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Symbolic computations of exact solutions to nonlinear integrable differential equations

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Title:
Symbolic computations of exact solutions to nonlinear integrable differential equations
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Book
Language:
English
Creator:
Grupcev, Vladimir
Publisher:
University of South Florida
Place of Publication:
Tampa, Fla.
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Subjects

Subjects / Keywords:
The tanh method
PDE
KdV
Solitary wave
Wave equation
Dissertations, Academic -- Mathematics -- Masters -- USF   ( lcsh )
Genre:
bibliography   ( marcgt )
theses   ( marcgt )
non-fiction   ( marcgt )

Notes

Abstract:
ABSTRACT: In this thesis, first the tanh method, a method for obtaining exact traveling wave solutions to nonlinear differential equations, is introduced and described. Then the method is applied to two classes of Nonlinear Partial Differential Equations. The first one is a system of two (1 + 1)-dimensional nonlinear Korteweg-de Vries (KdV) type equations. The second one is a (3 + 1)-dimensional nonlinear wave equation. At the end, a few graphic representations of the obtained solitary wave solutions are provided, in correspondence to different values of the parameters used in the equations.
Thesis:
Thesis (M.A.)--University of South Florida, 2007.
Bibliography:
Includes bibliographical references.
System Details:
System requirements: World Wide Web browser and PDF reader.
System Details:
Mode of access: World Wide Web.
Statement of Responsibility:
by Vladimir Grupcev.
General Note:
Title from PDF of title page.
General Note:
Document formatted into pages; contains 34 pages.

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Resource Identifier:
aleph - 001917289
oclc - 181590952
usfldc doi - E14-SFE0002025
usfldc handle - e14.2025
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SFS0026343:00001


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Symbolic computations of exact solutions to nonlinear integrable differential equations
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2007.
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ABSTRACT: In this thesis, first the tanh method, a method for obtaining exact traveling wave solutions to nonlinear differential equations, is introduced and described. Then the method is applied to two classes of Nonlinear Partial Differential Equations. The first one is a system of two (1 + 1)-dimensional nonlinear Korteweg-de Vries (KdV) type equations. The second one is a (3 + 1)-dimensional nonlinear wave equation. At the end, a few graphic representations of the obtained solitary wave solutions are provided, in correspondence to different values of the parameters used in the equations.
502
Thesis (M.A.)--University of South Florida, 2007.
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Includes bibliographical references.
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Text (Electronic thesis) in PDF format.
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System requirements: World Wide Web browser and PDF reader.
Mode of access: World Wide Web.
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Title from PDF of title page.
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Advisor: Wen-Xiu Ma, Ph.D.
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PDE.
KdV.
Solitary wave.
Wave equation.
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by VladimirGrupcev Athesissubmittedinpartialfulllment oftherequirementsforthedegreeof MasterofArts DepartmentofMathematics CollegeofArtsandSciences UniversityofSouthFlorida MajorProfessor:Wen-XiuMa,Ph.D. YunchengYou,Ph.D. AthanassiosG.Kartsatos,Ph.D. DateofApproval: April10,2007 Keywords:Thetanhmethod,PDE,KdV,Solitarywave,Waveequation cCopyright2007,VladimirGrupcev

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DierentialEquations VladimirGrupcev @t6@ @x+@3 @x3=@ @x+@3 @x3;@ @t6@ @x+@ @x+@3 @x3=k@ @x+@3 @x3; @t+@ @x+@ @x(r2)+@5 @x5=0;

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wherethesubscriptsdenotethepartialderivatives:ut(x;t)=@u(x;t)

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@t6@ @x+@3 @x3=@ @x+@3 @x3;@ @t6@ @x+@ @x+@3 @x3=k@ @x+@3 @x3; @t+@ @x+@ @x(r2)+@5 @x5=0;

