Research Article On the Harmonic Problem with Nonlinear ...downloads.hindawi.com/archive/2014/976520.pdf · In the present work, we deal with the harmonic problems in a bounded domain

Research ArticleOn the Harmonic Problem with NonlinearBoundary Integral Conditions

Saker Hacene

LMA Department of Mathematics Faculty of Sciences University of Badji Mokhtar PO Box 12 23000 Annaba Algeria

Correspondence should be addressed to Saker Hacene h sakeryahoofr

Received 18 November 2013 Revised 7 January 2014 Accepted 8 January 2014 Published 23 February 2014

Academic Editor Shamsul Qamar

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

In the present work we deal with the harmonic problems in a bounded domain of R2 with the nonlinear boundary integralconditions After applying the Boundary integral method a nonlinear boundary integral equation is obtained the existence anduniqueness of the solution will be a consequence of applying theory of monotone operators

1 Introduction

For the harmonic problem the simplest boundary conditionwe can impose specifies 119906 at all points on the boundary Γ andis known as the Dirichlet boundary condition The Dirichletproblem for the Laplace equation can easily be solved usingthe boundary integral equation [1] If the normal derivativeof 119906 that is 120597119906120597119899 where 119899 is the outward normal to theboundary Γ is specified at all points on the boundary Γ thatis the Neumann boundary condition with int

Γ(120597119906120597119899)119889119904 = 0

then given the value of 119906 at one point on Γ enables a uniquesolution to be obtained [1]

In this work we impose more general boundary condi-tions namely the nonlinear integral equation of Urysohntype [2 3]

Much attention has been paid to the resolution of bound-ary value problems for partial differential operators withnonlinear boundary conditions by the method of integralequations in many directions (see eg Atkinson and Chan-dler [4 5] and Ruotsalainen and Wendland [6])

Problems involving nonlinearities form a basis of math-ematical models of various steady-state phenomena andprocesses in mechanics physics and many other areas ofscience Among these is the steady-state heat transfer Alsosome electromagnetic problems contain nonlinearities inthe boundary conditions for instance problems where theelectrical conductivity of the boundary is variable [7] Furtherapplications arise in heat radiation and heat transfer [7 8]

In the present paper we look for the solution of the Lapla-cian equation with nonlinear data of the form

Δ119906 (119909) = 0 119909 isin Ω (1)

120597119906

120597119899

(119909) + int

Γ

119870 (119909 119910 119906 (119910)) 119889119904119910

= 119891 (119909) 119909 isin Γ

(2)

We recall that the nonlinear boundary integral operatordefined by

119860 (119909 119906 (119909)) = int

Γ

119870 (119909 119910 119906 (119910)) 119889119904119910 119909 isin Γ (3)

is the nonlinear integral operator of Urysohn typeIn (1) we assumeΩ is an open bounded region inR2 with

a smooth boundary Γ = 120597Ω and

119891 Γ 997888rarr R 119870 Γ times Γ times R 997888rarr R (4)

are given real value functionsBy the Green representation formula we formulate a

nonlinear integral equation on the boundary Γ of the domainΩ Under some assumptions on the Kernel of the nonlinearintegral equation of Urysohn 119870(119909 119910 119906) we prove the exis-tence and uniqueness of the solution

Hindawi Publishing CorporationInternational Journal of AnalysisVolume 2014 Article ID 976520 5 pageshttpdxdoiorg1011552014976520

2 International Journal of Analysis

11 Definitions and Notations

Definition 1 (see [1 9]) Let 119898 isin N one denotes by 119867119898

(Ω)

the Sobolev space

119867119898

(Ω) = 119906 isin 1198712

(Ω) 119863120572119906 isin 1198712

(Ω) |120572| le 119898 (5)

Definition 2 (see [1 9]) Let 119904 isin R one denotes by119867119904(R119899) the

Sobolev space

119867119904(R119899) = 119906 isin 119871

2(R119899) (1 +

10038161003816100381610038161205851003816100381610038161003816

2

)

1199042

|119865 [119906]| isin 1198712

(R119899)

(6)

and the associated norm

119906119867119904 = (int

R119899(1 +

10038161003816100381610038161205851003816100381610038161003816

2

)

