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64Pawe lKarczmarek


125i2ZZD22712722710x2720yd271d272fxyxy2D2whereD1D2arethequarter-planeandthewholecomplexplanerespectivelyispresentedWehavenotfoundintheliteratureanystudyoftheequationinwhichthesurfaceofintegrationisthe

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Document on Subject : "64Pawe lKarczmarek"— Transcript:

1 64Pawe lKarczmarek 1 (i)2ZZD2'(
64Pawe lKarczmarek 1 (i)2ZZD2'(1;2) (1�x)(2�y)d1d2=f(x;y);(x;y)2D2;whereD1;D2arethequarter-planeandthewholecomplexplane,respectively,ispresented.Wehavenotfoundintheliteratureanystudyoftheequationinwhichthesurfaceofintegrationisthehalf-plane.Inthispaper,weconsidertheequation1 (i)2ZZD'(1;2) (1�x)(2�y)d1d2=f(x;y);(1)where(x;y)2D=f(x;y):0Rez1;�1Imz1;z=x+iyg;f(x;y)isagivenfunctionand'(x;y)isanunknownfunction.2.FUNCTIONCLASSESLetusintroducefunctionclassesthatwillbeusedinthispaper.Wewrite'(x)2h(1);x&#x]TJ/;༔ ; .96; Tf;&#x 18.;͔ ;� Td;&#x[

2 000;0,ifthefunction'(t)='1+t 1&
000;0,ifthefunction'(t)='1+t 1�t;t2(�1;1);satis estheinequalityj'(t0)�'(t00)jKjt0�t00j;(2)whereK�0;01areconstantsindependentofthearrangementofthepointst0;t00ineachclosedintervalcontainedin(�1;1),andinaneighbourhoodofthepointt=�1thefollowingconditionissatis ed:'(t)='(t)jt+1j� ;0Re 1:Here'(t)isaHoldercontinuousfunctionontheinterval[�1;1),andlimt!1�0'(t)=limx!1'(x)=0:(3)Wewrite'(x)2h(1);�1x1,ifthefunction'(t)='i1+t 1�t;jtj=1;satis esinequality(2).Wewrite'(x;y)2h(1)h(1);x�0;�1y1,if

3 thefunction'(t1;t2)='1+t1 1�
thefunction'(t1;t2)='1+t1 1�t1;i1+t2 1�t2;(t1;t2)2(�1;1)L;L=ft2:jt2j=1g;satis estheinequality 66Pawe lKarczmarek 1 i1Z0p(1) 1+1d1=1 i1Z�1q(2) 2+id2=!;(10)thenthesolutionisgivenbytheformula'(x;y)=R(f;x;y)�p(x)+q(y)i p x+!i p x:(11)Proof.Wecanrewrite(1)intheform1 i1Z0d1 1�x1 i1Z�1'(1;2) 2�yd2=f(x;y);(12)or1 i1Z�1d2 2�y1 i1Z0'(1;2) 1�xd2=f(x;y):(13)Letusconsiderequation(12).Itcanberepresentedinthefollowingform1 i1Z0 1(1;y) 1�xd1=f(x;y);(14)whe

4 re 1(1;y)=1 i1Z�1'(1;&
re 1(1;y)=1 i1Z�1'(1;2) 2�yd2:(15)Letus ndthefunction 1(x;y)appearingin(14).Wesolve(14)inthefunctionclassh(1);0x1.By[8],weobtain 1(x;y)=x+1 p x1 i1Z0p 1 1+1f(1;y) 1�xd1+C3(y)i p x;whereC3(y);y2(�1;1),isanarbitraryfunctionofclassh(1):Next,solvingequation(15),weobtainthesolution'(x;y):'(x;y)=1 i1Z�1y+i 2+i 1(x;2) 2�yd2�C4(x)==1 (i)2(x+1)(y+i) p x1Z01Z�1p 1f(1;2) (1+1)(2+i)(1�x)(2�y)d1d2++(y+i)i ip x1Z�1C3(2) (2+i)(2�y)d2�C4(x);(16)whereC

5 4(x);0x1,isanarbitraryfunctionofclassh(1
4(x);0x1,isanarbitraryfunctionofclassh(1). 68Pawe lKarczmarek weget1 (i)21R01R�1R(f;1;2) (1�x)(2�y)d1d2=1 (i)21R01R�11 (i)2(1+1)(2+i) p 1(1�x)(2�y)1R01R�1p 01f(01;02) (01+1)(02+i)(01�1)(02�2)d01d02d1d2==1 (i)21R�11R�12+i (02+i)(2�y)(02�2)1 (i)21R01R0(1+1)p 01f(01;02) p 1(01+1)(01�1)(1�x)d01d1d02d2==1 (i)21R�11R�12+i (02+i)(