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Theparametersk,mandnsymbolizethewavenumbersinx,yandzdirections,respectively,and!isthefrequency,whichisassumedtobeafunctionofthewavenumbersk,mandn.Theintroductionofthenewvariablebringsthefollowingchanges:@ @t!cd d;@ @x!cd d;@ @t!cd din(1+1)dimensionalcase;@ @t!!d d;@ @x!kd d;(2.3)@ @y!md d;@ @z!nd din(1+3)dimensionalcase:

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d!(1~y2)d d~y;d2 d~y+(1~y2)d d~y;(2.4)::: where~y=tanh()=8<:tanh[c(xt)];tanh(kx+my+nz!t):

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@t6@ @x+@3 @x3=@ @x+@3 @x3;@ @t6@ @x+@ @x+@3 @x3=k@ @x+@3 @x3;(3.1)where,,,,,andkarerealconstants.Weintroduce=c(xt);(x;t)=U()and(x;t)=W():(3.2)Becauseoftheintroductionofthenewvariables,weobservethefollowingchanges@ @x=cdU d;@3 @x3=c3d3U d3;@ @t=cdU d;@ @x=cdW d;@3 @x3=c3d3W d3;@ @t=cdW d:(3.3)Ifweputtheseformulasintothesystem(3.1),wegetthefollowingsystem:cdU d6cUdU d+c3d3U d3=cdW d+c3d3W d3;cdW d6cWdW d+cdW d+c3d3W d3=kcdU d+c3d3W d3:(3.4)Now,letusdividethetwoequationsin(3.4)bycandintegratethemwithrespectto7

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d2=W+c2d2W d2;W3W2+W+c2d2W d2=kU+c2d2U d2:(3.5)Now,weintroduce~y=tanh()asanewindependentvariable.Wealsomakethefollowingsubstitutionsandproposethefollowingseriesexpansiontobeasolutionfor(x;t)and(x;t):(x;t)=U()=F(~y)=R1Xr1=0ar1~yr1;(x;t)=W()=G(~y)=R2Xr2=0br2~yr2:(3.6)Becauseofthesubstitutions,weobservethefollowingchanges:dU d=~y0dF d~y;d2U d2=~y00dF d~y+~y02d2F d~y2;dW d=~y0dG d~y;d2W d2=~y00dG d~y+~y02d2G d~y2:(3.7)Sincewearegoingtouse~y0and~y00,letusexpressthemintermsof~y:~y0=[tanh(~y)]0=1~y2;~y00=2~y0~y=2~y~y0=2~y(1~y2):(3.8)8

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d~y+(1~y2)d2F d~y2]=G+c2(1~y2)[2~ydG d~y+(1~y2)d2G d~y2];(3.9)G3G2+G+c2(1~y2)[2~ydG d~y+(1~y2)d2G d~y2]=kF+c2(1~y2)[2~ydF d~y+(1~y2)d2F d~y2]: d~ywhichyieldsthedegreeofR1+2.Thesecondoneis2c2~y3dG d~ywhichyieldsdegreeofR2+2.Nowletusconsiderthesecondequationof(3.9).Thenonlineartermofhighestorderis3G2anditsdegreeis2R2.Thereareseverallineartermsthatcanbeconsideredofhighestorder.Therstoneis2c2~y3dG d~yanditsdegreeisR2+2.Thesecondoneis2c2~y3dF d~yanditsdegreeisR1+2.Now,letusbalancethelineartermofhighestorderwiththenonlinearoneinbothequationsof(3.9)aswecanseewehavetwoalternatives:1.R1>R2or2.R2>R1.BothofthemgivethevaluesforR1=2andR2=2.Sonowknowingthis,(3.6)becomes:(x;t)=U()=F(~y)=2Xr1=0ar1~yr1=a2~y2+a1~y+a0;(x;t)=W()=G(~y)=2Xr2=0br2~yr2=b2~y2+b1~y+b0:(3.10)Substituting(3.10)into(3.9)andorganazingitinorderofthepowersof~y,wecanhavethefollowingcomputations.9

