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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Algebra
Discrete Dynamics in Nature and Society
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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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Complex AnalysisJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 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
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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
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
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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