205971199091

(

1

120583

120597

1205971199091

1198642) +

120597

1205971199093

(

1

120583

120597

1205971199093

1198642)

+ 119892 (1199091) 120578 (1199093) 1205791015840(119905) 119909

3lt 0 119905 gt 0

(10)

1198642

1003816100381610038161003816119905lt0

= 0 (11)

1198642

10038161003816100381610038161199093=+0

= 120593(1)(1199091 119905) (12)

In the last system we consider known additional infor-mation (8) as a boundary condition for solving the directproblem in the area 119909

3lt 0 (air) This fact enables us to

restrict the numerical solution of the inverse problem for theminimum possible size of the area in the plane 119909

3gt 0

If the coefficients of (10) do not depend on the variable1199091[11] then applying the Fourier transform 119865

1199091[sdot] to (10)ndash

(12) and similar to (6)ndash(9) we write the final statement of theproblem

In the air domain 1199093

lt 0 we have the followingstatement of the direct problem

120576120592119905119905=

1

120583

12059211990931199093

minus

1205822

120583

120592 + 119892120582120578 (1199093) 1205791015840(119905) 119909

3lt 0

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0

120592 (0 119905) = 119891(1)(119905)

(13)

In the earth domain 1199093gt 0 we have the following

statement of the direct problem

120592119905119905+

120590

120576

120592119905=

1

120583120576

12059211990931199093

minus

1205822

120583120576

119892120582120575 (1199093 119905) 119909

3gt 0 119909

3isin 1198771

(14)

120592|119905lt0

= 0 120592119905

1003816100381610038161003816119905lt0

= 0 (15)

1

120583

1205921199093(0 119905) = 119891

(2)(119905) (16)

120592 (0 119905) = 119891(1)(119905) (17)

Here 120582 is a Fourier parameter and 120592(119909 119905) = 1198651199091[1198642(1199091

0 1199093 119905)] 119891

(1)(119905) = 119865

1199091[120593(1)(1199091 119905)] and 119891

(2)(119905) = 119865

1199091[(120597

120597119905)120593(2)(1199091 119905)] are Fourier images

Direct Problem By the known values of 120576 120583 and 120590 find120592(1199093 119905) as the solution of the mixed problem (14)ndash(16)

Inverse Problem Find 120590(1199093) and 120592(119909

3 119905) from (14) to (16) for

given 119891(1)(119905) with fixed 120582 = 120582

0

To study conditional stability of the inverse geoelectricproblem it is convenient to use the integral formulation

Now we introduce the following notations 119887(1199093) =

1120583120576(1199093) and 119886(119909

3) = 120590(119909

3)120576(1199093) and change the variables

and functions

119911 = 119911 (1199093) = int

1199093

0

radic120583120576 (120585)119889120585 1199093= 120596 (119911)

119886 (119911) =

120590 (120596 (119911))

120576 (120596 (119911))

119887 (119911) =

1

120583120576 (120596 (119911))

119906 (119911 119905) = 120592 (120596 (119911) 119905) 1198950= minus119892 (120582)radic

120583

120576 (0)

(18)

Then (14)ndash(16) can be written in the form

119906119905119905(119911 119905) = 119906

119911119911(119911 119905) minus 119886 (119911) 119906

119905(119911 119905)

minus

1198871015840(119911)

119887 (119911)

119906119911(119911 119905) minus (120582119887 (119911))

2119906 (119911 119905)

(19)

119906|119905lt0

= 0 119906119905|119905lt0

= 0 (20)

1

120583

119906119911(0 119905) = 119891

(2)(119905) (21)

119906 (0 119905) = 119891(1)(119905) (22)

In the future we will get (19) without the derivative 119906119911 for

this we assume that

119906 (119911 119905) = 119866 (119911) V (119911 119905) (23)

Now we calculate derivatives as follows

119906119911= 1198661015840V + 119866V

119911

119906119911119911= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911

119906119905= 119866V119905 119906

119905119905= 119866V119905119905

(24)

Substituting (24) into (19) we obtain

119866V119905119905= 11986610158401015840V + 21198661015840V

119911+ 119866V119911119911minus 119886 (119911) 119866V

119905

minus

1198871015840

119887

(1198661015840V + 119866V

119911) minus (120582119887)

2119866V

(25)

Grouped together we obtain

V119905119905= V119911119911+ (2

1198661015840

119866

minus

1198871015840

119887

) V119911

minus 119886 (119911) V119905+ (

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2) V

(26)

Put that

2

1198661015840

119866

minus

1198871015840

119887

= 0 (27)

119892 (119911) =

11986610158401015840

119866

minus

1198871015840

119887

1198661015840

119866

minus (120582119887)2 (28)

4 Mathematical Problems in Engineering

Finally we have

V119905119905= V119911119911minus 119886 (119911) V

119905+ 119892 (119911) V

V|119905lt0

= 0 V119905|119905lt0

= 0

1

120583

V119911(0 119905) = 119891

(2)(119905)

V (0 119905) = 119891(1)(119905)

(29)

From (27) we have

1198661015840

119866

=

1198871015840

2119887

(ln119866)1015840 = (lnradic119887)1015840

ln119866 = lnradic119887 + ln 119878 (0) 119878 (0) = 1

119866 (119911) = radic119887 (119911)

(30)

Thus the function 119892(119911) is uniquely determined from (28)by the formula (30)

3 Statement of the Problem with the Data onthe Characteristics

In the domain Δ(119897) = (119911 119905)|0 lt |119911| lt 119905 lt 119897 we consider theinverse problem with data on the characteristics [11]

