FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER
50 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThusAsD 1−α R(ζ, T )v = 2 π D1−α( ∞ ∑k=12π(v, sin kx))k 2 sin kx.− ζD 1−α sin kx = d [D 1−α sin kx ] =dx= d [ (n −α sin nx − απ )]= n 1−α[ cos nx cos απdx22we obtainD 1−α R(ζ, T )v = 2 ∞∑ n 1−α (v, sin kx)[cos kx cos αππk2k=1= 2 απ∞∑ (v, sin kx)cosπ 2 k 2 cos kx + 2 απ∞∑sin− ζπ 2ThuswhereIndeed,k=1AR(ζ, T )v = 2 πk=1D 1−α R(ζ, T )v = A ˜R(ζ, T )v + B ˜R(ζ, T )v,Av = sin απ2Bv = cos απ2sinαπ2k=1∞∑(v, sin kx) sin kx,k=1∞∑(v, sin kx) cos kx.k=1∞∑k 1−α[ ∑∞ (v, sin kx)k 2 − ζk=1= 2 απ∞∑sinπ 2k=1BR(ζ, T )v = 2 απ∞∑cos k 1−α( ∑∞π 2k=11−α (v, sin kx)kk 2 ,− ζk=1(v, sin jx)k 2 − ζ= 2 απcosπ 2 ,[AR+BR] 2n+1 v = [ (AR) 2n+1 + BA 2n R 2n+1] v.+ sinαπ2 sin nx ],απ]+ sin =21−α (v, sin kx)kk 2 .− ζ]sin jx sin kx sin kx =)sin jx, sin jx cos kx =The fact that the operators A, R(ζ, T ) are permutable, is checked directly:AR(ζ, T )v = 2 απ∞∑cos k 1−α( ∑∞ (v, sin jx))π 2j 2 sin jx, sin kx sin kx =− ζ= 2 πsinαπ2k=1∞∑k=1j=11−α (v, sin kx)kk 2 − ζsin kx,
Boundary Value Problems for Differential Equations of Fractional Order 51R(ζ, T )Av = 2 απ∞∑ (Av, sin nx)cosπ 2 n 2 sin nx =− ζn=1∞∑= 2 ∞ π sin ∑n 1−α (v, sin kx) sin nxk=1n 2 sin nx =− ζn=1= 2 sin απ∞∑ n 1−α (v, sin kx)π 2 n 2 sin nx =− ζ= AR(ζ, T )v,n=1i.e.,AR(ζ, T )v = R(ζ, T )Av.Now show the nilpotence of the operator BR(ζ, T ):[ ] 2v BR(ζ, T ) = BR(ζ, T )BR(ζ, T )v == 2 π[ 2=π= 2 πcosαπ2cosαπ2∞∑k=1cosαπ2( 2π∞∑k=1] 2 ∞ ∑k=11−α (BR(ζ, T )v, sin kx)kk 2 − ζcosαπ2∞∑k=1cos kx =(v,sin jx)j 2 −ζcos jx, sin kx )k 2 − ζ( ∞∑ (v,sin jx)k 1−α j 2 −ζk=1cos kx =cos jx, sin kx ) cos kxk 2 = 0.− ζAccording to Theorem 1.4.1 [16], for the eigenvalues of the problem we haveµ k = n 2 + ̂λ 0m + ̂λ 1m .Calculate ̂λ∫0m = trAp n . As P n = 12iπR(ζ, T ) dζ, p n is an integral operatorwith the kernelΓ nP n (x, y) = 2 sin nx sin ny,πi.e.P n v = 2 π∫ π0sin nx sin nyv(y) dy = 2 (v, sin nx) sin nx.πWe obtainAp n v = 2 απ ∑sin k 1−α (v, sin nx)(sin nx, sin kx) sin kx =π 2= 2 απsinπ 2 (v, sin nx)n1−α sin nx.Find the eigenvalues of the operator AP n . AP n v = λv, that is,2 απsinπ 2 n1−α (v, sin nx) = vλ.
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Boundary Value Problems for Differential Equations of Fractional Order 51R(ζ, T )Av = 2 απ∞∑ (Av, sin nx)cosπ 2 n 2 sin nx =− ζn=1∞∑= 2 ∞ π sin ∑n 1−α (v, sin kx) sin nxk=1n 2 sin nx =− ζn=1= 2 sin απ∞∑ n 1−α (v, sin kx)π 2 n 2 sin nx =− ζ= AR(ζ, T )v,n=1i.e.,AR(ζ, T )v = R(ζ, T )Av.Now show the nilpotence of the operator BR(ζ, T ):[ ] 2v BR(ζ, T ) = BR(ζ, T )BR(ζ, T )v == 2 π[ 2=π= 2 πcosαπ2cosαπ2∞∑k=1cosαπ2( 2π∞∑k=1] 2 ∞ ∑k=11−α (BR(ζ, T )v, sin kx)kk 2 − ζcosαπ2∞∑k=1cos kx =(v,sin jx)j 2 −ζcos jx, sin kx )k 2 − ζ( ∞∑ (v,sin jx)k 1−α j 2 −ζk=1cos kx =cos jx, sin kx ) cos kxk 2 = 0.− ζAccording to Theorem 1.4.1 [16], for the eigenvalues of the problem we haveµ k = n 2 + ̂λ 0m + ̂λ 1m .Calculate ̂λ∫0m = trAp n . As P n = 12iπR(ζ, T ) dζ, p n is an integral operatorwith the kernelΓ nP n (x, y) = 2 sin nx sin ny,πi.e.P n v = 2 π∫ π0sin nx sin nyv(y) dy = 2 (v, sin nx) sin nx.πWe obtainAp n v = 2 απ ∑sin k 1−α (v, sin nx)(sin nx, sin kx) sin kx =π 2= 2 απsinπ 2 (v, sin nx)n1−α sin nx.Find the eigenvalues of the operator AP n . AP n v = λv, that is,2 απsinπ 2 n1−α (v, sin nx) = vλ.
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