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Comparingthecoecientsofeachpowerof~yin(3.11),weget:6a2c23a22=6c2b2;2c2a16a1a2=2c2b1;a26a0a23a218c2a2=b28c2b2;(3.13)a12c2a16a0a1=2c2b1+b1;3a20a0+2c2a2=2c2b2+b0:Comparingthecoecientsofeachpowerof~y,in(3.12)weget:3b22+6c2b2=6c2a2;6b1b2+2c2b1=2c2a1;b26b0b23b21+b28c2b2=ka28c2a2;(3.14)b16b0b1+b12c2b1=ka12c2a1;b03b20+b0+2c2b2=ka0+2c2a2:Aswecansee,in(3.13)and(3.14),wehaveasystemof10equationsand8variables,10

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Ifwecrossmultiplytheexpressionsin(3.16)anddividetheresultinequationbyb2,wegetthefollowingequation:(a2

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Goingbacktoa2=x2b2,weget:a2=2c2(x2) Now,fromthesecondequationsof(3.13)and(3.14)weexpressc2,equatetherighthandsidesofthetworepresentationforc2,crossmultiply,dividebyb2andthendividebyb1.Asaresultoftheseoperations,weget:x2x21+(x2)x1+=0;(3.21) wherea1

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6(x1+2c2x12c2+) Ifwenowpluga1=x1b1intothefourthequationin(3.14),weget:b16b0b1+b12c2b1=kx1b12c2x1b1; 6(+2c2kx1+2c2x1) Bynow,wehavea0,b0,a2andb2asfunctionsofandc2.Pluggingthesevaluesfora0,b0,a2andb2intothethirdequationof(3.13)andsolvethatfora1,wegeta1asafunctionofandc2:a1=s 3c2u1 whereu1=(x2)[(8c2)x1+(2c2)x26c2x1x2]: 3c2u2

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Wearegoingtousethepreviouslycomputedvaluesofx2,a2andb2presentedin(3.18),(3.20)and(3.19)respectivelyandplugthemintothethirdequationsin(3.13)and(3.14).Thisiswhatwegetasaresultofthat:from(3.13):2 (d2 3+12422d1 3)2(d2 3+12422d1 3+6d1 3(d2 312++42+2d1 34c2d2 3+48c216c228c2d1 3+24c2d1 3+6d1 3))=0;(3.28)14

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(d2 312+42+2d1 3)4(d2 312+42+2d1 36d1 3)(60k485+4k5d1 3+1443c222+85322c24++2kd2 34+4d1 34+123d132123d1 32+242c23d2 3+2d2 334d1 34+108k2222d2 33+7232+12965c222++324k224+6kd2 322522321642+216427232+5223+8k6243h+1222h+243h1222h+72c23d1 32163c22d1 3+36kd1 322+9kd2 32++93d2 393d2 3+363d1 32162c24d1 3363d1 354kd1 323+4kd1 32h+42d1 3h42d1 3h6kd1 32h202d1 3224k3d1 3432k23+132k238644c22++963c23+108321083236k2h+36k3h++12k3h483c2h+1444c2h+kd2 3h+2d2 3h2d2 3h24c23d1 3h72d2 38c22d2 32+202d1 32+72d2 39kd2 32+30k22d1 3)=0:(3.29) whered=36+1082+83+12h;h=p

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@t+@ @x+@ @x(r2)+@5 @x5=0;(4.1) wherer2=(@2 Becauseofthepreviousintroduction,weobservethefollowingchanges@ @t=!dU d;@ @x=kdU d;@ @y=mdU d;@ @z=ndU d;@2 @x2=k2d2U d2;@2 @y2=m2d2U d2;@2 @z2=n2d2U d2;(4.3)@3 @x3=k3d3U d3;@3 @x@y2=km2d3U d3;@3 @x@z2=kn2d3U d3;@5 @x5=k5d5U d5: d+kUdU d+k(k2+m2+n2)d3U d3+k5d5U d5=0:(4.4) Nowweintegrateoncethewholeequationandweget!U+1 2kU2+k(k2+m2+n2)d2U d2+k5d4U d4=0:(4.5)17