V119905119905(119911 119905) = V

119911119911(119911 119905) minus 119875V (119911 119905) (119911 119905) isin Δ119897 (31)

V (119911 119911) = 119878 (119911) 0 le 119911 le 119897 (32)

V (0 119905) = 119891 (119905) 0 le 119905 le 2119897 (33)

V119911(0 119905) = 120593 (119905) 0 le 119905 le 2119897 (34)

Here

119875V (119911 119905) = 119886 (119911) V119905(119911 119905) + 119892 (119911) V (119911 119905)

119891 (119905) = 119891(1)(119905) 120593 (119905) = 120583119891

(2)(119905)

(35)

We deem that 119886(119911) is an unknown function and thefunction 119892(119911) is to be known

Function 119878(119911) is a solution to Volterra integral equationof the second kind

119878 (119911) =

1

2

1205740minus

1

2

int

119911

0

119886 (120585) 119878 (120585) 119889120585 119911 isin (0 119897) (36)

Inverting the operator (12059721205971199052) minus (12059721205971199112) in (31) and

taking into account (33) and (34) we obtain

V (119911 119905) = Φ (119911 119905) + 119860119905119911[119875V] (119911 119905) isin Δ (119897) (37)

Here we use the following notations

Φ (119911 119905) =

1

2

[119891 (119905 + 119911) + 119891 (119905 minus 119911)] +

1

2

int

119905+119911

119905minus119911

120593 (120585) 119889120585

119860119905119911[V] =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

V (120585 120591) 119889120591 119889120585

(38)

Differentiating (37) with respect to 119905 we obtain

V119905(119911 119905)

= Φ119905(119911 119905) +

1

2

int

119911

0

[119875V (120585 119905 + 119911 minus 120585) minus 119875V (120585 119905 minus 119911 + 120585)] 119889120585

(39)

Put 119905 = 119911 + 0 in (37) and use condition (32) then we have

119878 (119911) = Φ (119911 119911 + 0) + 119860119911+0119911

[119875V] (40)

Differentiating both sides of the resulting equality withrespect to 119911 gives

1198781015840(119911) = Φ

1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585 (41)

It is not difficult to see that the function 119902(119911) = [119878(119911)]minus1

satisfies Volterra integral equation of the second kind

119902 (119911) = 120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585 120574 =

1205740

2

(42)

Taking this into account and the relation 119886(119911) =

21198781015840(119911)119878(119911) we get

119886 (119911) = 2 [Φ1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585]

sdot [120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585]

(43)

Thus we obtain a closed system of integral equations (37)(39) (42) and (43)

We write this system in vector form as follows

Υ = 119865 + 119870 (Υ) (44)

where

Υ (119911 119905) = (Υ1 Υ2 Υ3 Υ4)119879

119865 (119911 119905) = (1198651 1198652 1198653 1198654)119879

119870 (Υ) = (1198701(Υ) 119870

2(Υ) 119870

3(Υ) 119870

4(Υ))119879

Υ1(119911 119905) = V (119911 119905) Υ

2(119911 119905) = V

119905(119911 119905)

Υ3(119911) = 119902 (119911) Υ

4(119911) = 119886 (119911)

1198651(119911 119905) = Φ (119911 119905) 119865

2(119911 119905) = Φ

119905(119911 119905)

1198653= 1205740

minus1 119865

4(119911) = 120594 (119911)

(45)

Mathematical Problems in Engineering 5

where 120594(119911) = 2120574minus1Φ1015840(119911 119911 + 0)

1198701(Υ) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ (120585 120591) 119889120591 119889120585 (119911 119905) isin Δ (119897)

1198702(Υ) =

1

2

int

119911

0

[119875Υ (120585 119905 + 119911 minus 120585) minus 119875Υ (120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ) =

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585

1198704(Υ) = Φ

1015840(119911 119911 + 0) int

119911

0

Υ4(120585) Υ3(120585) 119889120585

+ 2int

119911

0

119875Υ (120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585)

(46)

Here

119875Υ (119911 119905) = Υ4(119911) sdot Υ

2(119911 119905) minus 119892 (119911) Υ

1(119911 119905) (47)

We deem that Υ = (Υ1 Υ2 Υ3 Υ4) isin 1198712(119897) if

Υ119895(119911 119905) isin 119871

2(Δ (119897)) 119895 = 1 2

Υ119895(119911) isin 119871

2(0 119897) 119895 = 3 4

(48)

Let Υ(119895) = (Υ(119895)1(119911 119905) Υ

(119895)

2(119911 119905) Υ

(119895)

3(119911) Υ

(119895)

4(119911))

119879

119895 = 1 2We define the scalar product and the norm as follows

⟨Υ(1) Υ(2)⟩

=

2

sum

119896=1

int

119897

0

int

2119897minus119911

119911

Υ(1)

119896(119911 119905) Υ

(2)

119896(119911 119905) 119889119905 119889119911

+

4

sum

119896=3

int

119897

0

Υ(1)

119896(119911) Υ(2)

119896(119911) 119889119911

Υ2= ⟨Υ Υ⟩

(49)

Inverse Problem Find vector Υ isin 1198712(119897) from (44) for given

119865 isin 1198712(119897)

4 Conditional Stability

Studying 1198671 conditional stability is similar to that in [12]

where it was done for the inverse acoustic problemWe suppose 1198862

1198712(0119897)= 1198721 11989121198712(0119897)