6 2�y)(02�2)&
2�y)(02�2)f(x;02)+1 (i)21R01R01+1 p 1(01�1)(1�x)d1p 01f(01;02) 01+1d01d02d2==1 (i)21R11R�1(2+i)f(x;02) (02+i)(2�y)(02�2)d02d2==f(x;y)+1 (i)21R�1 1R�12+i (2�y)(02�2)d2!f(x;02) 02+id02=f(x;y):NowwesubstitutethefunctionC1(x);x�0appearingin(7)into(1).Weobtain1 (i)21R01R�1C1(1) (1�x)(2�y)d1d2=1 i1R0C1(1) 1�xd11 i1R�1d2 

7 7;2�y=0:Finally,substitutingC2(y)i p
7;2�y=0:Finally,substitutingC2(y)i p x;�1y1into(1)weget1 (i)21R01R�1C2(2)i p 1(1�x)(2�y)d1d2=1 i1R�1C2(2)i 2�y1 i1R0d1 p 1(1�x)d2=0:Theabovecalculationsjustifyformula(7).Nowweproveformula(11).Tothisend,wesubstitute(7)intoconditions(8),(9).Wederive1 i1Z�1R(f;x;2) 2+id2+1 i1Z�1C1(x) 2+id2+1 i1Z�1C2(2)i p x(2+i)d2=p(x);(21)1 i1Z0R(f;1;y) 1+1d1+1 i1Z0C1(1) 1+1d1+1 i1Z0C2(y)i p 1(1+1)=q(y):(22)Since1 i1Z�1R(f;x;2

8 ) 2+id2=0;1 i1Z�1C1(x)
) 2+id2=0;1 i1Z�1C1(x) 2+id2=�C1(x);1 i1Z0R(f;1;y) 1+1d1=0;C2(y)i i1Z0d1 p 1(1+1)=C2(y); 70Pawe lKarczmarek Thenthesolutionof(1)inthefunctionclassh(1)h(1)hastheform'(x;y)=ip 2(x+1) p x(x+2)(y+1+i):4.SOLUTIONINTHECLASSh(0;1)h(1)Theorem4.1.Letf(x;y)2h(0;1)h(1)satisfycondition(6).Thenasolution'(x;y)of(1)inthefunctionclassh(0;1)h(1),satisfyingtherelations'(x;1)=0;x2[0;1);(23)and1 i1Z0'(1;y) 1+1d1=i(y+i) (i)21Z01Z�1f(1;2)d1d2 p 1(1+1)(2+i)(2�y);(24)isgivenbythefollowingformula:'(x;y)=

9 p x(y+i) (i)21Z01Z�1f(1;
p x(y+i) (i)21Z01Z�1f(1;2)d1d2 p 1(2+i)(1�x)(2�y):(25)Proof.AsintheproofofTheorem3.1,equation(1)canberewritteninform(14).Solving(14)inthefunctionclassh(0;1),weobtain(cf.[8]) 1(x;y)=p x i1Z0f(1;y) p 1(1�x)d1;(26)withthecondition1 i1Z0 1(1;y) 1+1d1=1 1Z0f(1;y) p 1(1+1)d1:(27)Substituting(15)into(27),wederive1 (i)21Z01Z�1'(1;2) (1+1)(2�y)d1d2=1 1Z0f(1;y) p 1(1+1)d1:(28)Multiplyingeachsideof(28)by1 y+i,onaccountof(23),usingHilberttransform[8],and

10 2;nallymultiplyingbothsidesoftheequation
2;nallymultiplyingbothsidesoftheequationbyy+i,weobtaincondition(24).Nowwesolve(15).By[8],thereis'(x;y)=y+i i1Z�1 1(x;2) (2+i)(2�y)d2==p x(y+i) (i)21Z01Z�1f(1;2)d1d2 p 1(2+i)(1�x)(2�y): 72Pawe lKarczmarek [9]M.Sheshko,SingularIntegralEquationswithCauchyandHilbertKernelsandTheirsAp-proximatedSolutions,TheLearnedSocietyoftheCatholicUniversityofLublin,Lublin,2003.Pawe lKarczmarekpawelk@kul.plTheJohnPaulIICatholicUniversityofLublinInstituteofMathematicsandComputerScienceAl.Rac lawickie14,20-950LublinReceived:June25,2007.Revised:July11,2007.Accepted:July12,