119904

|119865 [119906]|2119889120585)

12

(7)

with 119865[sdot] the Fourier transform

Definition 3 (see [1 9]) Let Ω sub R119899 a bounded domain andΓ = 120597Ω one defined

119867119904(Ω) = 119906 | Ω 119906 isin 119867

119904(R119899) 119904 isin R

119867119904(Γ) =

119906|Γ

119906 isin 119867119904+(12)

(R119899) 119904 gt 0

1198712

(Γ) 119904 = 0

(119867minus119904

(Γ))1015840

(dual space) 119904 lt 0

(8)

Definition 4 (see [1 9]) The Fichera trace spaces 119867119904(Γ) for

0 lt 119904 lt 1 is defined to be the completion of

1198620

119904(Γ) = 120593 isin 119862

0(Γ)

1003817100381710038171003817120593

1003817100381710038171003817119867119904(Γ)

lt infin (9)

with respect to the norm

119906119867119904(Γ)

= 1199062

1198712(Γ)

+ ∬

Γ

1003816100381610038161003816119906 (119909) minus 119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

12

(10)

2 The Boundary Integral Method

21 Representative Formula and Boundary Operator Weintroduce the fundamental solution of the Laplacian operatorin the plane defined by

119864 (119909 119910) =

1

2120587

log 1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (11)

We first consider some standard boundary integral operatorsFor 119909 isin Ω the single layer potential is

119878Ω

119906 (119909) = minus int

Γ

119864 (119909 119910) 119906 (119910) 119889119904119910 (12)

and the double layer potential is

119863Ω

119906 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910 (13)

Using Greenrsquos identity for harmonic functions we get

119906 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910

minus int

Γ

120597119906 (119910)

120597119899119910

119864 (119909 119910) 119889119904119910

(14)

for 119909 isin Ω which can be written as

119906 (119909) = 119863Ω

119906 (119909) + 119878Ω

120597119906 (119909)

120597119899

for 119909 isin Ω (15)

Sending in (15) 119909 rarr Γ The continuity of the simple layerpotential 119878

Ωand the jump relation of the double layer poten-

tial 119863Ω we can write the integral equation on the boundary

as follows

119906 (119909) minus 119863119906 (119909) = 119878

120597119906 (119909)

120597119899

119909 isin Γ (16)

where

119878

120597119906 (119909)

120597119899

= minus2 int

Γ

119864 (119909 119910)

120597119906 (119910)

120597119899

119889119904119910 119909 isin Γ

119863119906 (119909) = 2 int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910 119909 isin Γ

(17)

Clearly if 119906 isin 1198671(Ω) is the solution of (1) then the Cauchy

data 119906|Γand 120597119906120597119899|

Γsatisfies the integral equation (16)

Then the boundary conditions

120597119906

120597119899

(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (18)

yield

119906 (119909) minus 119863119906 (119909) = minus119878 (119860 (119909 119906 (119909))) + 119878119891 (119909) 119909 isin Γ

(19)

Equation (19) can be written as

(119868 minus 119863) 119906 (119909) + 119878 (119860 (119909 119906 (119909))) = 119878119891 (119909) 119909 isin Γ (20)

Conversely if 119906|Γsolves (20) then the solution of (1) can be

given by the representation formula (15) and will satisfy

120597119906

120597119899

(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (21)

due to (20) For studying the solvability of the nonlinearequation (20) we give some assumptions to be made here

(H1) We assume a diam(Ω) lt 1(H2) The Kernel 119870(sdot sdot sdot) of the Urysohn operator is a

Caratheodory function [3](H3) We assume that 120597119870(119909 119910 119906)120597119906 is measurable satisfy-

ing

0 lt 119886 le

120597119870 (119909 119910 119906)

120597119906

le 119887 lt +infin (22)

for some constants 119886 and 119887

International Journal of Analysis 3

Remark 5 (1) The operator 119878 may have eigenfunctions [1]then (H1) ensures that the integral operator