+ 1198911015840

2

1198712(0119897)= 1198722

1205932

1198712(0119897)= 1198723 and 1198922

1198712(0119897)le 1198724to be known

We define sum(1198971198721 119886lowast) as the class of possible solutions

of the inverse problem namely 119886(119911) isin sum(1198971198721 119886lowast) if 119886(119911)

satisfies the following conditions

(1) 119886(119911) isin 1198671(0 119897) cap 119862

1(0 119897)

(2) 1198861198671(0119897)

le 1198721

(3) 0 lt 119886lowastle 119886(119911) 119909 isin (0 119897)

We also define119865(119897119872211987231198724 1198960) as the class of possible

initial data namely 119891 isin 119865(119897 119876 1198960) if 119891 satisfies the following

conditions

(1) 119891 isin 1198671(0 2119897)

(2) 1198911198671(02119897)

le 1198722

(3) 119891(+0) = 1198960 1205931198671(02119897)

le 1198723

Suppose that for 119891(1) 119891(2) isin 119865(119897119872211987231198724 1198960) there

exist 119886(1) and 119886(2) from sum(1198971198721 119886lowast) which solve the inverse

problem

V(119895)119905119905(119911 119905) = V(119895)

119911119911(119911 119905) minus 119875V(119895) (119911 119905) (119911 119905) isin Δ (119897)

V(119895) (119911 119911) = 119878(119895) (119911) 0 le 119911 le 119897

V(119895) (0 119905) = 119891(119895) (119905) V119911(0 119905) = 120593

(119895)(119905) 0 le 119905 le 2119897

(50)

for 119895 = 1 2 respectivelyHere

119875V(119895) (119911 119905) = 119886(119895) (119911) V(119895)119905(119911 119905) + 119892 (119911) V(119895) (119911 119905)

119891(119895)(119905) = 119891

(119895)

(1)(119905) 120593

(119895)(119905) = 120583119891

(119895)

(2)(119905)

(51)

We deem that the function 119892(119911) is known and 119886(119895)(119911) isunknown 119895 = 1 2 We write the early resulting closed systemin the vector form as follows

Υ(119895)= 119865(119895)+ 119870 (Υ

(119895)) 119895 = 1 2 (52)

where

Υ(119895)= (Υ(119895)

1 Υ(119895)

2 Υ(119895)

3 Υ(119895)

4)

119879

Υ(119895)

1(119911 119905) = V(119895) (119911 119905) Υ

(119895)

2(119911 119905) = V(119895)

119905(119911 119905)

(53)

Υ(119895)

3(119911) = 119902 (119911) Υ

(119895)

4(119911) = 119886

(119895)(119911)

119865(119895)= (119865(119895)

1 119865(119895)

2 119865(119895)

3 119865(119895)

4)

119879

119895 = 1 2

119865(119895)

1(119911 119905) = Φ

(119895)(119911 119905) 119865

(119895)

2(119911 119905) = Φ

(119895)

119905(119911 119905)

1198653= 119895minus1 119865

4(119911) = 120594

(119895)(119911) 119895 = 1 2

(54)

1198701(Υ(119895)) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ(119895)(120585 120591) 119889120591 119889120585 (119911 119897) isin Δ (119897)

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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

4 Mathematical Problems in Engineering

Finally we have

V119905119905= V119911119911minus 119886 (119911) V

119905+ 119892 (119911) V

V|119905lt0

= 0 V119905|119905lt0

= 0

1

120583

V119911(0 119905) = 119891

(2)(119905)

V (0 119905) = 119891(1)(119905)

(29)

From (27) we have

1198661015840

119866

=

1198871015840

2119887

(ln119866)1015840 = (lnradic119887)1015840

ln119866 = lnradic119887 + ln 119878 (0) 119878 (0) = 1

119866 (119911) = radic119887 (119911)

(30)

Thus the function 119892(119911) is uniquely determined from (28)by the formula (30)

3 Statement of the Problem with the Data onthe Characteristics

In the domain Δ(119897) = (119911 119905)|0 lt |119911| lt 119905 lt 119897 we consider theinverse problem with data on the characteristics [11]

V119905119905(119911 119905) = V

119911119911(119911 119905) minus 119875V (119911 119905) (119911 119905) isin Δ119897 (31)

V (119911 119911) = 119878 (119911) 0 le 119911 le 119897 (32)

V (0 119905) = 119891 (119905) 0 le 119905 le 2119897 (33)

V119911(0 119905) = 120593 (119905) 0 le 119905 le 2119897 (34)

Here

119875V (119911 119905) = 119886 (119911) V119905(119911 119905) + 119892 (119911) V (119911 119905)

119891 (119905) = 119891(1)(119905) 120593 (119905) = 120583119891

(2)(119905)

(35)

We deem that 119886(119911) is an unknown function and thefunction 119892(119911) is to be known

Function 119878(119911) is a solution to Volterra integral equationof the second kind

119878 (119911) =

1

2

1205740minus

1

2

int

119911

0

119886 (120585) 119878 (120585) 119889120585 119911 isin (0 119897) (36)

Inverting the operator (12059721205971199052) minus (12059721205971199112) in (31) and

taking into account (33) and (34) we obtain

V (119911 119905) = Φ (119911 119905) + 119860119905119911[119875V] (119911 119905) isin Δ (119897) (37)

Here we use the following notations

Φ (119911 119905) =

1

2

[119891 (119905 + 119911) + 119891 (119905 minus 119911)] +

1

2

int

119905+119911

119905minus119911

120593 (120585) 119889120585

119860119905119911[V] =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

V (120585 120591) 119889120591 119889120585

(38)

Differentiating (37) with respect to 119905 we obtain

V119905(119911 119905)

= Φ119905(119911 119905) +

1

2

int

119911

0

[119875V (120585 119905 + 119911 minus 120585) minus 119875V (120585 119905 minus 119911 + 120585)] 119889120585

