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,