119878 119867119904(Γ) 997888rarr 119867

119904+1(Γ) (23)

is an isomorphism for every 119904 isin R and

(119878120583 120583) ge 1198881003817100381710038171003817120583

1003817100381710038171003817

2

119867minus12

(24)

for all 120583 isin 119867minus12 with some positive constant 119888 gt 0 [1] By

(sdot sdot) we denote the 1198712(Γ) scalar product

(2) The Kernel 119870(sdot sdot sdot) is a Caratheodory function (H2)that is 119870(sdot sdot 119906) is measurable for all 119906 isin R and 119870(119909 119910 sdot) iscontinuous for almost all 119909 119910 isin Γ

(3) The assumption (H3) implies that the Nemytskiioperator

119860 1198712

(Γ) 997888rarr 1198712

(Γ) (25)

is Lipschitz continuous and strongly monotonous such that

(119860119906 minus 119860V 119906 minus V) le 119887 mes (Γ) 119906 minus V20

(119860119906 minus 119860V 119906 minus V) ge 119886 mes (Γ) 119906 minus V20

(26)

for all 119906 V isin 1198712(Γ)

Theorem6 Let assumptions (H1) (H2) and (H3) holdThenfor every 119891 isin 119867

minus12 the nonlinear boundary integral equation(20) has a unique solution in 119867

12(Γ)

Proof The proof follows from the well-known theorem byBrowder and Minty on monotone operators [6 10]

Since the simple layer potential operator on Γ

119878 119867minus12

(Γ) 997888rarr 11986712

(Γ) (27)

is an isomorphism it is sufficient to consider the uniquesolvability of the following equation

119861119906 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ

(28)

We will prove that the operator

119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (29)

is continuous and strongly monotonous

(i) In the first we show that 119861 is continuous

It is clear from the continuity of the mapping propertiesof the simple and double layer operators that

119878minus1

(119868 minus 119863) 11986712

(Γ) 997888rarr 119867minus12

(Γ) (30)

is continuous And from (H3)

119860 11986712

(Γ) 997888rarr 119867minus12

(Γ) (31)

is continuous Hence the boundary integral operator

119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (32)

is continuous

(ii) In the second we show that 119861 is strongly monotoneoperator

The function 120583 isin 119867minus12

(Γ) defined by

120583 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) (33)

for 119906(119909) isin 11986712

(Γ) is the normal derivative of the harmonicfunction

119908 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910

minus int

Γ

120583 (119910) 119864 (119909 119910) 119889119904119910

(34)

for 119909 isin Ω this means that 119908 satisfies the problem

Δ119908 (119909) = 0 119909 isin Ω

119908 (119909) = 119906 (119909) 119909 isin Γ

(35)

Then Greenrsquos theorem yields

(119878minus1

(119868 minus 119863) 119906 119906) = int

Γ

120583119906 119889119904 = int

Γ

120597119908

120597119899

119906119889119904

= int

Γ

120597119908

120597119899

119908 119889119904 = int

Ω

(nabla119908)2119889119909

(36)

Hence for all 119906 V isin 11986712

(Γ)

(119878minus1

(119868 minus 119863) (119906 minus V) 119906 minus V)

= int

Ω

(nabla (1199081

minus 1199082))2

119889119909 =10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

(37)

where (1199081

minus 1199082) denotes the harmonic function correspond-

ing to the Cauchy data 119906 minus V and 119878minus1

(119868 minus 119863)(119906 minus V)On the other hand we note that there exists (]

1minus ]2) isin

119867minus12

(Γ) such that

119878 (]1

minus ]2) = 119906 minus V (38)

on Γ [1] Hence for all 119909 isin Ω we have

119878Ω

(]1

minus ]2) = 1199081

minus 1199082 (39)

The simple layer potential

119878Ω

119867119904(Γ) 997888rarr 119867

119904+(32)(Ω) (40)

is continuous for all 119904 isin R [1] Hence for 119904 = minus32 we find10038171003817100381710038171199081

minus 1199082

10038171003817100381710038171198712(Ω)

le 1198881

1003817100381710038171003817]1

minus ]2

1003817100381710038171003817119867minus32(Γ)

le 1198882

119906 minus V119867minus12(Γ)

le 1198883

119906 minus V0(41)

for some positive constants 1198881 1198882 and 119888

3

Hence we have

119906 minus V0 ge

1

1198883

10038171003817100381710038171199081

minus 1199082

10038171003817100381710038171198712(Ω)