(39)

Put 119905 = 119911 + 0 in (37) and use condition (32) then we have

119878 (119911) = Φ (119911 119911 + 0) + 119860119911+0119911

[119875V] (40)

Differentiating both sides of the resulting equality withrespect to 119911 gives

1198781015840(119911) = Φ

1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585 (41)

It is not difficult to see that the function 119902(119911) = [119878(119911)]minus1

satisfies Volterra integral equation of the second kind

119902 (119911) = 120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585 120574 =

1205740

2

(42)

Taking this into account and the relation 119886(119911) =

21198781015840(119911)119878(119911) we get

119886 (119911) = 2 [Φ1015840(119911 119911 + 0) + int

119911

0

119875V (120585 2119911 minus 120585) 119889120585]

sdot [120574minus1+

1

2

int

119911

0

119886 (120585) 119902 (120585) 119889120585]

(43)

Thus we obtain a closed system of integral equations (37)(39) (42) and (43)

We write this system in vector form as follows

Υ = 119865 + 119870 (Υ) (44)

where

Υ (119911 119905) = (Υ1 Υ2 Υ3 Υ4)119879

119865 (119911 119905) = (1198651 1198652 1198653 1198654)119879

119870 (Υ) = (1198701(Υ) 119870

2(Υ) 119870

3(Υ) 119870

4(Υ))119879

Υ1(119911 119905) = V (119911 119905) Υ

2(119911 119905) = V

119905(119911 119905)

Υ3(119911) = 119902 (119911) Υ

4(119911) = 119886 (119911)

1198651(119911 119905) = Φ (119911 119905) 119865

2(119911 119905) = Φ

119905(119911 119905)

1198653= 1205740

minus1 119865

4(119911) = 120594 (119911)

(45)

Mathematical Problems in Engineering 5

where 120594(119911) = 2120574minus1Φ1015840(119911 119911 + 0)

1198701(Υ) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ (120585 120591) 119889120591 119889120585 (119911 119905) isin Δ (119897)

1198702(Υ) =

1

2

int

119911

0

[119875Υ (120585 119905 + 119911 minus 120585) minus 119875Υ (120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ) =

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585

1198704(Υ) = Φ

1015840(119911 119911 + 0) int

119911

0

Υ4(120585) Υ3(120585) 119889120585

+ 2int

119911

0

119875Υ (120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585)

(46)

Here

119875Υ (119911 119905) = Υ4(119911) sdot Υ

2(119911 119905) minus 119892 (119911) Υ

1(119911 119905) (47)

We deem that Υ = (Υ1 Υ2 Υ3 Υ4) isin 1198712(119897) if

Υ119895(119911 119905) isin 119871

2(Δ (119897)) 119895 = 1 2

Υ119895(119911) isin 119871

2(0 119897) 119895 = 3 4

(48)

Let Υ(119895) = (Υ(119895)1(119911 119905) Υ

(119895)

2(119911 119905) Υ

(119895)

3(119911) Υ

(119895)

4(119911))

119879

119895 = 1 2We define the scalar product and the norm as follows

⟨Υ(1) Υ(2)⟩

=

2

sum

119896=1

int

119897

0

int

2119897minus119911

119911

Υ(1)

119896(119911 119905) Υ

(2)

119896(119911 119905) 119889119905 119889119911

+

4

sum

119896=3

int

119897

0

Υ(1)

119896(119911) Υ(2)

119896(119911) 119889119911

Υ2= ⟨Υ Υ⟩

(49)

Inverse Problem Find vector Υ isin 1198712(119897) from (44) for given

119865 isin 1198712(119897)

4 Conditional Stability

Studying 1198671 conditional stability is similar to that in [12]

where it was done for the inverse acoustic problemWe suppose 1198862

1198712(0119897)= 1198721 11989121198712(0119897)

+ 1198911015840

2

1198712(0119897)= 1198722

1205932

1198712(0119897)= 1198723 and 1198922

1198712(0119897)le 1198724to be known

We define sum(1198971198721 119886lowast) as the class of possible solutions

of the inverse problem namely 119886(119911) isin sum(1198971198721 119886lowast) if 119886(119911)

satisfies the following conditions

(1) 119886(119911) isin 1198671(0 119897) cap 119862

1(0 119897)

(2) 1198861198671(0119897)

le 1198721

(3) 0 lt 119886lowastle 119886(119911) 119909 isin (0 119897)

We also define119865(119897119872211987231198724 1198960) as the class of possible

initial data namely 119891 isin 119865(119897 119876 1198960) if 119891 satisfies the following

conditions

(1) 119891 isin 1198671(0 2119897)

(2) 1198911198671(02119897)

le 1198722

(3) 119891(+0) = 1198960 1205931198671(02119897)

le 1198723

Suppose that for 119891(1) 119891(2) isin 119865(119897119872211987231198724 1198960) there

exist 119886(1) and 119886(2) from sum(1198971198721 119886lowast) which solve the inverse

problem

V(119895)119905119905(119911 119905) = V(119895)

119911119911(119911 119905) minus 119875V(119895) (119911 119905) (119911 119905) isin Δ (119897)

V(119895) (119911 119911) = 119878(119895) (119911) 0 le 119911 le 119897

V(119895) (0 119905) = 119891(119895) (119905) V119911(0 119905) = 120593

(119895)(119905) 0 le 119905 le 2119897

(50)

for 119895 = 1 2 respectivelyHere

119875V(119895) (119911 119905) = 119886(119895) (119911) V(119895)119905(119911 119905) + 119892 (119911) V(119895) (119911 119905)