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Thisnewintroductionyieldsthefollowingnecessarychanges:dU d=~y0dS d~y;d2U d2=~y00dS d~y2;d3U d3=~y000dS d~y+3~y00~y0d2S d~y2+~y03d3S d~y3;(4.7)d4U d4=~y(4)dS d~y+4~y0~y000d2S d~y2+6~y02~y00d3S d~y3+~y04d4S d~y4: 2kS2+[k(k2+m2+n2)~y00+k5~yiv]dS d~y+[k(k2+m2+n2)~y02+k5(4~y0~y000+3~y002)]d2S d~y2+6k5~y02~y00d3S d~y3(4.8)+k5~y04d4S d~y4=0:

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2kS2++[24k5~y5+(2km2+2k340k5+2kn2)~y3+(2km22kn22k3++16k5)~y]dS d~y+[36k5~y6+(km2+k3+kn280k5)~y4+(52k52km22kn22k3)~y28k5+km2+k3+kn2]d2S d~y2++[12k5~y+36k5~y336k5~y5+12k5~y7]d3S d~y3++[4k5~y2+6k5~y44k5~y6+k5~y8+k5]d4S d~y4=0:(4.10) Letustakealookatthelinearandnonlineartermofhighestorderin(4.10).Wecanseethattherearemorethanonelineartermthatcanbeconsideredofhighestorderbutallofthemyieldthesamedegree.Consideroneofthem,say24k5~y5dS d~y.FromtheproposedsolutionforS(~y)in(4.6)itfollowsthatthetermofhighestorderinS(~y)hasadegreeofR,thusdS d~yhasadegreeofthehighesttermofR1.Thereforetheabovechosenlineartermofhighestorderin(4.10)hasadegreeofR+4.TheNonlineartermofhighestorder(andtheonlynonlinearterm)in(4.10)is1 2kS2.KnowingtheproposedsolutionforS(~y)in(4.6),itisobviousthatthedegreeofthenonlineartermis2R.BalancingtheLineartermofhighestorderwiththeNonlineartermofhighestorder,weget2R=R+4whichyieldsR=4.Sonow,bysubstitutingthisvalueforRintheequation(4.6),wegetthefollowingforthesolutionofS(~y):U()=S(~y)=4Xr=0ar~yr=a4~y4+a3~y3+a2~y2+a1~y+a0:(4.11) NowwendthederivativesofSwithregardsto~ydS d~y=4a4~y3+3a3~y2+2a2~y+a1;d2S d~y2=12a4~y2+6a3~y+2a2d3S d~y3=24a4~y+6a3;d4S d~y4=24a4:(4.12)19

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2ka24)~y8+(ka4a3+360k5a3)~y7+(1 2ka23+20km2a4+20k3a4+ka4a22080k5a4+20kn2a4+120k5a2)~y6+(12kn2a3+24k5a1+ka4a1+ka3a2+12km2a3+12k3a3816k5a3)~y5+(1 2ka22+1696k5a432k3a4+6km2a2240k5a2!a4+6k3a232kn2a4+ka4a0+6kn2a2+ka3a132km2a4)~y4+(40k5a1+2km2a118kn2a3+576k5a318km2a3+ka3a0+ka2a1++2kn2a1+2k3a118k3a3!a3)~y3+(ka2a0480k5a4+1 2ka21+12kn2a48km2a2+12km2a4+12k3a4++136k5a28k3a28kn2a2!a2)~y2+(6kn2a3+6k3a3+ka1a0!a12k3a1+16k5a12km2a1120k5a3++6km2a32kn2a1)~y+2km2a2+2kn2a2+2k3a216k5a2+24k5a4!a0+1 2ka20=0:(4.13) Comparingthecoecientsofeachpowerof~yinbothsides,wecangetthefollowingresult:From~y8840k5a4+1 2ka24=0(4.14) whichgivesusthesolutionfora4:a4=1680k4ora4=0:

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2ka23+20km2a4+20k3a4+ka4a22080k5a4+20kn2a4+120k5a2=0;(4.16) andknowingthevaluesfora4anda3,wegetthevalueofa2:a2=280 13m2280 13k2+2240k4280 13n2: Fromthis,thevalueofa1isa1=0: 2ka22+1696k5a432k3a4+6km2a2240k5a2!a4+6k3a232kn2a4+ka4a0+6kn2a2+ka3a132km2a4=0;(4.18) andwethenexpressa0asafunctionof!:a0=1 507(507!k37280k4n27280m2k431k431m4 k4:(4.19) From~y3weget0=0whichisTautology.21

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2ka21=0:(4.20) Whenweusethepreviouslycomputedvaluesfora4,a3,a2,a1anda0,then!cancancelout.Sothisequationleadstosomerestrictionsfortheconstantk.Factoringtheresultingequationforkandpluggingthepreviouslycomputedvaluesfora4,a3,a2,a1anda0,wesetallfactorsasfollows:280 6591k3;(52k4+n2+k2+m2);(4.21)(270402k81612k61612k4n21612m2k4+31k4++62m2k2+62k2n2+62m2n2+31m4+31n4) Wearegoingtousethesecondfactortoobtainrealsolutionsfork.Aftersolvingthefollowingequationfork:52k4+n2+k2+m2=0; 52q 52q 52q 52q 52q

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2ka20=0:(4.25) Now,everyvariableinthisequationiseitherknownorisafunctionof!,andthusinthisequationtheonlyunknownvariableis!.Since(4.25)isaquadraticequationof!,wegettwosolutionsfor!:!=1 507k3p whereh1=1377958400k12n23+451360m6k4+14423136k8m42+11532m2k2n4++14423136k8n42+28846272k10m22+1354080k6n4+1354080m2k8++1354080k4n4m2+28846272k8m22n2+1354080k4n2m4++2708160k6n2m2+451360k4n6+11532k4m2n2+1354080k8n2++1354080m4k61377958400k12m23+28846272k10n22++11532m4k2n2+961m8+961n8+310670563844k16+961k8++14423136k1221377958400k143+5766k4m4+3844k6m2+3844k6n2++5766k4n4+451360k10+3844m6k2+5766m4n4+3844m6n2++3844k2n6+3844n6m2; 507k3p

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507k4(7280k4n27280m2k431k431m462m2k262k2n2)1 507k4(31n462m2n2+2649922k87280k6);(4.27) where!andkhavethevaluespreviouslycomputed.So,nowbysubstitutingthepreviouslycomputedvaluesforallvariables,wegetthesolutionfor(4.1):(x;y;z;t)=1680k4tanh4(kx+my+nz!t)++h280 13(k2+m2+n2)+2240k4itanh2(kx+my+nz!t)++a0;(4.28) wherea0,!andkhavethepreviouslycomputedvalues.Ifweusetheothervaluefora4,whichis0,thenwegetthefollowingresultfortheothervariablesfromcomperingthecoecientsinfrontofeachpowerof~yin(4.13):a3=0;a2=0;a1=0:24

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2ka0)=0; 2ka0: 13(k2+m2+n2)+2240k4itanh2(kx+my+nz!t)+a0;

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Figure1:=1;=1;=30;=2;=2;=2;k=126

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Figure3:=5;=5;=10;=2;=2;=10;k=127

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@t6@ @x+@3 @x3=@ @x+@3 @x3+@5 @x5;@ @t6@ @x+@ @x+@3 @x3=k@ @x+@3 @x3+@5 @x5;