(42)


Then with (28) and (37) we get

(119861119906 minus 119861V 119906 minus V) = (119878minus1


+ (119860119906 minus 119860V 119906 minus V)

=10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ (119860119906 minus 119860V 119906 minus V)

(43)

and with (26) we get the inequality

(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ 119886mes (Γ) 119906 minus V20

(44)

hence with (42) we have

(119861119906 minus 119861V 119906 minus V)

ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

ge min1

119886mes (Γ)

1198882

3

times (10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

)

ge min1

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198671(Ω)

ge 1198884

119906 minus V211986712(Γ)

(45)

by the trace theorem [1 9] which completes the proof

Now we prove the regularity of the solution of thenonlinear boundary integral equation (20)

Theorem 7 For all 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 the unique

solution of the nonlinear boundary integral equation (20)belongs to the space 119867

119904(Γ)

In the proof of this theorem we will need the followinglemma

Lemma 8 For every 119906 isin 119867119904(Γ) 0 le 119904 le 1 one has 119860119906 isin

119867119904(Γ) and the mapping 119860 119867

119904(Γ) rarr 119867

119904(Γ) is bounded

Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871

2(Γ) has already been proved

For 119904 = 1 119906 isin 1198671(Γ) 119906 is an absolutely continuous

function By assumption (H3) the function119860119906(119909) is Lipschitzcontinuous Hence 119860119906(119909) is also absolutely continuous func-tion

It remains to prove the case 0 lt 119904 lt 1 by the assumption(H3) and due to the definition of the Sobolev space inDefinition 4 we have

∬

Γ

1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

le 1198872(mes (Γ))

21199062

119867119904(Γ)

(46)

which completes the proof of Lemma 8

Proof of Theorem 7 Let 119878119891 isin 119867119904(Γ) 12 le 119904 le 32 be given

By Theorem 6 there exists a unique solution 119906 isin 11986712

(Γ) ofthe nonlinear boundary integral equation

(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)

Lemma 8 ensure that

119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)

therefore

(119868 minus 119863) 119906 isin 119867119904(Γ) (49)

This implies together with the Fredholm property of thedouble layer potential operator that 119906 isin 119867

119904(Γ) 12 le 119904 le 32

Example 9 Here we give an example to illustrate thetheoretical results We consider the harmonic problems

Δ119906 (119909) = 0 119909 isin Ω

120597119906

120597119899

(119909) + int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910

= 119891 (119909) 119909 isin Γ

(50)

where the nonlinear boundary integral equation of Urysohntype is defined by

119860119906 (119909) = int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ

(51)

and the domain is

Ω = 119909 = (1199091 1199092) 1199092

1+ 1199092

2lt 1199032

lt

1

4

(52)

Clearly the nonlinearity satisfies our assumptions (1198671) (1198672)

and (1198673) such that

diam (Ω) = 2119903 lt 1 (53)

The Kernel (2119906(119910) + sin 119906(119910)) of the nonlinear boundaryintegral equation of Urysohn type is a Caratheodory func-tion And

120597119870 (119909 119910 119906)

120597119906

= 2 + cos 119906 (119910) (54)

is measurable satisfying

1 le

120597 (2119906 (119910) + sin 119906 (119910))

120597119906

le 3 lt +infin (55)

implying that the Nemytskii operator

119860 1198712

(Γ) 997888rarr 1198712

(Γ) (56)


2120587119903119906 minus V20

le (119860119906 minus 119860V 119906 minus V) le 6120587119903119906 minus V20

(57)

for all 119906 V isin 1198712(Γ)


Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

The author would like to thank the referee for his verycareful reading of the paper and his detailed comments andvaluable suggestions which improved both the content andthe presentation of this paper

References

[1] G C Hsiao and W Wendland Boundary Integral EquationsApplied Mathematical Sciences Springer Berlin Germany2008

[2] A jafarian Z Esmailzadeh and L Khoshbakhti ldquoA numericalmethod for solving nonlinear integral equation in the Urysohnformrdquo Applied Mathematical Sciences vol 7 no 28 pp 1375ndash1385 2013