119891(119895)(119905) = 119891

(119895)

(1)(119905) 120593

(119895)(119905) = 120583119891

(119895)

(2)(119905)

(51)

We deem that the function 119892(119911) is known and 119886(119895)(119911) isunknown 119895 = 1 2 We write the early resulting closed systemin the vector form as follows

Υ(119895)= 119865(119895)+ 119870 (Υ

(119895)) 119895 = 1 2 (52)

where

Υ(119895)= (Υ(119895)

1 Υ(119895)

2 Υ(119895)

3 Υ(119895)

4)

119879

Υ(119895)

1(119911 119905) = V(119895) (119911 119905) Υ

(119895)

2(119911 119905) = V(119895)

119905(119911 119905)

(53)

Υ(119895)

3(119911) = 119902 (119911) Υ

(119895)

4(119911) = 119886

(119895)(119911)

119865(119895)= (119865(119895)

1 119865(119895)

2 119865(119895)

3 119865(119895)

4)

119879

119895 = 1 2

119865(119895)

1(119911 119905) = Φ

(119895)(119911 119905) 119865

(119895)

2(119911 119905) = Φ

(119895)

119905(119911 119905)

1198653= 119895minus1 119865

4(119911) = 120594

(119895)(119911) 119895 = 1 2

(54)

1198701(Υ(119895)) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ(119895)(120585 120591) 119889120591 119889120585 (119911 119897) isin Δ (119897)

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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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

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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

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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

where 120594(119911) = 2120574minus1Φ1015840(119911 119911 + 0)

1198701(Υ) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ (120585 120591) 119889120591 119889120585 (119911 119905) isin Δ (119897)

1198702(Υ) =

1

2

int

119911

0

[119875Υ (120585 119905 + 119911 minus 120585) minus 119875Υ (120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ) =

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585

1198704(Υ) = Φ

1015840(119911 119911 + 0) int

119911

0

Υ4(120585) Υ3(120585) 119889120585

+ 2int

119911

0

119875Υ (120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ4(120585) Υ3(120585) 119889120585)

(46)

Here

119875Υ (119911 119905) = Υ4(119911) sdot Υ

2(119911 119905) minus 119892 (119911) Υ

1(119911 119905) (47)

We deem that Υ = (Υ1 Υ2 Υ3 Υ4) isin 1198712(119897) if

Υ119895(119911 119905) isin 119871

2(Δ (119897)) 119895 = 1 2

Υ119895(119911) isin 119871

2(0 119897) 119895 = 3 4

(48)

Let Υ(119895) = (Υ(119895)1(119911 119905) Υ

(119895)

2(119911 119905) Υ

(119895)

3(119911) Υ

(119895)

4(119911))

119879

119895 = 1 2We define the scalar product and the norm as follows

⟨Υ(1) Υ(2)⟩

=

2

sum

119896=1

int

119897

0

int

2119897minus119911

119911

Υ(1)

119896(119911 119905) Υ

(2)

119896(119911 119905) 119889119905 119889119911

+

4

sum

119896=3

int

119897

0

Υ(1)

119896(119911) Υ(2)

119896(119911) 119889119911

Υ2= ⟨Υ Υ⟩

(49)

Inverse Problem Find vector Υ isin 1198712(119897) from (44) for given

119865 isin 1198712(119897)

4 Conditional Stability

Studying 1198671 conditional stability is similar to that in [12]

where it was done for the inverse acoustic problemWe suppose 1198862

1198712(0119897)= 1198721 11989121198712(0119897)

+ 1198911015840

2

1198712(0119897)= 1198722

1205932

1198712(0119897)= 1198723 and 1198922

1198712(0119897)le 1198724to be known

We define sum(1198971198721 119886lowast) as the class of possible solutions

of the inverse problem namely 119886(119911) isin sum(1198971198721 119886lowast) if 119886(119911)

satisfies the following conditions

(1) 119886(119911) isin 1198671(0 119897) cap 119862

1(0 119897)

(2) 1198861198671(0119897)

le 1198721

(3) 0 lt 119886lowastle 119886(119911) 119909 isin (0 119897)

We also define119865(119897119872211987231198724 1198960) as the class of possible

initial data namely 119891 isin 119865(119897 119876 1198960) if 119891 satisfies the following

conditions

(1) 119891 isin 1198671(0 2119897)

(2) 1198911198671(02119897)

le 1198722

(3) 119891(+0) = 1198960 1205931198671(02119897)

le 1198723

Suppose that for 119891(1) 119891(2) isin 119865(119897119872211987231198724 1198960) there

exist 119886(1) and 119886(2) from sum(1198971198721 119886lowast) which solve the inverse

problem

V(119895)119905119905(119911 119905) = V(119895)

119911119911(119911 119905) minus 119875V(119895) (119911 119905) (119911 119905) isin Δ (119897)

V(119895) (119911 119911) = 119878(119895) (119911) 0 le 119911 le 119897

V(119895) (0 119905) = 119891(119895) (119905) V119911(0 119905) = 120593

(119895)(119905) 0 le 119905 le 2119897

(50)

for 119895 = 1 2 respectivelyHere

119875V(119895) (119911 119905) = 119886(119895) (119911) V(119895)119905(119911 119905) + 119892 (119911) V(119895) (119911 119905)

119891(119895)(119905) = 119891

(119895)

(1)(119905) 120593

(119895)(119905) = 120583119891

(119895)

(2)(119905)

(51)

We deem that the function 119892(119911) is known and 119886(119895)(119911) isunknown 119895 = 1 2 We write the early resulting closed systemin the vector form as follows