[3] M Krasnoselrsquoskii Topological Methods in the Theory of Nonlin-ear Integral Equations Macmillan New York NY USA 1964

[4] K Atkinson and G Chandler ldquoBoundary integral equationmethods for solving Laplacersquos equation with nonlinear bound-ary conditionsrdquo Mathematics of Computation vol 55 no 192pp 451ndash472 1990

[5] K Atkinson The Numerical Solution of Integral Equations ofthe Second Kind Cambridge University Press Cambridge UK1997

[6] K Ruotsalainen and W Wendland ldquoOn the boundary ele-ment method for some nonlinear boundary value problemsrdquoNumerische Mathematik vol 53 no 3 pp 299ndash314 1988

[7] R Bialecki and A J Nowak ldquoBoundary value problems in heatconduction with nonlinear material and nonlinear boundaryconditionsrdquo Applied Mathematical Modelling vol 5 no 6 pp417ndash421 1981

[8] C A Brebbia J C F Telles and L CWrobelBoundary ElementTechniques Springer Berlin Germany 1984

[9] R A Adams Sobolev Spaces Academic Press New York NYUSA 1975

[10] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

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Stochastic AnalysisInternational Journal of


11 Definitions and Notations

Definition 1 (see [1 9]) Let 119898 isin N one denotes by 119867119898

(Ω)

the Sobolev space

119867119898

(Ω) = 119906 isin 1198712

(Ω) 119863120572119906 isin 1198712

(Ω) |120572| le 119898 (5)

Definition 2 (see [1 9]) Let 119904 isin R one denotes by119867119904(R119899) the

Sobolev space

119867119904(R119899) = 119906 isin 119871

2(R119899) (1 +

10038161003816100381610038161205851003816100381610038161003816

2

)

1199042

|119865 [119906]| isin 1198712

(R119899)

(6)

and the associated norm

119906119867119904 = (int

R119899(1 +

10038161003816100381610038161205851003816100381610038161003816

2

)

119904

|119865 [119906]|2119889120585)

12

(7)

with 119865[sdot] the Fourier transform

Definition 3 (see [1 9]) Let Ω sub R119899 a bounded domain andΓ = 120597Ω one defined

119867119904(Ω) = 119906 | Ω 119906 isin 119867

119904(R119899) 119904 isin R

119867119904(Γ) =

119906|Γ

119906 isin 119867119904+(12)

(R119899) 119904 gt 0

1198712

(Γ) 119904 = 0

(119867minus119904

(Γ))1015840

(dual space) 119904 lt 0

(8)

Definition 4 (see [1 9]) The Fichera trace spaces 119867119904(Γ) for

0 lt 119904 lt 1 is defined to be the completion of

1198620

119904(Γ) = 120593 isin 119862

0(Γ)

1003817100381710038171003817120593

1003817100381710038171003817119867119904(Γ)

lt infin (9)

with respect to the norm

119906119867119904(Γ)

= 1199062

1198712(Γ)

+ ∬

Γ

1003816100381610038161003816119906 (119909) minus 119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

12

(10)

2 The Boundary Integral Method

21 Representative Formula and Boundary Operator Weintroduce the fundamental solution of the Laplacian operatorin the plane defined by

119864 (119909 119910) =

1

2120587

log 1003816100381610038161003816119909 minus 119910

1003816100381610038161003816 (11)

We first consider some standard boundary integral operatorsFor 119909 isin Ω the single layer potential is

119878Ω

119906 (119909) = minus int

Γ

119864 (119909 119910) 119906 (119910) 119889119904119910 (12)

and the double layer potential is

119863Ω

119906 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910 (13)

Using Greenrsquos identity for harmonic functions we get

119906 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910

minus int

Γ

120597119906 (119910)

120597119899119910

119864 (119909 119910) 119889119904119910

(14)

for 119909 isin Ω which can be written as

119906 (119909) = 119863Ω

119906 (119909) + 119878Ω

120597119906 (119909)

120597119899

for 119909 isin Ω (15)

Sending in (15) 119909 rarr Γ The continuity of the simple layerpotential 119878

Ωand the jump relation of the double layer poten-

tial 119863Ω we can write the integral equation on the boundary

as follows

119906 (119909) minus 119863119906 (119909) = 119878

120597119906 (119909)

120597119899

119909 isin Γ (16)

where

119878

120597119906 (119909)

120597119899

= minus2 int

Γ

119864 (119909 119910)

120597119906 (119910)

120597119899

119889119904119910 119909 isin Γ

119863119906 (119909) = 2 int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910 119909 isin Γ

(17)

Clearly if 119906 isin 1198671(Ω) is the solution of (1) then the Cauchy

data 119906|Γand 120597119906120597119899|

Γsatisfies the integral equation (16)

Then the boundary conditions

120597119906

120597119899

(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (18)

yield

119906 (119909) minus 119863119906 (119909) = minus119878 (119860 (119909 119906 (119909))) + 119878119891 (119909) 119909 isin Γ

(19)

Equation (19) can be written as

(119868 minus 119863) 119906 (119909) + 119878 (119860 (119909 119906 (119909))) = 119878119891 (119909) 119909 isin Γ (20)

Conversely if 119906|Γsolves (20) then the solution of (1) can be

given by the representation formula (15) and will satisfy

120597119906

120597119899

(119909) = minus119860 (119909 119906 (119909)) + 119891 (119909) (21)

due to (20) For studying the solvability of the nonlinearequation (20) we give some assumptions to be made here

(H1) We assume a diam(Ω) lt 1(H2) The Kernel 119870(sdot sdot sdot) of the Urysohn operator is a

Caratheodory function [3](H3) We assume that 120597119870(119909 119910 119906)120597119906 is measurable satisfy-

ing

0 lt 119886 le

120597119870 (119909 119910 119906)

120597119906

le 119887 lt +infin (22)

for some constants 119886 and 119887



119878 119867119904(Γ) 997888rarr 119867

119904+1(Γ) (23)


(119878120583 120583) ge 1198881003817100381710038171003817120583

1003817100381710038171003817

2

119867minus12

(24)





119860 1198712

(Γ) 997888rarr 1198712

(Γ) (25)




(26)

for all 119906 V isin 1198712(Γ)



12(Γ)



119878 119867minus12

(Γ) 997888rarr 11986712

(Γ) (27)


119861119906 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ

(28)


119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (29)




119878minus1

(119868 minus 119863) 11986712

(Γ) 997888rarr 119867minus12

(Γ) (30)


119860 11986712

(Γ) 997888rarr 119867minus12

(Γ) (31)


119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (32)

is continuous



(Γ) defined by

120583 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) (33)

for 119906(119909) isin 11986712


119908 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910

minus int

Γ

120583 (119910) 119864 (119909 119910) 119889119904119910

(34)


Δ119908 (119909) = 0 119909 isin Ω

119908 (119909) = 119906 (119909) 119909 isin Γ

(35)


(119878minus1

(119868 minus 119863) 119906 119906) = int

Γ

120583119906 119889119904 = int

Γ

120597119908

120597119899

119906119889119904

= int

Γ

120597119908

120597119899

119908 119889119904 = int

Ω

(nabla119908)2119889119909

(36)


(Γ)

(119878minus1


= int

Ω

(nabla (1199081

minus 1199082))2

119889119909 =10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

(37)

where (1199081




1minus ]2) isin

119867minus12

(Γ) such that

119878 (]1

minus ]2) = 119906 minus V (38)


119878Ω

(]1

minus ]2) = 1199081

minus 1199082 (39)


119878Ω

119867119904(Γ) 997888rarr 119867

119904+(32)(Ω) (40)


minus 1199082

10038171003817100381710038171198712(Ω)

le 1198881

1003817100381710038171003817]1

minus ]2

1003817100381710038171003817119867minus32(Γ)

le 1198882


le 1198883

119906 minus V0(41)


3

Hence we have

119906 minus V0 ge

1

1198883

10038171003817100381710038171199081

minus 1199082

10038171003817100381710038171198712(Ω)

(42)





+ (119860119906 minus 119860V 119906 minus V)

=10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ (119860119906 minus 119860V 119906 minus V)

(43)


(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ 119886mes (Γ) 119906 minus V20

(44)