Υ(119895)= 119865(119895)+ 119870 (Υ

(119895)) 119895 = 1 2 (52)

where

Υ(119895)= (Υ(119895)

1 Υ(119895)

2 Υ(119895)

3 Υ(119895)

4)

119879

Υ(119895)

1(119911 119905) = V(119895) (119911 119905) Υ

(119895)

2(119911 119905) = V(119895)

119905(119911 119905)

(53)

Υ(119895)

3(119911) = 119902 (119911) Υ

(119895)

4(119911) = 119886

(119895)(119911)

119865(119895)= (119865(119895)

1 119865(119895)

2 119865(119895)

3 119865(119895)

4)

119879

119895 = 1 2

119865(119895)

1(119911 119905) = Φ

(119895)(119911 119905) 119865

(119895)

2(119911 119905) = Φ

(119895)

119905(119911 119905)

1198653= 119895minus1 119865

4(119911) = 120594

(119895)(119911) 119895 = 1 2

(54)

1198701(Υ(119895)) =

1

2

int

119911

0

int

119905+119911minus120585

119905minus119911+120585

119875Υ(119895)(120585 120591) 119889120591 119889120585 (119911 119897) isin Δ (119897)

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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

6 Mathematical Problems in Engineering

1198702(Υ(119895))

=

1

2

int

119911

0

[119875Υ(119895)(120585 119905 + 119911 minus 120585) minus 119875Υ

(119895)(120585 119905 minus 119911 + 120585)] 119889120585

1198703(Υ(119895)) =

1

2

int

119911

0

Υ(119895)

4Υ(119895)

3119889120585

1198704(Υ(119895)) = Φ

1015840(119895)

(119911 119911 + 0) int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585

+ 2int

119911

0

119875Υ(119895)(120585 2119911 minus 120585) 119889120585

sdot (120574minus1+

1

2

int

119911

0

Υ(119895)

4(120585) Υ(119895)

3(120585) 119889120585)

11989711988611988711989011989711989011990255 (55)

Here we denote 119875Υ(119895)(119911 119905) = Υ4(119911)Υ2(119911 119905) minus 119892(119911)Υ

1(119911 119905)

Theorem 1 Suppose that for 119865(119895) isin 1198712(119897) 119895 = 1 2 there exist

Υ(119895)isin 1198712(Δ(119897)) as the solution of the inverse problem as follows

Υ(119895)(119911 119905) = 119865

(119895)(119911 119905) + 119870 (Υ

(119895)) 119895 = 1 2 (119911 119905) isin Δ (119897)

(56)

Then

10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le 119862

10038171003817100381710038171003817119891(1)minus 119891(2)10038171003817100381710038171003817

2

1198671(02119897) (57)

where

119862 = 119862 (1198971198721119872211987231198724 1198960) (58)

Proof We introduce

Υ (119909 119905) = (Υ1(119911 119905) Υ

2(119911 119905) Υ

3(119911) Υ

4(119911))

= Υ(1)(119911 119905) minus Υ

(2)(119911 119905)

119865 (119911 119905) = 119865(1)(119911 119905) minus 119865

(2)(119911 119905)

(59)

Then from (52) it follows that

Υ (119911 119905) = 119865 (119911 119905) minus 119870 (Υ) (119911 119905) isin Δ (119897) (60)

In the vector equation (60) we estimate each componentseparately taking into account the obvious inequalities asfollows

(119886 + 119887 + 119888)2le 3 (119886

2+ 1198872+ 1198882)

(radic119886 + radic119887)

2

le 2119886 + 2119887

(61)

for 119886 ge 0 119887 ge 0We obtain the chain of the inequalities

10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816le

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times[

[

radicint

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

119889120585

+radicint

119911

0

10038161003816100381610038161003816Υ2(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585]

]

(62)

Using the obvious inequality we get10038161003816100381610038161003816Υ1(119911 119905)

10038161003816100381610038161003816

2

le 3

100381610038161003816100381610038161198651(119911 119905)

10038161003816100381610038161003816

2

+

3

2

int

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

[

10038161003816100381610038161003816Υ(1)

1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(1)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

] 119889120585

+

3

2

int

119911

0

10038161003816100381610038161003816Υ(2)

3(120585)

10038161003816100381610038161003816

2

119889120585

times int

119911

0

10038161003816100381610038161003816Υ1(120585 119905 + 119911 minus 120585)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(120585 119905 minus 119911 + 120585)

10038161003816100381610038161003816

2

119889120585

(63)

Turning to the earlier introduced norms we have10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))119879

+

3

2

int

119911

0

int

2119897minus120585

120585

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

Mathematical Problems in Engineering 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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 7

+

10038161003816100381610038161003816Υ(1)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840

+ int

120585

0

10038161003816100381610038161003816Υ2

3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840

times int

120585

0

[

10038161003816100381610038161003816Υ1(1205851015840 120591 + 120585 minus 120585

1015840)

10038161003816100381610038161003816

2

+

10038161003816100381610038161003816Υ(2)

1(1205851015840 120591 minus 120585 + 120585

1015840)

10038161003816100381610038161003816

2

] 1198891205851015840119889120591 119889120585

le 3

100381710038171003817100381710038171198651

10038171003817100381710038171003817

2

1198712(Δ(119897119911))

+ 12Υ2

lowastint

119911

0

int

120585

0

10038161003816100381610038161003816Υ3(1205851015840)

10038161003816100381610038161003816

2

1198891205851015840119889120585 + 12Υ

2

lowastint

119911

0

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897120585))

119889120585

(64)