(119861119906 minus 119861V 119906 minus V)

ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

ge min1

119886mes (Γ)

1198882

3

times (10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

)

ge min1

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198671(Ω)

ge 1198884

119906 minus V211986712(Γ)

(45)





119904(Γ)



119867119904(Γ) and the mapping 119860 119867

119904(Γ) rarr 119867


Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871





∬

Γ

1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

le 1198872(mes (Γ))

21199062

119867119904(Γ)

(46)





(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)

Lemma 8 ensure that

119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)

therefore

(119868 minus 119863) 119906 isin 119867119904(Γ) (49)


119904(Γ) 12 le 119904 le 32


Δ119906 (119909) = 0 119909 isin Ω

120597119906

120597119899

(119909) + int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910

= 119891 (119909) 119909 isin Γ

(50)


119860119906 (119909) = int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ

(51)

and the domain is

Ω = 119909 = (1199091 1199092) 1199092

1+ 1199092

2lt 1199032

lt

1

4

(52)



diam (Ω) = 2119903 lt 1 (53)


120597119870 (119909 119910 119906)

120597119906

= 2 + cos 119906 (119910) (54)


1 le

120597 (2119906 (119910) + sin 119906 (119910))

120597119906

le 3 lt +infin (55)


119860 1198712

(Γ) 997888rarr 1198712

(Γ) (56)


2120587119903119906 minus V20


(57)

for all 119906 V isin 1198712(Γ)




Acknowledgment


References


















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119878 119867119904(Γ) 997888rarr 119867

119904+1(Γ) (23)


(119878120583 120583) ge 1198881003817100381710038171003817120583

1003817100381710038171003817

2

119867minus12

(24)





119860 1198712

(Γ) 997888rarr 1198712

(Γ) (25)




(26)

for all 119906 V isin 1198712(Γ)



12(Γ)



119878 119867minus12

(Γ) 997888rarr 11986712

(Γ) (27)


119861119906 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) + 119860 (119909 119906 (119909)) = 119891 (119909) 119909 isin Γ

(28)


119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (29)




119878minus1

(119868 minus 119863) 11986712

(Γ) 997888rarr 119867minus12

(Γ) (30)


119860 11986712

(Γ) 997888rarr 119867minus12

(Γ) (31)


119861 11986712

(Γ) 997888rarr 119867minus12

(Γ) (32)

is continuous



(Γ) defined by

120583 (119909) = 119878minus1

(119868 minus 119863) 119906 (119909) (33)

for 119906(119909) isin 11986712


119908 (119909) = int

Γ

119906 (119910)

120597

120597119899119910

119864 (119909 119910) 119889119904119910

minus int

Γ

120583 (119910) 119864 (119909 119910) 119889119904119910

(34)


Δ119908 (119909) = 0 119909 isin Ω

119908 (119909) = 119906 (119909) 119909 isin Γ

(35)


(119878minus1

(119868 minus 119863) 119906 119906) = int

Γ

120583119906 119889119904 = int

Γ

120597119908

120597119899

119906119889119904

= int

Γ

120597119908

120597119899

119908 119889119904 = int

Ω

(nabla119908)2119889119909

(36)


(Γ)

(119878minus1


= int

Ω

(nabla (1199081

minus 1199082))2

119889119909 =10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

(37)

where (1199081




1minus ]2) isin

119867minus12

(Γ) such that

119878 (]1

minus ]2) = 119906 minus V (38)


119878Ω

(]1

minus ]2) = 1199081

minus 1199082 (39)


119878Ω

119867119904(Γ) 997888rarr 119867

119904+(32)(Ω) (40)


minus 1199082

10038171003817100381710038171198712(Ω)

le 1198881

1003817100381710038171003817]1

minus ]2

1003817100381710038171003817119867minus32(Γ)

le 1198882


le 1198883

119906 minus V0(41)


3

Hence we have

119906 minus V0 ge

1

1198883

10038171003817100381710038171199081

minus 1199082

10038171003817100381710038171198712(Ω)

(42)





+ (119860119906 minus 119860V 119906 minus V)

=10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ (119860119906 minus 119860V 119906 minus V)

(43)


(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ 119886mes (Γ) 119906 minus V20

(44)


(119861119906 minus 119861V 119906 minus V)

ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

ge min1

119886mes (Γ)

1198882

3

times (10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

)

ge min1

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198671(Ω)

ge 1198884

119906 minus V211986712(Γ)

(45)





119904(Γ)



119867119904(Γ) and the mapping 119860 119867

119904(Γ) rarr 119867


Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871





∬

Γ

1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

le 1198872(mes (Γ))

21199062

119867119904(Γ)

(46)





(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)

Lemma 8 ensure that

119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)

therefore

(119868 minus 119863) 119906 isin 119867119904(Γ) (49)


119904(Γ) 12 le 119904 le 32


Δ119906 (119909) = 0 119909 isin Ω

120597119906

120597119899

(119909) + int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910

= 119891 (119909) 119909 isin Γ

(50)


119860119906 (119909) = int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ

(51)

and the domain is

Ω = 119909 = (1199091 1199092) 1199092

1+ 1199092

2lt 1199032

lt

1

4

(52)



diam (Ω) = 2119903 lt 1 (53)


120597119870 (119909 119910 119906)

120597119906

= 2 + cos 119906 (119910) (54)


1 le

120597 (2119906 (119910) + sin 119906 (119910))

120597119906

le 3 lt +infin (55)


119860 1198712

(Γ) 997888rarr 1198712

(Γ) (56)


2120587119903119906 minus V20


(57)

for all 119906 V isin 1198712(Γ)




Acknowledgment


References


















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













+ (119860119906 minus 119860V 119906 minus V)

=10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ (119860119906 minus 119860V 119906 minus V)

(43)


(119861119906 minus 119861V 119906 minus V) ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+ 119886mes (Γ) 119906 minus V20

(44)


(119861119906 minus 119861V 119906 minus V)

ge10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

ge min1

119886mes (Γ)

1198882

3

times (10038161003816100381610038161199081

minus 1199082

1003816100381610038161003816

2

1198671(Ω)

+10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198712(Ω)

)

ge min1

119886mes (Γ)

1198882

3

10038171003817100381710038171199081

minus 1199082

1003817100381710038171003817

2

1198671(Ω)

ge 1198884

119906 minus V211986712(Γ)

(45)





119904(Γ)



119867119904(Γ) and the mapping 119860 119867

119904(Γ) rarr 119867


Proof For 119904 = 0 119906 isin 1198670(Γ) = 119871





∬

Γ

1003816100381610038161003816119860119906 (119909) minus 119860119906 (119910)

1003816100381610038161003816

2

1003816100381610038161003816119909 minus 119910

1003816100381610038161003816

1+2119904119889119904119909

119889119904119910

le 1198872(mes (Γ))

21199062

119867119904(Γ)

(46)





(119868 minus 119863) 119906 + 119878119860119906 = 119878119891 (47)

Lemma 8 ensure that

119878119891 minus 119878119860119906 isin 119867119904(Γ) (48)

therefore

(119868 minus 119863) 119906 isin 119867119904(Γ) (49)


119904(Γ) 12 le 119904 le 32


Δ119906 (119909) = 0 119909 isin Ω

120597119906

120597119899

(119909) + int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910

= 119891 (119909) 119909 isin Γ

(50)


119860119906 (119909) = int

Γ

(2119906 (119910) + sin 119906 (119910)) 119889119904119910 119909 isin Γ

(51)

and the domain is

Ω = 119909 = (1199091 1199092) 1199092

1+ 1199092

2lt 1199032

lt

1

4

(52)



diam (Ω) = 2119903 lt 1 (53)


120597119870 (119909 119910 119906)

120597119906

= 2 + cos 119906 (119910) (54)


1 le

120597 (2119906 (119910) + sin 119906 (119910))

120597119906

le 3 lt +infin (55)


119860 1198712

(Γ) 997888rarr 1198712

(Γ) (56)


2120587119903119906 minus V20


(57)

for all 119906 V isin 1198712(Γ)




Acknowledgment


References


















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Acknowledgment


References


















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article On the Harmonic Problem with Nonlinear ...downloads.hindawi.com/archive/2014/976520.pdf · In the present work, we deal with the harmonic problems in a bounded domain