Here

Υlowast= max 1003817100381710038171003817

1003817Υ(1)10038171003817100381710038171003817

10038171003817100381710038171003817Υ(2)10038171003817100381710038171003817 (65)

We estimate the second component of (60)

10038161003816100381610038161003816Υ2(119911)

10038161003816100381610038161003816le

1

2

int

119911

0

10038161003816100381610038161003816Υ(1)

3(120585) Υ2(120585)

10038161003816100381610038161003816119889120585

+

1

2

int

119911

0

10038161003816100381610038161003816Υ(2)

2(120585) Υ3(120585)

10038161003816100381610038161003816119889120585

le

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(1)

3(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ2(120585)

10038161003816100381610038161003816

2

119889120585

+

1

2

radicint

119911

0

10038161003816100381610038161003816Υ(2)

2(120585)

10038161003816100381610038161003816

2

119889120585radicint

119911

0

10038161003816100381610038161003816Υ3(120585)

10038161003816100381610038161003816

2

119889120585

(66)

Then we have

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0119911)le

1

2

Υ2

lowastint

119911

0

[

10038171003817100381710038171003817Υ2

10038171003817100381710038171003817

2

1198712(0120585)+

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (67)

We estimate the third component of (60) and we have

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0119897)le

1

4

1198722int

119911

0

[

10038171003817100381710038171003817Υ3

10038171003817100381710038171003817

2

1198712(0120585)+1198723

10038171003817100381710038171003817Υ4

10038171003817100381710038171003817

2

1198712(0120585)] 119889120585 (68)

Finally for the fourth component of (60) we get theestimate

10038161003816100381610038161003816Υ4(119911)

10038161003816100381610038161003816le

100381610038161003816100381610038161198654(119911)

10038161003816100381610038161003816+

4

sum

119894=1

119908119894

1199081(119911) = 2

100381610038161003816100381610038161003816

(119891(1))

1015840

(119911)

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199082(119911) =

100381610038161003816100381610038161198706(Υ(1)) minus 1198706(Υ(2))

10038161003816100381610038161003816

1199083=

100381610038161003816100381610038161198702(Υ(1))

100381610038161003816100381610038161199082(119911)

1199084=

100381610038161003816100381610038161198704(Υ(2))

10038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(1)) minus 1198702(Υ(2))

10038161003816100381610038161003816

1199085=

100381610038161003816100381610038161003816

(119891(1))

1015840

minus (119891(2))

100381610038161003816100381610038161003816

100381610038161003816100381610038161198702(Υ(2))

10038161003816100381610038161003816

1198706(Υ) = int

119911

0

Υ3(120585)Φ1(120585 2119911 minus 120585) 119889120585

(69)

Estimating each term119908119894(119911) and substituting into (69) and

using the obvious inequality

(

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816)

2

le 4

4

sum

119896=1

1003816100381610038161003816119887119896

1003816100381610038161003816

2

(70)

we obtain10038171003817100381710038171003817Υ4(119911)

100381710038171003817100381710038171198712(0119911)

le ]0int

119911

0

[1198911015840(2120585)]

2

119889120585 +

1

2

]1

10038171003817100381710038171003817Υ1

100381710038171003817100381710038171198712(Δ(119897119911))

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ2

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]3(120585)

10038171003817100381710038171003817Υ3

100381710038171003817100381710038171198712(0120585)

119889120585

+ int

119911

0

]4(120585)

10038171003817100381710038171003817Υ4

100381710038171003817100381710038171198712(0120585)

119889120585

(71)

Now we combine all the obtained estimates for the fourcomponents (60) and denote for convenience

1205951(119911) =

10038171003817100381710038171003817Υ1

10038171003817100381710038171003817

2

1198712(Δ(119897119911)) 119911 isin (0 119897) (72)

and then

120595 (119911) = 1205951(119911) + 120595

2(119911) + 120595

3(119911) + 120595

4(119911) (73)

and for function 120595 we obtain the following estimate

120595 (119911) le 120578 + int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 (74)

where 120578 = 120578(Υ2lowast ]1 ]2 ]3 ]4)

Introduce a new function

] (119911) = 120578lowast+ int

119911

0

4

sum

119894=1

120574119894(120585) 120595119894(120585) 119889120585 120578 lt 120578

lowast (75)

where 120578lowastis constant

Then 120595(119911) le ](119911)

]1015840 (119911) =4

sum

119894=1

120574119894(119911) Υ119895(119911) le ] (119911)

4

sum

119894=1

120574119894(119911)

]1015840 (119911)] (119911)

le

4

sum

119894=1

120574119894(119911)

(76)

8 Mathematical Problems in Engineering

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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

Applying the Gronwall inequality we obtain

120595 (119911) le ] (119911) le ] (0) expint119911

0

4

sum

119894=1

120574119894(120585) 119889120585

int

119911

0

4

sum

119894=1

120574119894(120585) 119889120585 le 25Υ

2

lowasttimes 119911 + 12Υ

2

lowast

10038171003817100381710038171003817119891(1)10038171003817100381710038171003817

2

1198712(02119897)

+ 12Υ4

lowast+ 12Υ

2

lowast(12 + Υ

2

lowastsdot 119911)

(77)

Then from (77) we obtain10038171003817100381710038171003817Υ(1)minus Υ(2)10038171003817100381710038171003817

2

le

10038171003817100381710038171003817119891(1)minus 119891(2)100381710038171003817100381710038171198671(02119897)

(78)

where the constant 119862 gt 0 is given by (58)An explicit expression for the constant as a result of

successive computations is given by

119862 = [6119897 + 61198721(

4

1198962

0

+ Υ4

lowast)(1 + 12Υ

2

lowast119897)]

timesexpΥ2lowast[24119897 + 8119872

2(

4

1198962

0

+ Υ4

lowast)(1198723+ 36Υ

2

lowast119897)

+ 61198722Φ2+ 81198724Υ4

lowast]

(79)

5 Conclusions

The conditional stability of the inverse problem for thegeoelectric equation has been investigated For studying weconsider the integral formulation of the inverse geoelectricproblem The estimation of the conditional stability of theinverse problem solution has been obtained or rather lowerchanges in input data imply lower changes in the solution(of the numericalmethod)When determining the additionalinformation the device errors are possible That is why thisresearch is important for experimental studies with usage ofground penetrating radarsThe inlet data belongs to the class119865(119897119872

211987231198724 1198960) while the solution belongs to the class

sum(1198971198721 119886lowast)

Conflict of Interests

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

Acknowledgment

The work was supported by Ministry of Education andScience of the Republic of Kazakhstan (Grant no 139 (69) ot04022014)

References

[1] S I Kabanikhin Inverse and III-Posed Problems Theory andApplications de Gruyter Berlin Germany 2011

[2] A S Blagoveshchenskii and V M Babic ldquoOn a local methodof solution of a nonstationary inverse problem for a nonho-mogeneous stringrdquo in Mathematical Questions in the Theory ofWave Diffraction vol 115 of Proceedings of the Steklov Institute ofMathematics in the Academy of Sciences of the USSR pp 30ndash41American Mathematical Society Providence RI USA 1974

[3] V G Romanov Inverse Problems of Mathematical Physics VNUScience Press Utrecht The Netherlands 1987

[4] S I Kabanikhin andA Lorenzi Identification Problems ofWavephenomena VSP Utrecht The Netherlands 1999

[5] V G Romanov and M Yamamoto ldquoMultidimensional inversehyperbolic problem with impulse input and a single boundarymeasurementrdquo Journal of Inverse and Ill-Posed Problems vol 7no 6 pp 573ndash588 1999

[6] V Baranov and G Kunetz ldquoSynthetic seismograms with multi-plenreflections theory and numerical experiencerdquo GeophysicalProspecting vol 8 pp 315ndash325 1969

[7] A Bamberger G Chavent C Hemon and P Lailly ldquoInversionof normal incidence seismogramsrdquoGeophysics vol 47 no 5 pp757ndash770 1982

[8] A Bamberger G Chavent and P Lailly ldquoAbout the stabilityof the inverse problem in 1-D wave equationsmdashapplications tothe interpretation of seismic profilesrdquo Applied Mathematics andOptimization vol 5 no 1 pp 1ndash47 1979

[9] S He and S I Kabanikhin ldquoAn optimization approach toa three-dimensional acoustic inverse problem in the timedomainrdquo Journal of Mathematical Physics vol 36 no 8 pp4028ndash4043 1995

[10] J S Azamatov and S I Kabanikhin ldquoVolterra operator equa-tions 119871

2-theoryrdquo Journal of Inverse and Ill-Posed Problems vol

7 no 6 pp 487ndash510 1999[11] V G Romanov and S I Kabanikhin Inverse Problems for

Maxwellrsquos Equations VSP Utrecht The Netherlands 1994[12] S I Kabanikhin K T Iskakov and M Yamamoto ldquoH1-

conditional stability with explicit Lipshitz constant for a one-dimensional inverse acoustic problemrdquo Journal of Inverse andIll-Posed Problems vol 9 no 3 pp 249ndash267 2001

[13] S I Kabanikhin and K T Iskakov ldquoJustification of the steepestdescent method in an integral formulation of an inverse prob-lem for a hyperbolic equationrdquo Siberian Mathematical journalvol 42 no 3 pp 478ndash494 2001

[14] S I Kabanikhin and K T Iskakov Optimization Methodsof Coefficient Inverse Problems Solution NGU NovosibirskRussia 2001

[15] A L Bukhgeim and M V Klibanov ldquoUniqueness in thelarge of a class of multidimensional inverse problemsrdquo SovietMathematics Doklady vol 17 pp 244ndash247 1981

[16] M V Klibanov ldquoCarleman estimates for global uniqueness sta-bility and numerical methods for coefficient inverse problemsrdquoJournal of Inverse and Ill-Posed Problems vol 21 no 4 pp 477ndash560 2013

[17] M V Klibanov ldquoUniqueness of the solution of two inverseproblems for a Maxwell systemrdquo Computational Mathematicsand Mathematical Physics vol 26 no 7 pp 67ndash73 1986

[18] A V Kuzhuget L Beilina M V Klibanov A Sullivan LNguyen and M A Fiddy ldquoBlind backscattering experimentaldata collected in the field and an approximately globallyconvergent inverse algorithmrdquo Inverse Problems vol 28 no 9Article ID 095007 2012

[19] L Beilina andM V Klibanov Approximate Global Convergenceand Adaptivity for Coefficient Inverse Problems Springer NewYork NY USA 2012

Mathematical Problems in Engineering 9

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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

[20] A L Karchevsky M V Klibanov L Nguyen N Pantong andA Sullivan ldquoThe Krein method and the globally convergentmethod for experimental datardquo Applied Numerical Mathemat-ics vol 74 pp 111ndash127 2013

[21] S I Kabanikhin D B Nurseitov M A Shishlenin and BB Sholpanbaev ldquoInverse problems for the ground penetratingradarrdquo Journal of Inverse and Ill-Posed Problems vol 21 no 6pp 885ndash892 2013

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